Super 10 Mock Tests for KVPY SA for Class 11 2nd Edition Disha

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Contents TREND ANALYSIS OF KVPY - STREAM SA (2014-2019)

i-iv

KVPY SOLVED PAPER 2019 - STREAM SA

2019-1-20

KVPY SOLVED PAPER 2018 - STREAM SA

2018-1-24

MOCK TESTS 1. Mock Test - 1

1–8

2. Mock Test - 2

9–18

3. Mock Test - 3

19–26

4. Mock Test - 4

27–36

5. Mock Test - 5

37–46

6. Mock Test - 6

47–56

7. Mock Test - 7

57–66

8. Mock Test - 8

67–76

9. Mock Test - 9

77–86

10. Mock Test - 10

87–96

Mock Test Solutions (1-10)

97-220

EBD_7839

Sets, Functions and Number Systems

Algebra

Sequences and Series

Lines and Triangles

Quadrilaterals

Circles and Conic Sections

Mensuration

I

II

III

IV

V

VI

VII

Permutations and Combinations

TOTAL NO. OF QUESTIONS

IX

VIII Statistics and Probability

Unit Name

Unit No.

2015

15

1

2

1

1

2

3

5

5

1

1

1

2

15

3

3

1

1

1

1

2

3

5

3

1

1

Part Part Part Part I II I II

2014

2017

(i)

15

1

1

3

1

2

1

6

5

1

1

1

2

15

2

2

1

1

1

1

3

4

5

1

1

1

1

1

Part Part Part Part I II I II

2016

MATHEMATICS

15

2 (Q. 6, 12)

1 (Q. 7)

2 (1, 10)

1 (Q. 15)

4 (Q. 8, 9, 11, 14)

1 (Q. 3)

3 (Q. 2, 4, 5)

1 (Q.13)

P-1

2018

5

1 (Q. 62)

1 (Q. 63)

1 (Q. 64)

1 (Q. 65)

1 (Q. 61)

P-II

TREND ANALYSIS OF KVPY - Stream SA (2014-2019)

2 (Q. 63, 64) 1 (Q. 61)

3 (Q. 7, 10, 13) 4 (Q. 2, 3, 9, 12)

15

1 (Q. 8)

1 (Q. 14)

1 (Q. 1)

1 (Q. 11)

5

1 (Q. 62)

2 (Q. 5, 6) 1 (Q. 65)

2 (Q. 4, 15)

P-II

P-1

2019

1 9

Work, Energy and Power

Rotational Motion

Gravitation

Mechanical Properties of Solids and Fluids

Heat and Thermodynamics

Oscillations and Waves

Light, Electricity & Magnetism, Atoms & Nuclei

IV

V

VI

VII

VIII

IX

X 15

2

2

Laws of Motion

III

1

Kinematics

II

TOTAL NO. OF QUESTIONS

2014

2015

2016

2017

5

2

1

1

1

15

6

1

4

1

1

1

1

5

2

1

1

1

(ii)

15

8

1

1

1

1

2

1

5

1

1

3

15

8

2

1

1

3

5

2

1

1

1

15

8

3

2

2

5

1

2

1

1

Part II

2018

Part Part Part Part Part Part Part Part Part I II I II I II I II I

Physical World and Measurements

Unit Name

I

Unit No.

PHYSICS Part I

2019

1 (Q. 69)

1 (Q. 70)

1 (Q. 66)

1 (Q. 67)

Part II

15

5

8 (Q. 16, 19, 20, 1 (Q. 68) 21, 24, 25, 26, 30)

1 (Q. 27)

2 (Q. 18, 28)

1 (Q. 22)

2 (Q. 23, 29)

1 (Q. 17)

TREND ANALYSIS OF KVPY - Stream SA (2014-2019)

EBD_7839

2

15

2

15

1

3

States of Matter & Thermodynamics

Equilibrium, Redox Reactions & Chemical Kinetics

s & p Block Elements

G.O.C & Hydrocarbons

Analytical Chemistry, Biomolecules, Metals-Non Metals & Acid -Base

TOTAL NO. OF QUESTIONS

II

III

IV

V

VI

2018

15

5

6

5

3

2

4

1

2

1

5

(iii)

5

2

1

2

15

5

1

2

2

5

5

2

1

1

1

5

1

7

5

1

1

1

2

15

2

5

3

1

4

5

2

1

2

Part Part Part Part Part Part Part Part Part Part I II I II I II I II I II

I

2017

Basic Concepts of Chemistry, Structure of Atom, Chemical Bonding, Classification & Periodic Properties of Elements

2016

Unit Name

2015

Unit No.

2014

CHEMISTRY

Part I

2019 Part II

15

5

1 (Q. 74)

1 (Q. 75)

3 (Q. 32, 33, 34) 1 (Q. 40)

1 (Q. 71)

1 (Q. 72)

2 (Q. 35, 36)

1 (Q. 41)

2 (Q. 42, 45)

6 (Q. 31, 37, 1 (Q. 73) 38, 39, 43, 44)

TREND ANALYSIS OF KVPY - Stream SA (2014-2019)

1 1

1 5

Diversity of Living Organisms

Anatomy and Morphology of Living Organisms

Plant Physiology

Animal Physiology

Biochemistry and Cell Biology

Reproduction

Genetics and Evolution

Biotechnology

Ecology

Human Health and Microbes

II

III

IV

V

VI

VII

VIII

IX

X

TOTAL NO. OF QUESTIONS

2015

2016

2017

2018

15

2

1

1

9

1

1

2

2

15

3

1

1

2

2

4

5

2

1

1

1

(iv)

15

1

1

5

1

1

5

1

5

1

2

1

1

15

1

4

1

2

4

2

1

5

1

2

1

1

15

1

4

6

2

1

1

5

2

1

1

1

Part Part Part Part Part Part Part Part Part Part I II I II I II I II I II

I

2014

Unit Name

Unit No.

BIOLOGY Part I

1 (Q. 80)

5

1 (Q. 79)

4 (Q. 46, 50, 51, 54)

15

1 (Q. 77)

1 (Q. 60)

6 (Q. 47, 48, 49, 53, 56, 58)

2 (Q. 52, 57)

Part II

2 (Q. 76, 78)

2019

2 (Q. 55, 59)

TREND ANALYSIS OF KVPY - Stream SA (2014-2019)

EBD_7839

KISHORE VAIGYANIK PROTSAHAN YOJANA SOLVED PAPER 2019 STREAM : SA Time : 3 Hours

Maximum Marks : 100

INSTRUCTIONS 1. 2. 3. 4.

There are 80 questions in this paper. The question paper contains two parts; Part–I (1 Mark Questions) and Part–II (2 Marks Questions). There are four sections; Mathematics, Physics, Chemistry and Biology in each part. There are four options given with each question, only one of them is correct. For each incorrect answer 0.25 Mark in Part–I and 0.5 Mark in Part–II will be deducted. PART-I (1 MARK QUESTIONS) 3.

MATHEMATICS 1.

Let ABC be an equilateral triangle with side length a. Let R and r denote the radii of the circumcircle and the incircle of triangle ABC respectively. Then, as a function of a, the ratio

R r

(a) (b) (c) (d)

2.

strictly increases strictly decreases remains constant strictly increases for a < 1 and strictly decreases for a > 1 Let b be a non-zero real number. Suppose the 1 = 0 has two b

quadratic equation 2x2 + bx + distinct real roots. Then (a) b

1 b

5 2

(c) b2 – 3b > –2

(b) b

1 b

(d) b2 +

5 2

1 b2

0 be a real number, C denote a circle with circumference l, and T denote a triangle with perimeter l. Then (a) given any positive real number , we can choose C and T as above such that the ratio

Area(C ) is greater than Area(T )

(b) given any positive real number , we can choose C and T as above such that the ratio

Area(C ) is less than Area(T )

(c) given any C and T as above, the ratio

Area(C ) is independent of C and T Area(T ) (d) there exist real numbers a and b such that for any circle C and triangle T as above, we must have a < 15.

Area(C ) Q2 (b) Q1 < Q2 (c) M1Ql > M2Q2 (d) M1Q2 > M2Q1 Which one of the following schematic graphs best represents the variation of PV (in Joules) versus T (in Kelvin) of one mole of an ideal gas? (The dotted line represents PV = T.) (a)

30.

CHEMISTRY 31.

The hybridizations of N, C and O shown in the following compound

R

32.

N==C==O respectively, are (a) sp2, sp, sp2 (b) sp2, sp2, sp2 (c) sp2, sp, sp (d) sp, sp, sp2 The following compounds

33.

are (a) geometrical isomers (b) positional isomers (c) optical isomers (d) functional group isomers The major product of the following reaction

(b) PV (J)

PV (J)

T (K)

T (K)

(c)

(d) PV (J)

PV (J)

T ( K)

28.

29.

(a) 1% of Mumbai’s water needs (b) 10% of Mumbai’s water needs (c) 50% of Mumbai’s water needs (d) 100% of Mumbai’s water needs A mass M moving with a certain speed V collides elastically with another stationary mass m. After the collision the masses M and m move with speeds V and v respectively. All motion is in one dimension. Then (a) V = V + v (b) V = V + v (c) V = (V + v)/2 (d) v = V + V Four rays, 1, 2, 3 and 4 are incident normally on the face PQ of an isosceles prism PQR with apex angle Q = 120°. The refractive indices of the material of the prism for the above rays 1, 2, 3 and 4 are 1.85, 1.95, 2.05 and 2.15, respectively and the surrounding medium is air. Then the rays emerging from the face QR are (a) 4 only (b) 1 and 2 only (c) 3 and 4 only (d) 1, 2, 3 and 4

T (K)

Mumbai needs 1.4 l012 litres of water annually. Its effective surface area is 600 km 2 and it receives an average rainfall of 2.4 m annually. If 10% of this rain water is conserved it will meet approximately

Br Ph

1. excess alc. KOH

Br

2. NaNH 2 3. H3O+

is

KVPY SOLVED PAPER 2019 STREAM : SA

5

H

H

H

(b) Ph

concentration

(a) Ph

Br

Br (c) Ph

Br H

(d) Ph

H

0

Br H 34. IUPAC name of the following compound

O

35.

36.

37.

38.

39.

40.

41.

HO is (a) 1 -hydroxycyclohex-4-en-3-one (b) 1 -hydroxycyclohex-3-en-5-one (c) 3-hydroxycyclohex-5-en-1-one (d) 5-hydroxycyclohex-2-en-1 one In water-gas shift reaction, hydrogen gas is produced from the reaction of steam with (a) methane (b) coke (c) carbon monoxide (d) carbon dioxide Treatment with lime can remove hardness of water caused by (a) CaCl2 (b) CaSO4 (c) Ca(HCO3)2 (d) CaCO3 The most polarizable ion among the following is (a) F– (b) I– + (c) Na (d) Cl– For a multi-electron atom, the highest energy level among the following is (a) n = 5, l = 0, m = 0, s = +½ (b) n = 4, l = 2, m = 0, s = +½ (c) n = 4, l = 1, m = 0, s = +½ (d) n = 5, l = 1, m = 0, s = +½ The oxide which is neither acidic nor basic is (a) As 2 O 3 (b) Sb4 O10 (c) N2O (d) Na2O The element whose salts cannot be detected by flame test is (a) Mg (b) Na (c) Cu (d) Sr The plot of concentration of a reactant vs. time for a chemical reaction is shown below:

42.

43.

44.

45.

time

The order of this reaction with respect to the reactant is (a) 0 (b) 1 (c) 2 (d) not possible to determine from this plot During the free expansion of an ideal gas in an isolated chamber, (a) internal energy remains constant (b) internal energy decreases (c) work done on the system is negative (d) temperature increases The number of moles of water present in a spherical water droplet of radius 1.0 cm is [Given: density of water in the droplet = 1.0 g cm–3] 2 (a) (b) 18 27 2 (c) 24 (d) 9 Among the following, the correct statement about cathode ray discharge tube is (a) the electrical discharge can only be observed at high pressure and at low voltages. (b) in the absence of external electrical or magnetic field, cathode rays travel in straight lines. (c) the characteristics of cathode rays depend upon the material of electrodes. (d) the characteristics of cathode rays depend upon the gas present in the cathode ray tube. For a spontaneous process (a) enthalpy change of the system must be negative. (b) entropy change of the system must be positive. (c) entropy change of the surrounding must be positive. (d) en tr opy ch an ge of the system plus surrounding must be positive.

EBD_7839 6

46.

47.

48.

49.

50.

51.

52.

KVPY SOLVED PAPER 2019 STREAM : SA BIOLOGY Which one of the following is a CORRECT statement about primate evolution? (a) Chimpanzees and gorillas evolved from macaques (b) Humans and chimpanzees evolved from gorillas (c) Humans, chimpanz ees and gorill as evolved from a common ancestor (d) Humans and gorillas evolved from chimpanzees The crypts of Lieberkühn are found in which one of the following parts of the human digestive tract? (a) Oesophagus (b) Small intestine (c) Stomach (d) Rectum Removal of the pancreas impairs the breakdown of (a) lipids and carbohydrates only (b) lipids and proteins only (c) lipids, proteins and carbohydrates (d) proteins and carbohydrates only Microscopic examination of a blood smear reveals an abnormal increase in the number of granular cells with multiple nuclear lobes. Which one of the following cell types has increased in number? (a) Lymphocytes (b) Monocytes (c) Neutrophils (d) Thrombocytes Which one of the following genetic phenomena is represented by the blood group AB? (a) Codominance (b) Dominance (c) Overdominance (d) Semidominance The mode of speciation mediated by geographical isolation is referred to as (a) adaptive radiation (b) allopatric speciation (c) parapatric speciation (d) sympatric speciation Which one of the following metabolic conversions requires oxygen? (a) Glucose to pyruvate (b) Glucose to CO2 and ethanol

53.

54.

55.

56.

57.

58.

59.

60.

(c) Glucose to lactate (d) Glucose to CO2 and H2O Where are the proximal and distal convoluted tubules located within the human body? (a) Adrenal cortex (b) Adrenal medulla (c) Renal cortex (d) Renal medulla In a diploid organism, when the locus X is inactivated, transcription of the locus Y is triggered. Based on this observation, which one of the following statements is CORRECT? (a) X is dominant over Y (b) X is epistatic to Y (c) Y is dominant over X (d) Y is epistatic to X Which one of the following sequences represents the CORRECT taxonomical hierarchy? (a) Species, genus, family, order (b) Order, genus, family, species (c) Species, order, genus, family (d) Species, genus, order, family Which one of the following organs is NOT a site for the production of white blood cells? (a) Bone marrow (b) Kidney (c) Liver (d) Spleen Which one of the following anatomical structures is involved in guttation? (a) Cuticle (b) Hydathodes (c) Lenticels (d) Stomata Which one of the following parts of the eye is affected in cataract? (a) Cornea (b) Conjunctiva (c) Retina (d) Lens Which one of the following organisms is a bryophyte? (a) Liverwort (b) Volvox (c) Chlamdomonas (d) Fern During oogenesis in mammals, the second meiotic division occurs (a) before fertilisation (b) after implantation (c) before ovulation (d) after fertilisation

KVPY SOLVED PAPER 2019 STREAM : SA

7

PART-II (2 MARKS QUESTIONS)

and A = n

: fn

1

1 3(n 1)2 / 3

fn

Then (a) A = (b) A is a finite set (c) the complement of A in is nonempty, but finite (d) A and its complement in are both infinite 64. A prime number p is called special if there exist primes p1, p2, p3, p4 such that p = p1 + p2 = p3 – p4. The number of special primes is (a) 0 (b) 1 (c) more than one but finite (d) infinite 65. Let ABC be a triangle in which AB = BC. Let X be a point on AB such that AX : XB = AB : AX. If AC = AX, then the measure of ABC equals (a) 18° (b) 36° (c) 54° (d) 72° PHYSICS 66. A water-proof laser pointer of length 10 cm placed in a water tank rotates about a horizontal axis passing through its center of mass in a vertical plane as shown in the figure. The time period of rotation is 60 s. Assuming the water to be still and no reflections from the surface of the tank, the duration for which the light beam escapes the tank in one time period is close to (Refractive index of water =1.33)

10 cm 40 cm 30 cm

Laser

MATHEMATICS 61. Let a, b, c, d be distinct real numbers such that a, b are roots of x2 – 5cx – 6d = 0 and c, d are roots of x2 – 5ax – 6b = 0. Then b + d is (a) 180 (b) 162 (c) 144 (d) 126 62. Let S = {1, 2, 3, ..., 100). Suppose b and c are chosen at random from the set S. The probability that 4x2 + bx + c has equal roots is (a) 0.001 (b) 0.004 (c) 0.007 (d) 0.01 63. Let be the set of positive integers. For all n , let fn = (n + l)1/3 – n1/3

30 cm

(a) 8.13 s (b) 14.05 s (c) 16.27 s (d) 23.86 s 67. In an hour-glass approximately 100 grains of sand fall per second (starting from rest), and it takes 2 sec for each sand particle to reach the bottom of the hour-glass. If the average mass of each sand particle is 0.2 g then the average force exerted by the falling sand on the bottom of the hour-glass is close to. (a) 0.4 N (b) 0.8 N (c) 1.2 N (d) 1.6 N 68. A student uses the resistance of a known resistor (1 ) to calibrate a voltmeter and an ammeter using the circuits shown below. The student measures the ratio of the voltage to current to be l l03 in circuit (a) and 0.999 in circuit (b). From these measurements, the resistances (in ) of the voltmeter and ammeter are found to be close to:

1

(a)

102 and 10–2

1

(b)

103 and 10–3

(a) (b) (c) 10-2 and 102 (d) 10–3 and 103 69. A hot air balloon with a payload rises in the air. Assume that the balloon is spherical in shape with diameter of 11.7 m and the mass of the balloon and the payload (without the hot air inside) is 210 kg. Temperature and pressure of outside air are 27 °C and 1 atm = 105 N/m2 respectively. Molar mass of dry air is 30 g. The temperature of the hot air inside is close to, [The gas constant R = 8.31 J/K/mol] (a) 27 °C (b) 52 °C (c) 105 °C (d) 171 °C

EBD_7839 8 70.

KVPY SOLVED PAPER 2019 STREAM : SA A healthy adult of height 1.7 m has an average blood pressure (BP) of 100 mm of Hg. The heart is typically at a height of 1.3 m from the foot. Take the density of blood to be 103 kg/m3 and note that 100 mm of Hg is equivalent to 13.3 kPa (kilo Pascals). The ratio of BP in the foot region to that in the head region is close to. (a) one (b) two (c) three (d) four

75.

CHEMISTRY 71.

72.

PbO2 is obtained from (a) the reaction of PbO with HCl (b thermal decomposition of Pb(NO3)2 at 200 °C (c) the reaction of Pb3O4 with HNO3 (d) the reaction of Pb with air at room temperature For one mole of a van der Waals gas, the

PV at a fixed RT volume will certainly decrease if [Given: “a”, “b” are standard parameters for van der Waals gas] (a) “b” increases and “a” decreases at constant temperature (b) “b” decreases and “a” increases at constant temperature (c) temperature increases at constant “a” and “b” values (d) “b” increases at constant “‘a” and temperature The correct statements among the following i. E2s (H) > E2s (Li) > E2s (Na) > E2s (K) ii. The maximum number of electrons in the shell with principal quantum number n is equal to 2n2. iii. Extra stability of half-filled subshell is due to smaller exchange energy iv. Only two electrons, irrespective of their spin, may exist in the same orbital. are (a) i and ii (b) ii and iii (c) iii and iv (d) i and iv An organic compound contains 46.78% of a halogen X. When 2.00 g of this compound is heated with fuming HNO3 in the presence of AgNO3, 2.21 g AgX was formed. The halogen X is

(a) (c)

74.

(b)

C

(d) BIOLOGY

76.

compressibility factor Z

73.

[Given: atomic weight of Ag = 108, F = 19, Cl = 35.5, Bi = 80, I = 127] (a) F (b) Cl (c) Br (d) I An organic compound X with molecular formula C6H10, when treated with HBr, forms a gem dibromide. The compound X upon warming with HgSO4 and dil. H2SO4, produces a ketone which gives a positive iodoform test. The compound X is

77.

78.

A cell weighing 1 mg grows to double its initial mass before dividing into two daughter cells of equal mass. Assuming no death, at the end of 100 divisions what will be the ratio of the mass of the entire population of these cells to that of the mass of the Earth? Assume that mass of the Earth is 1024 kg and 210 is approximately equal to 1000. (a) 10–28 (b) 10–3 (c) 1 (d) 103 Papaya is a dioecious species with XY sexual genotype for male and XX for female. What will be the genotype of the embryos and endosperm nuclei after double fertilization? (a) 50% ovules would have XXX endosperm and XY embryo, while the other 50% would have XXY endosperm and XX embryo (b) 100% ovules would have XXX endosperm and XY embryo (c) 100% ovules would have XXY endosperm and XX embryo (d) 50% ovules would have XXX endosperm and XX embryo, while the other 50% would have XXY endosperm and XY embryo Solid and dotted lines represent the activities of pepsin and salivary amylase enzymes of the digestive tract, respectively. Which one of the following graphs best represents their activity vs pH?

KVPY SOLVED PAPER 2019 STREAM : SA

9 79. If the gene pool of the locus X in the human genome is 4, then what would be the highest possible number of genotypes in a large population? (a) 6 (b) 8 (c) 10 (d) 16 80. Match the plant hormones in Column I with their primary function in Column II. Column I Column II P. Abscisic acid (i) Promotes disease resistance Q. Ethylene (ii) Maintains seed dormancy R. Cytokinin (iii) Promotes seed germination S. Gibb ere llin (iv) Promotes fruit ripening (v) Inhibits leaf senescence Choose the CORRECT combination. (a) P - iii, Q - iv, R - i, S - ii (b) P - ii, Q - iv, R - v, S - iii (c) P - v, Q - iii, R- ii, S - i (d) P - iv, Q - ii, R - iii, S - v

Activity

(a)

1

5

10

pH

Activity

(b)

1

5

10

pH

Activity

(c)

1

5

10

pH

Activity

(d)

1

5

10

pH

ANS WER KEYS 1 2 3 4 5 6 7 8 9 10

(c) (c) (b) (a) (a) (b) (a) (c) (a) (d)

11 12 13 14 15 16 17 18 19 20

(c) (c) (d) (a) (b) (a) (a) (b) (c) (c)

21 22 23 24 25 26 27 28 29 30

Part-I 31 (d) 32 (d) 33 (a) 34 (c) 35 (a) 36 (d) 37 (a) 38 (b) 39 (d) 40 (c)

(a) (d) (a) (d) (c) (c) (b) (d) (c) (a)

41 42 43 44 45 46 47 48 49 50

(a) (a) (b) (b) (d) (c) (b) (c) (c) (a)

51 52 53 54 55 56 57 58 59 60

(b) (d) (c) (d) (a) (b) (b) (d) (a) (d)

61 62 63 64 65 66 67 68 69 70

Part-II 71 (c) 72 (a) 73 (a) 74 (b) (b) 75 76 (c) 77 (a) (b) 78 79 (c) 80 (c)

(c) (b) (a) (c) (d) (c) (d) (a) (c) (b)

EBD_7839 10

KVPY SOLVED PAPER 2019 STREAM : SA

HINTS & SOLUTIONS PART-I MATHEMATICS 1.

(c) In equilateral triangle centroid (G) and circumcentre are coincide A r=2 r = GD, R = AG

a

AG : GD = 2 : 1

Mn

a 10

G

R r

2.

4.

2

B

C

D

1 b

,0)

5.

b

0

8

0

b 0

3

0

range of f(b) when b = (f(2), ) = (–2, )

(– , 0)

(2, )

b – 3b > – 2 is correct So, b (– , 0) (2, ) is subset of solution set of b2 – 3b + 2 > 0 Also D is wrong as b

1

2

6.

(0, ) b2 (b) Let be the roots of p(x) = 0, then p( x ) ( x )( x )

a13 and

g ( x) ( x3 a13 )( x3 b13 ) g(x) = (x – a1)(x – b1) x2

a1x a12

x2

AC

BC 2

Area of

ABC

b1x b12

D 0 D 0 For exactly two values of x, g(x) = 0

AB 2

20

1 BC. AD 2 AB. AC

7.

1 AB.AC 2

25(AD) 15(20) AD = 12 AEDF is rectangle then, AD = EF = 12 (b) c2 = 2ab ...(i) a2 + c2 = 3b2 ...(ii) Substitute (i) in (ii) equation, we get, a2 + 2ab = 3b2 A (a + b)2 + 4b2 2a a

) b13

C

E

BC. AD

2

Let

n 110

n 1

AB = 15 & BC = 25

(2, )

)( x3

10

D

A

1.5

Let f(b) = b2 – 3b

( x3

2

(a) Here, BAC is a right angle triangle

F

Clearly options A and B are wrong

g ( x)

Mn

2(n 1) 2 2n

B 3

(b 2)(b2 2b 4) b b (

n 2a1 (n 1)d 2 n

n 1

(c) D > 0 (for real roots)

b2 4 2

3.

g(x) = x6 + ax3 + b and it is even degree polynomial. So, g(x) x R Hence, I and III are correct. (a) a1 = 2, d = 4

a = b, c = 2a A = 45° = B C and C = 90° B a (a) Sum of digits of N should be divisible by 9. In this case, the resulting number will always be divisible if c = 1, 2, 3, 5, 6 and 9. To make the resulting number divisible by 4, 7 and 8 as well N should be a multiple of LCM of (4, 7, 8) i.e. 56.

KVPY SOLVED PAPER 2019 STREAM : SA

8.

(c)

9.

(a)

10. (d)

11. (c)

Smallest positive integer which is multiple of 56 and whose sum of digits is a multiple of 9 is 504. N will be 504. Sum of digits = 9. Let x1 < x2 < ... < x11 x5 x6 Median of x1, x2 ,..., x10 2 x5 x6 Now replace x 11 by and then 2 arrange in order So, new set of numbers are x x x1, x2, x3, x4, x5, 5 6 , x6, x7, x8, x9, x10, 2 x5 x6 x6 Hence, median is 2 median decreases Let P(x) – 2x = a(x – ) (x – ) (x – ) (x – r) where r {1, 2, 3, 4} and r are distinct then, for any combination of r = 0, so P(x) = 2x no cubic polynomial possible. Total numbers of two digit number = 90 In each row from 11 to 19. 20 to 29, 30 to 39,...,90 to 99 i.e. every set of form {10 a + b; a = 1, 2, ....9; b = 0, 1, 2, ....., 9} has atleast one prime number so each number can be reduce in to prime number by changing atmost one digit. Hence, required number of almost prime two-digit number is 90. M C D N

P

L

B A K Let, 1 = Area of PAK = area of PBK [ AK = BK] Similarly, 2 = Ar( PBL) = Ar( PCL) 3 = Ar( PMC) = Ar( PMD) 4 = Ar( PMN) = Ar( PNA) Hence, ...(i) 1 + 4 = 25 + = 36 ...(ii) 1 2 ...(iii) 4 + 3 = 41 (ii) + (iii) – (i) + = 52 2 3 Area (PLCM) = 52

11 12. (c) x + y + z = 100 ...(i) and 6x + 4y + z = 200 ...(ii) From eqn. (i) – (ii) 5x + 3y = 100 ...(iii) No. of ordered pair (x, y) satisfying eqn. (iii) are (2, 30), (5,25), (8, 20), (11, 15) (14, 10), (17, 5), (20, 0) and for (x, y, z) using (i) equation (x, y, z) {(2, 30, 68), (5, 25, 70), (80, 20, 72) (11, 15, 74), (14, 10, 76), (17, 5, 78), (20, 20, 80)} No. of non negative integral solution = 7 13. (d) Given, N1 = 255 + 1, N2 = 165 = 5 × 3 × 11 N1 = (25)11 + 1 = (32)11 + 1 = (33 – 1)11 + 1 = 11C0 (33)11 – 11C1(33)32 + ...+ 11C1033 – 11C + 1 11 N1 is divisible by 33 Unit digit of N1 is 9 N1 is not divisible by 5. HCF of N1 and N2 = 33 14. (a) Let radius of circle be r and sides of be a, b, c Circumference of circle = l = perimeter of triangle 2 r

a b c

r

2 2

Area of circle (C) = r 2

2

2

4

area of (T) =

s( s a)(s b)( s c) where 2s = a + b + c = l. A.M. G.M. 1

( s a) ( s b) (s c ) 3

(s a)(s b)(s c )) 3

( s a )( s b )(s c )

s( s a )( s b)( s c) 2

Area of (C) Area of (T)

4

2

12 3

s 3

3

82 3 3

2

12 3

EBD_7839 12

KVPY SOLVED PAPER 2019 STREAM : SA Area of (C) Area (T)

3 3

Hence, for given positive real number , we can choose C and T such that the ratio of area of circle C to area of triangle T is greater than . 15. (b) It is given that b c 2

bc

b

1 2

c

1 2

1 4

(2b – 1) (2c – 1) = 1 2b – 1 = 1 and 2c – 1 = 1 or 2b – 1 = –1 and 2c – 1 = –1. b = 1 and c = 1 or b = 0 and c = 0 Numbers can be of form a11 to a00 where a = 1, 2, 3...,9 Number of three digit numbers = 18 PHYSICS 16. (a) In primary rainbow, two refraction and one TIR (1) Refraction of incident ray (2) TIR (3) Again refraction when rays come out of liquid drops

18. (b) Terminal speed is the maximum speed attained by an object as it falls through a fluid. When the object attains terminal speed, the net force on the object is zero. So acceleration a = 0 If we calculate the slope of given distance time graph between 0.3 m and 0.4 m then slope has approximately constant value, as the steel ball acquires constant terminal speed in liquid some time after falling into it. Terminal speed =

= 0.33 m/s 19. (c) Power delivered by the UPS battery is 1kVA i.e., 1000 V.A = 1000W When all the laptops connected directly to UPS then total power requirement 90 10 = 900W, So battery (UPS) can provide power to all laptops. If all laptops are used for 5 hours, then cost of electricity consumed as the cost of electricity is ` 5.00 per unit, =

(1)

(2)

(3)

In secondary rainbow, two refraction and two TIR. 17. (a) From question, 1 VSD =

11 = 1.1 mm 10

1 MSD = 1 mm Least count of vernier calliper L.C. = 1 VSD – 1 MSD LC = 1.1 mm – 1 mm = 0.1 mm

0.4 0.3 1 m/s = 0.3 3

900 5 3600 3.6 10 6

5 = ` 22.5

20. (c) Frosted glass is produced by coating on the surface and on its surface refractive index is different. Now, as a transparent adhesive tape is stuck on the glass surface, the refractive index of the upper coating is adjusted to be same as that of glass, so it becomes transparent. 21. (d) If refractive index of both prism parts are same, then the effects of dispersion are cancelled out and white light will appear as white light. Since R.I. of both parts will not be same when parts are filled with water at different temperatures, and given that the refractive index of water is inversely related to temperature, hence white light will not come out as white light. 22. (d) Since time period is 1.5 sec, so ball takes approximately 5.5 round in 8.3 seconds. Displacement in 5 round = 0

KVPY SOLVED PAPER 2019 STREAM : SA and in remaining 1/2 round, displacement = 2R = 2m Hence total displacement in 8.3 s is approximately 2m Since total time is 8.3 seconds so final remaining time is 8.3 –7.5 = 0.8 seconds =

=

t=

2 1.5

0.8

=

2 T

16 rad. = angle moved in last 0.8 sec 15 , so displacement 2R = 2m

13 4 3 R =2 3

4 3 R R0 i.e. R0 = 3 (2)1/3

[By volume conservation] Both nuclei separate to large distance, So, final, or total energy Ze 2

2 K

Ef =

=

2

2kZ 2 e2 1/3 2 4R

R0

kZ 2 e 2 R

1 21/3

0.63

kZ 2 e 2 R

Change in electrostatic energy

23. (a)

= Ei – Ef = 26. (d) If aM1 Block leaves contact at point P. Using W.E. theorem (A to P)

i.e.

1 mv2 2

g–

mg(R – h) = v=

2 g ( R h)

24. (c) All nuclei have nearly the same mass density. Density of nucleus (approx) =

M V

AmP 4 3 R 3

AmP 4 3 r0 A 3

=

3mP = constant (independent of mass 4 r03

no. A) 25. (a) Initially, electrostatic energy of nucleus = Ei =

kZ 2e 2 R

Now it breaks into two nuclei with radii R0 each

kZ 2 e 2 kZ 2 e 2 [1 – 0.63] = 0.37 R R

aM 2 then M1 hits the floor before M2

M1 g Q1 E M1 Q1E M1

g

M 2 g Q2 E M2

Q2 E M2

Q2 E Q1E > M2 M1

M1Q2 > M2Q1 27. (a) For ideal gas, PV = nRT or, PV = RT [ n = 1 mole given] Slope of PV Vs T graph is R PV = 8.314 T So with respect to PV = T PV = 8.314 T is having more slope so, graph (a) correctly depicts PV versus T graph. 28. (b) 10% of total rain water is conserved. Volume of water received annually V=A h

10 = 600 2.4 100

10 = 1.4 108 m3 100

[ A = 600 km2 = 600 106 m2 and h = 2.4 m] Mumbai needs = 1.4 1012 litres water annually = 1.4 109 m3

EBD_7839 14

KVPY SOLVED PAPER 2019 STREAM : SA For all rays angle of incidence on PR is 30°. For TIR at PR

1.4 108 % of water received = 100 1.4 109 = 10% i.e., 10% Mumbai’s water needs. 29. (d) According to question, all motion is in one dimension. M

m

V Before collision

;

M

V

sin

1

m

1

< 30°

1 or, 2

For elastic collision velocities after collision V1 =

(m1 m2 ) u1 (m1 m2 )

2m2 u2 m1 m2

V2 =

(m2 m1) u (m1 m2 ) 2

2m1 u m1 m2 1

(M (M

...(i)

or, V =

m) V m)

2M V ( M m) Subtracting (i) – (ii) and, v =

V

v=

[( M

...(ii)

m) 2M ]V ( M m)

Q

P

Br Ph

Br

Excess alc. KOH

Ph

Br

NaNH 2 H 3O

Ph—C

34. (d)

+

CH

O 1

6 5

2 3

4

5-Hydroxycyclohex-2-en-1-one

30° 60°

>2

CHEMISTRY

HO

120°

90°

< sin 30°

31. (a) R—N==C==O Nitrogen has 2 sigma bond and a lone pair of electron hybridisation is sp2. Carbon has 2 sigma bond hybridisation is sp. Oxygen has 1 sigma bond and two lone pair of electron hybridisation is sp2. NOTE: -bonding electron does not take part in hybridisation. 32. (d) an d compounds are functional group isomers. Functional group isomers are these the isomers which have same molecular formula but different functional group. 33. (a)

M m V v= V V – v = –V M m V +V=v 30. (c) For apex Q = 120° and normal incidence on PQ. For a ray to pass through QR it will get total internal reflection on PR i.e., ray will not directly go to QR.

30°

1

or,

So rays 3 ( = 2.05) and 4 ( = 2.15) will emerge out.

After collision

rest

1

30°

R

35. (c) In water gas shift reaction carbon monoxide reacts with steam at high temperature to produce carbondioxide and hydrogen. 36. (c) Lime can remove temporary hardness of water which is caused due to bicarbonate salt of Ca and Mg. 37. (b) As the size of anion increases polarizibility of anion increase. Among given anions I– has maximum size.

KVPY SOLVED PAPER 2019 STREAM : SA 38. (d) According to (n + l) rule, maximum is the (n + l) value higher is the energy level. If two level has same (n + l) value then the level which has higher value of n will have higher energy. 39. (c) As2O3 is amphoteric oxide. It reacts with both acids as well base. Sb4O10 is acidic oxide while Na2O is basic oxide. N2O is neutral oxide which neither react with acid nor with base. 40. (a) Magnesium salts cannot detected by flame test because magnesium has high ionisation enthalpy. Flame unable to provide sufficient energy to excite electron to next higher energy level. 41. (a) The rate of zero order reaction is independent of reactant concentration. As the reaction proceeds reactant concentration decrease with respect to time. 42. (a) Free expansion of an ideal gas in an isolated chamber is isothermal process. Internal energy remains constant durin g free expansion of an ideal gas. 43. (b) Volume of a spherical water droplet 4 3 4 r (1)3 3 3 4 cm3 V= 3 Mass of spherical water droplet 4 4 g = p×V= 1 3 3 18 g of water= 1 mol no. of moles of water in a spherical water droplet 4 2 = 3 18 27 44. (b) Cathode rays are independent of the nature of the gas and electrodes placed in discharge tube. They travel in a straight line in the absence of external electrical or magnetic field. 45. (d) For a spontaneous process, S = ( Ssystem + Ssurrounding) > 0

=

15 BIOLOGY 46. (c) According to the new genetic research-when combined with known fossils-the lineage that led to humans, chimps, and gorillas evolved from a common ancestor about 10 million years ago. 47. (b) The crypts of Lieberkuhn are simple, tubular glands which occur throughout the small intestine between the villi and secrete enzymes as well as mucus. 48. (c) The pancreas is a glandular organ. It is the part of the digestive system located in the abdomen and produces insulin and other important enzymes and hormones that help to break down the food. The enzymes include trypsin and chymotrypsin to digest proteins, amylase to break down carbohydrates and lipase, to break down fats into fatty acids and cholesterol. 49. (c) The correct answer will be Neutrophils. In this type of granules, granulocytes has very tiny granules and nucleus is multi lobed with lobed connected by thin strands of nuclear material. 50. (a) Codominance occurs when both alleles show dominance, as in the case of the AB blood type (IAIB) in humans. It is a relationship between two versions of a gene. Individuals receive one version of a gene, called an allele, from each parent. If the alleles are different, the dominant allele usually will be expressed, while the effect of the other allele, called recessive, is masked. 51. (b) Allopatric speciation, also referred to as geographic speciation or its earlier name, the dumbbell model, is a mode of speciation that occurs when biological populations of the same species become isolated from each other to an extent that prevents or interferes with gene flow. 52. (d) Aerobic respiration uses oxygen to break down glucose, amino acids and fatty acids and is the main way the body generates adenosine triphosphate (ATP), which supplies energy to the muscles.The products of this process are carbon dioxide and water.

EBD_7839 16 53. (c) The renal tubule is a long and convoluted structure that emerges from the glomerulus and can be divided into three parts based on function. The first part is called the proximal convoluted tubule (PCT) due to its proximity to the glomerulus which stays in the renal cortex. The second part is called the loop of Henle, or nephritic loop, because it forms a loop (with descending and ascending limbs) that goes through the renal medulla. The third part of the renal tubule is called the distal convoluted tubule (DCT) and this part is also restricted to the renal cortex. 54. (d) A gene is said to be epistatic when it's presence suppresses the effect of a gene at another locus, so here X is suppressing Y. Hence, X is epistatic to Y. 55. (a) Taxonomic hierarchy is a sequence of categories in a decreasing or increasing order from kingdom to species and vice versa. Kingdom is the highest rank followed by division, class, order, family, genus and species. 56. (b) In the human adult, the bone marrow produces all of the red blood cells, 60-70 percent of the white cells (i.e., the granulocytes), and all of the platelets. The reticuloendothelial tissues of the spleen, liver, lymph nodes, and other organs produce the monocytes (4-8 percent of the white cells). 57. (b) Hydathodes are involved in the process of guttation, in which positive xylem pressure (due to root pressure) causes liquid to exude from the pores. It mainly occur in the leaves of submerged aquatic plants. 58. (d) A cataract is a clouding of the normally clear lens of an eye. It often develop slowly and can affect one or both eyes. Symptoms may include faded colors, blurry or double vision, halos around light, trouble with bright lights, and trouble seeing at night. 59. (a) Bryophytes are a group of plant species that reproduce via. spores rather than flowers or seeds. It is consisting of three divisions of non-vascular land plants (embryophytes): the liverworts, hornworts and mosses.

KVPY SOLVED PAPER 2019 STREAM : SA

60.

Volvox, Chlamydomonas are grouped under algae and Ferns are a group of about 20,000 species of plants in the division Pteridophyta. (d) A primary oocyte begins the first meiotic division, but then arrests until later in life when it will finish this division in a developing follicle. This results in a secondary oocyte, which will complete meiosis if it is fertilized. So, the second meiotic division occurs after fertilisation. PART-II MATHEMATICS

61.

62.

(c) Since, a and b are the roots of the equation x2 – 5cx – 6d = 0 ...(i) Then, a + b = 5c and ab = –6d Since, c and d are the roots of the equation x2 – 5ax – 6b = 0 ...(ii) Then, c + d = 5a and cd = –6b Now, a + b + c + d = 5 (a + c) (a + c) + (b + d) = 5(a + c) (b + d) = 4 (a + c) And, (ab) (cd) = (– 6d) (–6b) (ac) (bd) = 36bd ac = 36 Since, a is the root of eqn. (i) and c is the root of the eqn. (ii), then a2 – 5ac – 6d = 0 ...(iii) c2 – 5ac – 6b = 0 ...(iv) From (iii) + (iv), (a + c)2 – 12 ac – 6(b + d) = 0 (a + c)2 – 12 × 36 – 6 × 4 (a + c) = 0 [ ac = 36 and (b + d) = 4 (a + c)] (a + c)2 – 24 (a + c) – 432 = 0 a + c = 36 or a + c = –12 when a + c = 36, then b + d = 4 (a + c) = 4 × 36 = 144 Hence, b + d = 144. (a) For equal r oots di scri mi n an t D = 0 b2 – 16c = 0. 2 b =4×4×c c has to be perfect square. Then, c can be any perfect square numbers from the given sets and the perfect square number in the set S = = 1, 4, 9, 16, 25, 36, 49, 64, 81, 100

KVPY SOLVED PAPER 2019 STREAM : SA

17 Ratio of length of sides of triangle > 0

Number of ordered pair (b, c) will be 10 Required probability

f n (n 1)1/3 n1/3 Vn N Rationalising fn, we get 1

fn

fn

1

fn

1

1 3(n 1)

1

fn

2/3

3n

1 3(n 1)

...(ii)

2/3

1

fn n

3(n 1)2/3

N

Hence, A = N 64. (b) If none, of p1, p2, p3, p4 is 2. All p1, p2, p3, p4 are odd Since, p is also a prime number. p1 + p2 + and p3 – p4 are both odd and hence cannot be prime. One of p1 or p2 (say p1) and p4 must be 2. p = 2 + p2 = p3 – 2. Above equation is satisfied only if p = 5, p2 = 3 and p3 = 7. Hence, only one special prime exists. 65. (b) Let ABC be a triangle, where AB = BC

2 5 1 4

cos

...(i)

2/3

36

PHYSICS 66. (c) Light will come out when the angle is less than critical angle ' C ' 10 cm C C 40 cm 30 cm

30 cm

1 sinc

Using sin c =

A

( BC ) 2 ( AC ) 2 2( AB )( BC )

2

2

cos

(n 1)2/3 (n)1/3 (n 1) 2/3 n 2/3

Clearly,

( AB ) 2

cos

r

63. (a)

5 1 2

0

0.001

La se

10 100 100

=

1 1.33

3 [ 4

water

= 1.33 given]

c = 50° (approx)

X

C

t = 2c [

AB

B

=

t

]

It is given that,

AX XB

AB AX

AX AB 2

1

t=

(say)

AB and XB AX

XB

1 0

2

AX 2

AB

AB

1

5 2

2c

2 50 180

AB

=

2 50 60 2 180

t = 16.67 s

16.27 s

2 50 180 = 2 T

EBD_7839 18 67.

KVPY SOLVED PAPER 2019 STREAM : SA (a) Velocity of sand particle just before striking the bottom v = u + at v = 0 + 10 2 = 20 m/s [ u = 0 and t = 2s given] Initial momentum, pi = mV = (0.2 10–3) 20 Final momentum, pf = 0 [ final velocity = 0] | p| = Pi Pf = 4 10–3 k m/s Rate of change of momentum exerted by falling sand

69.

| p| = force t

o

| p| 4 10 3 100 n = t 1 or, favg = 0.4 N (b) case – (a)

favg =

PM RT0

1

1 T0

I

Let, RV = resistance of voltmeter and RA = resistance of ammeter IRv I 1 RA

in

=

=

= RV(1 + RA) = 1000

or, RV(1 + RA) = 1000 case – (b)

210 V

M V

210 4 3 r 3

PM 210 3 = RTin 4 r3

PM 1 R T0

I

V = A

–

1 Tin

1 Tin

=

=

[

PV = nRT]

210 3 4 r3

210 3 R 4 r3

P M

210× 3×8.31× 7 × (2) 3 105 × 30× 4× 22× (11.7)3

0.0007

1 1 = – 0.0007 Tin 300

...(i)

or, Tin 107.22 °C 70.

(c) Given: Height of man = 1.7 m

I

A

Head

1

A

68.

(c) Here, mass of the balloon and payload, m = 210 kg Diameter of the balloon, d = 2r = 11.7 m Air pressure (outside), P = 105 N/m2 Temperature of outside air, T0 = 27°C = 300 k = density of air inside the balloon in = density of air outside the balloon o 210g + inVg = oVg

I 1 + RV

I

I 1 Rv V = I A

Rv =

Rv = 0.999 Rv 1

103 RV = 0.999 (1 + RV) RV = 999 Putting this value of RV in eqn (i) and solving, we get RA = 10 3

Heart 1.7 m

1.3 m

Feet

Height from heart to feet = 1.3 m Density of blood = 103 kg/m3 Pressure at the heart level = 100 mm of Hg = 13.3 kPa

KVPY SOLVED PAPER 2019 STREAM : SA

19

Total pressure on the feet of man

46.78 2 2.21 100

Pfoot = Pheart + gh = 13.3 + 103 10 1.3 = 26.3 kPa P hea d = P hea rt – gh = 13.3 – 10 3 (1.7 – 1.3) = 9.3 kPa 26.3 Pfoot = 9.3 Phead

10

0.42335 =

2Pb(NO3)2 + PbO2 + 2H2O

72. (b) Compressibility factor Z = Z=

PVm RT

Vm Vm

=

1 b 1 Vm

PVm RT

a RTVm

Br (X)

Let the atomic weight of the halogen (X) = x the atomic weight of AgX = 108 + x Substituting the value in the above equation we get,

HgSO4 H2SO4

... (i)

At. wt . of X × wt. of AgX ×100 = Mol. wt. of AgX × wt. of organic compound

2.21 100 2

Br

HBr

Considering the equation (i) the value of compressibility factor Z decreases as the value of ‘b’ decreases and the value of compressibility factor Z decreases as the value of ‘a’ increases at constant temperature and fixed volume. 73. (a) Correct statements are (i) and (ii) only. Extra stability of half filled subshell is due to greater exchange energy. In an orbital only two electrons of opposite spin may exist. 74. (c) Carius method of estimation of halogen: In this method a known weight of organic compound is heated with fuming nitric acid in presence of AgNO3. Halogen forms silver halide. Carbon and h ydrogen of th e compound are oxidised to CO2 and H2O. % of Halogen (X)

x 46.78 = 108 x

75. (d)

a RTVm

b

x 108 x

0.42335(108 + x) = x 45.7218 + 0.42335x = x x = 79.3 80 Halogen present in organic compound is Bromine.

3

CHEMISTRY 71. (c) Pb3O4 + 4HNO3

x 108 x

O

NaOH

O

H

–

Na+O

I2

O

+ CHI3

BIOLOGY 76. (c) Division in the cell is calculated by 2n so, After 100th division No. of cells = 2100 Total Numbers of cells = 2 Mass of one cell = 1 mg = 10–6 kg. Total mass of cells = 2100 × 10–6 kg { 210 = 103} 3 10 –6 24 = (10 ) × 10 kg = 10 kg Ratio of mass =

1024 1024

1

77. (d) After double fertilization when male gametes enter in embryo sac and fused with XX of female and XY of male then the genotypes, we receive i.e., 50% ovules having XXX endosperm and XX embryo, while the other 50% would have XXY endosperm and XY embryo.

EBD_7839 20

KVPY SOLVED PAPER 2019 STREAM : SA

78. (a) In the graph (A) indicating activity of pepsin at low pH i.e., 2.8. It will be highest and similarly activity of salivary amylase will be highest at the pH of 6.8. (B) is representing minimum activity and (C) and (D) graphs show constant activity of both enzyme at increasing pH. 79. (c) For locus 'x' total gene = 4 For example a, b, c and d Possible genotype aa

bb

cc

ab

bc

cd

ac

bd

ad

dd

Formula – for given gene alleles Total number = n Possibel genotype =

n n +1 2

4 4 1 2 = 2 (5) = 2 × 5 = 10 80. (b) ABA Maintain seed dormancy Ethylene Promote fruit ripening Cytokinin Inhibit leaf sensesence Gibberellin Promotes seed germination

KISHORE VAIGYANIK PROTSAHAN YOJANA SOLVED PAPER 2018 STREAM : SA Time : 3 Hours

Maximum Marks : 100

INSTRUCTIONS 1. 2. 3. 4.

There are 80 questions in this paper. The question paper contains two parts; Part–I (1 Mark Questions) and Part–II (2 Marks Questions). There are four sections; Mathematics, Physics, Chemistry and Biology in each part. There are four options given with each question, only one of them is correct. For each incorrect answer 0.25 Mark in Part–I and 0.5 Mark in Part–II will be deducted. PART-I (1 MARK QUESTIONS) MATHEMATICS

1.

2.

3.

4.

The number of pairs (a, b) of positive real numbers satisfying a4 + b4 < 1 and a2 + b2 > 1 is (a) 0 (b) 1 (c) 2 (d) More than 2 The number of real roots of the polynomial equation x4 – x2 + 2x – 1 = 0 is (a) 0 (b) 2 (c) 3 (d) 4 Suppose the sum of the first m terms of an arithmetic progression is n and the sum of its first n terms is m, where m ¹ n. Then, the sum of the first (m + n) terms of the arithmetic progression is (a) 1 – mn (b) mn – 5 (c) – (m + n) (d) m + n Consider the following two statements: I. Any pair of consistent linear equations in two variables must have a unique solution. II. There do not exist two consecutive integers, the sum of whose squares is 365.

5.

6. 7.

Then, (a) both I and II are true (b) both I and II are false (c) I is true and II is false (d) I is false and II is true The number of polynomials p(x) with integer coefficients such that curve y = p(x) passes through (2, 2) and (4, 5) is (a) 0 (b) 1 (c) more than 1 but finite (d) infinite The median of all 4-digit numbers that are divisible by 7 is (a) 5797 (b) 5498.5 (c) 5499.5 (d) 5490 A solid hemisphere is attached to the top of a cylinder, having the same radius as that of the cylinder. If the height of the cylinder were doubled (keeping both radii fixed), the volume of the entire system would have increased by 50%. By what percentage would the volume have increased if the radii of the hemisphere and the cylinder were doubled (keeping the height fixed)? (a) 300% (b) 400% (c) 500% (d) 600%

EBD_7839 2 8.

KVPY SOLVED PAPER 2018 STREAM : SA Consider a DPQR in which the relation QR2 + PR2 = 5 PQ2 holds. Let G be the points of intersection of medians PM and QN. Then ÐQGM is always (a) less than 45°

13.

14.

(b) obtuse (c) a right angle (d) acute and larger than 45° 9.

Then, as a, b, c vary, K can assume every value in the interval

10.

11.

æ 1 2ö (a) çè , ÷ø 4 3

æ 1 4ö (b) çè , ÷ø 2 5

æ3 ö (c) çè ,1÷ø 4

æ 4 5ö (d) çè , ÷ø 5 4

(

)

(c)

(| x 0 | +| y 0 | –1)

x 20 + y 20 – 1

2

Let PQR be an acute-angled triangle in which PQ < QR From the vertex Q draw the altitude QQ1, the angle bisector QQ2 and the median QQ3, with Q1, Q2, Q3 lying on PR. Then,

All the vertices of a rectangle are of the form (a, b) with a, b integers satisfying the equation (a – 8)2 – (b – 7 )2 = 5. Then, the perimeter of the rectangle is (b) 22

(c) 24

(d) 26

16.

(| x 0 | +| y0 |)

2

A block of wood is floating on water at 0°C with volume V0 above water. When the temperature of water increases from 0 to 10°C, the change in the volume of the block that is above water is best described schematically by the graph.

(a)

(b)

(c)

(d)

–1

Let PQR be a triangle is which PQ = 3. From the vertex R, draw the altitude RS to meet PQ at S. Assume that RS = equals (a)

12.

(d)

(d) S ³ 148

PHYSICS

(b) x 20 + y 02 – 1 2

(c) 140 £ S < 148

(a) 20

Let x 0, y 0 be fixed real numbers such that x02 + y02 > 1. If x, y are arbitrary real numbers such that x2 + y2 £ 1, then the minimum value of (x – x0)2 + (y – y0)2 is (a)

(b) 6 £ S < 140

(c) PQ1 < PQ3 < PQ2 (d) PQ3 < PQ1 < PQ2 15.

l+m+n . a+ b+ c

(a) S < 6

(a) PQ1 < PQ2 < PQ3 (b) PQ2 < PQ1 < PQ3

Let a, b, c be the side-lengths of a triangle and l, m, n be the lengths of its medians. Put K=

Let S be the sum of the digits of the number 152 × 518 in base 10. Then,

5

(b)

3 and PS = QR. Then, PR 6

(c)

7

(d)

8

A 100 mark examination was administered to a class of 50 students. Despite only integer marks being given, the average score of the class was 47.5. Then, the maximum number of students who could get marks more than the class average is (a) 25 (b) 35 (c) 45 (d) 49

17.

A very large block of ice of the size of a volleyball court and of uniform thickness of 8 m is floating on water. A person standing near its edge wishes to fetch a bucketful of water using a rope. The smallest length of rope required for this is about (a) 3.6 m

(b) 1.8 m (c) 0.9 m

(d) 0.4 m

KVPY SOLVED PAPER 2018 STREAM : SA 18. A box filled with water has a small hole on its side near the bottom. It is dropped from the top of a tower. As it falls, a camera attached on the side of the box records the shape of the water stream coming out of the hole. The resulting video will show (a) the water coming down forming a parabolic stream (b) the water going up forming a parabolic stream (c) the water coming out in a straight line (d) no water coming out 19. An earthen pitcher used in summer cools water in it essentially by evaporation of water from its porous surface. If a pitcher carries 4 kg of water and the rate of evaporation is 20 g per hour, temperature of water in it decreases by DT in two hours. The value of DT is close to (ratio of latent of evaporation to specific heat of water is 540°C) (a) 2.7°C (b) 4.2°C (c) 5.4°C (d) 10.8°C 20. Two plane mirrors are kept on a horizontal table making an angle q with each other as shown schematically in the figure. The angle q is such that any ray of light reflected after striking both the mirrors returns parallel to its incident path. For this to happen, the value of q should be

q

(a) 30° (b) 45° (c) 60° (d) 90° 21. A certain liquid has a melting point of –50°C and a boiling point of 150°C. A thermometer is designed with this liquid and its melting and boiling points are designated at 0°L and 100°L. The melting and boiling points of water on this scale are (a) 25°L and 75°L, respectively (b) 0°L and 100°L, respectively (c) 20°L and 70°L, respectively (d) 30°L and 80°L, respectively

3 22. One can define an alpha-volt (a-V) to be the energy acquired by an a-particle when it is accelerated by a potential of 1 V. For this problem, you may take a proton to be 2000 times heavier than an electron. Then, (a) 1 a- V = 1 eV/4000 (b) 1 a- V = 2 eV (c) 1 a- V = 8000 eV

(d) 1 a- V = 1 eV

23. In a particle accelerator, a current of 500 mA is carried by a proton beam in which each proton has a speed of 3 × 107 m/s. The cross-sectional area of the beam is 1.50 mm2. The charge density in this beam (in C/m3) is close to (a) 10–8 (b) 10–7 (c) 10–6 (d) 10–5 24. Which of the following is not true about the total lunar eclipse? (a) A lunar eclipse can occur on a new moon and full moon day (b) The lunar eclipse would occur roughly every month, if the orbits of earth and moon were perfectly coplanar (c) The moon appears red during the eclipse because the blue light is absorbed in earth’s atmosphere and red is transmitted (d) A lunar eclipse can occur only on a full moon day 25. Many exoplanets have been discovered by the transit method, where is one monitors, a dip in the intensity of the parent star as the exoplanet moves in front of it. The exoplanet has a radius R and the parent star has radius 100 R. If I0 is the intensity observed on earth due to the parent star, then as the exoplanet transits (a) the minimum observed intensity of the parent star is 0.9 I0 (b) the minimum observed intensity of the parent star is 0.99 I0 (c) the minimum observed intensity of the parent star is 0.999 I0 (d) the minimum observed intensity of the parent star is 0.9999 I0

EBD_7839 4

30.

D

31.

32.

B C

The cardboard is kept at a suitable distance behind a transparent empty glass of cylindrical shape. If the glass is now filled with water, one sees an inverted image of the pattern on the cardboard when looking through the glass. Ignoring magnification effects, the image would appear as A D C

(b)

C

B

D

(a)

A

(d)

If a ball is thrown at a velocity of 45 m/s in vertical upward direction, then what would be the velocity profile as function of height? (Assume, g = 10 m/s2)

(a)

(b)

(c)

(d)

CHEMISTRY

A

B

29.

(c)

C

28.

A B

27.

A steady current I is set up in a wire whose crosssectional area decreases in the direction of the flow of the current. Then, as we examine the narrowing region, (a) the current density decreases in value (b) the magnitude of the electric field increases (c) the current density remains constant (d) the average speed of the moving charges remains constant Select the correct statement about rainbow. (a) We can see a rainbow in the western sky in the late afternoon (b) The double rainbow has red on the inside and violet in the outside (c) A rainbow has an arc shape, since the earth is round (d) A rainbow on the moon is violet on the inside and red on the outside Remote sensing satellites move in an orbit that is at an average height of about 500 km from the surface of the earth. The camera onboard one such satellite has a screen of area A on which the images captured by it are formed. If the focal length of the camera lens is 50 cm, then the terrestrial area that can be observed from the satellite is close to (a) 2 × 103 A (b) 106 A (c) 1012 A (d) 4 × 1012 A Letters A, B, C and D are written on a cardboard as shown in the figure below.

D

26.

KVPY SOLVED PAPER 2018 STREAM : SA

33.

The number of water molecules in 250 mL of water is closest to [Given, density of water is 1.0 g mL–1; Avogadro’s number = 6.023 × 1023] (a) 83.6 × 1023 (b) 13.9 × 1023 23 (c) 1.5 × 10 (d) 33.6 × 1023 Among the following, the correct statement is: (a) pH decreases when solid ammonium chloride is added to a dilute aqueous solution of NH3 (b) pH decreases when solid sodium acetate is added to a dilute aqueous solution of acetic acid (c) pH decreases when solid NaCl is added to a dilute aqueous solution of NaOH (d) pH decreases when solid sodium oxalate is added to a dilute aqueous solution of oxalic acid The solubility of BaSO4 in pure water (in gL–1) is closest to [Given; Ksp for BaSO4 is 1.0 × 10–10 at 25°C. Molecular weight of BaSO4 is 233 g mol–1] (a) 10 × 10–5 (b) 10 × 10–3 (c) 2.3 × 10–5 (d) 2.3 × 10–3

KVPY SOLVED PAPER 2018 STREAM : SA 34. Among the following, the incorrect statement is (a) No two electrons in an atom can have the same set of four quantum numbers (b) The maximum number of electrons in the shell with principal quantum number, n is equal to n2 + 2 (c) Electrons in an orbital must, have opposite spin (d) In the ground state, atomic orbitals are filled in the order of their increasing energies 35. A container of volume 2.24 L can withstand a maximum pressure of 2 atm at 298 K before exploding. The maximum amount of nitrogen (in g) that can be safely put in this container at this temperature is closest to (a) 2.8 (b) 5.6 (c) 1.4 (d) 4.2 36. The compound shown below O

5 39. The chlorine atom of the following compound

Cl Cl

Br Y

is (a) Both are chiral (b) Both are achiral (c) X is chiral and Y is achiral (d) X is achiral and Y is chiral 38. The most acidic proton and the strongest, nucleophilic nitrogen in the following compound O c b N N CH3 H H a N H respectively, are (a) Na – H; Nb (b) Nb – H; Nc a c (c) N – H; N (d) Nc – H; Na

a

Cl

that reacts most readily with AgNO3 to give a precipitate is (a) Cla (b) Clb (c) Clc (d) Cld 40. Among the following sets, the most stable ionic species are –

+

(a)

(b)

Br X

b

Cl O

NO 2

can be readily prepared by Friedel-Craft’s reaction between (a) benzene and 2-nitrobenzoyl chloride (b) benzyl chloride and nitrobenzene (c) nitrobenzene and benzoyl chloride (d) benzene and 2-nitrobenzyl chloride 37. The correct statement about the following compounds

d

c

and +

+

+

and

–

–

(c)

and

+

–

(d)

and

41. The correct order of energy of 2s-orbitals in H, Li, Na and K, is (a) K < Na < Li < H (b) Na < Li < K < H (c) Na < K < H < Li (d) H < Na < Li < K 42. The hybridisation of xenon atom in XeF4 is (a) sp3 (b) dsp 2 3 2 (c) sp d (d) d2 sp 3 43. The formal oxidation numbers of Cr and Cl in the ions Cr2O72– and ClO3–, respectively are (a) + 6 and +7 (b) +7 and +5 (c) +6 and +5 (d) +8 and +7 44. A filter paper soaked in salt X turns brown when exposed to HNO3 vapor. The salt X is (a) KCl (b) KBr (c) KI (d) K2SO4 45. The role of haemoglobin is to (a) store oxygen in muscles. (b) transport oxygen to different parts of the body. (c) convert CO to CO2. (d) convert CO2 into carbonic acid.

EBD_7839 6

KVPY SOLVED PAPER 2018 STREAM : SA BIOLOGY

46.

47.

48.

49.

50.

51.

52.

53.

54.

Which one of the following molecules is a secondary metabolite? (a) Ethanol (b) Lactate (c) Penicillin (d) Citric acid Lecithin is a (a) carbohydrate (b) phospholipid (c) nucleoside (d) protein The water potential (Y p) of pure water at standard temperature and atmospheric pressure is: (a) 0 (b) 0.5 (c) 1.0 (d) 2.0 Action potential in neurons is generated by a rapid influx of (a) chloride ions (b) potassium ions (c) calcium ions (d) sodium ions Erythropoietin is produced by (a) heart (b) kidney (c) bone marrow (d) adrenal gland Tendrils are modifications of (a) stem or leaf (b) stem only (c) leaf only (d) aerial roots only Which one of the following combinations of biomolecules is present in the ribosomes? (a) RNA, DNA and protein (b) RNA, lipids and DNA (c) RNA and protein (d) RNA and DNA Which one of the following proteins does not play a role in skeletal muscle contraction? (a) Actin (b) Myosin (c) Troponin (d) Microtubule Which one of the following reactions is catalysed by high-energy ultraviolet radiation in the stratosphere?

55.

56.

57.

58.

59.

60.

(a) O2 + O ¾® O3 (b) O2 ¾® O + O (c) O3 + O3 ¾® 3O2 (d) O + O ¾® O2 Which one of the following statements is true about trypsinogen? (a) It is activated by enterokinase. (b) It is activated by renin. (c) It is activated by pepsin. (d) It does not need activation. Which one of the following organisms respires through the skin? (a) Blue whale (b) Salamander (c) Platypus (d) Peacock Which one of the following human cells lacks a nucleus? (a) Neutrophil (b) Neuron (c) Mature erythrocyte (d) Keratinocyte The first enzyme that the food encounters in human digestive system is : (a) Pepsin (b) Trypsin (c) Chymotrypsin (d) Amylase Glycoproteins are formed in which one of the following organelles? (a) Peroxisome (b) Lysosome (c) Golgi apparatus (d) Mitochondria An example of nastic movement (external stimulus-dependent movement) in plants is (a) folding-up of the leaves of Mimosa pudica. (b) climbing of tendrils. (c) growth of roots from seeds. (d) growth of pollen tube towards the ovule.

PART-II (2 MARKS QUESTIONS) repetitions) such that the sum of any two adjacent m digits is odd. Then is equal to n (a) 9 (b) 12 (c) 15 (d) 18

MATHEMATICS 61.

What is the sum of all natural numbers n such that the product of the digits of n (in base 10) is equal to n2 – 10n – 36? (a) 12

62.

(b) 13

(c) 124

(d) 2612

Let m (respectively, n ) be the number of 5-digit integers obtained by using the digit 1, 2, 3, 4, 5 with repetitions (respectively, without

63.

The number of solid cones with integer radius and integer height each having its volume numerically equal to its total surface area is (a) 0

(b) 1

(c) 2

(d) infinite

KVPY SOLVED PAPER 2018 STREAM : SA 64. Let ABCD be a square. An arc of a circle with A as centre and AB as radius is drawn inside the square joining the points B and D. Points P on AB, S on AD, Q and R on arc BD are taken such that PQRS is a square. Further suppose that PQ area PQRS and RS are parallel to AC. Then, is area ABCD 1 1 1 2 (a) (b) (c) (d) 8 5 4 5 65. Suppose ABCD is a trapezium whose sides and heigh are integers and AB is parallel to CD. If the area of ABCD is 12 and the sides are distinct, then |AB – CD| (a) is 2 (b) is 4 (c) is 8 (d) cannot be determined from the data

7 69. A proton of mass m and charge e is projected from a very large distance towards an a-particle with velocity v. Initially a-particle is at rest, but it is free to move. If gravity is neglected, then the minimum separation along the straight line of their motion will be (a) e2/4pe0 mv2 (b) 5e2/4pe0 mv2 2 2 (c) 2e /4pe0 mv (d) 4e2/4pe0 mv2 70. A potential is given by V(x) = k(x + a) 2/2 for x < 0 and V(x) = k(x– a)2/2 for x > 0. The schematic variation of oscillation period T for a particle performing periodic motion in this potential as a function of its energy E is

T

(b)

(a)

PHYSICS 66. A coffee maker makes coffee by passing steam through a mixture of coffee powder, milk and water. If the steam is mixed at the rate of 50 g per minute in a mug containing 500 g of mixture, then it takes about t0 seconds to make coffee at 70° C when the initial temperature of the mixture is 25°C. The value of t0 is close to (ratio of latent heat of evaporation to specific heat of water is 540°C and specific heat of the mixture can be taken to be the same as that of water) (a) 30 (b) 45 (c) 60 (d) 90 67. A person in front of a mountain is beating a drum at the rate of 40 per minute and hears no distinct echo. If the person moves 90 m closer to the mountain, he has to beat the drum at 60 per minute to not hear any distinct echo.The speed of sound is (a) 320 ms–1 (b) 340 ms–1 –1 (c) 360 ms (d) 380 ms–1 68. A glass beaker is filled with water up to 5 cm. It is kept on top of a 2 cm thick glass slab. When a coin at the bottom of the glass slab is viewed at the normal incidence from above the beaker, its apparent depth from the water surface is d cm. Value of d is close to (the refractive indices of water and glass are 1.33 and 1.5, respectively) (a) 2.5 cm (b) 5.1 cm (c) 3.7 cm (d) 6.0 cm

T

E

E T

T

(c)

(d)

E

E

CHEMISTRY 71. Among the following, the species with identical bond order are (a) CO and O2– (b) O2– and CO 2 2– (c) O2 and B2 (d) CO and N+2 72. The quantity of heat (in J) required to raise the temperature of 1.0 kg of ethanol from 293.45 K to the boiling point and then change to liquid to vapour at that temperature is closest to [Given, boiling point of ethanol 351.45 K. Specific heat capacity of liquid ethanol 2.44 J g–1 k–1. Latent heat of vaporisation of ethanol 855 J g–1] (a) 142 × 102 (b) 9.97 × 102 5 (c) 142 × 10 (d) 9.97 × 105 73. A solution of 20.2 g of 1,2-dibromopropane in MeOH upon heating with excess Zn produces 3.58 g of an unsaturated compound X. The yield (%) of X is closest to [Atomic weight of Br is 80] (a) 18 (b) 85 (c) 89 (d) 30

EBD_7839 8 74.

KVPY SOLVED PAPER 2018 STREAM : SA The lower stability of ethyl anion compared to methyl anion and the higher stability of ethyl radical compared to methyl radical, respectively, are due to

number of different genotypes and phenotypes obtained respectively would be

(a) + I- effect of the methyl group in ethyl anion s ® p-orbital conjugation in ethyl radical

If the H+ concentration

(b) – I - effect of the methyl group in ethyl anion and s ® s* p-conjugation in ethyl radical

Consider the following vision defects listed in column I and II and the corrective measures in column III. Choose the correct combination.

(a) identical in BrB5 but non-identical in PCl5 (b) identical in BrB5 and identical in PCl5 (c) non-identical in BrB5 but identical in PCl5

76.

77.

1 2 3 4 5 6 7 8 9 10

(b) 0.25

(c) 0.50

(d) 0.75

A cross was carried out between two individuals heterozygous for two pairs of genes. Assuming segregation and independent assortment, the

(d ) (b) (c) (b) (a) (b) (c) (c) (c) (a)

11 12 13 14 15 16 17 18 19 20

(c) (d) (b) (a) (a) (a) (c) (d) (c) (d)

21 22 23 24 25 26 27 28 29 30

80.

Column I

Column II

Column III

P. Hypermetropia

(i) Nearsightedness

a. Convex lens

Q. Myopia

(ii) Farsightedness

b. Concave lens

(a) P–ii–b

(b) Q–i–b

(c) P–i–a

(d) Q–i–a

Which one of the following properties causes the plant tendrils to coil around a bamboo stick? (a) Tendril has spines.

If the genotypes determining the blood groups of a couple are IAIO and IAIB, then the probability of their first child having type O blood is: (a) 0

(d) 11 and 4

79.

The F-Br-F bond angles in BrF5 and the Cl-P-Cl bond angles in PCl5, respectively, are

BIOLOGY

(c) 9 and 4

of an aqueous solution is 0.001 M, then the pOH of the solution would be (a) 0.001 (b) 0.999 (c) 3 (d) 11

(d) + I- effect of the methyl group in ethyl anion and s ® s* conjugation in ethyl radical.

(d) non-identical in BrF5 and non-identical in PCl5

(b) 6 and 3

78.

(c) + I effect of the methyl group in both cases

75.

(a) 4 and 9

(b) The base of the tendril grows faster than the tip. (c) Part of the tendril in contact with the bamboo stick grows at a slower rate than the part away from it. (d) The tip of the tendril grows faster than the base.

ANS WER KEYS Part-I (a) 31 (a) 41 (a) (b) 32 (a) 42 (c) (d) 33 (d) 43 (c) (a) 34 (b) 44 (c) (d) 35 (d) 45 (b ) (b) 36 (a) 46 (c) (Non e) 3 7 (c) 47 (b ) (c) 38 (b) 48 (a) (d) 39 (a) 49 (d ) (a) 40 (d) 50 (b)

51 52 53 54 55 56 57 58 59 60

(a) (c) (d) (b) (a) (b) (c) (d) (c) (a)

61 62 63 64 65 66 67 68 69 70

Part-II (b) 71 (c) 72 (b) 73 (d) 74 (b) 75 (b) 76 (c) 77 (b) 78 (b) 79 (b) 80

(c) (d ) (b ) (a) (d ) (a) (c) (d ) (b ) (c)

HINTS & SOLUTIONS PART-I

(m – n)a + (m – n)(m + n –1)

MATHEMATICS 1.

Þ 2a + (m + n –1) d = –2

(d) We have, a4 + b4 < 1 and a2 + b2 > 1 The graphs of

x2 + y2

= 1 and

x4 + y4 =

Y

1

=

x2+y2=1

X

m+n [(2a + (m + n - 1)d ] 2

m+n (- 2) = -(m + n) 2

(b) (I) Statement I is false. Consistent Linear equations may have unique of infinite solutions. (II) Statement II is also false Q 132 + 142 = 365

5.

(a) Let P(x) = anxn + an –1xn–1 + an – 2xn–2

Y'

+ ...+ a1x + a0 a0, a1, a2 ...anÎI

It is clear from graph, there are many positive real numbers (a, b) satisfying a4 + b4 < 1 and a2 + b2 > 1

Given, P(2) = 2 and P(4) = 5 \2 = an 2n + an –1 2n –1 + an – 2 2n – 2

(b) Given,

+a1 2 + a0 ...(1)

x4 – x2 + 2x – 1 = 0 Þ 0 2 2 Þ (x – x + 1) (x + x – 1) = 0 x4 – (x

– 1)2 =

5 = an 4n + an –1 4n –1

Þ x2 – x + 1 = 0 or x2 + x – 1 = 0

+ an – 2 4n – 2 + ... + 4a1 + a0 ....(2)

Here, x2 – x + 1 = 0 has no real roots. 3.

[m ¹ n]

x4+y4=1 4.

2.

Sm + n =

\

are given below:

X'

d = – (m – n) 2

On subtracting eq. (1) from eq. (2), we get

and x2 + x – 1 = 0 has two real roots (c) Given, Sm = n and Sn = m

(

)

(

3 = an 4n - 2n + a n -1 4n -1 - 2n -1

m S m = [2a + (m - 1)d ] = n 2

...(i)

n Sn = [2a + (n - 1)d ] = m 2

...(ii)

On subtracting Eq. (ii) from Eq. (i), we get (m - n ) d 2a + [ m( m - 1) - n(n - 1)] = n - m 2 2

)

+... + 2a1 Clearly, LHS is odd number and RHS is even number. 6.

\ no polynomials exists. (b) Four digit numbers which are divisible by 7 are 1001, 1008, 1015,...,9996. Hence, total number of such numbers = 1286

EBD_7839 10

KVPY SOLVED PAPER 2018 STREAM : SA Median

When height of cylinder is doubled, then

th

2 volume of solid, V2 = 2pr 2h + pr 3 3

th

æNö æN ö ç ÷ observation + ç + 1÷ observation 2 è2 ø =è ø 2

2 2 pr 2 h + p r 3 V2 3 3 \ = = V1 2 pr 2 h + 2 pr 3 3

[\ N is even] Median

=

=

=

=

7.

æ 1286 ö çè ÷ 2 ø

th

æ 1286 ö + 1÷ observation + ç è 2 ø 2

th

2 r 3 =3Þh=r Þ 2 2 2 3 h+ r 3

observation

2h +

643th + 644th 2

When the radius is doubled, then volume of solid,

(1001 + (642)7) + (1001 + (643)7)

V3 = 4p r 2 h +

2 2 (1001) + 7 ( 642 + 643)

16p r 3 3

16 4h + r V3 3 = 4h + 8h \ = 2 V1 h+h h+ r 3

2

= 1001 + 4497.5 = 5498.5 (c) Let the height and radius of cylinder are h are r, respectively. \ volume of cylinder = pr2h

é r hù êëQ 3 = 2 úû

=6

Hence, volume is increased by 500%. (c) Let DPQR

8.

r

r

Given, QR2 + PR2 = 5PQ2 Median PM and QN intersect at G.

h

Þ PN = NR =

h

QM = MR =

r

1 PR 2

&

1 QR 2

P and volume of hemisphere =

N

2 3 pr 3

2 \volumeof solid,V1 = pr 2h + pr 3 3

G Q

M

R

KVPY SOLVED PAPER 2018 STREAM : SA

11 and median of DABC : AD = l, BE = m, CF = n AD is median,

2 1 QG = QN , GM = PM 3 3 2

æ2 ö æ1 ö Þ QG 2 + GM 2 = ç QN ÷ + ç PM ÷ è3 ø è3 ø

2

\ AD
c 3 3

4 ( l + m + n) > a + b + c 3

l + m+ n 3 > ...(2) a+b+c 4 \ From eqs. (1) and (2), we get Þ

=

1 é10 PQ 2 + 7QR 2 - 2PR 2 ù ê ú 9ë 4 û

=

1 é 2 5 PQ - PR + 7QR ê 9ê 4 ë

=

1 é 2QR2 + 7QR 2 ù 1 2 2 ê ú = QR = QM 9ë 4 4 û

(

2

2

)

2

Þ

ù ú úû

l + m+ n æ 3 ö Î ç ,1÷ a+b+c è4 ø

10. (a) Let A(x0, y0) Given x2 + y2 < 1 Let any arbitrary point B(x, y) Y(0,1) A(X0,Y0)

Þ QG 2 + GM 2 = QM 2

B (X,Y)

\ ÐQGM = 90o

9.

(c) Let in DABC BC = a, AC = b, AB = c

X'

A

1 O

E

F G B

D

C

Y' AB2 = (x – x0)2 + (y – y0)2

(1,0)

X

EBD_7839 12

KVPY SOLVED PAPER 2018 STREAM : SA Also AB2 = (OA – OB)2

(

Þ AB2 =

Taking log base 10 both side

)

log10 N = log10 9 + log10 520

2

x02 + y02 –1 [Q OB = 1]

\ Minimum value of (x – x0)2 + (y – y0) is

(

)

x02 + y02 - 1

2

= 2log10 3 + 20 log10 5 = 2 × 0.4771 + 20 × (1 – 0.3010) = 14 characters value

11. (c) Given, in DPQR, PQ = 3

Hence, the number has 15 digits S = Sum of digits of the number

Altitude RS = 3 Þ PS = QR

Now, the last digit of N is 5.

R

\ minimum value of S = 1 + 5 = 6

and maximum value of S = 9 × 14 + 5 = 126 + 5 = 131 \ 6 < S < 140

P

S

Q

14. (a) Given, PQR is an acute angle triangle.

In DSQR, QR2 = SR2 + SQ2

Þ PS = 2

( 3)

2

+ ( QP - PS )

Q 2

[Q SQ = PQ – PS] PS = 3 + ( 3 - PS ) 2

PS 2 = 3 + 9 - 6 PS + PS 2 Þ PS = 2 In DPRS, PR = PS + RS = ( 2) + 2

2

2

2

( 3)

2

2375 » 49 48 13. (b) Given number, N = 152 × 518 N = 32 × 52 × 518 N = 9 × 520

P

= 4+ 3

Þ PR = 7 12. (d) Total number of students = 50 Average marks of student = 47.5 \ Total marks of students = 50 × 47.5 = 2375 Now, the student get integer marks Hence, for maximum number of students who could get marks more than 47.5, we will divide total marks by 48. \

p

r 2

Q1Q2 Q3 q

R

\ PQ < QR

Þ ÐQRP < QPR Again, PQ3 =

...(1)

1 PR 2

Also, PQ2 : Q2 R = r : p (angle bisector theorem). æ r ö Þ PQ2 = ç ÷ PR èr+ pø

But, r < p Þ PQ2
0

n=

Case II: When the digits are not repeated. The possibility of arrangement is odd even odd even odd = 3 ´ 2 ´ 1´ 2 ´ 1 = 12

10 ± 100 + 144 2

Þ n = 12

n = 5 ± 61

(

) (

\ n Î - ¥,5 - 61 È 5 + 61, ¥

)

But n is positive integer. \ n > 13 When n is a two digit number then maximum product = 9 × 9 = 81 \ n2 - 10n - 36 £ 81

n2 - 10n - 117 £ 0 \ n Î [5 – 142,5 + 142] Here n is taken as two digit number \ n Î[13,17 ] = 13,14,15,16 \ product of digits = 3, 4, 5, 6 (respectively) When put n = 13 in the given product,

13 - 10 ´ 13 - 36 = 169 - 166 = 3 \ n = 13 satisfies 62. (c) Here, m is 5-digits number using digits 1, 2, 3, 4, 5 with repetition such that sum of two adjacent digits is odd and n is 5-digits number using digits 1, 2, 3, 4, 5 without repetition such that sum of any two adjacents digits is odd. Sum of two digits are odd if one is even and other is odd. Even = 2, 4 Odd = 1, 3, 5 Case I : When the digits are repeated. Two possibilities (a) odd even odd even odd = 3 ×2 × 3 × 2 × 3 = 108 2

19

\

m 180 = = 15 n 12

63. (b) Let height and radius of cone is h and r respectively such that h, r Î I Q volume of cone = surface area of cone 1 \ pr 2 h = prl + pr 2 3 1 Þ pr 2 h = pr h 2 + r 2 + pr 2 3

1 Þ rh = h2 + r 2 + r 3

[ r ¹ 0]

Þ rh - 3r = 3 h2 + r 2

Þ r 2 h2 + 9r 2 - 6hr 2 = 9h2 + 9r 2

(

)

Þ h 2 r 2 - 9 = 6 hr 2 Þ h =

æ r2 ö Þ h = 6ç 2 è r - 9 ÷ø

6r 2 r2 - 9

Þ h = 6+

54 r2 - 9

h and r are integers Here (r2 – 9) should be factor of 54. \ r2 – 9 = 1, 2, 3, 6, 9, 18, 27, 54. Þ r2 = 10, 11, 12, 15, 18, 27, 36, 63. \ r = 6 (only possible value) \ h = 6+

54 = 6 + 2 =8 36 – 9

\ r = 6, h = 8

EBD_7839 20

KVPY SOLVED PAPER 2018 STREAM : SA

64. (d) Given, ABCD is a square. D

C R

65. (b) We have, ABCD is a trapezium. AB is parallel to CD. Area of trapezium = 12

N

D

b

C

Q

S

h

M A

P

B

ÐCAP = ÐMAP = 45o ÐAMP = ÐMPQ = 90(alternate angles)

A

b

E

Þ

1 ´ h ( AB + CD ) = 12 2

a

B

24 h

Þ ÐMPA = 45o

Þ AB + CD =

\ AM = MP = QN

Sides and height of trapezium are integer.

PQRS is a square,

\ h is a factor of 24

Q MN = PQ = PS

h = 1, 2, 3, 4, 6, 8, 12, 24

Also PS = 2 PM = 2 AM

AB + CD = 24, 12, 8, 6, 4, 3, 2, 1 But AB + CD > h

Þ AN = AM + MN = 3 AM

AB + CD = 24, 12, 8, 6

In DANQ,

In DBEC

AQ2 =

AN2 +

QN2

BEC is a right angled triangle

Þ AB 2 = ( 3 AM ) + AM 2 [Q AQ = AB ]

\ h must be 3 or 4

Þ AB 2 = 10 AM 2

Þ AB + CD = 8

2

Now, \

=

PS2 = (2AM)2 = 4AM2

When h = 3, BE = 4, CE =5

Þ AE + BE + AE = 8

area of square PQRS area of square ABCD

Þ 2 AE = 8 - BE = 8 - 4

PS 2 4 AM 2 2 = = 2 2 AB 10 AM 5

\ AB = 4 + 2 = 6 Þ and CD = 2

Þ AE = 2

\ | AB - CD |=| 6 - 2 |= 4

KVPY SOLVED PAPER 2018 STREAM : SA

21

PHYSICS 66. (b) According to principle of calorimetry, Heat lost by steam = Heat gained by mixture i.e., Heat of condensation of steam + Heat given by water formed = Heat gained by mixture or, m.L + mswDT = M.smDT Þ mL + msw(100 – 70) = 500 × sw × (70 – 25) 500 ´ sw ´ 45 500 ´ 45 = L + 30sw æ L ö + 30 ÷ ç s è w ø 500 ´ 45 » 40g or, m = (540 + 30) i.e., 40 g steam is condensed in the process of heating mixture from 25°C to 70°C.

Þ m=

As per question, in 1 min, 50 g of steam is condensed. \ Time to condensed 40 g of ice, t0 =

40 ´ 60 s = 48s 50

Very close to answer option (b) 45 s. 67. (c) According to question, drummer does not hear any echo i.e., time between two successive wavefronts = time taken by a wavefront to return back to drummer.

Drummer

Mountain 2x = time interval between two successive v wavefronts. v = speed of sound, then

2 x 60 = v 40

... (ii)

Substituting value of x from eq. (i) in eq. (ii) 2x – 180 = v Þ

3 v - 180 = v 2

i.e., speed of sound, v = 360 ms–1 68. (b) Given: d1 = 5 cm, m1 = 1.33 d2 = 2 cm, m2 = 1.5 d1 and d2 are the thickness of slabs of medium with refractive index m1 and m2, respectively. using formula, d =

d1 d2 + + ..... μ1 μ2

Apparent depth, d =

5 2 + 1.33 1.5

= 5.088 cm = 5.1 cm 69. (b) According to energy conservation principle, Initial kinetic energy = Potential energy at minimum separation r 1 2 1 2e 2 mv = × 2 4pe 0 r

where, m = reduced mass of system

x

Case I,

2( x - 90) 60 = v 60

Case II,

... (i)

=

or,

m × 4m . m + 4m

1 æ m × 4m ö 2 1 2e2 v = × ç ÷ 2 è m + 4m ø 4pe0 r

or, minimum separation, r=

5e2 4pe0 mv 2

70. (b) Given, potential function for the oscillating particle

EBD_7839 22

KVPY SOLVED PAPER 2018 STREAM : SA ì k ( x + a )2 , ï ï 2 V (x) = í 2 ï k ( x - a) , ïî 2

The electronic configuration of O 22 - (18) is x0

1 B.O = (10 - 8) = 1 2

So, potential energy of the particle of mass m. ì km ( x + a)2 , ï ï 2 U ( x) = í 2 ï km ( x - a) , ïî 2

The electronic configuration of O -2 (17) is

x0

p*2p2x = p *2 p1y 1 3 B.O = (10 - 7) = = 1.5 2 2 The electronic configuration of B2 (10) is

dU ìkm ( x + a), x < 0 =í dx îkm ( x - a), x > 0

s1s 2s *1s 2 s 2s 2s *2s 2 p 2 p1x = p 2 py1

dU If = 0, when x = ± a. dx

Now,

d 2U dx 2

= km > 0

So, at x = ± a particle is in unstable equilibrium. Hence, for – a > x and x > a particle is unbounded. In region, – a £ x £ a, time period of particle reduces from a maximum. Hence, graph (b) correctly depicts variation of oscillating period T as a function of energy E. CHEMISTRY 1 71. (c) B.O. = ( Nb - Na ) 2 where, Nb = electrons in bonding orbitals Na = electrons in antibonding orbitals. The electronic configuration of CO (14) is

s1s s*1s 2 s 2s 2 s *2 s 2 2 pz2 p 2 px2 = p 2p2y \

1 2 B.O = [6 - 4] = = 1 2 2

1 6 B.O = (10 - 4) = = 3 2 2

Electronic configuration of N +2 (13) is s1s 2s *1s 2 s2 s 2s *2s 2 p2 p2x = p2 py2 s2p1z 1 5 B.O = [9 - 4] = = 2.5 2 2 Thus, option (c) is correct. 72. (d) Given, mass of ethanol = 1kg = 1000 g Latent heat of vaporisation of ethanol = 855 Jg–1 Specific heat capacity of ethanol = 2.44 J/gK–1 Heat, q = mcDT + heat of vaporisation = 1000 × 2.44 (351. 45 – 293.45 + 855 × 1000 J) = 9.97 × 105 J

73. (b)

Moles of 1, 2-dibromo propane =

20.2 = 0.01 mole 202

KVPY SOLVED PAPER 2018 STREAM : SA

23

358 Moles of prop-1-ene = = 0.085 mole 42

% yield =

0.085 ´ 100 = 85% 0.1

C

C

H

76. (a) The genotypes of offspring of parents having IAIO and IAIB blood groups are: A O

Genotype

¬¾® H

C

CH2

H

H

Hyper conjugation in ethyl radical

A B

I I

×

A A

A O

A B

B O

I I I I I I I I Phenotype Blood Group A Blood Group AB Blood Group A Blood Group B Offspring

From the above cross, it is shown that none of the offspring will be of blood group O. \ The probability of their first child having type O blood is zero.

H

H

I I

Parents

74. (a) The lower stability of ethyl anion (CH3CH2–) compared to methyl anion (CH3–) is because of + I-effect of methyl group of ethyl anion. The higher stability of ethyl radical compared to methyl radical is due to s and p-orbital conjugation is ethyl radical. H

BIOLOGY

77. (c) In the given question, both parents are heterozygous for two pairs of genes. This means the cross is a dihybrid cross. Let us assume a dihybrid cross, Pure breeding – Yellow round × Wrinkled green traits seeds seeds (YYRR) (yyrr)

75. (d)

F1

– Yellow round seeds (YyRr) Gametes YR Yr

Due to the presence of lone pair at axial position in BrF5, bond angles will not remain identical as in regular octahedron (i.e., 90°).

Cl Cl

P

Cl Cl

Cl

YR Yr

Trigonal bipyamidal

Trigonal bipyramidal geometry does not have all identical angles, as angle between equatorial bonds is 120° and axial-equatorial bond is 90°.

yR yr

yR

yr

YR

Yr

yR

yr

YYRR yellow round

YYRr yellow round

YyRR yellow round

YyRr yellow round

YYRr Yyrr yellow yellow round wrinkled YyRR YyRr yellow yellow round round

YyRr yellow round yyRR green round

Yyrr yellow wrinkled

YyRr Yyrr yellow yellow round wrinkled

yyRr green round

yyrr green wrinkled

yyRr green round

EBD_7839 24 The genotype ratio is 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1 The phenotypic ratio is 9 : 3 : 3 : 1 \ The number of different genotypes and phenotypes obtained would be 9 and 4, respectively. 78. (d) The H+ ion concentration of an aqueous solution is 0.001 M or 1 × 10–3 M Since we know that pH = – log[H+] Using this equation, by putting in the values pH = – log 10–3 = – (–3) log 10 = 3 pH = 3 We know that pOH = 14 – pH = 14 – 3 = 11 \ pOH = 11 79. (b) Hypermetropia or far sightedness is a vision condition in which nearby objects appear blur. It is corrected by using convex lens.

KVPY SOLVED PAPER 2018 STREAM : SA Myopia or near sightedness is a condition in which close objects apear clearly but far ones do not. It is corrected by using concave lens. 80. (c) The tendrils are sensitive to touch. When they come in contact with any support, the part of the tendril in contact with the object does not grow rapidly as the part of the tendril away from the object. This causes the tendril to circle around the object and thus cling to it. This process is known as positive thigmotropism. Thigmotropism is the directional response of a plant organ to touch or physical contact with a solid object. This differential response is generally caused by the induction of some differential growth.

1 Time : 3 Hours

Stream-SA

Maximum Marks : 100

INSTRUCTIONS 1. 2. 3. 4.

There are 80 questions in this paper. The question paper contains two parts; Part–I (1 Mark Questions) and Part–II (2 Marks Questions). There are four sections; Mathematics, Physics, Chemistry and Biology in each part. There are four options given with each question, only one of them is correct. For each incorrect answer 0.25 Mark in Part–I and 0.5 Mark in Part–II will be deducted. PART-I (1 MARK QUESTIONS) MATHEMATICS

1.

2.

3.

The number of points (x, y) having integral 2 coordinates satisfying the condition x + y 2 < 25 is (a) 69 (b) 80 (c) 81 (d) 77 Find the sum of all the real roots of the equation P (x) = 2 x 98 + 3 x 97 + 2 x 96 +... + 2 x + 3 = 0. (a) 7 (b) 3 (c) 2 (d) None of these Suppose P is an interior point of a triangle ABC such that the ratios

d ( A, BC ) d ( B, CA) d (C , AB) , , are all equal. d ( P, BC ) d ( P, CA) d ( P, AB ) Here d(X, Y Z) denotes the perpendicular distance from a point X to the line Y Z. Then the common value of these ratios is. 2 3 The mean of fifteen different natural numbers is 13. The maximum value for the second largest of these numbers is (a) 46 (b) 51 (c) 52 (d) 53 A sphere with centre A and radius 7 cm and a larger sphere with centre B touches each other, two common planks to both the spheres have an angle of 60º in between them (refer the figure). Then find the radius of larger sphere.

(a) 2

4.

5.

(b) 3

(c) 4

(d)

B

A

6.

7.

8.

(a) 30 3 (b) 21 (c) 20 3 (d) 30 Let S(M) denote the sum of the digits of a positive integer M written in base 10. Let N be the smallest positive integer such that S(N) = 2013. Then the value of S(5N + 2013) (a) 15 (b) 17 (c) 25 (d) None of these If three positive real numbers a, b, c are in A. P. such that abc = 4, then the minimum possible value of b is (a) 23/2 (b) 22/3 (c) 21/3 (d) 25/2 If the polynomial of the lowest degree with integer coefficients one of whose roots is 2 + 3 3 is f(x). The degree of f(x) is:

(a) 6

(b) 7

(c) 5

(d) 4

EBD_7839 2 9.

10.

KVPY-SA In a triangle ABC, D is a point on BC such that AD is the internal bisector of ÐA. Suppose ÐB = 2ÐC and CD = AB. Then ÐA = (a) 60° (b) 70° (c) 45° (d) 72° Number of pairs of positive integers a, b such a2 + b b2 + a and 2 are both integers is: 2 b -a a -b (a) 5 (b) 7 (c) 10 (d) 6 The diagonals AC an d BD of a cyclic quadrilateral ABCD meet at right angles in E. Let R is radius of circumscribing circle and EA2 + EB2 + EC2 + ED2 = n·R2. Then n is:

that 11.

(a) 4 12.

(b) 3

(c) 11

(d) 9

How many pairs of positive integers m, n satisfy

1 4 1 + = where n is an odd integer m n 12

less than 60? (a) 4 13.

14.

15.

(b) 7

PHYSICS 16. A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. If 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific gravity of mercury = 13.6) (a) 2.20 cm (b) 0.22 cm (c) 22 cm (d) 0.02 cm 17. Water flows through a tunnel in a reservoir of dam towards the turbine installed in the power plant. This is situated h m below the reservoir. If the ratio of cross-sectional areas of the tunnel at the reservoir and power station is h, find the speed of the water entering into the turbine. (a) h

(d) 3

Positive integers from 1 to 21 are arranged in 3 groups of 7 integers each, in some particular order. Then the highest possible mean of the medians of these 3 groups is (a) 16 (b) 12.5 (c) 11 (d) 14 ABCD is a square and E is the intersection point of the diagonals. If N is any point on AE. Consider the following statements: (I) AB2 – BN2 = AN · NC (II) AN2 + NC2 = 2 BN2 Which of the following is correct (a) Only I is true (b) Only II is true (c) I & II are true (d) I & II are false The perimeter of rhombus is 2p cm, the sum of its diagonals is m cm. Then area of rhombus is(a)

m2 - p 2 4

(b)

m2 - p 2 2

(c)

1 (m. p) 2

(d) m2 – p2

2

h +1

(b)

2gh h2 - 1

h2 - 1 h2 - 1 2gh 18. A body cools in 7 minute from 60°C to 40°C. What will be its temperature after the next 7 minute? The temperature of the surroundings is 10°C. Assume that Newton’s law of cooling holds good throughout the process. (a) 8°C (b) 24°C (c) 28°C (d) 32°C 19. A child running a temperature of 101°F is given an antipyrin which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation by the drug. Assume the evaporation mechanism to be the only way to which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about 580 cal g–1. (a) 4.31 g/min (b) 4.31 g/sec (c) 4.41 g/sec (d) 3.41 g/min 20. A person speaking normally produces a sound intensity of 40 dB at a distance of 1 m. If the threshold intensity for reasonable audibility is 20 dB, the maximum distance at which he can be heard clearly is (a) 4 m (b) 5 m (c) 10 m (d) 20 m (c) h

(c) 5

2gh

2gh

(d) h

MOCK TEST-1

3

21. In a lamp and scale arrangement to measure small deflection, the arrangement is shown in the figure SS¢ is the glass scale placed at a distance of 1 m from the plane mirror MM and I is the position of the light spot formed after reflection from the undeflected mirror MM. The mirror is deflected by 10° and comes to the deflected position M ‘M’. The distance moved by the spot on the scale (IR) is :

(a) 24.6 cm (b) 36.4 cm (c) 46.4 cm (d) 34.6 cm 22. A steady current flows in a metallic conductor of non-uniform cross-section. The quantity/ quantities constant along the length of the conductor is/are (a) current, electric field and drift speed (b) drift speed only (c) current and drift speed (d) current only 23. When white light passes through a hollow prism then, (a) There is no dispersion and no angular deviation. (b) There is dispersion but no deviation. (c) There is angular deviation but no dispersion. (d) There is dispersion as well as deviation. 24. A certain planet completes one rotation about its axis in time T. The weight of an object placed at the equator on the planet's surface is a fraction f (f is close to unity) of its weight recorded at a latitude of 60°. The density of the planet (assumed to be uniform) is given by : æ 4 - f ö 3p (a) ç è 1 - f ø÷ 4GT 2

æ 4 - f ö 3p (b) çè + ø÷ 1 f 4GT 2

æ 4 - 3 f ö 3p æ 4 - 2 f ö 3p (c) çè 1 - f ø÷ (d) çè - f ø÷ 2 1 4GT 3GT 2 25. A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F(t) = F0e–bt in the x direction. Its speed v(t) is depicted by which of the following curves?

(a)

(b)

(c)

(d)

26. A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is (a)

a gk

(b)

a 2 gk

2a a (d) gk 4 gk 27. A ballet dancer, dancing on a smooth floor is spinning about a vertical axis with her arms folded with an angular velocity of 20 rad/s. When she stretches her arms fully, the spinning speed decreases by 10 rad/s. If I is the initial moment of inertia of the dancer, the new moment of inertia is (a) 2I (b) 3I (c)

(c) I/2 (d) I/3 28. The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun ? (a) 1.319 parsec, 2.638 second of arc (b) 3.119 parsec, 6.238 second of arc (c) 1.913 parsec, 2.863 second of arc (d) 13.19 parsec, 26.38 second of arc 29. The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and that emitted by the north star has the maximum value at 350 nm. If these stars behave like black bodies, then the ratio of the surface temperature of the sun and north star is (a) 1.46 (b) 0.69 (c) 1.21 (d) 0.83

EBD_7839 4 30.

KVPY-SA When a block of iron floats in mercury at 0°C, a fraction k1 of its volume is submerged , with at a temperature 60°C, a fraction k2 is seen to be submerged. If the coefficient of volume expansion of iron is gFe and that of mercury is gHg, then the ratio k1 / k2 can be expressed as :

1 + 60 g Fe (a) 1 + 60 g Hg

1 – 60 g Fe (b) 1 + 60 g Hg

1 – 60 g Fe (c) 1 – 60 g Hg

(d)

36. 37.

38.

31.

32.

33.

35.

C

CH

Si4+ ions are isoelectronic.

CH

(b)

(c)

(d)

1 + 60 g Fe

The value of ionic radii of these ions would be in the order : (a) Na + > Mg 2+ > Al3+ > Si 4+ (b) Na + < Mg 2+ < Al3+ < Si 4+ (c) Na + > Mg 2+ > Al3+ < Si 4+ (d) Na + < Mg 2+ > Al3+ > Si 4+ The heat of combustion of C, S and CS2 are –393.3 kJ, –293.7 kJ, and –1108.76 kJ. What will be the heat of formation of CS2? (a) –128.06 kJ (b) +970 kJ (c) +1108.7 kJ (d) +12 kJ The IUPAC name of O CH3

34.

Mg2+, Al3+ and

(a)

1 + 60 g Hg

CHEMISTRY Na+,

Which has the highest e/m ratio ? (a) He2+ (b) H+ (c) He+ (d) D+ Which of the following compound is planar and non-polar? (a) XeO4 (b) SF4 (c) XeF4 (d) CF4 Which of the following will show aromatic behaviour ?

39. 40.

41.

42.

CH3 is:

C2H5 CH3 (a) 3-(Methylethyl) pentan-2-one (b) 3-(Methylethyl) pentan-4-one (c) 3-Ethyl-4-methylpentan-2-one (d) 3-Ethyl-2-methylpentan-4-one The correct order of increasing pH values of the aqueous solutions of baking soda, rock salt, washing soda and slaked lime is: (a) Baking Soda < Rock Salt < Washing Soda < Slaked lime (b) Rock Salt < Baking Soda < Washing Soda < Slaked lime (c) Slaked lime < Washing Soda < Rock Salt < Baking Soda (d) Washing Soda < Baking Soda < Rock Salt < Slaked lime The weight of oxalic acid required to neutralise 100 mL of normal NaOH is : (a) 6.3 g (b) 126 g (c) 530 g (d) 63 g

43. 44. 45.

The oxidant which cannot act as reducing agent is: (a) SO2 (b) NO2 (c) CO2 (d) C1O2 Number of moles of NaOH present in 2 litre of 0.5 M NaOH is : (a) 1.5 (b) 2.0 (c) 1.0 (d) 2.5 The value of Planck's constant is 6.63 × 10–34 Js. The velocity of light is 3.0 × 108 m s–1. Which value is closest to the wavelength in nanometers of a quantum of light with frequency of 8 × 1015 s–1 ? (a) 3 × 107 (b) 2 × 10–25 –18 (c) 5 × 10 (d) 4 × 101 The closed containers of the same capacity and at the same temperature are filled with 44 g of H2 in one and 44 g of CO2 in the other. If the pressure of carbon dioxide in the second container is 1atm, then pressure of hydrogen in the first container would be: (a) 1 atm (b) 10 atm (c) 22 atm (d) 44 atm Hyperconjugation involves overlap of the following orbitals: (a) s-s (b) s - p (c) p- p (d) p-p Which of the following metals burn in air at high temperature with the evolution of heat ? (a) Cu (b) Hg (c) Pb (d) Al The bond order in N +2 is: (a) 1.5 (b) 3.0 (c) 2.5 (d) 2.0 BIOLOGY

46.

Hormone which stimulates pancreas to secrete enzymes is: (a) Enterocrinin (b) CCK-PZ (c) Gastrin (d) Secretin

MOCK TEST-1 47. Which of the following accelerates extrinsic or tissue thromboplastin formation? (a) Factor VII (proconvertin) (b) Factor XI (plasma thromboplastin antecedent) (c) Heparin (d) Calcium 48. Two pigment system theory of photosynthesis was proposed by or concept of evidence for existence of two photosystems in photosynthesis was given by: (a) Hill (b) Blackmann (c) Emerson (d) Arnon 49. A person suffers a permanent occlusion of a blood vessel to the right side of the cerebrum. The left side of the brain is unaffected. Which of the following could be a direct permanent result of this stroke? 1. Right arm weakness. 2. Loss of left knee-jerk reflex. 3. Blindness in the left visual field without complete blindness in either eye. (a) Only 1 (b) Only 2 (c) Only 3 (d) Both 1 and 2 50. Aquatic photodiffraction produces zones: (a) Euphotic, disphotic and aphotic (b) Aphotic, euphotic and disphotic (c) Euphotic, aphotic and disphotic (d) Disphotic, aphotic and euphotic 51. You experience a rapid increase in heart rate, your muscles tense, and your breathing becomes shallow. These changes in your physiology are most likely mediated by: (a) Neurotransmitters (b) Hormones (c) Target cells (d) Receptors 52. Which of the following plant tissue culture shows totipotency? (a) Meristem (b) Collenchyma (c) Sieve tube (d) Xylem vessel 53. The controlling centre of autonomic nervous system is: (a) Cerebrum (b) Spinal cord (c) Cerebellum (d) Medulla oblongata

5 54. The acidic condition within the lysosome is maintained by: (a) Digestive enzymes synthesised on RER (b) Pumping Cl– ion out of lysosome (c) Pumping protons (H+) into the lysosome (d) All of the above 55. What is the role of a calcium ion (Ca2+) in muscle contraction? (a) It opens up a binding site for a myosin head on a thin filament. (b) It opens up a binding site for an actin molecule on a thick filament. (c) It binds to a myosin head. (d) It release from a myosin head initiates a power stroke. 56. Person who is habitual of drinking (alcoholics) is always short of which vitamin? (a) K (b) B (c) C (d) D 57. Take the odd one out. (a) Rabies, Influenza, AIDS (b) Amoebiasis, Giardiasis, Trypanosomiasis (c) Taeniasis, Ascariasis, Elephantiasis (d) Cancer, Tuberculosis, Tetanus 58. The process of migration of chloride ions from plasma to RBC and carbonate ions from RBC to plasma is: (a) Chloride shift (b) Ionic shift (c) Atomic shift (d) Na+ pump 59. Which of the following food chain produces a greater availability of food to next trophic level? (a) Diatoms ® Fish ® Large fish ® Birds ® Man (b) Green plant ® Rabbit ® Man (c) Green plant ® Grasshopper ® Frog ® Snake ® Peacock (d) Green plant ® Rabbit ® Wolf ® Man 60. The major lymphatic vessel in the body is the thoracic duct. It empties into: (a) liver (b) kidney (c) spleen (d) a vein other than the portal vein

EBD_7839 6

KVPY-SA PART-II (2 MARKS QUESTIONS) one fourth the circumference of the tube.

MATHEMATICS 61.

(a)

(c)

62.

63.

Find the angle q that the radius to the interface makes with the vertical in equilibrium position. If the whole is given a small displacement from its equilibrium position, and the resulting oscillations are simple harmonic. Find the time period of these oscillations.

The smallest positive integer n, so that 999999 · n = 111 …. 11 is: 1054 + 1

(b)

9(106 - 1)

1054 - 1 9(106 - 1)

1054 - 1

(d) None of these 9(106 + 1) There are four bus routes between A and B; and three bus routes between B and C. A man can travel round-trip in number of ways by bus from A to C via B. If he does not want to use a bus route more than once, in how many ways can he make round trip? (a) 72 (b) 144 (c) 14 (d) 19 A piece of paper is in the form of a sector, making an angle a. The paper is rolled to form a right circular cone of radius 5 cm and height 12 cm. Then the value of angle a is 10p 9p 5p 6p (b) (c) (d) 13 13 13 13 Area of the quadrilaterals formed by joining the mid points of the adjacent sides of a quadrilateral is ____ the area of given quadrilateral. (a) 1/8th (b) 1/2th (c) 1/4th (d) 3/4th If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form (a) a square (b) a rhombus (c) a rectangle (d) any other parallelogram

1.803R æ 1ö (a) tan -1 ç ÷ , 2p è 5ø g æ 1 ö 2p 1.803 R (b) cot -1 ç ÷ , è 5ø g -1 (c) cot ( 5) ,

65.

2 p 1.803 R3 g A particle is moving in a circle of radius r under the action of a force F = ar2 which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy = 0 for r = 0):

(d) tan -1 ( 5) ,

67.

(a)

64.

2p 1.803R 3 g

(a)

1 3 ar 2

(b)

5 3 ar 6

4 3 ar (d) ar3 3 A plane mirror is held at a height h above the bottom of an empty beaker. The beaker is now filled with water up to depth d. The general expression for the distance from a scratch at the bottom of the beaker to its image in terms of h and the depth d of water in the beaker is :

(c)

68.

PHYSICS 66.

Two non-viscous, incompressible and immiscible liquids of densities r and 1.5 r are poured into the two limbs of a circular tube of radius R and small cross section kept fixed in a vertical plane as shown in fig. Each liquid occupies

R O q

æ m ö (a) 2h – d ç è m - 1÷ø

(b) 2h –

d æ m - 1ö 2 çè m ÷ø

æ m - 1ö (c) 2h – d ç è m ÷ø

æ 2m - 1ö (d) 2h – d ç è m ÷ø

MOCK TEST-1

7

69. The xy - plane separates two media A and B of refractive indices m1 = 1.5 and m2 = 2. A ray of light travels from A to B. Its directions in the two media are given by unit vectors uˆ1 = aiˆ + bjˆ and uˆ2 = ciˆ + djˆ . Then : (a) (a / c) = (4 / 3) (b) (a / c) = (3 / 4) (c) (b / d) = (4 / 3) (d) (b / d) = (3 / 4) 70. A cylinder of mass Mc and sphere of mass Ms are placed at points A and B of two inclines, respectively (See Figure). If they roll on the incline without slipping such that their sin qc accelerations are the same, then the ratio sin q s is: 8 7

(a)

15 14

(b) (c)

8 7

(d)

15 14

MC

A

M

S

B

qC qS

C

3 2B(s) + O 2 (g) ¾¾ ® B2O3 (s); 2 DH = –1273 kJ per mol 1 ® H 2O(l); (ii) H 2 (g) + O 2 (g) ¾¾ 2 DH = –286 kJ per mol ® H 2 O(g); (iii) H 2 O(l) ¾¾ DH = 44 kJ per mol (iv) 2B (s) + 3H2 (g) ¾¾ ® B2H6 (g); DH = 36 kJ per mol (a) + 2035 kJ per mol (b) – 2035 kJ per mol (c) + 2167 kJ per mol (d) – 2167 kJ per mol 73. The Statue of Liberty is made of 2.0 × 105 lbs of copper sheets bolted to a framework (1lb = 454 g). How many atoms of copper are on the statue? (Atomic weight: Cu = 63.5g/mole). (a) 2.1 × 1027 (b) 8.6 × 1029 26 (c) 4.3 × 10 (d) 8.6 × 1026 74. Select the incorrect graph for velocity of e– in 1 an orbit vs. Z, and n : n

(i)

D

(a)

CHEMISTRY

v

+

v

n

71. In which case the N O2 will attack at the meta position CCl3

(b)

NO2

(c)

v

1/n

(d)

v n

Z (ii)

(i)

+

O–

NH 3

(iii)

(iv)

(a) i, ii, iii (b) (c) ii and iii only (d) 72. Diborane is a potential under goes combustion equation

ii, iv ii only rocket fuel which according to th e

B2 H 6 (g) + 3O 2 (g) ¾¾ ® B 2O 3 (s) + 3H 2O(g) Calculate the enthalpy change for th e combustion of diborane. Given

+

75.

D /H 2O

A (major product) is– (a) (b) (c) (d)

¾¾¾¾ ® A;

EBD_7839 8

76.

KVPY-SA BIOLOGY In an experiment, three bottles were filled with water from an aquatic ecosystem. This water contained tiny plants and animals of the ecosystem. The following experiments were done with the bottles.

79.

Bottle Condition Oxygen Oxygen Number measurement (mg/L) 1 2 3

77.

78.

1 2 3 4 5 6 7 8 9 10

Control

Done immediately Light for Done after one hour one hour Dark for Done after one hour one hour

11 12 13 14 15 16 17 18 19 20

(a) (d ) (d ) (c) (a) (b ) (c) (c) (a) (c)

21 22 23 24 25 26 27 28 29 30

Root Parenchyma

10 4

80.

Root Xylem

yp

ys

yp

yys

(a) 200 (b) –200 (c) 200 (d) 200

–190 220 –220 –220

–200 65 65 –65

5 –5 –5 –5

9

The gross primary productivity for this ecosystem is: (a) 1 mg/L/h (b) 5 mg/L/h (c) 6 mg/L/h (d) 14 mg/L/h Assume that the average amino acid residue has a molecular weight of 110. The DNA strand coding for a polypeptide chain of molecular weight 11,000 has a length of: (a) 300 nucleotides (b) 250 nucleotides (c) 500 nucleotides (d) 800 nucleotides Turgor pressure is required to maintain the shape of all cells except: (a) Meristematic cells (b) Root cells (c) Collenchyma cells (d) Lignified cells

(a) (d ) (b ) (b ) (b ) (d ) (b ) (a) (d ) (d )

On a warm summer’s day, the transpiration pull is the main force that drives water from root parenchyma into the root xylem. The table shows values of Yp (pressure potential) and Ys (solute potential) in root xylem and root parenchyma, in kPa. In which of the alternatives (a – d) would transpiration pull cause water to move from root parenchyma into the root xylem?

In grasshopper, rosy body colour is caused by a recessive mutation. The wild-type body colour is green. If the gene for body colour is on the X chromosome, what kind of progeny would be obtained from a mating between a rosy female and a wild-type male? (a) All the daughters will be green and all the sons will be rosy. (b) 50% daughters will be green and 50% sons will be rosy. (c) All offspring will be green irrespective of sex. (d) All offspring will be rosy irrespective of sex.

ANS W ER KEYS Part-I (b) 31 (a) 41 (d ) (d) 32 (a) 42 (c) (a) 33 (c) 43 (b ) (a) 34 (b) 44 (d ) (c) 35 (a) 45 (c) (b) 36 (b ) 46 (b ) (a) 37 (c) 47 (a) (a) 38 (b) 48 (c) (b) 39 (c) 49 (c) (a) 40 (c) 50 (a)

51 52 53 54 55 56 57 58 59 60

(a) (a) (d ) (c) (a) (b ) (d ) (a) (b ) (d )

61 62 63 64 65 66 67 68 69 70

Part-II (b ) 71 (a) 72 (a) 73 (b ) 74 (c) 75 (a) 76 (b ) 77 (c) 78 (a) 79 (d ) 80

(a) (b) (b) (d) (d) (c) (a) (d) (a) (a)

21 Time : 3 Hours

Stream-SA

Maximum Marks : 100

INSTRUCTIONS 1. 2. 3. 4.

There are 80 questions in this paper. The question paper contains two parts; Part–I (1 Mark Questions) and Part–II (2 Marks Questions). There are four sections; Mathematics, Physics, Chemistry and Biology in each part. There are four options given with each question, only one of them is correct. For each incorrect answer 0.25 Mark in Part–I and 0.5 Mark in Part–II will be deducted. PART-I (1 MARK QUESTIONS) MATHEMATICS

1.

2.

The vertex C is a right angle in the triangle ABC. If the points D an d E are taken on the hypotenuse, so that BC = BD and AC = AE, then, the sum of the perpendicular from D and E on AC and BC respectively is (a) DE (b) DA (c) D B (d) CA If 3x = 5y = 75z, then the value of (a) 0

3.

4.

5.

(b) 1

z (2 x + y ) is xy

(c) 2

(d) 3

(a) 6.

(b) 1 £ M £ 2 (c) 2 £ M £ 3 (d) 3 £ M £ 4 2 2 If x + y = 25, xy = 12, then x = (a) {3, 4} (b) {3, –3} (c) {3, 4, –3, –4} (d) {–3, –3}

7.

If

Find the values of k1 × k2, if p(x) = x + k1.x2011 + k2 leaves remainder 0, when divided by (x – 1) and leaves remainder 4 when divided by (x + 1). 2010

(a) 3 (b) –3 (c) –2 (d) 2 The mean of a group of eleven consecutive natural numbers is m. What will be the percentage change in the mean when next six consecutive natural numbers are included in the group? (a) m % (c)

m % 300

(b)

m % 3

(d)

300 % m

If a, b, c, d are positive real numbers such that a + b + c + d = 2, then M = (a + b) (c + d) satisfies the relation 0 £ M £1

1 1 1 1 1 1 are A. P., then æç + - ö÷ , , a b c èa b cø

æ 1 1 1 ö is equal to ç + - ÷ èb c aø

8.

9.

b 2 – ac

(a)

4 3 – ac b2

(b)

(c)

1 4 – ac b2

(d) None of these

a 2b 2c2

If a, b, c are distinct +ve real numbers and a2 + b2 + c2 = 1 then ab + bc + ca is (a) less than 1 (b) equal to 1 (c) greater than 1 (d) any real no. In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on line segment CD so that

EBD_7839 10

KVPY-SA

DF = 1 and GC = 2. Lines AG and FB intersect at a E. The area of DAEB is equal to . Find a. 14 (a) 7 5 (b) 7 2 (c) 6 2 (d) 6 8 10. If n = 1 + x, where x is a product of four consecutive positive integers, then which of the following not true ? (a) n is odd (b) n is prime (c) n is a perfect square (d) None of these 11. A circular metallic sheet is divided into two parts in such a way that each part can be folded into a cone. If the ratio of their curved surface areas is 1 : 2, then ratio of their volumes is: (a) 1 : 8

2 n +1

(b)

AD = DE

1 (ÐC – ÐB). 2 Then, which of the following is/are true? (a) Only (i) (b) Only (ii) (c) Both (i) and (ii) (d) None 15. The mean of three positive numbers is 10 more than the smallest of the numbers and 15 less than the largest of the three. If the median of the three numbers is 5, then the means of squares of the numbers is

(ii) ÐDAE =

(b) 1: 16

(c) 1: 10 (d) 2 : 3 12. In a parallelogram ABCD, a point P is taken on AD, such that AP : AD is 1 : n. If BP intersects AC at Q then AQ : AC is (a)

(i)

(a)

108

2 3

(b) 116

2 3

(c)

208

1 3

(d) 216

2 3

PHYSICS 16.

1 n -1

Two identical glass bulbs are interconnected by a thin glass tube. A gas is filled in these bulbs at NTP. If one bulb is placed on ice and another bulb is placed on hot bath, then the pressure of the gas become 1.5 times. The temperature of hot bath will be

1 2 (d) n +1 n -1 13. In the figure, OABCD is a sector of a circle. If

(c)

» » = CD » , then x = AB = BC Ice

A B

C

D

x

O (b) 120° (a) 105° (c) 150° (d) 144° 14. ABC is a right angled triangle at A and AB > AC. D bisects BC. DE is perpendicular to hypotenuse BC and meets the bisector of the right angle A in E. Consider the following statements.

17.

Hot bath

(a) 100º C (b) 182º C (c) 256º C (d) 546º C A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring such that it is half submerged in a liquid of density s at equilibrium position. The extension x 0 of the spring when it is in equilibrium is: (a)

Mg k

(b)

Mg æ LAs ö 1– k çè M ÷ø

(c)

Mg æ LAs ö 1– k çè 2 M ÷ø

(d)

Mg æ LAs ö 1+ k çè M ÷ø

MOCK TEST-2

11

18. A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 second for this and records 40 second for 20 oscillations. For this observation, which of the following statement (s) is (are) true? (I) Error DT in measuring T, the time period, is 0.05 second (II) Error DT in measuring T, the time period, is 1 second. (III) Percentage error in the determination of g is 5% (IV) Percentage error in the determination of g is 2.5% (a) I and II (b) II and III (c) I and III (d) IV and II 19. The ice storm in the province of Jammu strained many wires to the breaking point. In a particular situation, the transmission pylons are separated by 500 m of wire. The top grounding wire is 15° from the horizontal at the pylons, and has a diameter of 1.5 cm. The steel wire has a density of 7860 kg/m3. When ice (density 900 kg/m3) built up on the wire to a total diameter of 10.0 cm, the wire snapped. What was the breaking stress (force/unit area) in N/m2 in the wire at the breaking point? You may assume the ice has no strength. (a) 7.4 × 107 N/m2 (b) 4.5 × 108 N/m2 6 2 (c) 2.6 × 10 N/m (d) 1.15 × 107 N/m2 20. The dimensions of Stefan’s Boltzmann constant s can be written in terms of Plank’s constant h, Boltzmann constant kB and speed of light c as

21.

s = h a k Bb c g , then (a) a = 3, b = 4, g = –3 (b) a = 3, a = –4, a = 2 (c) a = –3, b = 4, g = –2 (d) a = 2, a = –3, a = –1 The v - x graph for a car in a race on a straight road is given. v(m/s)

(a)

200

x (m)

Identify the correct a – x graph

a(m/s 2)

0.64 100 200

x(m)

(b)

0.64 100

0.64

(c)

200

x(m)

a(m/s 2)

a(m/s 2) 0.64

100

200 x(m)

–0.64

(d)

100 200

x(m)

–0.64

22. Circular part in the centre of retina of human eye is called (a) Blind spot (b) Yellow spot (c) Red spot (d) None of these 23. A student performs on experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with on uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01 mm. Take g = 9.8 m s–2(exact). The Young’s modulus obtained from the reading is close to (a) (2.0 ± 0.3) × 1011 Nm–2 (b) (2.0 ± 0.2) × 1011 Nm–2 (c) (2.0 ± 0.1) × 1011 Nm–2 (d) (2.0 ± 0.05) × 1011 Nm–2 24. Three elephants A, B and C are moving along a straight line with constant speed in same direction as shown in figure. Speed of A is 5 m/s and speed of C is 10 m/s. Initially separation between A and B is ‘d’ and between B and C is also d. When ‘B’ catches ‘C’ separation between A and C becomes 3d. Then the speed of B will be – u

5 m/s A

10 m/s

B d

8

100

a(m/s2 )

C d

(a) 7.5 m/s (b) 15 m/s (c) 20 m/s (d) 5 m/s 25. An air bubble in a water tank rises from the bottom to the top. Which of the following statements are true :

EBD_7839 12

KVPY-SA (I) (II) (III) (IV) (a)

Bubble rises upwards because pressure at the bottom is less than that at the top. Bubble rises upwards because pressure at the bottom is greater than that at the top. As the bubble rises, its size increases. As the bubble rises, its size decreases. II and III (b) II and IV

(c) I and IV 26.

30.

(d) III and IV

In a certain region of space, the potential field depends on x and y-coordinates as V = (–x2 + y2). The corresponding electric field lines in xy-plane are correctly represented by

CHEMISTRY

y

y

31.

(a)

x

(d)

x

y y

(c)

x

(d)

x

32.

27.

28.

29.

The splitting of white light into several colours on passing through a glass prism is due to (a) Refraction (b) Reflection (c) Interference (d) Diffraction Two different masses m and 3m of an ideal gas are heated separately in a vessel of constant volume, the pressure P and absolute temperature T, graphs for these two cases are shown in the figure as A and B. The ratio of slopes of curves B to A is P B (a) 3 : 1 A (b) 1 : 3 3m (c) 9 : 1 m T (d) 1 : 9 The resistance of a wire is R. It is bent at the middle by 180° and both the ends are twisted together to make a shorter wire. The resistance of the new wire is (a) 2 R (b) R/2 (c) R/4 (d) R/8

A pulley of radius 2 m is rotated about its axis by a force F = (20t – 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg-m2 the number of rotations made by the pulley before its direction of motion is reversed, is: (a) more than 3 but less than 6 (b) more than 6 but less than 9 (c) more than 9 (d) less than 3

33.

34.

35.

Consider a real gas placed in a container. If the inter-molecular attractions are supposed to disappear suddenly which of the following would happen? (a) Pressure decreases (b) Pressure increases (c) Pressure remains unchanged (d) Gas collapse Conjugate acid of HF2- is (a) H+ (b) HF (c) F2 (d) H2F2 Two elements A and B on burning in air give corresponding oxides. Oxides of both A and B are soluble in water. The aqueous solution of oxide of A is alkaline and reacts with aqueous solution of oxide of B to give another compound. Identify A and B (a) A and B both are metals (b) A and B are non-metals (c) A is metal and B is non-metal (d) A is non-metal and B is metal Which of the following is strongest nucleophile? (a) Br– (b) : OH– (c) : CN (d) C6 H5O : Which of the following metal sulphides has maximum solubility in water? (a) CdS (Ksp = 36 × 10–30) (b) FeS (Ksp = 11 × 10–20) (c) HgS (Ksp = 32 × 10–54) (d) ZnS (Ksp = 11 × 10–22)

MOCK TEST-2

13

36. In the Kjeldahl’s method for estimation of nitrogen present in a soil sample, ammonia evolved from 0.75 g of sample neutralized 10 mL of 1 M H2SO4. The percentage of nitrogen in the soil is : (a) 37.33 (b) 45.33 (c) 35.33 (d) 43.33 37. Which of the following describes the best relationship between the methyl groups in the chair conformation of the substance shown below? CH3 CH3 (a) Trans (b) Anti (c) Gauche (d) Eclipsed 38. Which graph shows how the energy (E) of a photon of light is related to its wavelengths (l) ?

­ (a)

­

E

(b) l

®

l®

­ (c)

E

­

E

(d) l®

E

l®

39. Which of the underlined atoms in the molecules shown below have sp-hybridization? (u) CH2CHCH3 (v) CH2CCHCl (w) CH 3CH +2 (x) H — C ºº C — H (y) CH3CN (z) (CH3)2CNNH2 (a) x and z (b) x, y and z (c) u, w and x (d) v, x and y 40. The following quantum numbers are possible for how many orbital(s) n = 3, l = 2, m = +2 ? (a) 1 (b) 3 (c) 2 (d) 4 41. KO2 is used in oxygen cylinders in space and submarines because it:

(a) Absorbs CO2 and increases O2 content. (b) Eliminates moisture. (c) Absorbs CO2. (d) Produces ozone. 42. What is the IUPAC name of the compound ?

H Cl

CH3

(a) 2 - Chloro - 2 - butene (b) 3 - Chloro - 1 - butene (c) 3 - Methyl - 3 - chloropropene - 1 (d) 3 - Chloro - 3 - methyl - 1 - propene 43. Which molecular geometry is least likely to result from a trigonal bipyramidal electron geometry? (a) Trigonal planar (b) See-saw (c) Linear (d) T-shaped 44. Cyclohexene on reaction with cold alkaline KMnO4 forms, (a) trans-hexanediol (b) hexadiketone (c) cis-hexanediol (d) pentadiketone 45. Which of the following has unpaired electron(s)? (a) N2

(b) O-2

(c) N 22+

(d) O 22 BIOLOGY

46. Which is th e most abun dant gas in our atmosphere that cannot be utilised by plants directly in its atmospheric form and is, therefore, captured by certain bacteria that live symbiotically in the nodules of roots? (a) Oxygen (b) Nitrogen (c) Neon (d) Hydrogen 47. Phospholipids are important cell membrane constituents, because they: (a) contain glycerol. (b) can form bilayers in water. (c) combine covalently with proteins. (d) contain polar and non-polar portions.

EBD_7839 14 48.

49. 50.

51.

52.

53. 54.

KVPY-SA Allergens are non-infectious foreign substances that: (a) Increases the secretion of IgA (b) Increases the secretion of IgE (c) Increases the secretion of IgG (d) Increases the secretion of IgM Which of the following is not found in birds? (a) Hind limb (b) Fore limb (c) Pelvic girdle (d) Pectoral girdle A drop of each of the following, is placed separately on four slides. Which of them will not coagulate? (a) Whole blood from pulmonary vein (b) Blood plasma (c) Blood serum (d) Sample from the thoracic duct of lymphatic system Cancer cells are : (a) HeLa cells (b) CD4 cell (c) Memory cell (d) Plasma cell Which of the following options does not hold good regarding anaerobic respiration or fermentation? (a) Occurs inside the mitochondrion (b) Partial breakdown of glucose occurs (c) Net gain of only 2 ATP molecules (d) None of these Phloem parenchyma is absent in: (a) Dicots (b) Monocots (c) Gymnosperms (d) All of these Which among the following has highest water potential? (a) 1 M salt solution (b) 1 M glucose solution (c) Distilled water (d) None of the above

55. 56.

57.

58.

59.

The number of essential amino acids is: (a) 9 (b) 8 (c) 10 (d) 14 All are the products of anaerobic respiration except: (a) Ethyl alcohol (b) Lactic acid (c) Fumaric acid (d) Butyric acid Cell mediated immunity, also involved in the killing of the cancerous cells, is the responsibility of which of the following group of cells? (a) B- cells (b) CT- cells (c) HT- cells (d) Malignant cells A patient is generally advised to specially consume more meat, lentils, milk and eggs in diet only when he suffers from: (a) Scurvy (b) Kwashiorkar (c) Rickets (d) Anaemia The tumour inducing capacity of Agrobacterium tumefaciens is located in large extrachromosomal plasmid called: (a) Ri plasmid (b) Lambda phage (c) pBR322 (d) Ti plasmid

60. The process by which a lake ecosystem is altered by eutrophication involves several stages, each causing the next. Which of the following stages would occur second in the chain of causation? (a) Algal blooms occur. (b) Oxygen levels drop in deeper water. (c) Phosphorus input from sewage and agricultural runoff increases. (d) Respiratory demand from decomposers increases.

PART-II (2 MARKS QUESTIONS) MATHEMATICS 61. Suppose n is an integer such that the sum of digits of n is 2, and 1010 < n < 1011. Find the number of different values of n. (a) 9

(b) 10

(c) 11

(d) 12

62. The number of 3 letters words, with or without meaning which can be formed out of the letters of the word ‘NUMBER’. Statement I : When repetition of letters is not allowed is 120. Statement II : When repetition of letters is allowed is 216.

MOCK TEST-2

15

Choose the correct option.

(II) The net elongation of the spring is

(a) Only Statement I is correct

8pR3rg 3K

(b) Only Statement II is correct (c) Both I and II are correct (d) Both I and II are false 63. In a parallelogram ABCD, A point ‘K’ is taken on diagonal BD. When AK is extended, it intersects CD and BC at M and L respectively. If AK = 6 units, MK = 4 units, then LM is (a) 9 (b) 6 (c) 5

(d)

24

64. A conical cup contains water equivalent to

Perimeter of rectangle Perimeter of parallelogram is-

1 >1 II > IV > III (b) III > IV > II > I (c) II > IV > I > III (d) I > III > II > IV A mixture of O2 and gas "Y" (mol. mass 80) in the mole ratio a : b has a mean molecular mass 40. What would be mean molecular mass, if the gases are mixed in the ratio b : a under identical conditions? (Assume that gases are non-reacting) (a) 40 (b) 48 (c) 62 (d) 72 A gaseous mixture contains three gases A, B and C with a total number of moles of 10 and total pressure of 10 atm. The partial pressure of A and B are 3 atm and 1 atm respectively. If C has molecular weight of 2 g/mol then, the weight of C present in the mixture will be : (a) 8 g (b) 12 g (c) 3 g (d) 6 g Arrange in the order of increasing acidity.

CHEMISTRY 71.

The electronegativity difference between N and F is greater than that between N and H, yet the dipole moment of NH3 (1.5 D) is larger than that of NF3 (0.2D). This is because (a) In NH3, the atomic dipole and bond dipole are in the same direction, whereas in NF3 these are in opposite directions. (b) In NH3 as well as NF3, the atomic dipole and bond dipole are in opposite directions.

OH

NH2 (II)

(I)

OH (III)

(a) III < I < II (c) III < II < I

(b) I < III < II (d) II < I < III

MOCK TEST-2

17

(a) (a)

(b) (b)

(d)

(c)

(d) Hydrophilic end

Hydrophobic end Phospholipid

80. Photosynthetic features of four plants P, Q, R and S are depicted in the graphs below. Uptake of CO2

76. The nucleus of a human liver cell contains 6 picograms (pg) of DNA. What amount of DNA is likely to found in the nucleus of an actively dividing human skin cell at the end of interphase? (a) 12 pg (b) 6 pg (c) 24 pg (d) 18 pg 77. Three uniformly watered plants A, B and C were kept in 45% relative humidity, 45% relative humidity with blowing wind and 95% relative humidity respectively. Arrange these plants in the order (fastest to slowest) in which they dry up. (a) A= B, C (b) B, A, C (c) C, B, A (d) C, A = B 78. A scientist introduced bacterial plasmids into baker ’s yeast. However, the cells lost these plasmids over a period of time. Which of the following needs to be inserted into the bacterial plasmid to overcome this problem? (a) Centromere (b) Yeast origin of replication (c) Telomere (d) Bacterial origin of replication 79. A red blood corpuscle (RBC) was kept in a solution and treated so that it became inside out. What will be the polarity of the phospholipid bilayer in this cell?

(c)

P Q Leaf temperature

Uptake of CO2

BIOLOGY

R S

Concentration of CO2 within the intercellular space These plant species belong to: (a) P: C3 plant Q: C4 plant R: CAM plant S: shade plant (b) P: C4 plant Q: C3 plant R: C4 plant S: C3 plant (c) P: C4 plant Q: sun plant R: C3 plant S: CAM plant (d) P: C3 plant Q: C4 plant R: C3 plant S: C4 plant

EBD_7839 18

1 2 3 4 5 6 7 8 9 10

KVPY-SA

(a) (b ) (c) (d ) (a) (c) (a) (a) (a) (b )

11 12 13 14 15 16 17 18 19 20

(c) (c) (c) (c) (d ) (d ) (c) (c) (b ) (c)

21 22 23 24 25 26 27 28 29 30

ANS W ER KEYS Part-I (c) 31 (b) 41 (a) (b) 32 (d) 42 (b ) (b) 33 (c) 43 (a) (b) 34 (c) 44 (c) (a) 35 (b) 45 (b ) (b) 36 (a) 46 (b ) (a) 37 (c) 47 (d ) (a) 38 (d) 48 (b ) (c) 39 (d) 49 (b ) (a) 40 (a) 50 (c)

51 52 53 54 55 56 57 58 59 60

(a) (a) (b ) (c) (a) (c) (b ) (b ) (d ) (a)

61 62 63 64 65 66 67 68 69 70

Part-II (c) 71 (c) 72 (c) 73 (d ) 74 (c) 75 (a) 76 (d ) 77 (b ) 78 (b ) 79 (c) 80

(a) (c) (d) (b) (d) (a) (b) (b) (a) (b)

31 Time : 3 Hours

Stream-SA

Maximum Marks : 100

INSTRUCTIONS 1. 2. 3. 4.

There are 80 questions in this paper. The question paper contains two parts; Part–I (1 Mark Questions) and Part–II (2 Marks Questions). There are four sections; Mathematics, Physics, Chemistry and Biology in each part. There are four options given with each question, only one of them is correct. For each incorrect answer 0.25 Mark in Part–I and 0.5 Mark in Part–II will be deducted. PART-I (1 MARK QUESTIONS) MATHEMATICS

1. In an international convention, 5 participants from each of USA, China and Russia were arranged around a circle. In how many ways this arrangement can be made if participants from the same country are always together? (a) (3!)(5!)(5!)(5!) (b) (2!)(5!)(5!)(5!) (c) (3!)(5!)(5!)(5!) (d) None of these 2. Two numbers A and B are such that their HCF is 72 and when written in prime factorization form A = 2a3b7c while B = 2d3e5f where all a, b, c, d, e and f are distinct natural numbers. Find the minimum value of LCM of two numbers A and B. (a) 25345271 (b) 26345271 (c) 26335171 (d) 26355471 1 7 3. If, a, b, c are real numbers such that a + = ; b 3 1 b + = 4; then value of abc is: c (a) 0‘ (b) 4 (c) 1 (d) 2 4. Let ABC be a triangle in which AB = AC and ÐCAB = 90°. Suppose M and N are points on hypotenuse BC such that BM2 + CN2 = MN2. Then ÐMAN = (a) 45° (b) 60° (c) 30° (d) None of these

5. If the bisectors of two adjacent angles A and B of a quadrilateral ABCD intersect at a point O such that ÐC + ÐD = k ÐAOB, then find the value of k – 1 is 5 6 (a) 1 (b) (c) 2 (d) 3 2 6. A number N is divisible by 3 and 4 but not by 9 then which one of the following cannot be an integer? (a) N/6 (b) N/42 (c) N/18 (d) N/21 7. Consider the following situation : – (i) A set S = { 1, 2, 3, ….., 19, 20} i.e 1st 20 positive integers. (ii) Another set P is a subset of set “S” and minimum number of elements in set P is 2. (iii) HCF of any two elements of set P is not more than 2. What is the maximum number of elements that set P can have? (a) 9 (b) 10 (c) 11 (d) 13 8. Let T be the set of integers {3, 11, 19, 27... 451, 459, 467} and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in S is (a) 32 (b) 28 (c) 29 (d) 30

EBD_7839 20 9. In a Annual general meeting of XYZ company, participants are ‘2a’ number of executives, ‘b’ number of managers, ‘c’ number of senior managers and director of the company. The seating arrangement is made in such a way that all the executives are seated at the either end (‘a’ number of executives on one end), All the managers will be together while Director don’t want to sit next to an executive or a manager, in how many ways they can be arranged in a straight line? (a) {(a)!}{(a)!}(b!){(c+1)!}{cP2} (b) {(2a)!}(b!){(c+1)!}{cP2} (c) {(2a)!}(b!){(c)!}{cP2} (d) None of these 10. When you reverse the digits of the number 13, the number increases by 18. How many other two digit numbers increase by 18 when their digits are reversed? (a) 5 (b) 6 (c) 7 (d) 8 11. ABC is a triangle and AD is the median. If the coordinates of A are ( 4, 7, –8) and the coordinates of centroid of the triangle ABC are (1, 1, 1), what are the coordinates of D? 11ö æ 1 ö æ 1 (a) ç – , 2,11÷ (b) ç – , –2, ÷ è 2 ø è 2 2ø (c) (–1, 2, 11) (d) (–5, –11, 19) 12. x and y are two non-negative numbers such that 2x + y = 10. The sum of the maximum and minimum values of (x + y) is (a) 6 (b) 9 (c) 10 (d) 15 13. Three circles each of radius 1 touch one another externally and they lie between two parallel lines. The minimum possible distance between the lines is. (a) 2 + 3 (b) 3 + 3 1 (c) 4 (d) 2 + 3 14. In a small village, there are 87 families, of which 52 families have at most 2 children. In a rural development programme 20 families are to be chosen for assistance, of which at least 18 families must have at most 2 children. In how many ways can the choice be made? (a) 52C18 × 35C2 (b) 52C18 × 35C2 + 52C19 × 35C1 + 52C20 (c) 52C18 + 35C2 + 52C19 (d) 52C18 × 35C2 + 35C1 × 52C19

KVPY-SA 15. If x, y are real numbers such that x +1 y

3 is (a)

-3

x -1 y

0

= 24 then the value of (x + y)/(x – y)

(b) 1

(c) 2

(d) 3

PHYSICS 16.

17.

A stone weighs (10.0 ± 0.1) kg in air. The weight of the stone in water is (5.0 ± 0.1) kg. Find the maximum percentage error in the measurement of specific gravity. (a) 5% (b) 0.5% (c) 0.25% (d) 0.2% Which of the following graph cannot possibly represent one dimensional motion of a particle? x

x

(a)

(b)

t

t

speed

(c) 18.

19.

t

(d) All of these

While an aquarium is being filled with water, a motionless fish looks up vertically through the surface of the water at a monochromatic plane wave source of frequency f. If the index of refraction of water is m and water level rises at a rate of dh/dt, the shift in the frequency df/f, that the fish observes is (velocity of light is c) : m dh ( m - 1) dh / dt (a) (b) c dt c c dh / dt c dh (c) (d) ( m - 1) m dt Which of the following is/are correct. I. An object moving the earth under the influence of Earth’s gravitational force is in a state of “free-fall”. II. The dominant effect causing a slowdown in the rotation of the earth is the gravitational pull of other planets in the solar system. III. Generally the path of a projectile from the earth is parabolic but it is elliptical for projectiles going to a very large height. IV. The path of a projectile is independent of the gravitational force of earth. (a) I and II (b) II and III (c) I and III (d) I and IV

MOCK TEST-3

21

20. A calorimeter (of water equivalent 50 g) contains 250 g of water and 50 g of ice at 0°C. 30 g of water at 80°C is added to it. The final condition of the system will be : (a) the temperature of the system will be 4.2°C. (b) the temperature of the system will still be 0°C and the entire ice will melt. (c) the temperature will be 0°C and half of the ice will melt. (d) the temperature will be 0°C and 20 g of ice will left. 21. Two identical balls A and B each of mass 0.1 kg are attached to two identical massless springs. The spring mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle as shown in the figure. The pipe is fixed in a horizontal plane. The centres of the balls can move in a circle of radius 0.06 m. Each spring has a natural length of 0.06p m and force constant 0.1 N/m. Initially both the balls are displaced by an angle q = p/6 radian with respect to the diameter PQ of the circle and released from rest. The frequency of oscillation of the ball B is (a) p Hz 1 Hz (b) p (c) 2p Hz 1 Hz (d) 2p

bT 2 22. The emf is given by E = aT + and a = 2 -1 10µV/ºC2 , b = µV/ºC2. If temperature of cold 20 junction is zero find neutral temperature and temperature of inversion ? (a) 200°C and 400°C (b) 100°C and 200°C (c) 20°C and 40°C (d) 400°C and 20°C 23. Which of the following graphs correctly represents the relation between (ln E) and (ln T) where E is the amount of radiation emitted per unit time from unit area of a body and T is the absolute temperature (a)

ln E

(c)

ln E

ln E

(b) ln T

ln T ln E

ln T

(d)

ln T

24. An a particle is moving along a circle of radius R with a constant angular velocity w. Point A lies in the same plane at a distance 2R from the centre. Point A records magnetic field produced by a particle. If the minimum time interval between two successive times at which A records zero magnetic field is t, the angular speed w, in terms of t is (a) 2p/t (b) 2p/3t (c) p/3t (d) p/t 25. An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed state. The gas is then heated very slowly to temperature T2, pressure P2 and volume V2. During this process the piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the incorrect statement is

(a) If V2 = 2V1 and T2 = 3T1, then the energy 1 stored in the spring is P1V1 4 (b) If V2 = 2V1 and T2 = 3T1, then the change in internal energy is 3P1V1 (c) If V2 = 3V1 and T2 = 4T1, then the work done by the gas is P1V1 (d) If V2 = 3V1 and T2 = 4T1, then the heat sup17 plied to the gas is PV 6 1 1 26. Mark the correct statements. I. Gravitational potential at the centre of curvature of a thin hemispherical shell of radius R and mass M is equal to GM/R II. Gravitation field strength at a point lying on the axis at a distance of ‘a’ on a thin, uniform circular ring of radius R and mass GMR M is equal 2 ( a + R 2 )3 2 III. Newton's law of gravitation for gravitational force between two bodies is applicable only when bodies have sphrical symmetric distribution of mass (a) I and II (b) II and III (c) I and III (d) None of these

EBD_7839 22 27.

28.

KVPY-SA A solid homogeneous sphere of mass M and radius R is moving on a rough horizontal surface, partly rolling and partly sliding. During this kind of motion of the sphere (a) total kinetic energy is conserved (b) the angular momentum of the sphere about the point of contact with the plane is conserved (c) only the rotational kinetic energy about the centre of mass is conserved (d) angular momentum about the centre of mass is conserved. A large number of droplets, each of radius a, coalesce to form a bigger drop of radius b. Assume that the energy released in the process is converted into the kinetic energy of the drop. The velocity of the drop is (S = surface tension and r = density of drop) é3 S æ 1 1ö ù (a) ê ç - ÷ú ë r è a bø û

1/2

(b)

é 6 S æ 1 1ö ù ê r çè a - b ÷ø ú ë û

1/2

29.

30.

1/2

1/2

é 2 S æ 1 1ö ù é S æ 1 1ö ù (c) ê çè - ÷ø ú (d) ê çè - ÷ø ú ë r a b û ër a b û A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ? (a) 8.6 m/sec2, 45°28¢ (b) 6.8 m/sec2, 54°28¢ (c) 0.86 m/sec2, 54°28¢ (d) 0.68 m/sec2, 45°28¢ A diver looking up through the water sees the outside world contained in a circular horizon. 4 The refractive index of water is and the diver’ss 3 eyes are 15 cm below the surface of water. Then the radius of the circle is: (a) 15 ´ 3 ´ 5 cm (b) 15 ´ 3 7 cm

15 ´ 7 cm (c) 3

15 ´ 3 (d) cm 7

CHEMISTRY 31.

If one assumes linear structure instead of bent structure for water, then which one of the following properties cannot be explained? (a) The formation of intermolecular hydrogen bond in water.

32.

33.

34.

35.

36.

37.

38.

39.

(b) The high boiling point of water. (c) Solubility of polar compounds in water. (d) Ability of water to form coordinate covalent bond. 1, 2-Dibromocyclohexane on dehydrobromination gives : (a)

(b)

(c)

(d)

Br Electronic configuration of an atom is 2, 8, 1. Which of the following elements is similar with it in chemical reactivity? (a) K (b) Cl (c) N (d) Ar The number of secondary hydrogens in 2, 2-dimethylbutane is : (a) 8 (b) 6 (c) 4 (d) 2 Which of the following sets of quantum numbers is correct for an electron in 4f orbital ? (a) n = 4, l = 3, m = + 1, s = + ½ (b) n = 4, l = 4, m = – 4, s = – ½ (c) n = 4, l = 3, m = + 4, s = + ½ (d) n = 3, l = 2, m = – 2, s = + ½ Which solution has pH equal to 10 ? (a) 10–4 M KOH (b) 10–10 M KOH (c) 10–10 M HCl (d) 10–4 M HCl Bleaching powder is soluble in cold water giving a milky solution due to – (a) available chlorine (b) lime present in it (c) calcium carbonate formation (d) the absorption of carbon dioxide from atmosphere What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene ? (a) 2.8 kg (b) 6.4 kg (c) 9.6 kg (d) 96 kg The structure of the noble gas compound XeF4 is (a) square planar (b) distorted tetrahedral (c) tetrahedral (d) octahedral

MOCK TEST-3

23

40. Fire extinguishers contain H2SO4 and which one of the following? (a) NaHCO3 and Na2CO3 both (b) Na2CO3 only (c) NaHCO3 only (d) CaCO3 only 41. What happens when Al is added to KOH solution? (a) No reaction takes place (b) Oxygen is evolved (c) Water is produced (d) Hydrogen is evolved 42. Which pair of atoms form strongest ionic bond? (a) Al and As (b) Al and N (c) Al and Se (d) Al and O 43. In which of the following ways does the hydride ion tend to function ? (a) An electrophile (b) A nucleophile (c) A free radical (d) An acid 44. Le-Chatelier principle is not applicable to : (a) H 2 (g) + I2 (g) ‡ˆˆ ˆˆ† 2HI (g) (b) Fe (s) + S(s) ‡ˆˆ ˆˆ† FeS(s) (c) N 2 (g) + 3H 2 (g) ‡ˆˆ ˆˆ† 2NH3 (g) (d) N 2 (g) + O 2 (g) ‡ˆˆ ˆˆ† 2NO (g) 45. Which of the following would not rearrange to a more stable form? (a)

(b) H

(c)

(d)

+

+ +

BIOLOGY 46. In the genetic code dictionary, how many codons are used to code for all the 20 essential amino acids? (a) 60 (b) 20 (c) 64 (d) 61 47. Hemichordates have now been placed with the non-chordates, close to echinoderms, because (a) notochord is absent. (b) pharyngeal gill-slits are lacking. (c) dorsal nerve cord is absent. (d) heart is lacking.

48. In terms of being open or closed, what is the state of the mitral and tricuspid valves (left and right atrioventricular valves, respectively) at the end of the first heart sound? (a) Both are closed (b) Both are open (c) Mitral is closed, tricuspid is open (d) Mitral is open, tricuspid is closed 49. Complete failure of the anterior lobe of pituitary causes: (a) Conn’s disease (b) Acromegaly (c) Cushing’s disease (d) Simmond’s disease 50. During translation, proteins are synthesised by: (a) ribosomes using the information on DNA. (b) lysosome using the information on DNA. (c) ribosome using the information on mRNA. (d) lysosome using the information on mRNA. 51. Hydrolysis of maltose gives rise to (a) two molecules of glucose. (b) two molecules of galactose. (c) one molecule of glucose and one molecule of galactose. (d) one molecule of glucose and one molecule of fructose. 52. Chargaff’s rules of base pairing states that: (a) the ratio of purines to pyrimidines is roughly equal in all tested organisms. (b) the ratio of A to T is roughly equal in all tested organisms. (c) the ratio of A + T and G + C is roughly equal in all tested organisms. (d) Both (a) and (b) 53. In the Hardy-Weinberg equation, the term 2 pq represents the (a) overall gene frequency of the population. (b) frequency of both homozygous genotypes. (c) frequency of the heterozygous genotype. (d) allele frequencies of the population. 54. The glomerular filtrate consists of: (a) urea, sodium chloride, fibrinogen and water (b) glucose, amino acids, urea, oxytocin and calcitonin (c) Both (a) and (b) (d) urea, glucose, salts and water 55. Choose the mismatch pair. (a) Glutamic acid – Acidic (b) Lysine – Basic (c) Valine – Charged (d) Phenylalanine – Aromatic

EBD_7839 24 56.

57. 58.

KVPY-SA The distance between A and B genes is too long on a chromosome, the strength of linkage between them is: (a) More (b) Less (c) No relation to distance (d) Same irrespective of close or distant location A triple antigen (vaccine) is: (a) BCG (b) DPT (c) ATS (d) TAB Which of the following is not the function of HCl in stomach? (a) Breaking down proteins into peptones.

59.

60.

(b) Killing the bacteria ingested with food and drinks. (c) Promoting the formation of pepsin. (d) Softening fibrous food elements. Which of the following is a method for birth control? (a) IUDs (b) GIFT (c) HTF (d) IVF-ET Affinity of CO for haemoglobin as compared to O2 is: (a) 2 times (b) 20 times (c) 100 times (d) 200 times

PART-II (2 MARKS QUESTIONS) MATHEMATICS

65. The value of

61. ABCD is a parallelogram. A straight line is drawn through A and meets CB, CD produced in E, F. Then CB · CE + CD · CF = (a) AC2 + AE.AF (b) AB2 + AE.AF 2 (c) AC + AB.AF (d) AB2 + AB.AF 62. HCF of A & B is h1, a & C is h2 and that of B & C is h3. Find HCF of h1, h2 and h3. (a) 1 (b) h1 h2 h 3 (c) h1 + h2 + h3 (d) (h1 h2 h3 )2 63. In the given figure, a circle with centre B overlaps another circle with centre A and a square. The ratio of areas of P and Q is 5 : 4 and the area of Q 1 the area of circle B. The radii of circle A and 8 circle B are 10 cm and 8 cm respectively.

is

A

P B Q

7 cm

Find the area of the unshaded part of the figure. (Take p = 3.14) (a) 449.75 cm2 (b) 520.60 cm2 (c) 563.72 cm2 (d) 507.44 cm2 64. From the middle point C of an arc AB of a circle, a diameter CD is drawn and also a chord CE which meets the straight line AB in F. If a circle, drawn with centre C to bisect FE, meets BD in G, then EF = (a) BG (b) 2 BG 3 (c) BG (d) None of these 2

21 + 3 59 + 16 + 3 722 + 49 is (a) 4 (c) 6

(b) 5 (d) 8 PHYSICS

66.

67.

A proton, a deuteron and an a-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If rp, rd and ra denote respectively the radii of the trajectories of these particles, then (a) ra = rp < rd (b) ra > rd > rp (c) ra = rd > rp (d) rp = rd = ra A block of mass m is on an inclined plane of angle q. The coefficient of friction between the block and the plane is m and tan q > µ. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from P1 = mg(sin q – m cos q ) to P2 = mg(sin q + m cos q), the frictional force f versus P graph will look like f

f

(a)

(b)

f

(c)

f

(d)

MOCK TEST-3

25

68. A balloon of volume V, contains a gas whose density is to that of the air at the earth’s surface as 1 : 15. If the envelope of the balloon be of weight w but of negligible volume, find the acceleration with which it will begin to ascend. æ 7Vg s - w ö æ 2Vg s - w ö (a) ç ´ g (b) ç ´g ÷ è Vg s + w ø è Vg s + w ø÷ æ 14Vg s - w ö æ 14Vg s + w ö (c) ç ´ g (d) ç ´g è Vg s + w ø÷ è Vg s - w ø÷ 69. In the diagram below (not to scale), each of the loudspeakers emits a continuous sound of the same frequency. A microphone moved along the line PQ detects a series of maximum and minimum sound intensities. Which one of the following actions on its own, will lead to an increase in the distance between the maxima of sound intensity ? Q Loudspeaker

Loudspeaker P

(a) Decreasing the frequency of the sound emitted by the loudspeakers. (b) Increasing the frequency of the sound emitted by the loudspeakers. (c) Increasing the separation of the loudspeakers (d) Decreasing the distance of the loudspeakers from the line PQ. 70. A 0.5 kg block slides from the point A on horizontal track with an initial speed of 3 m/s towards a weightless horizontal spring of length 1 m and force constant 2 Nm–1. The part AB of the track is frictionless and the part BC has the coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distances AB and BD are 2 m and 2.14 m respectively, find the total distance through which the block moves before it comes to rest completely. [Take g = 10 m /s2] R (a) 2.42 m (b) 1.42 m mk R (c) 4.24 m A B C E C (d) 2.14 m 2m 2.14m

mg

CHEMISTRY 71. In forming (i) N 2 ® N +2 and (ii) O2 ® O2+ ; the electrons respectively are removed from : (a) (p*2pz) and (p*2py or p*2px) (b) (p 2pz) and (p2py or p2px) (c) (p2pz) and (p*2py or p*2px) (d) (p*2pz) and (p2py or p2px) 72. Arrange the following in decreasing order of stability.

I

II

III

(b) I > III > II (a) I > II > III (c) II > I > III (d) II > III > I 73. If degree of dissociation of pure water at 100 °C is 1.8 × 10–8, then the dissociation constant of water will be : (Density of H2O = 1 g/cc) (a) 1 × 10–12 (b) 1 × 10–14 (c) 1.8 × 10–12 (d) 1.8 × 10–14 – 74. E Values of some redox couples are given below. On the basis of these values choose the correct option. – E values: Br 2/Br – = + 1.90; Ag+ /Ag(s) = + 0.80; Cu2+ /Cu(s) = + 0.34; I2(s) /I– = + 0.54 (a) Cu will reduce Br – (b) Cu will reduce Ag (c) Cu will reduce I – (d) Cu will reduce Br2 75. Identify the unknown compounds. HNO H 2SO 4 (50 ° C)

Br FeBr3

Sn

3® 2 ® ® (C) ¾¾¾¾ (A) ¾¾¾ (B) ¾¾¾ HCl

(a) A ® Nitrobenzene, B ® Dinitrobenzene, C ® p-Bromoaniline (b) A ® C6H5SO3H2, B ® m-Benzenesulphonic acid, C ® m-Benzenesulphonate (c) A ® C6H5NO2, B ® m-Bromonitrobenzene, C ® m-Bromoaniline (d) A ® p-Nitrobenzene, B ® m-Trinitrobenzene, C ® m-Bromoaniline

EBD_7839 26

KVPY-SA

BIOLOGY 76. There are several types of enzyme catalysed reactions. In one type of enzyme catalysed reaction, in addition to the catalytic site to which the substrate (X) binds, the enzyme also has a site to which some other substance (Y) can bind. When Y binds to such an enzyme, the enzyme can still bind to the substrate but cannot convert it to the product. Which of the following will occur in such a case? (a) The affinity of the enzyme for the substrate will reduce. (b) Vmax of the reaction will decrease. (c) Y will alter the conformation of X. (d) The effect of Y can be overcome by increasing the concentration of X. 77. Match column-I (function) with column-II (types of enzymes) and select the correct option. Column-I Column-II (Function) (Types of enzymes) A. Enzyme catalyses I. Isomerases breakdown without addition of water. B. Enzyme catalyses II. Oxidoreductase the conversion of an aldose sugar to a ketose sugar. C. Enzyme catalyses III. Ligases transfer of electrons from one molecule to another. D. Enzyme catalyses IV. Lyases bonding of two components with the help of ATP.

1 2 3 4 5 6 7 8 9 10

(b ) (d ) (c) (a) (a) (c) (b ) (d ) (c) (b )

11 12 13 14 15 16 17 18 19 20

(b ) (d ) (a) (b ) (d ) (a) (d ) (a) (c) (d )

21 22 23 24 25 26 27 28 29 30

(a) A – I; B – IV; C – III; D – II (b) A – I; B – IV; C – II; D – III (c) A – IV; B – I; C – II; D – III (d) A – IV; B – I; C – III; D – II 78. If alveolar ventilation is 4200 mL/min, respiratory frequency is 12 breaths per minute and tidal volume is 500 mL, what is anatomical dead space ventilation? (a) 1800 mL/min.

(b) 6000 mL/min.

(c) 350 mL/min.

(d) 1200 mL/min.

79. Total population of 800 individuals which formed F2 population (9 : 3 : 3 : 1) of a cross between yellow round and green wrinkled. Find the number of plants with yellow and wrinkled seeds. (a) 150

(b) 400

(c) 800

(d) 300

80. Choose the correct option with appropriate medium of circulation and transport against each animal. Column-I

Column-II

A.

Hydra

(i)

Water surrounding the body

B.

Octopus

(ii) Haemolymph

C.

Prawn

(iii) Blood

(a) A – (iii)

(b) B – (iii)

(c) B – (ii)

(d) C – (i)

ANS W ER KEYS Part-I (b) 31 (c) 41 (d ) (a) 32 (b) 42 (d ) (d) 33 (a) 43 (b ) (b) 34 (d) 44 (b ) (d) 35 (a) 45 (c) (b) 36 (a) 46 (d ) (b) 37 (b) 47 (a) (b) 38 (c) 48 (a) (c) 39 (a) 49 (d ) (d) 40 (a) 50 (c)

51 52 53 54 55 56 57 58 59 60

(a) (d ) (c) (d ) (c) (b ) (b ) (a) (a) (d )

61 62 63 64 65 66 67 68 69 70

Part-II (a) 71 (a) 72 (d ) 73 (b ) 74 (b ) 75 (a) 76 (a) 77 (c) 78 (a) 79 (c) 80

(c) (c) (d) (d) (c) (b) (c) (a) (a) (b)

41 Time : 3 Hours

Stream-SA

Maximum Marks : 100

INSTRUCTIONS 1. 2. 3. 4.

There are 80 questions in this paper. The question paper contains two parts; Part–I (1 Mark Questions) and Part–II (2 Marks Questions). There are four sections; Mathematics, Physics, Chemistry and Biology in each part. There are four options given with each question, only one of them is correct. For each incorrect answer 0.25 Mark in Part–I and 0.5 Mark in Part–II will be deducted. PART-I (1 MARK QUESTIONS) MATHEMATICS

1.

2.

3.

4.

In an international convention, 5 participants from each of USA, China and Russia were arranged around a circle. In how many ways this arrangement can be made if no two participants from USA are together? (a) (9!)(10P5) (b) (9!)(10C5) (c) (3!)(5!)(5!)(5!) (d) None of these Find the smallest number that is divisible by all the numbers from 1 to 15 except one number. (a) 180180 (b) 120120 (c) 32760 (d) 27720 Given real numbers and b, if a =

x+3 , 4

2x + 1 7 , b < < 2a , find the range of 3 3 the value of x.

5.

6.

5 < x NH 2 2 2 2 (d) NO + NH = 1 mole (c) NO = NH 2

2

2

2

32. The combination of plots which does not represent isothermal expansion of an ideal gas is:

(A)

(B)

(C)

(D)

(a) (B) and (D) (b) (A) and (C) (c) (B) and (C) (d) (A) and (D) 33. When steam reacts with red hot coke to form CO2 and hydrogen : (a) Water acts as an oxidising agent. (b) Water acts as a reducing agent. (c) Carbon acts as an oxidising agent. (d) There is no oxidation or reduction. 34. Which of the following is not correct for electronic distribution in the ground state ?

(c)

– 36. A substance on treatment with dil.H2 SO 4 liberates a colourless gas which turns acidified dichromate solution green. The reaction indicates the presence of (a) CO 32- (b) S2 - (c) SO 32 - (d) NO 337. What is the correct order of electronegativity? (a) M– < M2– < M3– < M4– (b) M– > M2– > M3– > M4– (c) M– < M2– > M3– < M4– (d) M4– < M2– < M3– < M– 38. The difference in number of water molecules in Gypsum and Plaster of Paris is 1 (a) 5 (b) 2 (c) 1/2 (d) 1 2 OH

39. The IUPAC name of

40.

41.

(a) Co : [Ar] (b) Ni : [Ar] 42.

(c) Cu : [Ar] (d) All of the above 35. The non aromatic compound among the following is : (a)

(b)

S

(d)

43.

O OH

is:

(a) 2-Carboxyphenol (b) 2-Hydroxybenzoic acid (c) 1-Carboxy-2-hydroxybenzene (d) 2-Carboxy-1-hydroxybenzene The bond having the highest bond energy is : (a) C = C (b) C = S (c) C = O (d) P = N A buffer solution is prepared by mixing 0.1 M ammonia and 1.0 M ammonium chloride. At 298 K, the pKb of NH4OH is 5.0. The pH of the buffer is (a) 10.0 (b) 9.0 (c) 6.0 (d) 8.0 The correct order of increasing chemical reactivity is – (a) Zn < Fe < Mg < K (b) Fe < Mg < Zn < K (c) Fe < Mg < K < Zn (d) Fe < Zn < Mg < K The shape of methyl carbanion is similar to that of – (a) BF3 (b) NH3 (c) Methyl free radical (d) Methyl carbocation

EBD_7839 32 44.

45.

46.

47. 48.

49.

50.

KVPY-SA Calculate the standard enthalpy change (in kJ mol–1) for the reaction H2(g) + O2(g) ® H2O2(g), given that bond enthalpy of H–H, O=O, O–H and O–O (in kJ mol–1) are respectively 438, 498, 464 and 138. (a) – 130 (b) – 65 (c) + 130 (d) – 334 In pyrophosphoric acid, H4P2O7, number of s and dp – pp bonds are respectively (a) 8 and 2 (b) 6 and 2 (c) 12 and zero (d) 12 and 2 BIOLOGY Smoking destroys the cilia in the respiratory passageways. This (a) makes it harder to move air in and out of the lungs. (b) decreases the surface area for respiration. (c) slows blood flow through lung blood vessels. (d) makes it harder to keep the lungs clean. Cholesterol is required for the synthesis of: (a) Relaxin (b) Vitamin E (c) Oestradiol (d) All of these What pathway is taken by water and solutes when they travel through a nephron? (a) Glomerulus ® Bowman’s capsule ® proximal tubule ® loop of Henle ® distal tubule ® collecting ducts (b) Bowman’s capsule ® glomerulus ® distal tubule ® loop of Henle ® proximal tubule ® collecting ducts (c) Glomerulus ® Bowman’s capsule ® distal tubule ® loop of Henle ® proximal tubule ® collecting ducts (d) Glomerulus ® Bowman’s capsule ® proximal tubule® collecting ducts ® distal tubule ® loop of Henle Gram-negative cells stain pink because (a) they have specialised lipids in their cell walls. (b) their peptidoglycan layer is thin. (c) their peptidoglycan layer is thick. (d) they are receptive to antibiotics. Which one of the following is common to glycolysis as well as Krebs cycle in eukaryotes? (a) Substrate level phosphorylation (b) Photophosphorylation (c) Localisation in mitochondria (d) Production of FADH2

51.

52.

53.

54.

From the experiments carried out by Avery, MacLeod and McCarty by using various enzymes, which of the following results prominently proved that DNA is the transforming material? (a) DNA of heat killed ‘S’ + R type + DNAase ® non-virulent strain (b) DNA of heat killed ‘S’ + R type + RNAase ® virulent strain (c) DNA of heat killed ‘S’ + R type ® virulent strain (d) DNA of heat killed ‘S’ + R type + Protease ® virulent strain Match the column-I with column-II and choose the correct option. Column-I Column-II A. Human embryos I. Chemical evolution have gill B. Oparin and II. Stimulation Haldane experiment C. Miller and Urey III. Wings of bird and butterfly D. Analogous IV. Ontogeny repeats organs phylogeny (a) A – III; B – IV; C – II; D – I (b) A – II; B – I; C – IV; D – III (c) A – IV; B – I; C – II; D – III (d) A – IV; B – I; C – III; D – II Baculoviruses are excellent candidates for (a) species-specific, narrow spectrum pesticidal applications. (b) species-specific, broad spectrum pesticidal applications. (c) species-specific, narrow spectrum insecticidal applications. (d) species-specific, broad spectrum insecticidal applications. The most abundant structural polysaccharide is cellulose and the second largest structural polysaccharide is: (a) Chitin (b) Mannan (c) Glycogen (d) Hyaluronic acid

MOCK TEST-4

33

55. “A” cells start division and enter in “B” stage of meiotic division and get temporarily “C” at this stage, called “D”. Identify A, B, C and D. (a) A: Oogonia; B: Metaphase I; C: Arrested; D: Primary oocyte. (b) A: Oogonia; B: Anaphase I; C: Released; D: Secondary oocyte. (c) A: Oogonia; B: Prophase I; C: Arrested; D: Primary oocyte. (d) A: Oogonia; B: Telophase I; C: Released; D: Secondary oocyte. 56. Select the correct order of geological time scale of earth. (a) Palaeozoic ® Archaeozoic ® Coenozoic (b) Archaeozoic ® Palaeozoic ® Proterozoic (c) Palaeozoic ® Mesozoic ® Coenozoic (d) Mesozoic ® Archaeozoic ® Proterozoic 57. PS I occurs in: (a) Appressed part of granal thylakoids (b) Appressed and non-appressed part of granal thylakoids (c) Stroma (d) Stroma thylakoids and non-appressed part of granal thylakoids

58. Hormones that are secreted by one endocrine gland and control the activities of another endocrine glands are called: (a) Growth hormones (b) Steroid hormones (c) Tropic hormones (d) Peptide hormones 59. In terms of DNA and RNA structure, what is a nucleotide? (a) A nucleotide is a heterocyclic base. (b) A nucleotide is a sugar molecule covalently bonded to a heterocyclic base. (c) A nucleotide is a sugar molecule bonded to phosphate group and a heterocyclic base. (d) A nucleotide is a heterocyclic base bonded to phosphate group. 60. A human bone marrow cell, in prophase of mitosis, contains 46 chromosomes. How many chromatids does it contain altogether? (a) 46 (b) 92 (c) 23 (d) 23 or 46, depending when you look during prophase

PART-II (2 MARKS QUESTIONS) MATHEMATICS 61. PQRS is a rectangle in which PQ = 2 PS, T and U are mid points of PS and PQ respectively. QT and US intersect at V. The area of QRSV divided by area of PQT is9 7 6 8 (a) (b) (c) (d) 7 4 5 3 62. Let us define a set S such that any element of S can be expressed as sum of a two digit number and number formed by reversing that 2 digit number. What is the HCF of all the elements of S ? (a) 3 (b) 9 (c) 7 (d) 11 63. A semi-circle of diameter 1 unit sits at the top of a semi-circle of diameter 2 units. The shaded region inside the smaller semi-circle but outside the larger semi-circle is called a lune. The area of the lune is

(a)

p 3 6 4

(c)

3 p 4 12

(b)

3 p 4 24

(d)

3 p 4 8

64. Let ABC be a triangle and D be the mid-point of side BC. Suppose ÐDAB = ÐBCA and ÐDAC = 15°. Show that ÐADC is obtuse. Further, if O is the circumcentre of ADC, then DAOD is :

65.

(a) Equilateral

(b) Right angled

(c) Isosceles

(d) None of these

7+4 3 - 7-4 3 =

(a) 4

(b) 2

(c)

4 3 (d)

2 3

EBD_7839 34

KVPY-SA

PHYSICS 66. Figure shows a smooth spherical ball of mass m striking two identical equilateral triangular wedges, each of mass m. The velocity of ball at the instant of impact is v0. If e is the coefficient of restitution, then velocity of either wedge after impact is m v0 m

(a)

3 (1 + e)v0 5

5 (1 + e)v0 3

(b)

ev0 3 67. An upright U-tube manometer with its limbs 0.6m high and spaced 0.3m apart contains a liquid to a height of 0.4m in each limb. If the U-tube is rotated at 10 radians/second about a vertical axis at 0.1m from one limb. Choose the correct statement/s

3(1 + e)v0

(c)

(d)

10rad/s 0.4m

z2 z1

zmin 0.1m

é æT ö2ù A (a) ê1 - ç M ÷ ú ë è T ø û Mg

é æ T ö2ù A (b) ê1 - ç ÷ ú êë è TM ø úû Mg

éæ T ö 2 ù A (c) ê ç M ÷ - 1ú ëè T ø û Mg é æ T ö 2 ù Mg (d) ê ç M ÷ - 1ú ëè T ø û A

69. Two spheres P and Q of equal radii have densities r1 and r2 , respectively. The spheres are connected by a massless string and placed in liquids L1 and L2 of densities s1 and s2 and viscosities h1 and h2, respectively. They float in equilibrium with the sphere P in L1 and sphere Q in L2 and the string being taut (see figure). If ur sphere P alone in L2 has terminal velocity V P , ur and Q alone in L1 has terminal velocity V Q , then ur VP h ur = 1 (a) L1 h2 VQ

0.2m

I. The value of z1 = 0.324 II. The value of z2 = 0.477 III. The value of zmin = 0.273 IV. The value of z1 + z2 = 0.8 (a) I only (b) I and II only (c) III only (d) I, II, III and IV 68. A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to TM. If the Young’s modulus of the material of the wire is Y then (g = gravitational acceleration)

1 is equal to: Y

(b) (c) (d)

ur VP h ur = 2 h1 VQ

L2

ur ur V P .V Q > 0 ur ur V P .V Q = 0

70. Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and / or record time for different number of oscillations. The observations are shown in the table. Least count for length = 0.1 cm least count for time = 0.1 s

MOCK TEST-4

Student I

35

Total time Numbe r of Le ngth of the for (n) oscillations oscil lations pen dulum(cm) (n) (s) 64.0 8 128.0

Time pe riod (s) 16.0

II

64.0

4

64.0

16.0

III

20.0

4

36.0

9.0

If EI, EII and EIII are the percentage errors in g, æ Dg ö i.e., ç ´ 100 ÷ for students I, II and III, è g ø

respectively, then (a) (b) (c) (d)

EI = 0 EI is minimum EI = EII EII is minimum CHEMISTRY

71. Which of the following reagents convert propene to 1-propanol? (a) H2O, H2SO4 (b) Aqueous KOH (c) MgSO4, NaBH4/H2O (d) B2H6, H2O2, OH–

® SO2 ; DH = -298.2 kJ ... (i) 72. If S + O2 ¾¾ 1 SO 2 + O2 ¾¾ ® SO3 ; DH = – 98.7 kJ ... (ii) 2 SO3 + H 2 O ¾¾ ® H 2SO 4 ; DH = -130.2 kJ ... (iii) 1 H 2 + O 2 ¾¾ ® H 2O; DH = -287.3 kJ ... (iv) 2 Then the enthalpy of formation of H2SO4 at 298 K is (a) –814.4 kJ (b) –650.3 kJ (c) –320.5 kJ (d) –233.5 kJ 73. A metal oxide has the formula Z2O3. It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is (a) 27.9 (b) 159.6 (c) 79.8 (d) 55.8

74. Bond order normally gives idea of stability of a molecular species. All the molecules viz. H2, Li2 and B2 have the same bond order yet they are not equally stable. Their stability order is (a) H2 > B2 > Li2 (b) Li2 > H2 > B2 (c) Li2 > B2 > H2 (d) H2 > Li2 > B2 75. The stability of 2, 3 - dimethyl but - 2- ene is more than 2- butene. This can be explained in terms of (a) resonance (b) hyperconjugation (c) electromeric effect (d) inductive effect BIOLOGY 76. Analysis of a protein sequence from a recently mutated gene showed that the newly expressed protein had a completely different sequence from the wild-type protein. This was due to a: (a) Missense mutation (b) Nonsense mutation (c) Silent mutation (d) Frameshift mutation 77. A student while studying the anatomy of leaves of four specimens (A-D), observed the following characters: A. Reticulate venation and no bundle sheath B. Parallel venation and bundle sheath containing chloroplasts C. Reticulate venation and bundle sheath containing chloroplasts D. Parallel venation and bundle sheath without chloroplasts Identify the type of plants. (a) A and C are C4 dicots, while B and D are C4 monocots. (b) A and C are C3 dicots, while B and D are C3 monocots. (c) A and D are C 3 dicot and monocot, respectively, while B and C are C4 monocot and dicot, respectively. (d) A and D are C 4 dicot and monocot, respectively, while B and C are C3 monocot and dicot, respectively.

EBD_7839 36 78.

79.

1 2 3 4 5 6 7 8 9 10

KVPY-SA Because of the relatively high altitude of Antonito, Colorado, the town has a normal barometric pressure of about 600 mm Hg rather than 760 mm Hg as at sea level. The partial pressure of oxygen in Antonito’s air is approximately (a) 75 mm Hg (b) 126 mm Hg (c) 160 mm Hg (d) 760 mm Hg The following structures are ‘analogous’ from the evolutionary point of view except: (a) Sting of scorpion and honeybee (b) Wings of an owl and wings of a butterfly (c) Spines of a cactus and quills of a porcupine (d) Tendrils of pea and spines of barberry

(a) (d ) (a) (c) (b ) (b ) (c) (c) (d ) (a)

11 12 13 14 15 16 17 18 19 20

(a) (a) (b) (c) (c) (a) (b) (b) (b) (c)

21 22 23 24 25 26 27 28 29 30

80.

Match the items given in column-I with those in column-II and select the correct option given below. Column-I Column-II A. Tidal volume I. 2500 – 3000 mL B. Inspiratory II. 1100 – 1200 mL Reserve volume C. Expiratory III. 500 – 550 mL Reserve volume D. Residual volume IV. 1000 – 1100 mL (a) A – III; B – II; C – I; D – IV (b) A – III; B – I; C –IV; D – II (c) A – IV; B – III; C – II; D – I (d) A – I; B – IV; C – II; D – III

ANS W ER KEYS Part-I (c) 31 (c) 41 (d) (d ) 32 (a) 42 (d) (c) 33 (a) 43 (b) (b ) 34 (d ) 44 (a) (d ) 35 (d ) 45 (d ) (b ) 36 (c) 46 (d) (d ) 37 (b ) 47 (c) (c) 38 (d ) 48 (a) (c) 39 (b ) 49 (b) (c) 40 (c) 50 (a)

51 52 53 54 55 56 57 58 59 60

(a) (c) (c) (a) (c) (c) (d ) (c) (c) (b )

61 62 63 64 65 66 67 68 69 70

Part-II (d ) 71 (d ) 72 (b ) 73 (a) 74 (d ) 75 (a) 76 (d ) 77 (c) 78 (a) 79 (b ) 80

(d) (a) (d) (d) (b) (d) (c) (b) (d) (b)

51 Time : 3 Hours

Stream-SA

Maximum Marks : 100

INSTRUCTIONS 1. 2. 3. 4.

There are 80 questions in this paper. The question paper contains two parts; Part–I (1 Mark Questions) and Part–II (2 Marks Questions). There are four sections; Mathematics, Physics, Chemistry and Biology in each part. There are four options given with each question, only one of them is correct. For each incorrect answer 0.25 Mark in Part–I and 0.5 Mark in Part–II will be deducted. PART-I (1 MARK QUESTIONS) MATHEMATICS

1.

5.

Let A1, B1, C1, D1 be the midpoints of the sides of a convex quadrilateral ABCD and let A2, B2, C2, D 2 be the midpoints of the sides of the quadrilateral A 1B 1C 1D 1. If A 2B 2C 2D 2 is a rectangle with sides 4 and 6, and the product of the lengths of the diagonals of ABCD is denoted

é n ù by n then find ê ú , where [·] denotes the ë 100 û

mid-points of diagonals AC and BD respectively. Then PQ = (a)

2.

3.

4.

1 (AB + CD) 3

(b)

1 (AB – CD) 2

1 1 (AB – CD) (d) (AB + CD) 3 2 Let X be a four-digit positive integer such that the unit digit of X is prime and the product of all digits of X is also prime. How many such integers are possible? (a) 4 (b) 8 (c) 12 (d) 24 If number 368A37982B is divisible by 36, here A and B are digits then what could be the sum of all the values of A? (a) 12 (b) 21 (c) 24 (d) 16 Consider the set S = {2, 3, 4, ...., 2n + 1), where n is a positive integer larger than 2007. Define, X as the average of the odd integers in S and, Y as the average of the even integers in S. What is the value of X – Y? (c)

6.

greatest integer function. (a) 1 (b) 2 (c) 3 (d) 4 If unit digit of a! + b! + c! is 9 then find the value of {(a!)(b!)(c!)} (a) 8 (b) 5040 (c) 0 (d) 12 2 2 If x + xy + x = 14, y + xy + y = 28 and z = x + y, then the sum of all possible values of z is: (a) 7 (b) 6 (c) –1 (d) –6 N is a natural number such that 6000 < N < 8000. The sum of number N and a new number formed by reversing its digits is S. if S is P times sum of digits of number N then find the minimum value of P. (a) 330.25 (b) 335.50 (c) 332.75 (d) None of these

A trapezium ABCD in which P and Q are the

7.

8.

(a) 1

(b) n/2

(c) (n + 1)/2n

(d) 2008

EBD_7839 38 9.

KVPY-SA In figure, ABCD is a rectangle of 20 cm × 10 cm. A semi-circle is drawn with centre at O and radius

14.

rope which is wound round a wheel of diameter

10 2 cm and it passes through A and B. Find the area of the shaded region in the figure. 20 cm

D

77 cm. Given that the bucket ascends in 1 min. 28 seconds with a uniform speed of 1.1 m/sec, calculate the number of compolete revolutions

C

P A

the wheel makes in raising the bucket.

10 cm

(a) 38

(b) 40

B

(c) 45

(d) None of these

15. 10 2 cm

L

10.

11.

12.

13.

10 2 cm

O 20 2 cm

A bucket is raised from a well by means of a

M

(a) (p – 10) × 50

pö æ (b) ç 2 - ÷ ´ 100 è 2ø

pö æ (c) ç 3 - ÷ ´ 100 è 2ø

pö æ (d) ç 5 - ÷ ´ 100 è 2ø

(a) nx1 + b (c) n ( x1 )

é l1 l2 ù êl + l ú 1 ú where [n] denotes greatest value of ê 2 ë 10 û integer function of n. (a) 24 (b) 25 (c) 30 (d) 32

n -1

+a

(b) n ( x1 )

n -1

(d) n ( x1 )

n -1

+b

PHYSICS 16.

Find the remainder when x2018 + x2017 + x2016 +...+ x2 + x + 1 is divided by x + 1. (a) 1 (b) 0 (c) –1 (d) 2 N is a 99 digit number comprises of only 4’s and 8’s. If N is divisible by 72 then find the minimum value of sum of digits of the number N. (a) 360 (b) 369 (c) 387 (d) None of these If XY is a line parallel to side BC of DABC. BE || AC and CF || AB meets XY in E and F respectively. Then ar DABE = ? (a) ar DACY (b) ar DAYF (c) ar DACF (d) ar (Quad. BCFX) Let a, b are the roots of the equation l (x2 – x) + x + 5 = 0. If l1 and l2 are two values of l for which the roots a b 4 a, b are related by + = , then find the b a 5

If x1, x2, x3,....., xn are the roots of x n + ax + b = 0, then the value of (x1 – x2) (x1 – x3) (x1 – x4) ... (x1 – xn) is equal to

A particle which is simultaneously subjected to two perpendicular simple harmonic motions represented by; x = a1 cos wt and y = a2 cos 2 wt traces a curve given by:

(a)

y

y

a2

a2

a1

O

x

(b)

O a1

y

y

(c)

17.

a1

x

a2 O

x

(d)

a1

a2 O

x

For a particle executing simple harmonic motion, which of the following statement is not correct ? (a) The total energy of the particle always remains the same. (b) The restoring force of always directed towards a fixed point. (c) The restoring force is maximum at the extreme positions. (d) The acceleration of the particle is maximum at the equilibrium position.

MOCK TEST-5 18. There are two processes ABC and DEF. In which of the process is the amount of work done by the gas greater?

39 21. What do you conclude from the graph about the frequency of KE, PE and SHM? Energy

Total energy

A

B KE

PE 0

(a) ABC (b) DEF (c) Equal in both processes (d) It cannot be predicted 19. A person sitting at the rear end of the compartment throws a ball towards the front end. The ball follows a parabolic path. The train is moving with uniform velocity of 20 ms–1. A person standing outside on the ground also observes the ball. How will the maximum height (hm) attained and the ranges (R) seen by the thrower and the outside observer compare each other? (a) same hm different R (b) same h m and R (c) different h m same R (d) different h m and R 20. A river flow with a speed more than the maximum speed with which a person can swim in the still water. He intends to cross the river by shortest possible path (i.e., he wants to reach the point on the opposite bank which directly opposite to the starting point). Which of the following is correct statement? (I) He should start normal to the river bank (II) He should start in such a way that, he moves normal to the bank, relative to the bank. (III) He should start in a particular (calculated) direction making an obtuse angle with the direction of water current (IV) The man cannot cross the river, in that way (a) I only (b) II only (c) III only (d) IV only

T/4

2T/4

3T/4

4T/4

t

(a) Frequency of KE and PE is double the frequency of SHM (b) Frequency of KE and PE is four times the frequency SHM. (c) Frequency of PE is double the frequency of K.E. (d) Frequency of KE and PE is equal to the frequency of SHM. 22. Two wooden blocks A and B float in a liquid of density rL as shown. The distance L and H are shown. After some time, block B falls into the liquid, so that L decreases and H increases. If density of block B is rB, find the correct option. (a) rL = rB

B

(b) rL > rB

A

L H

(c) rL < rB

(d) Unpridictable 23. Trajectories are shown in figure are for three kicked footballs, ignoring the effect of the air on the footballs. If T1, T2 and T3 are their respective time of flights then: y 1

2 R 1.5 R 2R

(a) T1 > T3 (c) T2 =

T3 2

3 x

(b) T1 < T3 (d) T1 = T2 = T3

EBD_7839 40 24.

25.

KVPY-SA A ray of light travelling inside a rectangular glass block of refractive index 2 is incident on the glass-air surface at an angle of incidence of 45º. The refractive index of air is one. Under these conditions the ray will (a) emerge into the air without any deviation (b) be reflected back into the glass (c) be absorbed (d) emerge into the air with an angle of refraction equal to 90º A light ray falls on a square glass slab as shown in the diagram. The index of refraction of the glass, if total internal reflection is to occur at the vertical face, is equal to :

( 2 + 1) 2 5 (b) 2 3 (c) 2

28.

45° Incident ray

(a)

(a)

(c)

3 2 The period of oscillation of a simple pendulum is 2p L / g . Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the percentage error in the determination of g ? (a) 4% (b) 1% (c) 6% (d) 3% A spaceship is sent to investigate a planet of mass M and radius R. While hanging motionless in space at a distance 5R from the centre of the planet, the spaceship fires an instrument package with speed v 0 as shown in the figure. The package has mass m, which is much smaller than the mass of the spaceship. For what angle q will the package just graze the surface of the planet? v0

29.

(d)

26.

27.

æ 8GM ö -1 (c) sin ç 5 1 + 2 ÷ ç 5v 0 R ÷ø è æ 8GM ö (d) cos -1 ç 5 1 + 2 ÷ ç 5v 0 R ÷ø è An artificial satellite of mass m of a planet of mass M, revolves in a circular orbit whose radius is n times the radius R of the planet. In the process of motion, the satellite experiences a slight resistance due to cosmic dust. Assuming resistance force on satellite depends on velocity as F = av2 where ‘a’ is constant. Calculate how long the satellite will stay in orbit before it falls onto the planet’s surface?

m

R 5R

æ1 8GM ö (a) s in -1 ç 1 + ÷ ç5 5v 20 R ÷ø è æ1 8GM ö (b) cos -1 ç 1 + 2 ÷ ç5 5v 0 R ÷ø è

M

30.

m R

(

)

n -1

a GM mR

(

)

n +1

(b)

(d)

a GM

mR

(

)

n -1

a GM

m R

(

n -1

)

a GM

Two satelites A and B of equal mass move in the equatorial plane of the earth, close to earth’s surface. Satellite A moves in the same direction as that of the rotation of the earth while satellite B moves in the opposite direction. Calculate the ratio of the kinetic energy of B to that of A in the reference frame fixed to the earth. (g = 9.8 m s–2 and radius of the earth = 6.37 × 106 km) (a) 2.17 (b) 1.27 (c) 7.12 (d) 1.72 Two soap bubbles of different size are formed at two ends of a tube as shown in fig. If the stop cock is opened then which bulb shrinks ? T

r

R

(I) Smaller bubble shrinks and the bigger bubble expands. (II) Bigger bubble shrinks and the smaller bubble expands. (III) Both bubble shrinks. (IV) Both bubble expands. Choose the correct statement (a) I only (b) II only (c) III (d) IV

MOCK TEST-5

41 CHEMISTRY

31. Which of the following represents the correct order of stability of the given carbocations ? | + (a) F3C > F3C - C+ > CH3 | | + + (b) H3C > F3C - C + > F3C | | + + (c) F3C - C+ > F3C > H3C | | + + (d) F3C - C+ > H3C > F3C | 32. Caffiene has a molecular mass of 194. If it contains 28.9% by mass of nitrogen, number of atoms of nitrogen in one molecule of caffeine is: (a) 4 (b) 6 (c) 2 (d) 3 33. Global warming is mainly due to increase of: (a) Methane and nitrous oxide in atmosphere. (b) Methane and CO2 in atmosphere. (c) Methane and O3 in atmosphere. (d) Methane and CO in atmosphere. 34. Which compound (s) out of the following is/are

37. Tautomerism will be exhibited by (a) (CH3)2NH (b) (CH3)3CNO (c) R3CNO2 (d) RCH2NO2 38. The beans are cooked earlier in pressure cooker, because, (a) boiling point increases with increasing pressure. (b) boiling point decreases with increasing pressure. (c) internal energy is not lost while cooking in pressure cooker. (d) extra pressure of pressure cooker softens the beans. 39. The correct order for acid strength of compounds CHºCH, CH3–CºCH and CH2=CH2 is as follows: (a) CHºCH > CH2=CH2 > CH3–CºCH (b) CH3–CºCH > CHºCH > CH2=CH2 (c) CH3–CºCH > CH2=CH2 > HCºCH (d) HCºCH > CH3–CºCH > CH2=CH2

40. A solution contains 10 mL 0.1 N NaOH and 10 mL 0.05 N H2SO4, pH of this solution is : (a) less than 7 (b) 7 (c) zero (d) greater than 7 41. The maximum number of 90º angles between bond pair-bond pair of electrons is observed in not aromatic? (a) dsp2 hybridization (b) sp3d hybridization (c) dsp3 hybridization (d) sp3d2 hybridization ˆˆ† C + D + heat, has 42. The reaction A + B ‡ˆˆ – + + reached equilibrium. The reaction may be made (A) (B) (C) (D) to proceed forward by (a) (B), (C) and (D) (b) (C) and (D) (a) adding more C (b) adding more D (c) (B) (d) (A) and (C) (c) decreasing the temperature 35. What is the maximum number of electron in a (d) increasing the temperature subshell that can have the quantum numbers 43. Which of the following compounds contains 1°, n = 3 and l = 2? 2°, 3° as well as 4° carbon atoms ? (a) 2 (b) 5 (c) 6 (d) 10 (a) Neopentane 36. Given 3+ – 2+ o (b) 2-Methylpentane Fe (aq) + e ® Fe (aq); E = + 0.77 V 3+ – o (c) 2,3-Dimethylbutane Al (aq) + 3e ® Al(s); E = – 1.66 V – – o (d) 2,2,3-Trimethylpentane Br2(aq) + 2e ® 2Br (aq); E = + 1.09 V 44. H2O is polar, whereas BeF2 is not. It is because Considering the electrode potentials, which of (a) EN of F > EN of O. the following represents the correct order of (b) H2O involves H-bonding, whereas BeF2 is reducing power? a discrete molecule. 2+ 2+ (a) Fe < Al < Br (aq)(b) Br (aq) < Fe < Al (c) H 2O is linear and BeF2 is angular. (c) Al < Br - < Fe2+ (d) Al < Fe2+ < Br - (aq) (d) H2O is angular and BeF2 is linear.

EBD_7839 42 45.

KVPY-SA Unlike the other elements of its group carbon and silicon does not form MX2 type molecules because : (a) Energetically this is not possible. (b) They undergoes catenation. (c) They are non-metallic. (d) They do not contain d-orbital. BIOLOGY

46.

47.

48.

49.

50.

A segment of DNA molecule contains 200 guanine and 200 thymine bases. What will be the total number of nucleotides in this segment of DNA? (a) 400 (b) 200 (c) 800 (d) 100 If a grasshopper cell contained 200 units of DNA during G2, what would be the quantity of DNA in one of the grasshopper cells at the end of telophase II of meiosis? (a) 50 units (b) 100 units (c) Between 50 and 100 units (d) 200 units Which of the following is the correct group of vestigial organs in man? (a) Nictitating membrane, ear muscles, eyelids and coccyx. (b) Appendix, coccyx, ear muscles and elbow joint. (c) Wisdom tooth, coccyx, body hair and eyelids . (d) Wisdom tooth, body hairs, nictitating membrane and vermiform appendix. Oxidative phosphorylation refers to: (a) anaerobic production of ATP (b) the citric acid cycle production of ATP (c) production of ATP by chemiosmosis (d) alcoholic fermentation In tertiary structure of DNA, what is a histone octamer? (a) A complex consisting of eight positively charged histone proteins (two of each H2A, H2B, H3 and H4) that aid in the packaging of DNA.

51.

52.

53.

54.

(b) A complex consisting of eight negatively charged histone proteins (two of each H2A, H2B, H3 and H4) that aid in the packaging of DNA. (c) A complex consisting of nine positively charged histone proteins (H1 and two of each H2A, H2B, H3 and H4) that aid in the packaging of DNA. (d) A complex consisting of nine negatively charged histone proteins (H1 and two of each H2A, H2B, H3 and H4) that aid in the packaging of DNA. Rate of respiration is directly proportional to (a) Concentration of oxygen in blood (b) Concentration of carbon dioxide in blood (c) Oxygen in trachea (d) Diaphragm expansion Haem protein complexes that act as oxidising agents during plant respiration are known as: (a) Chlorophyll (b) Haemoglobin (c) Myoglobin (d) Cytochrome Which of the following pair is correctly matched? (a) Anaphase I: Homologous chromosomes are separated. (b) Metaphase I: Pairing of maternal and paternal chromosomes. (c) Interphase: A nuclear envelope encloses each haploid set of chromosomes. (d) Prophase I: Non-homologous chromosomes are separated. STD/VD/UTI are: (a) devices which are used to delay the pregnancy. (b) infections which are caused by food contamination. (c) diseases which are transmitted through sexual intercourse. (d) action plans and programmes to create awareness about various reproductive health related problems.

MOCK TEST-5

43

55. B-lymphocytes are primarily involved in:

58. Sucrose is composed of:

(a) Humoral immunity

(a) Glucose and fructose

(b) Autoimmune disorders

(b) Glucose and glycogen

(c) Graft rejection

(c) Two molecules of glucose

(d) Cell-mediated immunity

(d) Glycogen and fructose

56. The differences in D and L forms of sugars are based on (a) Spatial arrangement of carbon atom to which the functional group is attached. (b) Configuration of H and OH groups around the penultimate carbon. (c) If the plane of polarised light is rotated to left (levo), it is L form; and if shifted to right (dextro), it is D form. (d) Special arrangement of H and OH groups at the last carbon atom. 57. Splitting of water in photosynthesis is called:

59. The Miller-Urey abiotic synthesis experiment (and other subsequent, similar experiments) showed that (a) simple organic molecules can form spontaneously under conditions like those thought to prevail early in the earth’s history. (b) the earliest life forms introduced large amounts of oxygen to the atmosphere. (c) life can be created in a test tube. (d) long chains of DNA can form under abiotic conditions. 60. Life on earth is carbon based. Similar molecules could be formed with:

(a) Dark reaction

(b) Photolysis

(a) Potassium

(b) Aluminum

(c) Electron transfer

(d) Phototropism

(c) Iron

(d) Silicon

PART-II (2 MARKS QUESTIONS) MATHEMATICS A

61. In the figure, the two circles touch each other at C. The diameter AB. of the bigger circle is tangent to the smaller circle at D. If DE bisects ÐADC, find q

P Q E

H B

(a) 24°

(b) 38°

(c) 45°

(d) 52°

62. In the figure BE ^ AC. AD is any line from A to BC intersecting BE in H. Also P, Q and R are mid points of AH, AB and BC. Then find ÐPQR.

D

(a) 30º (b) 60º (c) 90º (d) cannot be determined

R

C

EBD_7839 44 63.

64.

KVPY-SA Let A = {q : sin(q) = tan(q)} and B = {q : cos(q) = 1} be two sets. Then : (a) A = B (b)

AË B

(c)

BË A

A plane is in level flight at constant speed and each of its two wings has an area of 25 m2. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m–3).

(d)

A Ì B and B - A ¹ f

(a) 440 kg

In the given figure ABCDE is a regular pentagon. BP is drawn parallel to AC and it meets DC extended at P. EQ is drawn parallel to AD and meets CD extended at Q. Then choose the correct option.

67.

68.

(b) 400 kg

(c) 4400 kg (d) 4044 kg Two positive charges of magnitude ‘q’ are placed, at the ends of a side (side 1) of a square of side ‘2a’. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge Q moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is (a) zero (b)

1 2 qQ æ 1 ö 1+ ÷ 4 pe 0 a çè 5ø

1 2qQ æ 2 ö (c) 4pe a çè1 ÷ 5ø 0

(a) ar (ABCDE) = ar DAPQ (b) ar DEDQ = ar DBAC 69.

(c) ar (ABCDE) = ar (PBEQ) (d) ar DBPC = ar DADE 65. If highest power of 7 in N! is k and that in (N + 1)! is K + 2 then how many values of N exist if N < 100. (a) 0 (b) 1 (c) 2 (d) 3 PHYSICS 66.

A car having a mass of 200 kg is rolling at a speed of 1 m/s towards a spring - stop system. If the spring is non - linear such that it develops 300 x2 N force for a deflection of x m. The maximum deceleration that the car A undergoes;

1 2qQ æ 1 ö (d) 4pe a çè1 ÷ 5ø 0 A long rectangular slab of transparent medium of thickness d is placed on a table with length parallel to x-axis and width parallel to y-axis. A ray of light is traveling along y-axis at origin. The refractive index µ of the medium varies x/d as, m = 1 + e .

The refractive index of air is 1. The value of x, where the ray intersects the upper surface of the slab-air boundary is

Glassy shaddy

d

q

x

(a) 1 m/s2

(b) 1.5 m/s2

(c) 2 m/s

(d) 2.5 m/s

2

2

(a) d ln 2 (c) 3d ln 2

(b) d ln 4 (d) none of these

MOCK TEST-5

45

70. A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the two equal end containing air at the same pressure P. When the tube is held at an angle of 60° with the vertical direction, the length of the air column above and below the mercury column are 46 cm and 44.5 cm respectively. Calculate the pressure P in centimeters of mercury. (The temperature of the system is kept at 30°C). (a) 75.4 cm (b) 57.4 cm (c) 70.4 cm (d) 44.5 cm

V(L)

(b) (22.4L, 273K)

T(K) V(L)

(30.6 L, 373K)

(c) (22.4L, 273K)

T(K)

V(L)

(d) (22.4L,

CHEMISTRY

273K)

(14.2 L, 373K) T(K)

71. The geometry of ammonia molecule can be best described as: (a) Nitrogen at one vertex of a regular tetrahedron, the other three vertices being occupied by three hydrogens. (b) Nitrogen at the centre of the tetrahedron, three of the vertices being occupied by three hydrogens. (c) Nitrogen at the centre of an equilateral triangle, three corners being occupied by three hydrogens. (d) Nitrogen at the junction of a T, three open ends being occupied by three hydrogens. 72. The oxidation states of sulphur in the anions SO 32 -, S2O24-, and S 2 O 62 - follow the order..

(a) S2O42– < SO32– < S2O62–

74. (A)

(B)

(C)

(D)

Decreasing order of rate of electrophilic aromatic substitution is: (a) A > B > C > D (b) A > C > B > D (c) B > A > C > D (d) B > C > A > D 75. The density (in g mL–1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (molar mass = 98 g mol–1) by mass will be (a) 1.45 (b) 1.64 (c) 1.88 (d) 1.22 BIOLOGY 76. Maltose can be broken down into glucose molecules by the enzyme maltase. Which of the following would slow the reaction rate?

(b) SO32– < S2O42– < S2O62– (c) S2O42– < S2O62– < SO32–

(a) Adding maltase

(d) S2O62– < S2O42– < SO32– 73. Which of the following volume (V) - temperature (T ) plots represents the behaviour of one mole of an ideal gas at one atmospheric pressure ? V(L)

(28.6 L, 373K)

(38.8 L, 373K)

(a) (22.4L, 273K)

(b) Adding maltose (c) Removing glucose (d) Diluting with water 77. From the cross AABBCC × aabbcc, how many different kinds of (i) F1 gametes, (ii) F2 genotypes and (iii) F2 phenotypes would be expected? (a) (i) 16, (ii) 24, (iii) 16 (b) (i) 8, (ii) 27, (iii) 8

T(K)

(c) (i) 8, (ii) 64, (iii) 16 (d) (i) 8, (ii) 32, (iii) 16

EBD_7839 46 78.

79.

KVPY-SA Tidal volume and Expiratory reserve volume of an athlete is 500 mL and 1000 mL respectively. What will be his Expiratory capacity if the residual volume is 1200 mL? (a) 1500 mL

(b) 1700 mL

(c) 2200 mL

(d) 2700 mL

If in a test tube, mRNA of human, tRNA of Rhizobium and ribosomes of sunflower are added along with sufficient amino acids and ATP, the polypeptide synthesised will be of the nature of: (a) body cells of human (c) Sunflower (d) Both (a) and (b)

(b ) (d ) (c) (c) (b ) (a) (b ) (a) (d ) (a)

11 12 13 14 15 16 17 18 19 20

(d ) (c) (b ) (b ) (c) (a) (d ) (b ) (a) (d )

Tay-Sachs disease, which is lethal, results from having the homozygous recessive condition of the responsible gene. Which one of the following statements is true? (a) Because homozygous recessive individuals die, the recessive allele will eventually be lost from the population. (b) Only homozygous dominant individuals will be able to survive and reproduce. (c) Heterozygous individuals will survive and may pass the recessive allele on to their offspring. (d) In the heterozygous condition, the dominant allele will overcome the recessive allele and only the dominant allele will be passed on to offspring.

(b) Rhizobium

1 2 3 4 5 6 7 8 9 10

80.

21 22 23 24 25 26 27 28 29 30

ANS W ER KEYS Part-I (a) 31 (b) 41 (d ) (b) 32 (a) 42 (c) (d) 33 (b) 43 (d ) (d) 34 (a) 44 (d ) (d) 35 (d) 45 (a) (d) 36 (d) 46 (c) (a) 37 (d) 47 (a) (a) 38 (a) 48 (d ) (b) 39 (d) 49 (c) (a) 40 (b) 50 (a)

51 52 53 54 55 56 57 58 59 60

(b ) (d ) (a) (c) (a) (b ) (b ) (a) (a) (d )

61 62 63 64 65 66 67 68 69 70

Part-II (a) 71 (c) 72 (b ) 73 (a) 74 (c) 75 (b ) 76 (c) 77 (d ) 78 (b ) 79 (a) 80

(b) (a) (c) (b) (d) (d ) (b) (a) (a) (c)

61 Time : 3 Hours

Stream-SA

Maximum Marks : 100

INSTRUCTIONS 1. 2. 3. 4.

There are 80 questions in this paper. The question paper contains two parts; Part–I (1 Mark Questions) and Part–II (2 Marks Questions). There are four sections; Mathematics, Physics, Chemistry and Biology in each part. There are four options given with each question, only one of them is correct. For each incorrect answer 0.25 Mark in Part–I and 0.5 Mark in Part–II will be deducted. PART-I (1 MARK QUESTIONS) MATHEMATICS

1.

4.

Let ABCD be a quadrilateral; X and Y be the midpoints of AC and BD respectively and the lines through X and Y respectively parallel to BD, AC meet in O. Let P, Q, R, S be the midpoints of AB, BC, CD, DA respectively. Consider the statements. (i)

quadrilaterals APOS and APXS have the same area.

(ii) the areas of the quadrilateral APOS, BQOP, CROQ, DSOR are all equal.

5.

Consider a set S = { 1, “2, 2, 2”2, ……. 1024}, consider another set P that is subset of S and has two distinct elements. If product of these two elements is more than or equal to 1024 the how many values of set P exist? (a) 55 (b) 110 (c) 140 (d) None of these A square ABCD is constructed inside a DPQR, having sides 10, 17, 21 as shown in figure. Find the approximation value of perimeter of square ABCD.

Then, which of the following is/are true?

2.

(a) Only (i)

(b) Only (ii)

(c) Only (i) and (ii)

(d) None

If a(b!) is completely divisible by 511 where a is a single digit no then find the minimum value of b. (a) 40

3.

(b) 45

(c) 50

(d) 46

If p = (k + 10)(k + 11)(k + 12)(k + 13)(k + 14) and q = p + 2 then which one of these is false for q. (a) q is even (b) when divided by 2 it leaves remainder 0 (c) q is divisible by 3 (d) q is not prime

(a) 28

(b) 23.2

(c) 25.4

(d) 28.8

EBD_7839 48 6.

7.

8.

9.

KVPY-SA Let S be the set of integers x such that: (i) 100 d” x d” 200 (ii) x is odd (iii) x is divisible by 3 but not by 7 How many elements does S contain? (a) 16 (b) 12 (c) 11 (d) 13 If a four digit number “abcd” is divisible by 4 and another four digit number “acbd” is divisible by 8 then what is the minimum value of a + b + c + d ? (a) 33 (b) 34 (c) 36 (d) None of these Suppose, the seed of any positive integer n is defined as follows: Seed(n) = n, if n < 10 = seed(s(n)), otherwise, where s(n) indicates the sum of digits of n. For example, seed(7) = 7, seed(248) = seed(2 + 4 + 8) = seed(14) = seed(1 + 4) = seed(5) = 5 etc. How many positive integers n, such that n < 500, will have seed(n) = 9? (a) 39 (b) 72 (c) 81 (d) 55 AB is a line segment of 4 cm length. P is the midpoint of AB. Circles are drawn with A, P and B as centre and radii AP = PB. The area of shaded portion (in cm2) is

D A

C

P

B

(a)

6 3

(b) 2p - 6 3

(c)

2p - 3 3

(d) 3 3

10. If f(x) is a polynomial of degree 5 with leading coefficient 2009. Also f(1) = 1, f(2) = 3, f(3) = 5, f(4) = 7, f(5) = 9. What is the value of f(6). (a) 322191 (b) 11 (c) 312331 (d) 241091

11.

N is a 75 digit number that is formed by writing all the two digit numbers in which ten’s digit is greater than unit’s digit, in increasing order one after the other. What is remainder when n divided by 16? (a) 5 (b) 7 (c) 9 (d) None of these 12. ABCD is a parallelogram, X and Y are midpoints of BC and CD respectively. Then choose the correct option. (a) ar DAXY =

5 ar (ABCD) 8

(b) ar DAXY =

1 ar (ABCD) 2

(c) ar DAXY =

3 ar (ABCD) 8

3 ar (ABCD) 4 13. Find all real numbers a for which the equation x2 + (a “ 2)x + 1 = 3|x| has exactly three distinct real solutions in x. (a) 1, 2 (b) 2, 3 (c) 1, 3 (d) 2, 4 14. There are two concentric circles such that the area of the outer circle is four times the area of the inner circle . Let A, B and C be three distinct points on the perimeter of the outer circle such that AB and AC are tangents to the inner circle. If the area of the outer circle is 12 square centimeters then find the area (in square centimeters) of the triangle ABC .

(d) ar DAXY =

(a)

9 3 p

(b)

8 3 p

(c)

6 3 p

(d) None of these

15. If x, y Î[ 0,10] , then number of solutions (x, y) of the inequation 3sec (a) 2 (c) 4

2

x -1

9 y 2 - 6 y + 2 £ 1 is

(b) 1 (d) infinite

MOCK TEST-6

49

16. 5 moles of nitrogen gas are enclosed in an adiabatic cylindrical vessel. The piston itself is a rigid light cylindrical container containing 3 moles of Helium gas. There is a heater which gives out a power 100 cal to the nitrogen gas. A power of 30 cal is transferred to Helium through the bottom surface of the piston. The rate of increment of temperature of the nitrogen gas assuming that the piston moves slowly : (a) 2K/sec

He N2

(b) 4K/sec (c) 6K/sec

(d) 8K/sec 17. A force (F) acting on a body is dependent on its displacement s as F µ s -1/3 . Therefore, the power delivered by the force varies with its displacement as: (a)

s 2/3

(b)

(c)

s -5/3

(d) s0

s1/2

18. A horizontally oriented copper rod of length l is rotated about a vertical axis passing through its middle. Calculate the rotated frequency at which the rod ruptures. Breaking or rupture strength of copper takes as s and density of copper r. (a)

1 8s 2p rl 2

(b)

(c)

1 2s 3p rl 2

(d) None of these

1 3s 4p rl 2

19. A pulley is in the form of a disc of mass M and radius R. In following figure two masses M1 and M2 are connected by a light inextensible string which passes over the pulley. Assuming that the string does not slip over the pulley, the angular momentum of system at the instant shown, about axis of rotation of pulley is

1 ù é êë M 2 + M1 + k M úû vR then

///////////////////////////

find the value of k.

PHYSICS

M = mass of disc shaped pulley and R = radius of the pulley. (a) 1 (b) 2 (c) 4 (d) 5

O

M2 v

M1

20. A ball is dropped from a certain height on a horizontal floor. The coefficient of restitution 1 . The 2 displacement time graph of the ball will be

between the ball and the floor is

(a)

(b)

(c)

(d)

21. Consider the following statementsI. Isotopes are atoms having same number of protons but different number of neutrons. II. Isotopes are atoms having same number of neutrons but different number of protons. III. Isotopes are atoms having same number of protons and neutrons. Choose the correct statement. (a) I and II (b) II only (c) I only (d) I, II and III 22. Express which of the following set ups can be used to verify Ohm’s law? A

(a)

(b)

V

A

V

V

A

(c)

(d) V

A

EBD_7839 50 23.

24.

25.

26.

KVPY-SA A satisfactory photographic print is obtained when the exposure time is 10 sec at a distance of 2 m from a 60 cd lamp. The time of exposure required for the same quality print at a distance of 4m from a 120 cd lamp is (a) 5 sec (b) 10 sec (c) 15 sec (d) 20 sec Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will I. Keep floating at the same distance between them II. Move towards each other III. Move away from each other IV. Will becomes stationary Choose the correct statement. (a) I only (b) II only (c) III and II (d) IV and I A defective eye cannot see close objects clearly because their image is formed (a) On the eye lens (b) Between eye lens and retina (c) on the retina (d) Beyond retina A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length l. The system is rotated about the other end of the spring with an angular velocity w, in gravity free space. The increase in length of the spring is (a)

(a) absolute error same for both, relative error greater for MA and lesser for MB. (b) absolute error same for both, relative error greater for MB and lesser for MA. (c) relative error same for both, absolute error greater for MA and lesser for MB. (d) relative error same for both, absolute error greater for MB and lesser for MA. 28.

Suppose the drift velocity nd in a material varied with the applied electric field E as nd µ E . Then v – I graph for a wire made of such a material is best given by : v

v

(a)

(b) I

I

v

v

(d)

(c) I

29.

I

A jet plane flying at a constant velocity v at a height h = 8 km, is being tracked by a radar R located at O directly below the line of flight, if the angle q is decreasing at the rate of 0.025 rad/ s, the velocity of the plane when q = 60° is :

mw 2 l k mw 2 l

(b)

( k - mw ) 2

mw 2 l

(c)

27.

( k + mw ) 2

(d) None of these Two masses MA and MB (MA < MB) are weighed using same weighing machine. Absolute error and relative error in two measurement are (Assume only systematic errors are involved)

(a) 1440 km/h (c) 1920 km/h 30.

(b) 960 km/h (d) 480 km/h

What is the total kinetic energy of 2g of nitrogen at 300 K? Given : molecular weight of nitrogen = 28. (a) 267.1 × 107 erg (c) 762.1 × 104 erg

(b) 672.1 × 107 erg (d) 267.1 × 104 erg

MOCK TEST-6

51 CHEMISTRY

31. The spectrum of He is expected to be similar to that of : (a) H (b) Li+ (c) Na (d) He+ 32. Which of the following compounds is not aromatic? (a)

(b) ++

(c)

N H

–

33. Among the following which compound will show the highest lattice energy ? (a) KF (b) NaF (c) CsF (d) RbF 34. Which of the following explanation is best for not placing hydrogen with alkali metals or halogen? (a) The ionization energy of hydrogen is high for group of alkali metals or halogen. (b) Hydrogen can form compounds. (c) Hydrogen is much lighter element than the alkali metals or halogens. (d) Hydrogen atom does not contain any neutron. 35. An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be – (a) Calcium (b) Carbon (c) Silicon (d) Iron 36. The rate of reaction for which of the following is not affected by pressure: (a) PCl3 + Cl2 ‡ˆˆ ˆˆ† PCl5 (b) N 2 + 3H 2 ‡ˆˆ ˆˆ† 2NH 3 ˆˆ† 2NO (c) N 2 + O2 ‡ˆˆ (d)

ˆˆ† 2SO 3 2SO 2 + O 2 ‡ˆˆ

(a) d (b) f (c) g (d) p 38. The screening effect of d-electron is (a) equal to p-electron. (b) much more than p-electron. (c) same as ¦-electrons. (d) less than p-electrons. 39. Which of the following does not have a tetrahedral structure? (a) BH–4

(d)

N

37. The orbital having m = – 2 should not be present in the following sub-shell:

(b) BH3

(c) NH +4 (d) CH4

40. The plot which does not represent isothermal expansion of an ideal gas is:

(a)

P

P

(b) Vm

1/Vm

PVm

(d) U

(c) P

Vm

41. Which of the following substances consists of only one element? (a) Marble (b) Sand (c) Diamond (d) Glass 42. What will be the molarity of a solution containing 5 g of sodium hydroxide in 250 mL solution? (a) 0.5 (b) 1.0 (c) 2.0 (d) 0.1 43. The presence or absence of hydroxyl group on which carbon atom of sugar differentiates RNA and DNA? (a) 1st (b) 2nd (c) 3rd (d) 4 th 44. Which of the following is acidic salt? (a) Pb(OH)NO3

(b) Cu(OH)NO3

(c) KHSO3

(d) None of these

EBD_7839 52 45.

KVPY-SA The oxidation number of Cr in CrO5 which has the following structure is: O

(a) + 4

O || Cr

53.

O

O

O

(b) + 5

(c) + 6

(d) + 3

54.

BIOLOGY 46.

47.

48.

49.

50.

51.

52.

The normal pH of the arterial blood is: (a) 6.8 (b) 7.8 (c) 7.4 (d) 8.8 Which of the following factors influence the flowering in plants? (a) Acidity of soil (b) Amount of green pigment (c) Photoperiod (d) Water in the soil It is said that the Taj Mahal may be destroyed due to: (a) Food in Yamuna river (b) Decomposition of marble as a result of high temperature (c) Air pollutants released from oil refinery of Mathura (d) All of the above Glycolysis cannot occur in the absence of: (a) Ca (b) Mg (c) Mn (d) Co Which of the following gland is not involved in the process of digestion? (a) Liver (b) Pancreas (c) Salivary gland (d) Adrenal gland Which of the following is false? (a) Cardiac muscle is striated. (b) Smooth muscle does not contain actin. (c) Skeletal muscle is considered voluntary. (d) Smooth muscle is found in the iris of the eye and the walls of the bladder. The compound acting as an oxygen store in skeletal muscle is: (a) Myoglobin (b) Haemoglobin (c) ATP (d) Myokinase

The bacterium Bacillus thuringiensis can withstand heating, dryness, and toxic chemicals that would kill most other bacteria. This indicates that it is probably able to form: (a) Pseudopodia

(b) Endotoxins

(c) Endospores

(d) Pili

The direction of the conduction of food through phloem is from (a) bottom to upwards (b) top to bottom. (c) leaves to roots. (d) None of the above

55.

The structure of glucose and galactose are same except with respect to: (a) First carbon atom (b) Second carbon atom (c) Third carbon atom (d) Fourth carbon atom

56.

Bamboo and grasses elongate by the activity of: (a) Apical meristem (b) Lateral meristem (c) Secondary meristem

57.

(d) Intercalary meristem Amylopsin acts upon (a) polypeptide in acidic medium. (b) polysaccharide in acidic medium. (c) polysaccharide in alkaline medium. (d) polypeptide in alkaline medium.

58.

Which of the following is a micronutrient? (a) Copper

59.

(c) Minerals 60.

(b) Calcium

(c) Phosphorus (d) Magnesium Endocrine glands produce or action of endocrine glands is mediated through: (a) Hormones (b) Enzymes (d) Vitamins

Which one of the following cell organelles is found only in plants? (a) Golgi complex (c) Plastids

(b) Mitochondria (d) Ribosomes

MOCK TEST-6

53 PART-II (2 MARKS QUESTIONS) MATHEMATICS

61. Three circles each of radius r units are drawn inside an equilateral triangle of side a units, such that each circle touches the other two and two sides of the triangle as shown in the figure, (P, Q and R are the centres of the three circles). Then relation between r and a is (a)

a = 2 ( 3 +1)r

(b) a = ( 3 +1)r

(c)

a = ( 3 + 2)r

(d) a = 2 ( 3 + 2)r

62. Square ABPQ and ADRS are drawn on the sides AB and AD of a parallelogram ABCD. Then

64. ABCD is a trapezium with AB || CD. M and N are points on AD and BC such that MN || CD. If area of ABNM is half the area of ABCD, then which of the relation is true. (a) MN2

= AB2 + CD2

(b) 2MN2 = AB + CD (c) 2MN2 = AB2 + CD2 (d) None of the above 65. Lets define a set S = { Pi | set of all 3 digit natural number that has odd number of factors}. Find the highest power of 12 in the product of all the elements of set S. (a) 16

(b) 19

(c) 20

(d) 25 PHYSICS

66. Spherical particles of pollen are shaken up in water and allowed to settle. The depth of water is 2 × 10–2 m. What is the diameter of the largest particles remaining in suspension one hour later? [Density of pollen = 1.8 × 103 kg m–3. Viscosity of water = 1 × 10 –2 poise and density of water, 1 × 10–5 kg m–3.] (a) 5.34 mm

(a)

ÐSAQ = ÐABC

(b)

ÐSRD = ÐADC

(c)

ÐSAD = ÐABC

(d) None of these 63. Let P = {q : sin q – cos q = 2 cos q} and Q = {q : sin q + cos q = 2 sin q} be two sets. Then (a) P Ì Q and Q – P ¹ Æ (b) Q Ë P (c) P Ë Q (d) P = Q

(b) 1.77 mm

(c) 7.11 mm (d) 3.54 mm 67. The density of a sphere is measured by measuring its mass and diameter. If, it is known that the maximum percentage errors in the measurement are 2% and 3%, then find the maximum percentage error in the measurement of density? (a) 15% (b) 18% (c) 9% (d) 11% 68. Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and /or record time for different number of oscillations. The observations are shown in the table. Least count for length = 0.1 cm Least count for time = 0.1 s

EBD_7839 54

KVPY-SA

Student Length of the No. of Total time Time pendulum oscillations for (n) period (cm) (n) oscillations (s) (s) I 64.0 8 128.0 16.0 II 64.0 4 64.0 16.0 III 20.0 4 36.0 9.0

70.

The reflecting surface is represented by the equation 2x = y2 as shown in the figure. A ray travelling horizontal becomes vertical after reflection. The co-ordinates of the point of incidence are :

If EI, EII and EIII are the percentage errors in g, i.e., æ Dg ö çè g ´ 100÷ø for students I, II and III, respectively, then

69.

(a) EI = 0 (b) EI is minimum (c) EI = EII (d) EII is maximum A horizontally insulated cylindrical vessel of length 2l is separated by a thin insulating piston into two equal parts each of which contains n moles of an ideal monoatomic gas at temperature T. The piston is connected to the end faces of the vessel by undeformed springs of force constant k each. The left part in the contact with a thermostat (a device which maintains a constant temperature). When an amount of heat Q is supplied to the gas in the right part, the piston is displaced to the left by a distance x = l/2. Determine the heat Q’ given away at the temperature T to the thermostat by the left part of the piston.

2L

(a) (1/2, 1) (c) (1/2, 1/2)

CHEMISTRY 71.

k

n,T

n,T

(a)

5 Q ' = Q - kl 2 + 3nRT 2

(b)

5 Q ' = Q + kl 2 - 3nRT 2

(c)

5 Q ' = Q - kl 2 - 3nRT 2

(d)

5 Q ' = Q + kl 2 + 3nRT 2

Q

The hydrocarbon which can react with sodium in liquid ammonia is : (a) CH 3CH 2 CH 2C º CCH 2CH 2 CH3 (b) CH 3CH 2C º CH (c) CH 3CH = CHCH3 (d) CH 3CH 2C º CCH 2CH3

72.

73.

k

(b) (1, 1/2) (d) None

The mass of BaCO3 produced when excess CO2 is bubbled through a solution of 0.205 mol Ba(OH)2 is (a) 81 g (b) 40.5 g (c) 20.25 g (d) 162 g From the following bond energies: H – H bond energy: 431.37 kJ mol–1 C = C bond energy: 606.10 kJ mol–1 C – C bond energy: 336.49 kJ mol–1 C – H bond energy: 410.50 kJ mol–1 Enthalpy for the reaction, H H H H | | | | C = C + H - H ¾¾ ® H - C- C - H | | | | H H H H

will be: (a) – 243.6 kJ mol–1 (c) 553.0 kJ mol–1

(b) –120.0 kJ mol–1 (d) 1523.6 kJ mol–1

MOCK TEST-6

55

74. The van der Waal’s constant ‘a’ for four gases P, Q, R and S are 4.17, 3.59, 6.71 and 3.8 atm L2 mol–2 respectively. Therefore, the ascending order of their liquefaction is: (a) R < P < S < Q (b) Q < S < R < P (c) Q < S < P < R (d) R < P < Q < S 75. Which of the following structures correspond to the product expected, when excess of C6H6 reacts with CH2Cl2 in presence of anhydrous AlCl3? (a)

(b)

(a) A – III; B – V; C – I; D – II; E – IV (b) A – I; B – II; C – III; D – IV; E – V (c) A – II; B – III; C – IV; D – V; E – I (d) A – IV; B – I; C – V; D – II; E – III 77. Which of the following statements are incorrect? (i)

A dominant allele determines the phenotype when paired with a recessive allele.

(ii) A recesive allele is weaker than a dominant allele. (iii) A recessive allele shows its effects when paired with a dominant allele.

CH Cl

(iv) A dominant allele is always better for an organism.

CHCl2

(a) (i), (ii) and (iv) (b) (ii), (iii) and (iv)

(c)

Cl

(c) (i), (ii) and (iii)

C

(d) (i), (iii) and (iv) 78. Mark the odd pair out.

Cl

(a) Totipotency: Parenchyma (d)

(b) Secondary growth: Cambium

CH2

(c) Dermal cells: Trichomes BIOLOGY 76. Match the terms given in column-I with their definition given in column-II and choose the correct option. Column-I (Terms)

(d) Periderm: Collenchyma 79. Which parental phenotypes would produce offspring with blood group phenotypes in the expected ratio of 1 type A : 1 type B?

Column-II (Definition)

Blood group of mother

Blood group of father

A. Semi-circular canal I. Spiral organ of Corti

(a)

A

B

B. Vestibule

(b)

AB

AB

(c)

AB

B

(d)

AB

O

II. Fluid found in the scala vestibuli and scala tympani

C. Cochlea

III. Evaluates rotational motion

D. Perilymph

IV. Fluid found within the organ of Corti

E. Endolymph

V. Responds to gravity and movements of the head

80. Bacteriorhodopsin absorbs a quantum of light by P 570 which is transformed into P 412 accompanied by release of H+. It is also known that F0-F1 system generates ATP through the movement of H + . What should be the appropriate diagram for a photosynthetic Halobacterium halobium?

EBD_7839 56

KVPY-SA

(a)

(b)

(c)

1 2 3 4 5 6 7 8 9 10

(d)

(c) (b ) (a) (b ) (b ) (d ) (a) (d ) (c) (d )

11 12 13 14 15 16 17 18 19 20

(a) (c) (c) (a) (c) (a) (d ) (a) (b ) (c)

21 22 23 24 25 26 27 28 29 30

ANS W ER KEYS Part-I (c) 31 (b) 41 (c) (a) 32 (a) 42 (a) (d) 33 (b) 43 (b ) (b) 34 (c) 44 (c) (d) 35 (a) 45 (c) (b) 36 (c) 46 (c) (a) 37 (d) 47 (c) (c) 38 (d) 48 (c) (b) 39 (b) 49 (b ) (a) 40 (d) 50 (d )

51 52 53 54 55 56 57 58 59 60

(b ) (a) (c) (c) (d ) (d ) (c) (a) (a) (c)

61 62 63 64 65 66 67 68 69 70

Part-II (a) 71 (a) 72 (d ) 73 (c) 74 (b ) 75 (d ) 76 (d ) 77 (b ) 78 (c) 79 (a) 80

(b) (b) (b) (c) (d) (a) (b) (d) (d) (a)

71 Time : 3 Hours

Stream-SA

Maximum Marks : 100

INSTRUCTIONS 1. 2. 3. 4.

There are 80 questions in this paper. The question paper contains two parts; Part I (1 Mark Questions) and Part II (2 Marks Questions). There are four sections; Mathematics, Physics, Chemistry and Biology in each part. There are four options given with each question, only one of them is correct. For each incorrect answer 0.25 Mark in Part–I and 0.5 Mark in Part–II will be deducted. PART-I (1 MARK QUESTIONS) MATHEMATICS

1.

3.

4.

5.

(b) n is odd

If p, q , r are + ve and are in A. P., the roots of quadratic equation px2 + qx + r = 0 are all real for (a)

2.

(a) n is even

r -7 ³ 4 3 p

p

(b) r - 7 ³ 4 3

(c) all p and r (d) no p and r ab is a two digit number less than 50 such that (ab)2 = c0cb where a,b,c are distinct digits not equal to zero. How many values of “ab” exist. (a) 2 (b) 1 (c) 4 (d) 5 What should be the value of n such that A = 31 + 32 + …. + 3n is a perfect square? (a) 112 (b) 144 (c) 225 (d) None of these x and y are two non-negative numbers such that 2x + y = 10. The sum of the maximum and minimum values of (x + y) is (a) 6 (b) 9 (c) 10 (d) 15 Each of the numbers x1, x2. . . xn, n > 4, is equal to 1 or –1. Suppose, x1 x2 x3 x 4 + x2 x3 x 4x 5 + x3x4x5x6 + .... + xn-3xn-2xn-1xn + xn-2xn-1xnx1 + xn-1xnx1x2 + xnx1x2x3 = 0, then,

(c) n is an odd multiple of 3 (d) n is prime 6.

In the given figure, ABCD is a quadrilateral in which diagonals AC and BD intersect at a point E Then

D

A N E M B

C

(a) Area DAED × Area DBEC = Area DABE × Area DCDE (b) Area DABE × Area DBEC = Area DAED × Area DCDE (c) Area DBEC × Area DCDE = Area DAED × Area DABE (d) None of these

EBD_7839 58 7.

KVPY-SA In the diagram O is the center of a circle. AE + EB = CE + ED. OP ^ B and OQ ^ CD then true relation between OP and OQ is :

Q

O

B

E P A

8.

12.

The diameter of one of the bases of a truncated cone is 100 mm. If the diameter of this base is increased by 21% such that it still remains a truncated cone with the height and the other base unchanged, the volume also increases by 21%. The radius of other base is (a) 65 (b) 55 (c) 45 (d) 35 13. Suppose A1, A2, ......, A30 are thirty sets each having 5 elements and B1, B2, ......, Bn are n sets

C

(a) OP > OQ (b) OP < OQ 1 (c) OP = OQ (d) OP = OQ 2 In the given figure, DABC is a right angled triangle semicircles are drawn on AB, AC and BC as diameters. It is given that AB = 3 cm and AC = 4 cm. Find the area of shaded region.

30

n

i =1

j =1

each with 3 elements. Let U Aj = U Bj = S and each element of S belongs to exactly 10 of the Ai's and exactly 9 of the Bj's. Then n is equal to (a) 15 (b) 3 (c) 45 (d) 35 14. Consider n is a two digit number such that the ratio of the number to the sum of its digit is maximum. How many such numbers exist? (a) 4 (b) 9 (c) 12 (d) 11 15. Let X and Y be two non-empty sets. Let f : X ® Y be a function. For A Ì X and

9.

10.

11.

(a) 12 cm2 (b) 6 cm2 2 (c) 9 cm (d) 15 cm2 In DABC, AB = AC, P and Q are points on AC and AB respectively such that BC = BP = PQ = AQ. Then, ÐAQP is equal to (use p =180º) 2p 3p (a) (b) 7 7 4p 5p (c) (d) 7 7 A drain cover is made from a square metal plate of side 40 cm having 441 holes of diameter 1 cm each drilled in it. Find the area of the remaining square plate. (a) 1250.5 cm2 (b) 1253.5 cm2 (c) 1240.2 cm2 (d) 1260.2 cm2 A and B have some amount with them. B gave some amount to A hence due to this change now A and B both have as much rupee as much they had paisa and as much paisa as much they had rupee. What can be the maximum total sum of money with them? (a) 199.98 (b) 188.98 (c) 198.97 (d) 198.98

B Ì Y,

f ( A)= { f ( x) : x Î A};

define

f -1 ( B)= {x Î X : f ( x) Î B}, then

(a)

f ( f -1 ( B )) = B

(b) f ( f -1 ( B)) Ì B

(c)

f -1 ( f ( A)) = A

(d) f -1 ( f ( A)) Ì A

PHYSICS 16.

Time graph between the displacement x and time t for a particle moving in a straight line is shown in fig. During the interval OA, AB, BC and CD, the acceleration of the particle is x A

B

C

t

O

(a) (b) (c) (d)

OA + – + –

D

AB 0 0 0 0

BC + + – –

CD + 0 + 0

MOCK TEST-7

59

17. In the determination of Young’s modulus 4 MLg ö æ çè Y = ÷ by using Searle’s method, a wire pld 2 ø

of length L = 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l = 0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement (a) due to the errors in the measurements of d and l are the same. (b) due to the error in the measurement of d is twice that due to the error in the measurement of l. (c) due to the error in the measurement of l is twice that due to the error in the measurement of d. (d) due to the error in the measurement of d is four times that due to the error in the measurement of l. 18. A rod of length l is in motion such that its ends A and B are moving along x-axis and y-axis dq = 2 rad/sec dt always. P is a fixed point on the rod.

respectively. It is given that y B

P q

19. A circular coil ABCD carrying a current i is placed in a uniform magnetic field. If the magnetic force r on the segment AB is F , the force on the remaining segment BCDA is A r (a) F i r (b) -F D B r (c) 3F r C (d) -3F 20. A catapult consists of two parallel rubber strings, each of lengths 10 cm and cross sectional area 10 mm2. When stretched by 5 cm, it can throw a stone of mass 100 g to a vertical height of 25 m. Determine Young’s modulus of elasticity of rubber. (a) 9.8 × 107 N/m2 (b) 8.9 × 107 N/m2 (c) 25 × 10–1 N/m2 (d) 2.5 × 10–1 N/m2 21. An exhausted chamber with nonconducting walls is connected through a valve to the atmosphere where the pressure is p0 and the temperature is T0 . The valve is opened slightly and air flows into the chamber until the pressure within the chamber is p0. Assuming the air to behave like an ideal gas with constant heat capacities, the final temperature of the air in the chamber is (a) g T0 (b) (T0 / g) (c) (1 / g – 1)T0 (d) (g / 1 – g) T0 22. A hollow sphere is held suspended. Sand is now poured into it in stages. The centre of gravity of the sphere with the sand

A M

x

Let M be the projection of P on x-axis. For the p time interval in which q changes from 0 to , 2 choose the correct statement.

(a) The speed of M is always directed towards right (b) M executes S.H.M. (c) M moves with constant speed (d) M moves with constant velocity

SAND

(a) rises continuously (b) remains unchanged in the process (c) first rises and then falls to the original position (d) first falls and then rises to the original position

EBD_7839 60 23.

24.

Velocity (m/s) [west]

25.

26.

KVPY-SA Adding detergents to water helps in removing dirty greasy stains. This is because I. It increases the oil-water surface tension II. It decreases the oil-water surface tension III. Viscosity of oil is high IV. Dirt is held suspended surrounded by detergent molecules (a) II and IV (b) I only (c) III and IV (d) IV only The pair of physical quantities that has the different dimensions is : (a) Reynolds number and coefficient of friction (b) Curie and frequency of a light wave (c) Latent heat and gravitational potential (d) Planck’s constant and torque The graph shown illustrates velocity versus time for two cars A and B constrained to move in a straight line. Both cars were at the same position at t = 0s. Consider the following statements. (1) Car A is travelling west and Car B is travelling east. (2) Car A overtakes Car B at t = 5 s. (3) Car A overtakes Car B at t = 10 s.

14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 0.0

27.

28.

29.

Car A

Car B 2.0

4.0

6.0

8.0 Time (s)

10.0

12.0

14.0

Which of the following is correct? (a) Only statement 1 (b) Only statement 2 (c) Only statement 3 (d) Statements 1 and 2 In the given figure, a smooth parabolic wire track lies in the xy-plane (vertical). The shape of track is defined by the equation y = x2. A ring of mass m which can slide freely on the wire track, is placed at the position A (1,1). The track is rotated with constant angular speed w such there is no relative slipping between the ring and the track. y The value of w is (a)

g /2

(b)

g

(c)

2g

(d) 2 g

w

O (0, 0)

x

30.

Read the following statements S1 : An object shall weigh more at pole than at equator when weighed by using a physical balance S2 : It shall weigh the same at pole and equator when weighed by using a physical balance S3 : It shall weigh the same at pole and equator when weighed by using a spring balance S4 : It shall weigh more at the pole than at equator when weighed using a spring balance Which of the above statements is/are correct (a) S1 and S2 (b) S1 and S4 (c) S2 and S3 (d) S2 and S4 A point object is kept in front of a plane mirror. The plane mirror is doing SHM of amplitude 2 cm. The plane mirror moves along the x - axis which is normal to the mirror. The amplitude of the mirror is such that the object is always in front of the mirror. The amplitude of SHM of the image is (a) 0 (b) 2 cm (c) 4 cm (d) 1 cm R A circular disc of radius R and thickness has 6 moment of inertia I about an axis passing through its centre perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter is 2I (a) I (b) 8 I I (c) (d) 5 10 A man is standing on an international space station, which is orbiting earth at an altitude 520 km with a constant speed 7.6 km/s. If the man’s weight is 50 kg, his acceleration is (a) 7.6 km/s2 (b) 7.6 m/s2 (c) 8.4 m/s2 (d) 10 m/s2 CHEMISTRY

31.

From a heated mixture of nitrogen, oxygen and carbon, two compounds (out of the many obtained) are isolated. The rates of diffusion of the two isolated compounds are almost identical. The two compounds are (a) N2O and CO2 (b) CO and NO (c) CO2 and NO2 (d) N2O and CO

MOCK TEST-7 32.

61

HSO-4 + OH - ® SO 42- + H 2O Which statement is correct about conjugate acid base pair?

(a) HSO-4 is conjugate acid of base SO24- . (b) HSO-4 is conjugate base of acid SO24- . (c) SO24- is conjugate acid of base HSO-4 . (d) Conjugate acid of SO24- is not possible. 33. Equivalent mass of H 3 PO 2 when it disproportionate into PH3 and H3PO3 is: (a) M (b) M/2 (c) M/4 (d) 3M/4 34. Which of the following electronic configuration of an atom has the lowest first ionisation enthalpy? (a) 1s2 2s2 2p3 (b) 1s2 2s2 2p6 3s1 2 2 6 (c) 1s 2s 2p (d) 1s2 2s2 2p5 35. In the following reaction

39. The size of species I, I+ and I– decreases in the order: (a) I+ > I– > I (b) I– > I > I+ – + (c) I > I > I (d) I > I+ > I– 40. Which of the following pairs of species have identical shapes? (b) PCl5 and BrF5 (a) NO+2 and NO2(c) XeF4 and ICl4 (d) TeCl4 and XeO4 41. A metal M readily forms its sulphate MSO4 which is water-soluble. It forms its oxide MO which becomes inert on heating. It forms an insoluble hyroxide M(OH)2 which is soluble in NaOH solution. Then M is (a) Mg (b) Ba (c) Ca (d) Be 42. In the isochoric process, DH for a system is equal to: (a) P. DV (b) P V (c) E + P.DV (d) D E 43. The IUPAC name of the following compound is

H SO

2 4 CH3 – CH2 – CH2 – CH3 ¾¾¾¾ ® 475 K

(a) (b) (c) (d)

CH3CH = CHCH3 predominates. CH2 = CHCH2CH3 predominates. Both are formed in equal amounts. The amount of production depends on the nature of catalyst. 36. The conjugate base of HBr is : (a) H +

(b) H 2 Br +

-

+

(c) Br (d) Br 37. Hydrogen bonding is maximum in : (a) C2H5OH (b) CH3OCH3 (c) (CH3)2 C = O (d) CH3CHO 38. Which of the following will have larger dipole moment?

(a) trans-2-chloro-3-iodo-2-pentene (b) cis-3-iodo-4-chloro-3-pentene (c) trans-3-iodo-4-chloro-3-pentene (d) cis-2-chloro-3-iodo-2-pentene 44. In the reaction Cr2 O 72- + 14 H + + 6I - ¾ ¾® 2Cr 3+ + 7H 2 O + 3I 2

Which element is reduced? (a) I (b) O (c) H (d) Cr 45. According to Einstein's photoelectric equation, the graph between kinetic energy of photoelectrons ejected and the frequency of the incident radiation is :

(a)

(b)

(a)

(b)

(c)

(d)

(c)

(d)

EBD_7839 62

KVPY-SA BIOLOGY

When dominant and recessive alleles express together, it is called: (a) Co-dominance (b) Dominance (c) Amphidominance (d) Pseudodominance 47. Stool of a person is whitish grey coloured due to malfunction of which of the following organ? (a) Pancreas (b) Spleen (c) Kidney (d) Liver 48. Secondary air pollutant is: (a) PAN (b) CO

53.

46.

49.

50.

51.

52.

(c) SO2 (d) Aerosols Which digestive processes take place in the mouth? (a) Both chemical and physical digestion. (b) Chemical digestion only. (c) Physical digestion only. (d) Neither chemical nor physical digestion. Which of the following characteristic distinguish arthropods from annelids and molluscs? (a) An external skeleton made of chitin (a polysaccharide) and protein rather than a shell made chiefly of mineral salts. (b) Subdivision of the legs into movable segments. (c) Distinct group of muscles, derived from many body segments, that move the separate parts of the exoskeleton. (d) All of the above Copper-T is a device that prevents: (a) Implantation of blastocyst (b) Ovulation (c) Fertilisation (d) Egg maturation A functional piece of mRNA has 66 codons. What is the maximum number of amino acids that could be present in the protein, coded for this mRNA? (a) 22 (b) 64 (c) 65 (d) 66

54.

55.

56.

57.

58.

59.

60.

In micturition: (a) Urethra relaxes (b) Ureter contracts (c) Ureter relaxes (d) Urethra contracts Thirty percent of the bases in a sample of DNA extracted from eukaryotic cells are adenine. What percentage of cytosine is present in this DNA? (a) 10% (b) 20% (c) 30% (d) 40% Which of the following is incorrectly matched? (a) Ribozyme: Proteinaceous in nature (b) Apoenzyme: The protein part of enzyme (c) Co-enzyme: Loosely attached organic cofactor of holoenzyme (d) Co-factor : Non-protein part of holoenzyme Hepatitis B is transmitted through: (a) Sneezing (b) Female Anopheles (c) Coughing (d) Blood transfusion At the instant following the second heart sound, which heart valves are open? (a) Atrioventricular valves only. (b) Both atrioventricular valves and semilunar valves. (c) All valves are closed. (d) Semilunar valves only. Hugo de Vries gave his mutation theory on organic evolution while working on: (a) Pisum sativum (b) Drosophila melanogaster (c) Oenothera lamarckiana (d) Althea rosea Source of somatostatin is the same as that of: (a) Insulin and glucagon (b) Vasopressin and oxytocin (c) Thyroxine and calcitonin (d) Somatotropin and prolactin DNA replication is (a) conservative and discontinuous. (b) semiconservative and semi-discontinuous. (c) semiconservative and discontinuous. (d) conservative.

MOCK TEST-7

63 PART-II (2 MARKS QUESTIONS) MATHEMATICS

61. If N is 100 consecutive even multiples of 3. If P is the smallest prime number divisible by N + 1 then which one of the following is true about P? (a) P is a single digit prime number (b) 10 < P < 25 (c) 25 < P < 50 (d) P > 100 62. If a, b are the roots of x 2 + px + q = 0 and

a b , are root b a of xn + 1 + (x + 1)n = 0, then n can be i + p ¹ 0 (a) 13 (b) 14 (c) 15 (d) None of these 63. What is the area of the shaded region shown below, if the radius of each circle is equal to the side of the hexagon, which in turn is equal to 6 cm, and A and B are the centres of the circles ? also of x 2 n + p n x n + q n = 0 and if

A (a)

B

6 ( 9 3 - 4p )

(b) 4 ( 9 3 - 4p )

(c) 6 ( 9 2 - 4p ) (d) None of these 64. In given figure ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Then (a) Area (ABCDE) = Area (DAPQ) (b) Area (ABCDE) = Area (DACQ) (c) Area (ABCDE) = 2 Area (DADQ) (d) None of these

65. Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares? (a) 2 (b) 4 (c) 0 (d) 1 PHYSICS 66. A double-pane window used for insulating a room thermally from outside consists of two glass sheets each of area 1m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state, the room glass inter-face and the glass -outdoor interface are at constant temperatures of 27°C and 0°C respectively. Calculate the rate of heat flow through the window pane. Given thermal conductivities of glass and air as 0.8 and 0.08 Wm –1 K –1 respectively. (a) 41.5 J/s (b) 80.3 J/s (c) 90 J/s (d) 90.6 J/s 67. In the arrangement shown in fig., the ends P and Q of an in extensible string move downwards with uniform speed u and v respectively. Pulley A and B are fixed. Find the velocity of mass M at the instant shown in the figure is

A

(a)

(u + v ) cos q 2

(b)

2(u + v ) cos q

E

B

(c) (u + v) / 2 cos q

P

C

D

Q

(d) u / cosq

EBD_7839 64 68.

KVPY-SA time interval. Which of the following graphs correctly describes (x1 – x2) as a function of time ‘t’?

A rod PQ of length l is pivoted at an end P and freely rotated in a horizontal plane at an angular speed w about a vertical axis passing through P. If coefficient of linear expansion of material of rod is a, find the percentage change in its angular velocity if temperature of system is increased by DT is

(x1 – x2)

(x1 – x2)

(a)

(a) (a D T ´ 100)%

(b)

t

O (x1 – x2)

(b) (2a D T ´100)%

O

t

(x1 – x2)

(c) (3a D T ´ 100) ´100% (c)

(d) (4a D T ´100)% 69. The angle of incidence and refraction of a monochromatic ray of light of wavelength l at an air-glass interface are i and r, respectively. A parallel beam of light with a small spread dl in wavelength about a mean wavelength l is refracted at the same air-glass interface. The refractive index µ of glass depends on the wavelength l as m(l) = a +

(a)

71.

72.

æ 2b ö ç 3 ÷ dl èl ø

(c)

æ 2b tan r ö ç 3 ÷ dl è al + bl ø

(d)

æ (2b ( a + b / l 2 ) sin i ö ç ÷ dl ç ÷ l3 è ø

O

t

When 3, 3-dimethyl-2-butanol is heated with H2SO4 the major product obtained is (a) cis and trans isomers of 2, 3-dimethyl-2butene (b) 3, 3-dimethyl-1-butene (c) 2, 3-dimethyl-2-butene (d) 2, 3-dimethyl-1-butene Among the following carbocations +

(I) Ph2 CCH 2 Me

æ sin i ö ç 3 ÷ dl è l cos r ø

(b)

(d)

CHEMISTRY

b

, where a and b l2 are constants. Then the angular spread in the angle of refraction of the beam is :

t

O

+

(II) PhCH 2 CH 2 CHPh +

(III) Ph2CH CHMe +

70.

A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive xdirection with a constant speed. The position of the first body is given by x1(t) after time ‘t’; and that of the second body by x2(t) after the same

(IV) Ph2C(Me) CH 2

73.

The order of stability is (a) IV > II > I > III (b) I > II > III > IV (c) II > I > IV > III (d) I> IV > III > II Four diatomic species are listed below in different sequences. Which of these presents the correct order of their increasing bond order ? (a) O-2 < NO < C22- < He+2 (b) NO < C22 - < O 2- < He 2+ (c) C 22 - < He +2 < NO < O -2 (d) He+2 < O 2- < NO < C 22 -

MOCK TEST-7

65

74. For the following three reactions (i), (ii) and (iii), equilibrium constants are given: (i) CO(g) + H 2O(g) ƒ CO2 (g) + H 2 (g); K1 (ii) CH 4 (g) + H 2O(g) ƒ CO(g) + 3H 2 (g); K2 (iii) CH 4 (g) + 2H 2O(g) ƒ CO2 (g) + 4H 2 (g); K3 (a) K1 K 2 = K3

(b) K 2 K 3 = K1

(c) K3 = K1 K2

(d) K3 .K23 = K12

75. The wave number of the spectral line in the emission spectrum of hydrogen will be equal to

8 times the Rydberg’s constant if the electron 9 jumps from .......... (a) n = 3 to n = 1 (b) n = 10 to n = 1 (c) n = 9 to n = 1 (d) n = 2 to n = 2 BIOLOGY 76. At rest, most people require an alveolar ventilation for about 4 lt./minute. Suppose an emphysema victim with a tidal volume of 0.5 lt. has suffered alveolar damage that has produced a respiratory dead space of 0.3 lt. What will be that person’s frequency of breathing at rest? (a) 5 breaths/min (b) 8 breaths/min (c) 20 breaths/min (d) 50 breaths/min 77. In a plant, genes A, B and C control the pod colour, height of plant and pod shape, respectively. Study the following genotypes and corresponding phenotypes: AA ® Green pod aa ® Yellow pod Aa ® Purple pod B_ ® Tall plant bb ® Dwarf plant C_ ®Full pod shape cc ® Constricted pod shape If AabbCC and aaBbCc are crossed, what proportion of plants would be tall, have purple and full pod in the progeny? (a) 25% (b) 50% (c) 75% (d) 100%

78. The steps in catalytic cycle of an enzyme action are given in random order. (i) The enzyme releases the products. Now enzyme is free to bind another substrate. (ii) The active sites, now in close proximity of substrate breaks the chemical bonds of substrate and forms E-P complex. (iii) Binding of substrate induces the enzyme to alter its shape fitting more tightly around the substrate. (iv) The substrate binds to the active site of enzyme (i.e., fitting into the active site). (a) (i), (ii), (iii), (iv) (b) (iv), (iii), (ii), (i) (c) (i), (iii), (ii), (iv) (d) (i), (ii), (iv), (iii) 79. Which one of th e following groups of structures/organs have similar function? (a) Typhlosole in earthworm, intestinal villi in rat and contractile vacuole in Amoeba. (b) Nephridia in earthworm, Malpighian tubules in cockroach and urinary tubules in rat. (c) Antennae of cockroach, tympanum of frog and clitellum of earthworm. (d) Incisors of rat, gizzard (proventriculus) of cockroach and tube feet of starfish. 80. Match column-I with column-II and select the correct answer using the codes given below. Column-I Column-II A. Triglyceride I. Animal hormones B. Membrane lipid II. Coating on leaves and stems C. Steroid III. Phospholipids D. Wax IV. Fat stored in the form of droplets (a) A – IV; B – III; C – I; D – II (b) A – II; B – III; C – IV; D – I (c) A – III; B – IV; C – I; D – II (d) A – IV; B – I; C – II; D – III

EBD_7839 66

1 2 3 4 5 6 7 8 9 10

KVPY-SA

(b ) (b ) (d ) (d ) (c) (a) (d ) (b ) (d ) (b )

11 12 13 14 15 16 17 18 19 20

(c) (b ) (c) (b ) (b ) (b ) (a) (b ) (b ) (a)

21 22 23 24 25 26 27 28 29 30

ANS W ER KEYS Part-I (a) 31 (a) 41 (d ) (d) 32 (a) 42 (d ) (d) 33 (d) 43 (a) (d) 34 (b) 44 (d ) (c) 35 (a) 45 (c) (c) 36 (c) 46 (a) (d) 37 (a) 47 (d ) (c) 38 (c) 48 (a) (c) 39 (b) 49 (a) (c) 40 (c) 50 (d )

51 52 53 54 55 56 57 58 59 60

(a) (c) (a) (b ) (a) (d ) (c) (c) (a) (c)

61 62 63 64 65 66 67 68 69 70

Part-II (d ) 71 (b ) 72 (a) 73 (a) 74 (d ) 75 (a) 76 (c) 77 (b ) 78 (c) 79 (b ) 80

(c) (b ) (d) (c) (a) (c) (a) (b) (b) (a)

81 Time : 3 Hours

Stream-SA

Maximum Marks : 100

INSTRUCTIONS 1. 2. 3. 4.

There are 80 questions in this paper. The question paper contains two parts; Part–I (1 Mark Questions) and Part–II (2 Marks Questions). There are four sections; Mathematics, Physics, Chemistry and Biology in each part. There are four options given with each question, only one of them is correct. For each incorrect answer 0.25 Mark in Part–I and 0.5 Mark in Part–II will be deducted. PART-I (1 MARK QUESTIONS) MATHEMATICS

1.

æ Pö p In a DPQR , ÐR = . If tan çè ÷ø 2 2 æ Qö tan ç ÷ è 2ø

2.

3.

4.

(a) 2r = x + t (b) 2t = r + s (c) 2s = r + t

and

are roots of the equation

ax 2 + bx + c = 0 (a ¹ 0), then (a) a + b = c (b) b + c = 0 (c) a + c = b (d) b = c If N is a two digit number such that last two digit of N3 is 44 then what is the summation of all the possible values of N? (a) 78 (b) 80 (c) 72 (d) 86 How many values of natural number N exist such that both N and N + 3293 are perfect squares? (a) 0 (b) 1 (c) 9 (d) None of these If (k, k) is the set of solution of system of equations

5.

6.

(d) r + s + t = 0 Let a, b, c be distinct digits. Consider a two digit number ‘ab’ and three digits ‘ccb’ both defined under the usual decimal number system. If (ab)² = ccb and ccb > 300, then the value of b is: (a) 1 (b) 0 (c) 5 (d) 6 A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Then (a) Area (DABCD) = Area (DACF) (b) Area (DADF) = Area (ABFC) (c) Area (DABF) = Area (DACF) (d) None of these A

B

ax + by + (t – s) = 0 an d bx + ay

E

+ (s – r) = 0, a ¹ b. Then which of the following must be true.

D

C

F

EBD_7839 68 7.

KVPY-SA In the figure, O is the centre of the circle BCD. ABC and EDC are straight lines, BC= DC and ÐAED = 70°, Find ÐBOD

AB = 4 units and CD = 2 units, what is the area, in sq. units of the largest circle? (a) 128p 11.

8.

(a) 40° (b) 70° (c) 80° (d) 90° In the figure shown, two identical circles of radii r are drawn. If the centre of the circles lie on each other circumference, what is the area of shaded region?

12.

13.

9.

(a)

2 2 3 2 pr r 3 2

(c)

1 2 pr 3

(d)

2pr 2 pr 2

3 2 r 2

(c)

5 AC2 4

14.

(b) 5 AC2 (d) 6AC2

AD is a diameter of a circle. Two more circles pass through A and intersect AD in B and C respectively, such that AB and AC are diameters of these circles and AD > AC > AB. If the circumference of the middle circle is average of the circumference of the other two, then given

(c) 48À (d) 16p If cost price of 40 pens is more than cost of 15 copies but less than cost of 16 copies. What can be the minimum total cost of a pen and a copy if price of both pen and copy is an integer (a) 16

(b) 18

(c) 20

(d) 21

A hollow cube is inscribed in a hollow sphere, such that all the corners of cube touches sphere. Then the ratio of air between sphere and cube to the air in cube is (a)

3p - 2 2

(b)

2 3p - 2

(c)

3p - 8 8

(d)

8 3p - 8

Let S = {x Î R : x ³ 0 and

2 | x - 3 | + x ( x - 6) + 6 = 0 . Then S :

2

If ABC is a triangle right angled at B and M, N are the mid-points of AB and BC, then 4(AN2 + CM2) is equal to : (a) 4 AC2

10.

(b)

(b) 64À

15.

(a) contains exactly one element. (b) contains exactly two elements. (c) contains exactly four elements. (d) is an empty set. How many two digit numbers exist such that the product of two digit number and two digit number formed by reversing its digits end with 5. (a) 2 (b) 3 (c) 4 (d) 5 If two roots of the equation

( p - 1)( x 2 + x + 1)2 - ( p + 1)( x 4 + x 2 + 1) = 0 1- x , then are real and distinct and f ( x ) = 1+ x æ æ 1 öö f ( f ( x)) + f ç f ç ÷ ÷ is equal to è è x øø (a) p (b) –p (c) 2p

(d) –2p

MOCK TEST-8

69 PHYSICS

16. One end of uniform wire of length L and of weight W is attached rigidly to a point in the roof and a weight W1 is suspended from its lower end. If s is the area of cross section of the wire, the stress æ 3L ö in the wire at a height ç ÷ from its lower end is è 4ø (a)

W1 s

(b)

Wù é êëW1 + 4 úû s

W1 + W 3W ù é (d) êëW1 + 4 úû / s s 17. Two plates identical in size, one of black and rough surface (B1) and the other smooth and polished (A2 ) are interconnected by a thin horizontal pipe with a mercury pellet at the centre. Two more plates A1 (identical to A2) and B2 (identical to B1 ) are heated to the same temperature and placed closed to the plates B1, and A2 as shown in the fig. The mercury pellet

(c)

P

A1 B1

A2 B2

(a) moves to the right (b) moves to the left (c) remains stationary (d) starts oscillating left and right 18. Three equal masses (each m) are placed at the corners of an equilateral triangle of side ‘a’. Then the escape velocity of an object from the circumcentre P of triangle is :

(a)

2 3 Gm a

(b)

3 Gm a

(c)

6 3 Gm a

(d)

3 3 Gm a

19. Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Select the correct statement : I. If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm II. If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm III. If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm IV. If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm (a) I only (b) II and III (c) III and IV (d) I and IV 20. In an ideal transformer the number of turns of primary and secondary coil is given as 100 and 300 respectively. If the power input is 60 W, the power output is (a) 100 W (b) 300 W (c) 180 W (d) 60 W 21. A cylindrical solid of length L and radius a is having varying resistivity given by r = r0x, where r0 is a positive constant and x is measured from left end of solid. The cell shown in the figure is having emf V and negligible internal resistance. The magnitude of electric field as a function of x is best described by

EBD_7839 70

KVPY-SA r = r 0x

24.

x V

(a) (c)

22.

23.

2V 2

L

V

2V

x

x

(b)

x

(d) None of these

r0 L2

An electron moves along the line PQ as shown which lies in the same plane as a circular loop of conducting wire as shown in figure. What will be the direction of the induced current initially in the loop when electron comes closer to loop? loop P

L2 Consider a moving charged particle in a region of magnetic field. Which of the following statements are correct ? (a) If v is parallel to B, then path of particle is spiral. (b) If v is perpendicular to B, then path of particle is a circle. (c) If v has a component along B, then path of particle is zig-zag. (d) If v is along B, then path of particle is a circle. Two masses m1 and m2 are connected by a massless spring of spring constant k and unstretched length l. The masses are placed on a frictionless straight channel, which are consider our x-axis. They are initially at x = 0 and x = l respectively. At t = 0, a velocity v0 is suddenly imparted to the first particle. At a later time t, the centre of mass of the two masses is at:

(a) x =

m2l m1 + m2

(b) x =

m1l mvt + 2 0 m1 + m2 m1 + m2

(c) x =

m2l m vt + 2 0 m1 + m1 m1 + m2

m2l m1v0t (d) x = m + m + m + m 1 2 1 2

25.

26.

Q

(a) Anticlockwise (b) Clockwise (c) Direction can not be predicted (d) No current will be induced Choose the incorrect statement: (a) Torque depends on transverse component of force. (b) Angular momentum depends on transverse component of linear momentum. (c) Radial component of linear momentum does not contribute to the angular momentum. (d) None of these A set of n identical cubical blocks lies at rest parallel to each other along a line on a smooth horizontal surface. The separation between the near surfaces of any two adjacent blocks is L. The block of one end is given a speed v towards the next one at time t = 0. All collisions are completely inelastic, then ; (n - 1) L (a) the last block starts moving at t = v (b) the last block starts moving at n ( n + 1) L t= 2v (c) the centre of mass of the system will have a final speed v (d) the centre of mass of the system will have a

v n 27. The figure shows a spherical hollow inside a lead sphere of radius R, the surface of the hollow passes through the centre of the sphere and "touches" the right side of the sphere. The mass of the sphere before hollowing was M. With what gravitational force does the hollowed out

final speed

MOCK TEST-8

71

lead sphere attract a small sphere of mass m that lies at a distance 'd' from the centre of the lead sphere, on the straight line connecting the centres of the sphere and of the hollow :

m R d

(a)

(b)

(c)

(d)

GMm é 1 ù 12 ê 2ú d ê R ö ú æ 8 ç1 ÷ êë d ø úû 2 è GMm é 1 ù 1+ 2 ê 2ú d ê R ö ú æ 4 ç1 ÷ êë è 2 d ø úû

CHEMISTRY 31. The compound which contains both ionic and covalent bond is: (a) KCl (b) KCN (c) CH4 (d) H2 32. Geometrical isomerism is possible in :

GMm é 1 ù 12 ê 2ú d ê R ö ú æ 4 ç1 ÷ êë è 2d ø úû

(a) CH3CH(CH3 )CH 2CH 2CH3

GMm é 1 ù 1+ 2 ê 2ú d ê R ö ú æ 8 ç1 ÷ êë d ø úû 2 è

(b) CH3CH=CHCH3 (c) CH3CH=CH 2

28. In the given pressure-temperature diagram, for water, which point indicates triple point? D

(a) A

(c) P (d) E

B water

Pressure (P)

(b) C

ice C

A

steam

P F

If equal amount of heat is supplied to the two spheres, then (a) temperature of A will be greater than B (b) temperature of B will be greater than A (c) their temperature will be equal (d) can’t be predicted 30. An aeroplane flying at a constant speed releases a bomb. As the bomb moves away from the aeroplane, it will (a) always be vertically below the aeroplane on ly i f th e aer opl an e was flyi n g horizontally (b) always be vertically below the aeroplane only if the aeroplane was flying at an angle of 45° to the horizontal (c) always be vertically below the aeroplane (d) gradually fall behind the aeroplane if the aeroplane was flying horizontally.

E Temp. (T)

29. Consider two identical iron spheres, one of which lies on a thermally insulating plate, while the other hangs from thermally insulating thread. The spheres are arranged so that their centres of mass are at the same height level (see figure).

(d) ClH2C - CH2Cl 33. The number of hydrogen atoms present in 25.6 g of sucrose (C12H22O11) which has a molar mass of 342.3 g is: (a) 22 × 1023

(b) 9.91 × 1023

(c) 11 × 1023 34. For the reaction

(d) 44 × 1023

ˆˆ† 2NH (g), DH = ? N 2 (g) + 3H 2 (g) ‡ˆˆ 3

(a) DE + 2RT (b) DE –2RT (c) DE – RT (d) None of these 35. The number of moles of KMnO4 needed to react completely with one mole of ferrous oxalate in acidic solution are : (a)

3 5

(b)

2 5

(c)

4 5

(d) 1

EBD_7839 72 36.

37.

KVPY-SA Which of the following has the maximum number of unpaired electrons? (a) Mn (b) Ti (c) V (d) Al Consider the following reaction :

43.

ˆˆ† Z (g) . At equilibrium, it was X (g) + Y (g) ‡ˆˆ

Water + H2SO4 + NH3 ¾¾¾® NH4 + HSO4–

38.

Which of the following is correct for above equation? (a) HSO–4 is a weak conjugate acid of H2SO4. (b) NH3 is a weak base. + (c) NH4 is a weak conjugate acid of NH3.

44.

(d) H2SO4 is a weak acid.

45.

In which of the following reactions, the underlined substance has been reduced? (b) CuO + 2HCl ¾® CuCl2 + H2O 46.

(d) C + 4HNO3 ¾® CO2 + 2H2O + 4NO2

41.

42.

Which step is chain propagation step in the following mechanism? hv (a) Cl 2 ¾¾ ® Clg + Clg (b) Clg + CH 4 ¾¾ ® g CH3 + HCl (c) Clg + Clg ¾¾ ® Cl 2 g g (d) CH3 + Cl ¾¾ ® CH3 Cl Aluminium is usually found in +3 oxidation state. In contrast, thallium exists in +1 and + 3 oxidation states. This is due to: (a) Inert pair effect (b) Diagonal relationship (c) Lattice effect (d) Lanthanoid contraction Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements O, S, F and Cl ? (a) Cl < F < O < S

(b) O < S < F < Cl

(c) F < S < O < Cl

(d) S < O < Cl < F

When a metal is added to dilute HCl solution, there is no evolution of gas. Metal is – (a) K

(b) Na

(c) Ag

(b) 8

(c) 64

(d) 6.4

Which of the following oxides is not expected to react with sodium hydroxide? (a) CaO (b) SiO2 (c) BeO (d) B2O3 BIOLOGY

(c) 4H2O + 3Fe ¾® 4H2 + Fe3O4

40.

1 1 observed that [X] = [Y] = [Z] . What is 2 2 value of [Z] (in mol L–1) at equilibrium? (a) 2 × 10–4 (b) 10–4 4 (c) 2 × 10 (d) 104 Molecular weight of a gas that diffuses twice as rapidly as the gas with molecular weight 64 is: (a) 16

(a) CO + CuO ¾® CO2 + Cu

39.

At 550K, the Kc for the following reaction is 104 mol–1 L

(d) Zn

47.

48.

49.

50.

51.

Maltose consists of which one of the following? (a) b- glucose and b-galactose (b) a- glucose and a-fructose (c) a-sucrose and b-glucose (d) Glucose and glucose Which one of the following shortens when a muscle fibre contracts? (a) Thick filament (b) Sarcomere (c) Troponin (d) Thin filament Bacteria having single flagellum on both sides: (a) Amphitrichous (b) Cephalotrichous (c) Peritrichous (d) Lophotrichous Meals which are rich in fat are not digested in the intestine in absence of: (a) Pepsin (b) Enterokinase (c) Insulin (d) Steapsin Sinks are related to: (a) Transport of minerals (b) Stomata (c) Enzymes (d) Phytochrome Pectin is: (a) Monosaccharides (b) Homopolysaccharides (c) Mucopolysaccharides (d) Heteropolysaccharides

MOCK TEST-8

73

52. The most common indicator organism of polluted water is: (a) E. coli (b) P. typhi (c) Vibrio (d) Entamoeba 53. Pentoses and hexoses are the most common: (a) Disaccharides (b) Monosaccharides (c) Oilgosaccharides (d) Polysaccharides 54. Which of the following elements are involved in muscle contraction? (a) Ca2+ and Mg2+ (b) Ca2+ and Na+ (c) Na+ and K+ (d) Mg2+ and K+ 55. In dicotyledonous stem, fascicular cambium is a meristematic tissue. It is an example of which of the following meristem? (a) Lateral (b) Secondary (c) Apical (d) Intercalary

56. Which of the following is not a steroid hormone? (a) Androgen (b) Aldosterone (c) Estrogen (d) Relaxin 57. Which of the following hormones can replace vernalisation? (a) Ethylene (b) Gibberellins (c) Cytokinins (d) Auxins 58. Which of the following conversions takes place in the blood clotting pathways? (a) Conversion of vitamin K to prothrombin. (b) Conversion of fibrin to fibrinogen. (c) Conversion of thrombin to prothrombin. (d) None of the above 59. Contraction of gall bladder is induced by: (a) Gastrin (b) Cholecystokinin (c) Secretin (d) Enterogastrone 60. Smallest cell of human body is: (a) Erythrocyte (b) Monocyte (c) Neuron (d) Blood platelets

PART-II (2 MARKS QUESTIONS) MATHEMATICS 61.

If a8 and 8a is completely divisible by 50! Then

which one of the following is true about ‘highest value of a’? (a) 10 < a < 14 (b) 14 < a < 16 (c) 16 < a < 18 (d) 18 < a < 20 p p 62. Let - < q < – . Suppose a1 and b1 are the 6 12 roots of the equation x2 – 2x sec a + 1 = 0 and a2 and b2 are the roots of the equation x2 + 2x tan q – 1 = 0. If a1 > b1 and a2 > b2, then a1 + b2 equals (a) 2 (sec q – tan q)

(b) 2 sec q

(c) –2 tan q (d) 0 63. The sum of the area of two circles, which touch each other externally is 153p. If the sum of their radii is 15, then ratio of the areas of smaller to the larger circle is

64. Through the vertex A of a parallelogram ABCD, line AEF is drawn to meet BC at E and DC produced at F. Then which of the following is true. (a) The triangles BEF and DCE are equal in area (b) Area (DBEF) =

1 Area (DABD) 2

(c) Area (DDCE) =

1 Area (DABF) 2

(d) Area (DBEF) = 2 Area (DDCE) 65. For a positive integer n, let pn denote the product of the digits of n, and sn denote the sum of the digits of n. the number of integers between 10 and 1000 for which pn + sn = n is

(a) 1 : 2

(b) 1 : 4

(a) 81

(b) 16

(c) 1 : 6

(d) 1 : 5

(c) 18

(d) 9

EBD_7839 74

KVPY-SA PHYSICS

66.

70.

A vertical cylinder contains a movable frictionless piston separating the former into two portions. The two portions are filled with ideal gases, and at equilibrium, the ratio of the volumes of upper to the lower portions is 5 : 3. When the temperature of the system is doubled, the ratio becomes 3 : 2. Then the ratio of number of molecules in the upper to the lower portions is (a) 35 : 26

(b) 26 : 35

Two identical uniform rectangular blocks (with longest side l) and a solid sphere of radius R are to be balanced at the edge of a heavy table such that the centre of the sphere remains at the maximum possible horizontal distance from the vertical edge of the table without toppling as indicated in the figure. If the mass of each block M is M and of the sphere is , the maximum 2 distance x that can be achieved is:

(c) 5 : 6 (d) 2 : 3 67. Steel ruptures when a shear of 3.5 × 108 N m–2 is applied. The force needed to punch a 1 cm diameter hole in a steel sheet 0.3 cm thick is nearly: (a) 1.4 × 104 N (b) 2.7 × 104 N (c) 3.3 × 104 N 68.

69.

(d) 1.1 × 104 N

A uniformly tapering conical wire is made from a material of Young's modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values R and 3R, respectively. The upper end of the wire is fixed to a rigid support and a mass M is suspended from its lower end. The equilibrium extended length, of this wire, would equal : (a)

æ 2 Mg ö L ç1 + è 9 pYR 2 ÷ø

(b)

æ 1 Mg ö L ç1 + è 9 pYR 2 ÷ø

(c)

æ 1 Mg ö L ç1 + è 3 pYR 2 ÷ø

(d)

æ 2 Mg ö L ç1 + è 3 pYR 2 ÷ø

Two 20 kg cannon balls are chained together and fired horizontally with a velocity of 200 m/s from the top of a 30 m wall. The chain breaks during the flight of the cannon balls and one of them strikes the ground at t = 1.5s, at a distance of 250 m from the foot of the wall, and 5m to the right of the line of fire. Determine the position of the other cannon ball at that instant. Neglect the resistance of air. (a) (20,550) (b) (550,20) (c) (2,55) (d) (55,2)

(a)

8l 15

(b)

5l 6

(c)

3l +R 4

(d)

7l +R 15

CHEMISTRY 71.

Find the product of the given reaction CH3 CH3 OH CH3 CH3

(a)

(b) CH3 (c)

CH3 H3C

(d)

H3C

H+ ¾ ¾¾® D

MOCK TEST-8

75

72. 1 c.c. N2O at NTP contains : (a)

1.8 ´ 10 22 atoms 224

(b)

6.02 ´ 10 23 molecules 22400

(c) 1.32 ´ 10 23 electrons 224 (d) All of the above 73. 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is : (a) Six

(b) Zero

(c) Two (d) Four 74. Which of the following compounds of elements in group IV would you expect to be most ionic in character? (a) CCl4 (b) SiCl4 (c) PbCl 2

77. Sex determination in grasshoppers, humans, and Drosophila is similar because (a) females are hemizygous. (b) males have one X chromosome and females have two X chromosomes. (c) all males always have one Y chromosome in all three species. (d) the ratio of autosomes to sex chromosomes is the same in all three organisms. 78. Which of the following pair is not correctly matched? (a) Rods – Twilight vision (b) Ciliary body – Iris (c) Retina – Optic chiasma (d) Vitreous humour – Posterior compartment 79. Match the genetic phenomena given in column-I with their respective ratios given in column-II and select the correct option.

(d) PbCl 4

75. The boiling point of p-nitrophenol is higher than that of o-nitrophenol because (a) NO2 group at p-position behaves in a different way from that at o-position. (b) intramolecular hydrogen bonding exists in p-nitrophenol. (c) there is intermolecular hydrogen bonding in p-nitrophenol. (d) p-nitrophenol has a higher molecular weight than o-nitrophenol.

Column-I

Column-II

(Genetic phenomena)

(Ratios)

A.

Inhibitory gene ratio

I.

9:3:4

B.

Complementary gene

II.

1:1:1:1

C.

Recessive epistasis

III.

12 : 3 : 1

IV.

13 : 3

V.

9: 7

ratio ratio D.

Dihybrid test cross ratio

E.

Dominant epistasis ratio

BIOLOGY

(a) A – V; B – IV; C – III; D – II; E – I

76. Which of the following are not used in the conversion by pyruvate to acetyl CoA? (i) Oxidative dehydrogenation (ii) Oxidative dehydration (iii) Oxidative phosphorylation (iv) Oxidative decarboxylation (a) (i), (ii) and (iii) (b) (i) and (ii) (c) (ii) and (iv) (d) (i) and (iii)

(b) A – IV; B – V; C – I; D – II; E – III (c) A – I; B – II; C – IV; D – III; E – V (d) A – II; B – I; C – IV; D – V; E – III 80. The concentration of OH– ions in a solution with the H+ ions concentration of 1.3 ´ 10–4 M is: (a) 7.7 ´ 10–4 M (b) 1.3 ´ 10–4 M (c) 2.6 ´ 10–8 M (d) 7.7 ´ 10–11 M

EBD_7839 76

1 2 3 4 5 6 7 8 9 10

KVPY-SA

(a) (a) (d ) (c) (a) (b ) (c) (a) (b ) (d )

11 12 13 14 15 16 17 18 19 20

(b ) (a) (b ) (b ) (a) (c) (c) (c) (b ) (d )

21 22 23 24 25 26 27 28 29 30

ANS W ER KEYS Part-I (a) 31 (b ) 41 (b ) (b) 32 (b) 42 (c) (d) 33 (b) 43 (a) (a) 34 (b) 44 (a) (a) 35 (a) 45 (a) (d) 36 (a) 46 (d ) (a) 37 (c) 47 (b ) (c) 38 (c) 48 (a) (b) 39 (b) 49 (d ) (c) 40 (a) 50 (a)

51 52 53 54 55 56 57 58 59 60

(d ) (a) (b ) (a) (a) (d ) (b ) (d ) (b ) (a)

61 62 63 64 65 66 67 68 69 70

Part-II (b ) 71 (c) 72 (b ) 73 (a) 74 (d ) 75 (a) 76 (c) 77 (c) 78 (b ) 79 (a) 80

(d) (d) (d) (c) (c) (a) (b) (c) (b) (d)

91 Time : 3 Hours

Stream-SA

Maximum Marks : 100

INSTRUCTIONS 1. 2. 3. 4.

There are 80 questions in this paper. The question paper contains two parts; Part–I (1 Mark Questions) and Part–II (2 Marks Questions). There are four sections; Mathematics, Physics, Chemistry and Biology in each part. There are four options given with each question, only one of them is correct. For each incorrect answer 0.25 Mark in Part–I and 0.5 Mark in Part–II will be deducted. PART-I (1 MARK QUESTIONS) MATHEMATICS

1.

2.

3.

4.

If p(x) = x3 – 3x2 + 2x + 5 and p(a) = p(b) = p(c) = 0, then the value of (2 – a) (2 – b) (2 – c) is (a) 5 (b) 3 (c) 9 (d) 2 How many 2 digit numbers exist such that sum of the squares of the numbers and the number formed by reversing its digits ends with 3. (a) 6 (b) 1 (c) 12 (d) 8 If p and q are distinct integers such that 2005 + p = q2 and 2005 + q = p2 then the product p.q is: (a) 2004 (b) 2009 (c) 2005 (d) – 2005 In figure DABC is equilateral triangle. Find height of DABC if the distances from the point O to the sides are 4, 5 and 6 respectively.

6.

(a) 122 (b) 47 (c) 99 (d) None of these A triangle has side lengths 4, 6, 8. A tangent is drawn to the incircle parallel to side 4 cutting other two sides at M and N, then the length of MN is

10 20 5 4 (b) (c) (d) 9 9 3 3 In the given fig., OB is the perpendicular bisector of the line segment DE, FA ^ OB and FE

(a) 7.

intersects OB at the point C. D

F

O

(a) 15 (c) 11 5.

A

C

B

E

(b) 14 (d) None of these

(a)

1 1 1 + = OA OC OB

(b)

1 1 + = OC2 OA OB

If 100! = K ( 2!) ( 5!) ( 7!) Then find the maxi-mum possible value of x + y + z

(c)

1 1 1 = 2+ 2 OA OC OB

(d)

1 1 2 + = OA OB OC

x

y

2

EBD_7839 78 8.

KVPY-SA If the remaining 9 are drinking water from a pond, then the difference between the number of deer who are grazing and those who are playing is a multiple of

In the diagram, the circle and the square have the same centre O and equal areas. The circle has radius 1 and intersects one side of the square at P and Q. The length of PQ is

(a) 4 12.

(b) 6

(c) 8

(d) 9

Find square root of 5 + 10 + 15 + 6 æ æ 3ö æ5öö (a) ç 1 + ç ÷ + ç ÷ ÷ ç è 2ø è 2 ø ÷ø è

9.

(a) 4 - p

(b) 1

(c)

(d)

2

æ æ 3ö æ5öö (b) ç 1 + ç ÷ - ç ÷ ÷ ç è 2ø è 2 ø ÷ø è 4-p

(a)

1 2 cm 4

(b)

1 2 cm 8

1 1 cm2 (d) cm2 16 32 PQ is a chord of length 8 cm of a circle of radius 5cm. The tangents at P and Q intersects at a point T. Then the length of TP is (c)

10.

(a)

20 cm 3

(b) 24.5 cm

20 cm (d) 12 cm 3 Half of a herd of deer are grazing in the field and three fourth of the remaining are playing nearby. (c)

11.

æ æ1ö æ3ö æ5öö (c) ç ç ÷ + ç ÷ + ç ÷ ÷ ç è2ø è2ø è 2 ø ÷ø è

In the figure, the area of square ABCD is 4 cm2 and E is mid point of AB, F, G, H and K are the mid points of DE, CF, DG and CH respectively. The area of DKDC is

æ æ 3ö æ5öö (d) ç 1 + ç ÷ + ç ÷ ÷ ç è 4ø è 2 ø ÷ø è

13.

If a number N = x. yz where x, y and z are single digits with not more than one digit is zero. Then which of the following will be always be an integer? (a) 37 N (b) 355 N (c) 396 N (d) 497 N

14.

If highest power of 7 in N! is k and that in (N + 3)! is K + 2 then how many values of N exist if N < 100. (a) 2 (b) 4 (c) 5 (d) 6

15.

In a particular batch of Pioneer career Kolkata there are 4 boys and certain number of girls. In every mock test only 5 students including at least 3 boys can appear. If different group of students write the Mock exam every time, if number of times test conducted is 66 then find the total number of students in the class. (a) 5 (c) 9

(b) 6 (d) None of these

MOCK TEST-9

79 PHYSICS

v0

16. A cylindrical drum, pushed along by a board rolls forward on the ground. There is no slipping at any contact. Find the distance moved by the man who is pushing the board, when axis of the cylinder covers a distance L.

wind

q

L

(a) 2 L (b) L (c) L / 2 (d) 4 L 17. An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be: (a) 1.7% (b) 2.7% (c) 4.4% (d) 2.27% 18. An experiment on the take-off performance of an aeroplane shown that the acceleration varies as shown in the figure, and that 12 s to take-off from a rest position. The distance along the runway covered by the aeroplane is

(a) 210 m (b) 2100 m (c) 21000 m (d) None of these 19. A boy throws a ball upward with velocity v0 = 20 m/s making an angle q with the vertical. The wind imparts a horizontal acceleration of 4 m/s2 to the left. The angle at which the ball must be thrown so that the ball returns to the boy’s hand is (g = 10 m/s2)

(a) tan–1(1.2) (b) tan –1(0.2) –1 (c) tan (2) (d) tan –1(0.4) 20. Carbon monoxide is carried around a closed cycle abc in which bc is an isothermal process as shown in the figure . The gas absorbs 7000 J of heat as its temperture increases from 300 K to 1000 K in going from a to b. P b

P2 P1 O

a

c V1

V2

V

The quantity of heat rejected by the gas during the process ca is (a) 4200 J (b) 5000 J (c) 9000 J (d) 9800 J 21. A biconvex lens of focal length f forms a circular image of sun of radius r in focal plane. Then :

(a)

p r2 µ f

(b)

p r2 µ f 2

(c) If lower half part is covered by black sheet, then area of the image is equal to

p r2 2

(d) If f is doubled, intensity will increase

EBD_7839 80 22.

KVPY-SA A mass is at the centre of a square, with four masses at the corners as shown.

5M

M

(A)

2M

3M

3M

5M

M

2M

5M

3M

2M

3M

2M

(c)

M

(B)

M

(C)

v(cm)

2M

(D)

5M 2M M M Rank the choices according to the magnitude of the gravitational force on the centre mass. (a) FA = FB < FC = FD (b) FA > FB < FC < FD (c) FA = FB > FC = FD (d) None of these 23.

24.

25.

A ball B is suspended from a string of length l attached to a cart A, which may roll on a frictionless surface. Initially the cart is at rest and the ball is given a horizontal velocity v0. Determine the velocity of B as it reaches the maximum height. (a)

n0 m A + mB

(b)

mB n 0 m A + mB

(c)

mB n 0 m A - mB

B

v(cm)

mB

A body travels uniformly a distance of (13.8 ± 0.2) m in a time (4.0 ± 0.3) s. Calculate its velocity with error limits. What is percentage error in velocity? (a) (3.5 + 3.1) m/sec, + 10% (b) (3.5 + 1.3) m/sec, + 9% (c) (3.5 + 0.31) m/sec, + 9% (d) (5.3 + 0.31) m/sec, + 9% For a particle executing S.H.M. the displacement x is given by x = A cos wt. Identify the graph which represents the variation of potential energy (P.E.) as a function of time t and displacement x.

v0

O

28.

29.

(a) I, III (b) II, IV (c) II, III (d) I, IV A 50 kg woman is on a large swing (generally seen in fairs) of radius 9 m that rotates in a vertical circle 6 rev/min. What is the magnitude of her weight when she has moved halfway up ? (a) 521 N (b) 512 N (c) 251 N (d) 215 N Let the x-z plane be the boundary between two transparent media. Medium 1 in z ³ 0 has a refractive index of 2 and medium 2 with z < 0 has a refractive index of 3. A ray of light in medium 1 given by the vector

r A = 6 3iˆ + 8 3 ˆj - 10kˆ is incident on the plane of

v(cm) (b)

O u(cm)

27.

O u(cm)

Cart A mA

(d) n0 ( m A + mB ) The radius of the earth is reduced by 4%. The mass of the earth remains unchanged. What will be the change in escape velocity? (a) Increased by 2% (b) Decreased by 4% (c) Increased by 6% (d) Decreased by 8% A student measures the focal length of a convex lens by putting an object pin at a distance ‘u’ from the lens and measuring the distance ‘v’ of the image pin. The graph between ‘u’ and ‘v’ plotted by the student should look like

(a)

(d) O u(cm)

26.

v(cm)

u(cm)

separation. The angle of refraction in medium 2 is: (a) 45° (b) 60° (c) 75° (d) 30°

MOCK TEST-9

81

30. Determine the minimum coefficient of friction between a thin rod and a floor at which a person can slowly lift the rod from the floor without slipping, to the vertical position, applying to its end a force always perpendicular to its length. (a)

1

(b)

2 2

(c)

(d)

2

OH OH

(c)

1

H H

2

H H

2 2 OH H

CHEMISTRY 31. Which of the following pairs of compounds illustrate the law of multiple proportions ? (a) H2O and Na2O (b) MgO and Na2O (c) Na2O and BaO (d) SnCl2 and SnCl4 32. Which bonds are formed by a carbon atom with sp2-hybridisation? (a) 4p-bonds (b) 2p-bonds and 2s-bonds (c) 1p-bond and 3s-bonds (d) 4s-bonds 33. Solubility of MX2 type electrolyte is 0.5 × 10–4 mol/litre. The value of Ksp of the electrolyte is (a) 5 × 10–13 (b) 25 × 10–10 –13 (c) 1.25 × 10 (d) 5 × 1012 34. Which of the following sub-shells will be filled by the electron after complete filling up of the orbitals of the third principal shell? (a) 4 s (b) 4 f (c) 4 d (d) 4 p 35. Which of the following conformers for ethylene glycol is most stable? OH OH

H

(d) HO H

H H

36. Which of the following behaves as both oxidising and reducing agents ? (a) H2SO4 (b) SO2 (c) H2O (d) HNO3 37. The molecule which does not exhibit dipole moment is : (a) NH3 (b) CHCl3 (c) H2O (d) CCl4 38. Among the following four structures I to IV,

CH3 C2H5 –– CH –– C3H7, (I) O

CH3

CH3 –– C –– CH –– C2H5, (II) H

CH3 +

H –– C , C2H5 –– CH –– C2H5

(a) H

H H OH

H

H

H

H

(b)

OH

H (III)

(IV)

It is true that, (a) only I and II are chiral compounds. (b) only III is a chiral compound. (c) only II and IV are chiral compounds. (d) all four are chiral compounds. 39. In the reaction, CO32– + H2O ® HCO3– + OH– water is a (a) Bronsted acid (b) Bronsted base (c) Conjugate acid (d) Conjugate base

EBD_7839 82

KVPY-SA

40.

The correct order of increasing thermal stability of K2CO3, MgCO3, CaCO3 and BeCO3 is (a) BeCO3< MgCO3 < CaCO3 < K2CO3 (b) MgCO3 < BeCO3 < CaCO3 < K2CO3 (c) K2CO3 < MgCO3 < CaCO3 < BeCO3 (d) BeCO3 < MgCO3 < K2CO3 < CaCO3

41.

The reactivities of iron, magnesium, sodium and zinc towards water are in the following order –

47.

(a) Fe > Mg > Na > Zn (b) Zn > Na > Mg > Fe (c) Na > Mg > Zn > Fe 42.

43.

44.

45.

(d) Mg > Na > Fe > Zn The IUPAC name of the compound CH3CH = CHC º CH is (a) Pent-l-yn-3-ene (b) Pent-4-yn-2-ene (c) Pent-3-en-1-yne (d) Pent-2-en-4-yne The two functional groups present in a typical carbohydrate are: (a) – CHO and – COOH and – OH (b) (c) – OH and – CHO (d) – OH and – COOH A sample of Na2CO3.H2O weighing 0.62 g is added to 100 mL of 0.1 N H2SO4 solution. The resulting solution would be: (a) Acidic (b) Neutral (c) Alkaline (d) None of the above Which one of the following conformations of cyclohexane is chiral? (a) Boat (b) Twist boat (c) Rigid (d) Chair

48.

49.

50.

BIOLOGY 46.

Sugars are not as good as fats as a source of energy for cellular respiration, because sugars (a) produce toxic amino groups when broken down. (b) contain more hydrogen. (c) usually bypass glycolysis and the Krebs cycle. (d) contain fewer hydrogen atoms and electrons.

51.

Match column-I with column-II and select the correct option from the codes given below. Column-I Column-II A. Stele I. Innermost layer of cortex B. Endodermis II. Suberin C. Casparian strip III. All the tissues exterior to vascular cambium D. Bark IV. All the tissues inner to endodermis (a) A – IV; B – I; C – II; D – III (b) A – III; B – II; C – I; D – IV (c) A – I; B – II; C – III; D – IV (d) A – IV; B – II; C – I; D – III Which of the following will not make minerals more available to plants? (a) Increasing the rainfall in a wet forested area. (b) Raising the pH of a very acidic soil. (c) Tillering a packed-down or water-logged soil. (d) Introducing fungi that can form mycorrhiza into a soil that lacks them. Which of the following is an amino acid, that is metabolised by the brain? (a) Alanine (b) Histidine (c) Glycine (d) Glutamic acid Which of the following hormone is a derivative of amino acid? (a) Oestrogen (b) Epinephrine (c) Progesterone (d) Prostaglandin Under which of the following circumstances will insulin be secreted? (a) The blood sugar level in the liver is low. (b) The blood sugar level in the hepatic portal vein is low. (c) The blood sugar level in the islets of Langerhans is high. (d) The glycogen level in the skeletal muscle is high.

MOCK TEST-9

83

52. A student brought home a strange animal which he found outside under a rock. It had moist skin, a complete digestive tract, a ventral nerve cord, and had gone through torsion. Identify the phylum of the animal. (a) Porifera (b) Annelida (c) Mollusca (d) Echinodermata 53. What is the most efficient region of water absorption in roots? (a) Root cap (b) Growing point (c) Zone of elongation (d) Zone of differentiation 54. ELISA is used to detect viruses, where (a) DNA probes are required. (b) Southern blotting is done. (c) alkaline phosphatase is the key reagent. (d) catalase is the key reagent. 55. Which of the following roles, given in below statements, are played by pioneer species on a barren rock? (i) It helps in dissolving rocks. (ii) It helps in weathering. (iii) It helps in soil formation. (a) (i) and (ii) (b) (i) and (iii)

(c) (ii) and (iii) (d) (i), (ii) and (iii) 56. The cell in the human body invaded by the human immuno-deficiency virus (HIV) is: (a) T-helper cell (b) Erythrocyte (c) B-cell (d) Macrophage 57. Protein is used as respiratory substrate only when: (a) Carbohydrates are absent (b) Fats are absent (c) Both carbohydrates and fats are exhausted (d) Fats and carbohydrates are abundant 58. Disintegration of the platelets gives rise to which factor, required for normal blood coagulation? (a) Thromboplastin (b) Hageman factor (c) Fibrinogen (d) None of these 59. Small proteins produced by vertebrate cells in response to viral infections inhibiting viral multiplication are known as: (a) Lipoproteins (b) Immunoglobulins (c) Interferons (d) Antitoxins 60. Which of the following enzyme is used in case of fungus to release DNA along with other macromolecules from the cells? (a) Lysozyme (b) Cellulase (c) Chitinase (d) Amylase

PART-II (2 MARKS QUESTIONS) MATHEMATICS 61. In given figure, CD || AE and CY || BA. Then (a) Area (DAXY) = Area (DCXY)

62. For positive real numbers x and y, let log y

f (x, y) = x 2 . If the sum of the solutions of the equation 4096 f ( f (x, x), x) = x13 can be expressed

(b) Area (ABCY) = 3 Area (DAXY) (c) Area (DCBX) = Area (DAXY) (d) None of these

as

m ( where m, n are coprime numbers), then n

(m – 10n) is (a) 1

(b) 2

(c) –1

(d) None of these

EBD_7839 84

KVPY-SA

63. The medians BE and CF of a DABC intersect at G. Then choose the correct option. (a) ar (DGBC) = ar of quad. AFGE (b) ar (DGBC) = ar (c) ar (DGBC) = ar (d) ar (DBGC) = ar (BACG) 64. A right circular cylinder has its height equal to two times its radius. It is inscribed in a right circular cone having its diameter equal to 10 cm and height 12 cm, and the axes of both the cylinder and the cone coincide. Then, the volume (in cm 3) of the cylinder is approximately (a) 107.5 (b) 118.6 (c) 127.5 (d) 128.7 65. The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is: (a) 15.8 (b) 14.0 (c) 16.8 (d) 16.0

68.

(a) The wave C is ahead by a phase angle of p and the wave B lags behind by a phase 2 p angle 2 (b) The wave C is lag behind by a phase angle p of and the wave B is ahead by a phase 2 p angle 2 (c) The wave C is ahead by a phase angle of p and the wave B lags behind by a phase angle p (d) The wave C lags behind by a phase angle of p and the wave B is ahead by a phase angle p Following are expressions for four plane simple harmonic waves (i)

æ xö y1 = A cos 2p ç n1t + ÷ l1 ø è

(ii)

æ ö x + p÷ y2 = A cos 2p ç n1t + l1 è ø

(iii)

æ xö y3 = A cos 2p ç n2 t + ÷ l2 ø è

PHYSICS 66. The distance between object and its real image in convex lens is D, and magnification is m. The focal length of the lens is: (a) (c)

D 1+ m

mD

(1 + m)

æ xö (iv) y4 = A cos 2p ç n2t - ÷ l2 ø è

D (b) 1 - m 2

(d)

The pairs of waves which will produce destructive interference and stationary waves respectively in a medium, are (a) (iii, iv), (i, ii) (b) (i, iii), (ii, iv) (c) (i, iv), (ii, iii) (d) (i, ii), (iii, iv)

mD

(1 - m)

2

67. The figure shows three progressive waves A, B and C with their phases expressed with respect to the wave A. What can be concluded from the figure? y B A C wt wt = p/2 wt = p wt = 3p/2

69.

A rectangular glass slab ABCD, of refractive index m1, is immersed in water of refractive index µ2(µ1 >µ2). A ray of light is incident at the surface AB of the slab as shown. The maximum value of the angle of incidence amax such that the ray comes out only from the other surface CD is given by

MOCK TEST-9

85

(a)

ém æ m sin -1 ê 1 cos ç sin -1 2 m m1 è ëê 2

(b)

é æ 1 öù sin -1 êm1 cos ç sin -1 ÷ú m 2 ø úû êë è

(c)

öù ÷ú ø ûú

-1 æ

m ö sin ç 1 ÷ è m2 ø

æm ö sin -1 ç 2 ÷ è m1 ø 70. A body initially at 80°C cools to 64°C in 5 minutes and to 52°C in 5 minutes. What will be the temperature after 15 minutes and what is the temperature of surroundings?

(d)

(a) 16°C, 43°C (c) 43°C, 16°C

(b) 6°C, 3°C (d) 3°C, 6°C

CHEMISTRY 71. In which of the following arrangements, the order is NOT according to the property indicated against it? (a) Li < Na < K < Rb : Increasing metallic radius (b) I < Br < F < Cl : Increasing electron gain enthalpy (with negative sign) (c) B < C < N < O : Increasing first ionisation enthalpy (d) Al

3+

< Mg

2+

+

-

< Na < F :

Increasing ionic size 72. 2, 2, 6, 6-tetramethyl cyclohexanol is treated with an acid. An alkene is formed. The structure of the alkene is

(a)

(b)

(c)

(d)

73. Calculate the total pressure in a 10.0 L cylinder which contains 0.4 g helium, 1.6 g oxygen and 1.4 g nitrogen at 27 °C. (a) 0.492 atm

(b) 49.2 atm

(c) 4.52 atm

(d) 0.0492 atm

74. The enthalpy changes for the following processes are listed below :

Cl2(g) ® 2Cl(g),

242.3 kJ mol–1

I2(g) ® 2I(g),

151.0 kJ mol–1

ICl(g) ® I(g) + Cl(g),

211.3 kJ mol–1

I2(s) ® I2(g),

62.76 kJ mol–1

Given that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of formation for ICl(g) is : (a) +16.8 kJ mol–1

(b) +244.8 kJ mol–1

(c) –14.6 kJ mol–1

(d) –16.8 kJ mol–1

75. An unknown compound A has a molecular formula C4H6. When A is treated with an excess of Br2 a new substance B with formula C4H6Br4 is formed. A forms a white precipitate with ammonical silver nitrate solution. A may be (a) 1-Butyne (b) 2-Butyne (c) 1-Butene (d) 2-Butene BIOLOGY 76. The sequence given below represents: Blue green algae ® Crustose lichen ® Foliose lichen ® Mosses ® Shrubs ® Dicot trees (a) Food chain (b) Ecological succession (c) Ecological trend (d) Food pyramid

EBD_7839 86 77.

78.

KVPY-SA An actively dividing bacterial culture is grown in a medium containing radioactive adenine (A*). After all the adenine is labelled, the bacteria are transferred to a medium containing nonradioactive adenine (A). Following one round of DNA replication in the nonradioactive medium, the DNA is analysed. Which of the following sequences could represent this DNA? (a) A*A*TTGA*TC TTAACTAG (b) A*ATTGA*TC TTA*A*CTAG (c) AATTGATC TTAACTAG (d) A*A*TTGA*TC TTA*A*CTA*G Homologous structures among organisms provide evidence for evolution in that these structures are (a) different in different animals, but are modifications of the same basic structure. (b) similar in function, but of different basic structure. (c) all shown in the fossil record. (d) all produced by the same gene.

79.

80.

After which of the following treatments (i – iii), would an enzyme still be expected to have activity? (i) Protease treatment. (ii) Heating almost to the point of denaturation and then cooling once only. (iii) Freezing and then thawing once only. (a) (i) only (b) (iii) only (c) (i) and (ii) only (d) (ii) and (iii) only Calvin and colleagues determined the pathway of carbohydrate synthesis in plants by studying the incorporation of radioactive carbon dioxide into biological compounds. Suppose that photosynthesis is proceeding at a steady pace in a typical experiment with the lights on, and carbon dioxide is being combined with ribulose-bisphosphate (RuBP) to produce 3-phosphoglycerate (3PG). Then suddenly the source of carbon dioxide is eliminated. What changes in the concentrations of 3PG and RuBP would occur? (a) 3PG levels rise, RuBP levels fall. (b) 3PG levels fall, RuBP levels rise. (c) 3PG levels rise, RuBP levels rise. (d) 3PG levels rise, RuBP levels stay the same.

ART - 1 : BIOLOGY 1 2 3 4 5 6 7 8 9 10

(a) (d ) (c) (a) (d ) (b ) (d ) (d ) (b ) (c)

11 12 13 14 15 16 17 18 19 20

(d ) (a) (c) (d ) (d ) (a) (b ) (a) (d ) (d )

21 22 23 24 25 26 27 28 29 30

ANS W ER KEYS Part-I (b) 31 (d ) 41 (c) (a) 32 (c) 42 (c) (b) 33 (a) 43 (c) (a) 34 (d) 44 (b ) (c) 35 (a) 45 (b ) (c) 36 (b) 46 (d ) (a) 37 (d) 47 (a) (a) 38 (a) 48 (a) (a) 39 (a) 49 (d ) (a) 40 (a) 50 (b )

51 52 53 54 55 56 57 58 59 60

(c) (c) (d ) (c) (d) (a) (c) (a) (c) (c)

61 62 63 64 65 66 67 68 69 70

Part-II (c) 71 (a) 72 (a) 73 (c) 74 (b ) 75 (c) 76 (b ) 77 (d ) 78 (a) 79 (c) 80

(c) (a) (a) (a) (a) (b) (a) (a) (b) (b)

10 1 Time : 3 Hours

Stream-SA

Maximum Marks : 100

INSTRUCTIONS 1. 2. 3. 4.

There are 80 questions in this paper. The question paper contains two parts; Part–I (1 Mark Questions) and Part–II (2 Marks Questions). There are four sections; Mathematics, Physics, Chemistry and Biology in each part. There are four options given with each question, only one of them is correct. For each incorrect answer 0.25 Mark in Part–I and 0.5 Mark in Part–II will be deducted. PART-I (1 MARK QUESTIONS) MATHEMATICS

1.

2.

3.

4.

If

f (n + 1) =

2 f (n ) + 1 , n = 1, 2, ...... 2

a nd

f (1) = 2, then f (101) = ____________. (a) 53 (b) 52 (c) 51 (d) 50 When 648 is added or 739 subtracted from a number N result in both the cases is a perfect cube then which one of the following is correct about N? (i) N is an even number (ii) N is divisible by 3 (iii) N is divisible by 5 (a) Only (i) & (ii) (b) Only (i) & (iii) (c) Only (ii) & (iii) (d) None of these The numbers a, b, c are the digits of a three digit number which satisfy 49a + 7b + c = 286. Then, the value of 100a + 10b + c is (a) 556 (b) 256 (c) 356 (d) 286 In a trapezium ABCD, P and Q are points on AD and BC respectively such that

DP CQ 1 = = . AP BQ 3

If the diagonal BD intersect PQ at E, find the

5.

6.

ratio of ar (DDPE) to (trapezium ABCD) (a) 1 : 8 (b) 1 : 16 (c) 1 : 32 (d) 1 : 27 N is the product of first 100 multiple of K. If N is divisible by 10100 then find the minimum number of zeros at the end of N (a) 24 (b) 124 (c) 97 (d) 121 Let X be any point on the side BC of a triangle ABC. If XM, XN are drawn parallel to BA and CA meeting CA. BA in M, N respectively, MN meets CB produced in T. Then

(a) TB2 = TX × TC (c) TX2 = TB × TC

(b) (d)

TC2 = TX × TX TX2 = 2(TB × TC)

EBD_7839 88 7.

KVPY-SA Let ABC be an acute angled triangle and CD be the altitude through C. If AB = 8 and CD = 6, then the distance between the mid-points of AD and BC (i.e. EF) is

time of flight increased by 30 minutes, then the duration of flight is

12.

13.

8.

(a) 3

(b) 5

(c) 6

(d) 4

14.

A circle is inscribed in a square and the square is circumscribed by another circle. What is the ratio of the areas of the inner circle to the outer circle?

9.

(a) 1 : 2

(b) 1: 2

(c)

(d) 1: 3

2 :4

15.

P, Q and R are respectively the mid–points of sides BC, CA and AB of a triangle ABC. PR and BQ meet at X. CR and PQ meet at Y. Then (a) XY=

1 BC 3

(b) XY =

(a) 1 hr.

(b) 2 hrs.

(c) 3 hrs.

(d) 4 hrs.

Find 6 root of 99 + 70 2 th

(a) 1 +

2

(b) 2 +

3

(c) 3 +

5

(d) 2 +

5

What is the last 3 digit number of 1223334444 …… 1000 digit. (a) 212 (b) 414 (c) 373 (d) 323 Let n! = 1 × 2 × 3 × … × n for integer n. If p = 1! + (2 × 2!) + (3 × 3!) + … + (10 × 10!), then p + 2 when divided by 11! Leaves a remainder of (a) 10 (b) 0 (c) 7 (d) 1 In how many ways Ram can distribute 40 apples in his six children named A, B, C, D, E and F such that A gets two more than B, C gets 3 more than F and D gets five less than E and every one must have atleast one fruit (a) 101 (b) 91 (c) 96 (d) 136

2 BC 3

PHYSICS 16.

A river is flowing with a velocity of 1 m/s towards east directions. When the boat runs with a velocity of 3 m/s relative to the river is the direction of the river flow, the flag on the boat flutter in north direction. If the boat runs with the same speed but in north direction relative to river, the flag flutters towards north-east direction. The actual velocity of the wind should

10.

1 2 BC (d) XY = BC 4 5 In the figure, ABCD is a semicircle. ÐCAD =

11.

(a) 25° (b) 40° (c) 45° (d) 50° In a flight of 600 km an aeroplane was slowed

(a) 4 $i + 6 $j

(b) 6 $i + 4 $j

down due to a bad weather. If the average speed

(c) 4 $i – 6 $j

(d) 6 $i – 4 $j

(c)

XY =

for the trip was reduced by 200 km/h and the

be : ( $i ® east direction and $j ® north direction).

MOCK TEST-10

89

17. A U-tube is of non uniform cross-section. The area of cross-sections of two sides of tube are A and 2A (see fig.). It contains non-viscous liquid of mass m. The liquid is displaced slightly and free to oscillate. Its time period of oscillations is m (a) T = 2p 3rgA m (b) T = 2p 2rgA

(c) T = 2p

A

2A

m

m rgA

(d) None of these 18. Variation of radiant energy emitted by Sun, filament of tungsten lamp and welding arc as a function of its wavelength is shown in figure. Which of the following option is the correct match ?

(a) Sun-T3 , tungsten filament - T1, welding arc - T2 (b) Sun-T2 , tungsten filament - T1, welding arc - T3 (c) Sun-T3 , tungsten filament - T2, welding arc - T1 (d) Sun-T1 , tungsten filament - T2, welding arc - T3 19. A system of two identical rods (L-shaped) of mass m and length l are resting on a peg P as shown in the figure. If the system is displaced in its plane by a small angle q, find the period of oscillations

(a) 2p

2l 3g

(b) 2p

2 2l 3g

(c) 2p

2l 3g

(d) 3p

l 3g

20. Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is : (a) Zero (b) 1% (c) 3% (d) 6% 21. Consider the following statements : A gas can be liquefied by increasing the pressure 1. above the critical pressure only 2. Only when the temperature of the enclosed gas is below the critical temperature 3. Only when the volume of the enclosed gas is below the critical volume Which of the statements given above is / are correct : (a) 1 and 2 (b) 2 only (c) 3 only (d) 2 and 3 22. The period of oscillation of a simple pendulum is T = 2p

L . Measured value of L is 20.0 cm g

known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution. The accuracy in the determination of g is : (a) 1% (b) 5% (c) 2% (d) 3%

EBD_7839 90 23.

KVPY-SA A wire is bent in the form of circle of radius 2 m. Resistance per unit length of wire is 1/p W/m. Battery of 6V is connected between A and B. ÐAOB = 90°. Find the current through the battery.

26.

(a) 8 A

3v 2 (d) None of these 27. The refractive index of a material of a prism of angles 45°, 45°, 90° is 1.5. The path of the ray of light incident normally on the hypotenuse side is shown in figure

I

B

(b) 4 A (c) 3 A

A

(d) 9 A 24.

6V

A black coloured solid sphere of radius R and mass M is inside a cavity with vacuum inside. The walls of the cavity are maintained at temperature T0. The initial temperature of the sphere is 3T0. If the specific heat of the material of the sphere varies as aT3 per unit mass with the temperature T of the sphere, where a is a constant, then the time taken for the sphere to cool down to temperature 2T0 will be (s is Stefan Boltzmann constant) (a)

Ma

æ 3ö ln ç ÷ 4pR s è 2 ø 2

Ma

æ 16 ö ln ç ÷ (c) 2 16pR s è 3 ø 25.

(b)

Velocity of the centre of a small cylinder is v. There is no slipping anywhere. The angular velocity of the centre of the larger cylinder is (a) 2v R (b) v v 2R (c)

(a)

(c) 28.

Ma

æ 16 ö ln ç ÷ 4pR s è 3 ø 2

æ 3ö ln ç ÷ (d) 2 16pR s è 2 ø

C

B

A

A 90°

(d)

45°

C

B

45°

45°

C

45°

C

v2

(b) a2

a2

v2

v2

(c)

(d) a2

29.

45°

90°

(a)

Heater R

It is connected in a combinations with a resistance of 10 ohms and a resistance R to a 100 volts mains as shown in the figure. What should be the value of R so that the heater operates with a power of 62.5 watts. (a) 25 W (b) 0.05 W (c) 5 W (d) 65 W

45°

(b)

45°

45°

v2

Ma

100V

B

A 90°

A graph of the square of the velocity against the square of the acceleration of a given simple harmonic motion is

A heater is designed to operate with a power of 1000 watts in a 100 volt line. 10W

B

A 90°

a2

Which one of the following statements is true? (a) Both light and sound waves in air are transverse (b) The sound waves in air are longitudinal while the light waves are transverse (c) Both light and sound waves in air are longitudinal (d) Both light and sound waves can travel in vacuum

MOCK TEST-10

91

30. A box of mass 1 kg is mounted with two cylinders each of mass 1 kg, moment of inertia 0.5 kg m2 and radius 1m as shown in figure. Cylinders are mounted on their control axis of rotation and this system is placed on a rough horizontal surface. The rear cylinder is connected to battery operated motor which provides a torque of 100N-m to this cylinder via a belt as shown. If sufficient friction is present between cylinder and horizontal surface for pure rolling, find acceleration of the vehicle in m/s2. (Neglect mass of motor, belt and other accessories of vehicle). (a) 20

Electric motor

m/s2

(b) 10 m/s2 (c)

m

25 m/s2

(d) 30 m/s2 CHEMISTRY 31. The maximum number of molecules are present in (a) 15 L of H2 gas at STP (b) 5 L of N2 gas at STP (c) 0.5 g of H2 gas (d) 10 g of O2 gas 32. Which of the following will be most easily attacked by an electrophile? OH

Cl

(a)

(a) NO2

(b) O2

(c) N2

(d) N2O

35. Which of the following pair will give displacement reaction? (a) ZnSO4 solution and Aluminium metal. (b) MgCl2 solution and aluminium metal. (c) FeSO4 solution and silver metal. (d) AgNO3 solution and copper metal. 36. Which of the following is not aromatic? (a) Benzene (b) Naphthalene (c) Pyridine (d) 1,3,5 Heptatriene 37. For which one of the following sets of four quantum numbers, an electron will have the highest energy? n

l

m

s

(a) 3

2

1

1/2

(b) 4

2

–1

1/2

(c) 4

1

0

–1/2

(d) 5

0

0

–1/2

38. I, II, and III are three isotherms, respectively, at T1, T2 and T3. Temperature will be in order:

(b)

CH3 (c)

34. When PbO2 reacts with conc. HNO3 the gas evolved is:

P

(d) I

33. Which of the following statement is incorrect? (a) Alkali metal hydroxide are hygroscopic. (b) Dissolution of alkali metal hydroxide is endothermic. (c) Aqueous solution of alkali metal hydroxides are strongly basic. (d) Alkali metal hydroxides form ionic crystals.

II III V3

V2

V1

(a) T1 = T2 = T3

(b) T1 < T2 < T3

(c) T1 > T2 > T3

(d) T1 > T2 = T3

EBD_7839 92 39.

KVPY-SA Standard reduction potentials of the half reactions are given below : F2(g) + 2e– ® 2F– (aq); E° = + 2.85 V Cl2(g) + 2e– ® 2Cl–(aq); E° = + 1.36 V Br2(l) + 2e– ® 2Br–(aq); E° = + 1.06 V I2(s) + 2e– ® 2I–(aq); E° = + 0.53 V The strongest oxidising and reducing agents respectively are :

40.

(a) F2 and I–

(b) Br2 and Cl–

(c) Cl2 and Br–

(d) Cl2 and I2

0.45 g acid of molecular weight 90 is neutralised by 20 mL of 0.5N caustic potash. The basicity of acid is: (a) 1

41.

(b) 2

46.

47.

48.

(d) 4

Which of the following oxides is amphoteric in character? (a) SnO2 (c) CO2

42.

(c) 3

BIOLOGY

(b) SiO 2 (d) CaO

The highest boiling point is expected for :

49.

(a) iso-octane (b) n-octane (c) 2, 2, 3, 3-tetramethylbutane (d) n-butane 43.

44.

45.

Pressure of a mixture of 4 g of O2 and 2 g of H2 confined in a bulb of 1 litre at 0 °C is (a) 25.22 atm

(b) 31.20 atm

(c) 45.21 atm

(d) 15.21 atm

Diacidic base is: (a) CH2(OH)2

(b) Ca(OH)2

(c) CH3CH(OH)2

(d) All of these

50.

51.

For which of the following process, DS is negative? (a) H2 (g) ¾¾ ® 2 H(g) (b) 2SO3 (g) ¾¾ ® 2SO2(g) + O2 (g) compressed

(c) N 2 (4L) ¾¾¾¾¾® N 2 (2L) (d) C(diamond) ¾¾ ® C (graphite)

52.

The branches of the nodal tissue, which give rise to minute fibres throughout the ventricular musculature of the respective sides are called: (a) SA node (b) AV node (c) Purkinje fibre (d) bundle of His Peroxisomes are rich in: (a) DNA (b) RNA (c) Catalytic enzymes (d) Oxidative enzymes The threshold of a neuron is: (a) the amount of inhibitory neurotransmitter required to inhibit an action potential. (b) the membrane voltage at which an axon potential will be suppressed. (c) the amount of excitatory neurotransmitter required to elicit an action potential. (d) the membrane voltage at which the membrane potential develops into an action potential. The lacunae in vascular bundles of monocot stems are: (a) Metaxylem (b) Mucilage canal (c) Lysigenous water cavity (d) Large-sized vessel Nutrient enrichment of water body is: (a) Eutrophication (b) Stratification (c) Biomagnification (d) None of these Energy pyramids are used to represent energy transfer in an ecosystem because energy is _______ at each trophic level. (a) Gained (b) Lost (c) Conserved (d) Either conserved or gained Vaccines are (a) treated bacteria or viruses or one of their proteins. (b) MHC (major histocompatibility complex) proteins. (c) curative medicines. (d) monoclonal antibodies.

MOCK TEST-10

93

53. The rate of photosynthesis of a freshwater plant is measured using five spectral colours. Which sequence of colours would give an increasing photosynthetic response? Largest response Smallest (a) Blue Green Yellow Orange Red (b) Green Yellow Orange Red Blue (c) Red Orange Yellow Green Blue (d) Yellow Green Orange Blue Red 54. All but one of the following chemicals are neurotransmitters that function in the human brain. Select the exception. (a) Dopamine (b) Glycine (c) Atropine (d) Glutamic acid 55. Purkinje fibres are present in: (a) Left auricle (b) Right auricle (c) Ventricular myocardium (d) SA node 56. Injections of which hormone are sometimes given to strengthen contractions of the uterus during childbirth?

(a) Adrenocorticotropic hormone (ACTH) (b) Thyroxine (c) Oxytocin (d) Insulin 57. When a piece of bread is chewed, it tastes sweet because (a) the sugar contents are drawn out. (b) saliva converts starch into maltose. (c) the taste buds are stimulated by chewing. (d) None of the above 58. Resting membrane potential will be greater in: (a) Large size fibre (b) Small size fibre (c) Both (a) and (b) (d) None of these 59. Substance that accumulates in a fatigued muscle is: (a) Pyruvic acid (b) Carbon dioxide (c) ADP (d) Lactic acid 60. Inhaled air passes through which of the following in the last? (a) Bronchiole (b) Larynx (c) Pharynx (d) Trachea

PART-II (2 MARKS QUESTIONS) MATHEMATICS 61. ABCD is a parallelogram. Any line through A cuts DC at a point P and BC produced at Q.

63. In the figure ABC is a triangle, D is mid point of AB. P is any point on BC. Line CQ is drawn parallel to PD to intersect AB at Q. PQ is joined. Then choose the correct option.

Then, (a) Area (DBPC) = Area (DDPQ) (b) Area (DBPC) =

1 Area (DACB) 4

(c) Area (DDPQ) =

1 Area (DADP) 4

(d) None of these

2 p - 2ü ì 62. If A = íx : - £ x £ ý, 5 5 þ î B = {y : – 1 £ y £ 1|} and f (x) = cos (5x + 2), then the mapping f : A ® B is (a) one-one but not onto (b) onto but not one-one (c) both one-one and onto (d) neither one-one nor onto

(a) ar (DBPQ) =

1 ar (DABC) 4

(b) ar (DBPQ) =

2 ar (DABC) 3

(c) ar (DBPQ) =

1 ar (DABC) 3

(d) ar (DBPQ) =

1 ar (DABC) 2

EBD_7839 94 64.

65.

KVPY-SA There is a right circular cone of height h and vertical angle 60º. A sphere when placed inside the cone, it touches the curved surface and the base of the cone. The volume of sphere is (a)

4 3 ph 3

(b)

4 ph3 27

(c)

4 3 ph 9

(d)

4 3 ph 81

In a set of 2n distinct observations, each of the observations below the median of all the observations is increased by 5 and each of the remaining observations is decreased by 3. Then the mean of the new set of observations: (a) increases by 1 (b) decreases by 1 (c) decreases by 2 (d) increases by 2

68.

Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m2. The radii of bubbles A and B are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04 N/m. Find the ratio nB / nA where nA and nB are the number of moles of air in bubbles A and B, respectively. [ Neglect the effect of gravity]. (a) 2 (b) 9 (c) 8 (d) 6

69.

A beaker contains a fluid of density r kg / m3, specific heat S J/kgºC and viscosity h. The beaker is filled upto height h. To estimate the rate of heat transfer per unit area (Q/A) by convection when beaker is put on a hot plate, a student proposes that it should depend on h ,

PHYSICS 66.

67.

The time taken by the earth to travel over half its orbit, remote from the sun, separated by the minor axis is about 2 days more than half the year, then the eccentricity of the orbit is (a) 1/30 (b) 1/60 (c) 1/15 (d) 1/70 A 2 m wide truck is moving with a uniform speed v0 = 8 m/s along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed v when the truck is 4 m away from him. The minimum value of v so that he can cross the road safely is

(a) 2.62 m/s (c) 3.57 m/s

(b) 4.6 m/s (d) 1.414 m/s

æ S Dq ö and æ 1 ö when Dq (in ºC) is the çè ÷ çè rg ø÷ h ø difference in the temperature between the bottom and top of the fluid. In that situation the correct option for (Q/A) is: æ S Dq ö æ 1 ö (a) h ç è h ÷ø çè rg ø÷

(c) 70.

S Dq hh

æ S Dq ö æ 1 ö (b) ç è hh ÷ø èç rg ø÷

(d) h

S Dq h

Two particles of masses m1 and m2 moving in coplanar parabolas round the sun, collide at right angles and coalesce when their common distance from sun is R. The subsequent path of the combined particles is an ellipse of major axis (a)

( m1 + m2 ) 2 R 2 m1m2

(b)

( m1 - m2 ) 2 R 2m1 m2

(c)

( m1 + m2 ) 2 R m1m2

(d)

( m1 + m2 ) 2 R 3m1 m2

MOCK TEST-10

95 CHEMISTRY

71. In hydrogen atom, energy of first excited state is –3.4 eV. Find out KE of the same orbit of hydrogen atom: (a) + 3.4 eV (b) + 6.8 eV (c) – 13.6 eV (d) + 13.6 eV 72. The addition of Br2 to (E)-but-2-ene gives: (a) (R,R)-2,3-dibromobutane (b) (S,S)-2,3-dibromobutane (c) (R,S)-2,3-dibromobutane (d) A mixture of (R,R) and (S,S)-2,3-dibromobutane 73. Consider the following changes : ® M (g) (1) M ( s ) ¾¾ ® M 2 + ( g ) + 2e (2) M ( s ) ¾¾ ® M + ( g ) + e(3) M ( g ) ¾¾ + ® M 2+ ( g ) + e (4) M ( g ) ¾¾

® M 2+ ( g ) + 2e (5) M ( g ) ¾¾

The second ionization energy of M could be calculated from the energy values associated with : (a) 1 + 3 + 4 (b) 2 – 1 + 3 (c) 1 + 5 (d) 5 – 3 74. The oxidation states of Cr in [Cr(H2O)6]Cl3, [Cr(C6H6)2], and K2[Cr(CN)2(O)2(O2)(NH3)] respectively are : (a) +3, +4, and +6 (b) +3, +2, and +4 (c) +3, 0, and +6 (d) +3, 0, and + 4 75. Which one is a nucleophilic substitution reaction among the following ? (a) CH3 – CH = CH2 + H2O H+

¾¾¾ ® CH3 - CH - CH 3 | OH

(b) RCHO + R¢MgX ¾¾® R – CH – R¢ | OH

CH 3 | (c) CH3 – CH2–CH–CH2Br + NH3 ¾¾®

CH3 | ¾¾® CH3 – CH2 – CH – CH2NH2 ® CH3CH (OH) CN (d) CH3CHO + HCN ¾¾

BIOLOGY 76. Plant cell ‘A’ has an osmotic pressure of 12 atm and is immersed in solution of 10 atm osmotic pressure. Another cell ‘B’ has 10 atm osmotic pressure and is immersed in solution of osmotic pressure of 8 atm. Both the cells are allowed to come to equilibrium, then removed from their solution and brought in intimate contact. Assuming that there is no external influencing force, what will be the result? (a) There will be a net flow of water from ‘A’to ‘B’. (b) There will be net flow of water from ‘B’ to ‘A’. (c) There will be no net flow of water. (d) Water will freely pass from ‘A’ to ‘B’ but not from ‘B’ to ‘A’. 77. A female with normal genotype (XX, 44) showed the presence of testis determining factor (TDF) gene on the X chromosome. This is most likely a result of: (a) X chromosome inactivation (b) Dosage compensation effect (c) Mutation (d) Meiotic recombination 78. Two alleles are found in Calico cats for coat colour and are located on the X chromosome. One allele is responsible for black colour while the other for orange colour. Which of the following is correct about the phenotype of Calico cats? (a) All male cats will have the same coat colour. (b) Female cats can have mosaic pattern. (c) All females will show intermediate coat colour. (d) All female cats born to mothers with black coat colour will have black coat colour.

EBD_7839 96

KVPY-SA

79. The critical day lengths for 4 plants are as follows: Plant A – 15.5 hrs Plant B – 15.5 hrs Plant C – 10.0 hrs Plant D – 9.5 hrs Plant A flowers when it receives 8.5 or more hours of darkness. Plant B flowers when it receives a minimum of 15.5 hrs of light. Plant C flowers when it receives less than 10 hrs of light. Plant D flowers when it receives less than 9.5 hrs of light. Which one is a long day plant? (a) Plant A (b) Plant A and C (c) Plant D (d) Plant B

1 2 3 4 5 6 7 8 9 10

(b ) (c) (a) (c) (b ) (c) (b ) (a) (c) (b )

11 12 13 14 15 16 17 18 19 20

(a) (a) (d) (d) (b) (a) (a) (a) (b) (d)

21 22 23 24 25 26 27 28 29 30

80.

Which of the following pathways depicts the generalised glycolytic scheme most accurately?

(a) Glucose ® a ® b ® c ® d ® f ® g ® h ® Pyruvate e

(b) Glucose « a « b « c « d « f « g « h « Pyruvate e

(c) Glucose ® a « b ® c « d « f « g « h ® Pyruvate e

(d) Pyruvate ® Glucose ® a ® b ® c h ¬ g ¬ f ¬ e ¬ d

ANS W ER KEYS Part-I (d ) 31 (a) 41 (a) (d ) 32 (b) 42 (b) (a) 33 (b ) 43 (a) (c) 34 (b ) 44 (b) 35 (d ) 45 (c) (c) (b ) 36 (d ) 46 (c) (a) 37 (b ) 47 (d) (d ) 38 (c) 48 (d) (b ) 39 (a) 49 (c) (a) 40 (b ) 50 (a)

51 52 53 54 55 56 57 58 59 60

(b ) (a) (b ) (c) (c) (c) (b ) (a) (d ) (a)

61 62 63 64 65 66 67 68 69 70

Part-II (a) 71 (c) 72 (d ) 73 (d ) 74 (a) 75 (b ) 76 (c) 77 (d ) 78 (d ) 79 (a) 80

(a) (c) (d) (c) (c) (b) (d) (b) (d) (c)

HINTS & SOLUTIONS

MOCK TEST-1 So, ÐABP = ÐDEP and hence, AB is parallel to DE. Similarly, BC is parallel to EF and CA is parallel to DF.

PART – I MATHEMATICS 1.

2.

3.

(a) Since x2 + y2 < 25 and x and y are integers, the possible values of x and y Î (0, ± 1, ± 2, ± 3, ± 4). Thus x and y can be chosen in 9 ways each and (x, y) can be chosen in 9 × 9 = 81 ways. However we have to exclude cases ( ± 3, ± 4), ( ± 4, ± 3) and ( ± 4, ± 4) (i.e.) 3 × 4 = 12 cases. Hence the number of permissible values = 81 – 12 = 69. (d) p (x) = (2 x + 3) (x 97 + x 96 +... + 1). = (2 x + 3) (x + 1) (x 96 + x 94 +.. + x 2 + 1) Also, x 96 + x 94 +... + x 2 + 1 > 1 " x Î R. So required real roots are –1 and –3/2 Sum of roots is –5/2 (b) Let AP, BP, CP when extended, meet the sides BC, CA, AB in D, E, F respectively. Draw AK, PL perpendicular to BC with K, L on BC. A

F

P

AD BE CF = = = 3, and consequently PD PE PF each of the given ratios is also equal to 3.

So,

x1 + ... + x15 = 13 15 x1 + x2 + x3 + ... + x15 = 15 × 13 = 195 In order to set the second largest and largest, first thirteen natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13

4.

(b) Given,

5.

13 ´ 14 2 Þ x14 + x15 = 195 – 91 = 104 Now, from the options, we have x14 = 51 and x15 = 53 Now, second largest is 51. (b)

E

B K LD C d ( A, BC ) AK AD = = . Now, d ( P, BC ) PL PD Similarly,

BD AE AF DC = = = DC EC FB BD Hence BD 2 = CD 2 or which is same as BD = CD. Thus, D is the midpoint of BC. Similarly, E, F are the midpoints of CA and AB respectively. We infer that AD, BE, CF are indeed the medians of the triangle ABC and hence P is the centroid of the triangle.

Using these, we obtain

So, x14 + x15 = 195 –

d ( B, CA) BE d (C, AB) CF = = . and d ( P, CA) PE d (P, AB) PF

So, we obtain

AD BE CF = = , and hence PD PE PF

AP BP CP = = . PD PE PF

C

AP BP = and ÐAPB = ÐDPE, it follows PD PE that triangles APB and DPE are similar.

From

60 A

B F 30º r1 D

G

r2 E

EBD_7839 98

KVPY-SA r1 ® radius of smaller sphere r2 ® radius of bigger sphere ÐBAD = 30°, r1 = 7 cm

Substituting two expressions for into (ii),we get:

(

= 2r1 sin 30º AB = AF + BF 2r1 = AF + r1 (Q BF = r1) AF = r1 = 7 cm also, AC = 2r2 AG + GC = 2r2 Þ AF + FG + r2 = 2r2 6.

7.

8.

Or equivalently,

9

)

9.

...(ii)

3 = x - 2 now squaring both the sides we get 3

(

9 = x² - 2 - 2 3 3 = x² - 2 - 2 2 x - 2

= x² + 2 - 2 x 2

)

x 6 - 6 x 4 - 6 x3 + 12 x 2 - 36 x + 1 = 0, which is the required polynomial of least degree and required degree is 6. (d) Draw the angle bisector BE of ÐABC to meet AC in E. Join ED. b b

a

And cubing both the side we get x3 = 2 2 + 6 3 3 + 3 2 3 9 + 3

)

A

...(i)

x2 = 2 + 2 2 3 3 + 3 9

(

Squaring both sides yields

(a) Let x = 2 + 3 3 . Then squaring both the sides we get

3

(

3

x3 + 6 x - 3 = 2 3x 2 + 2 .

Þ r2 = 7 + 14 Þ r2 = 21 cm (d) The number of digits of the numbers N cannot be less than 224 in which case the sum will be less than 223 × 9 = 2007. Moreover, if the number has 224 digits, the first digit cannot be less than 6 as the sum of digits would be less than 6 + 223 × 9 = 2013. Now, if the first digit is 6, the rest 223 digits would have to be 9 giving the sum of digits as 2013 and as there is no number less than this number whose sum of digits can be 2013, the least number N = 6999…9, with 9 appearing 223 times. Therefore, 5N + 2013 = 5 × 6999 … 9 + 2013 = 5 × (7000 … 0 1) + 2013 = 35000 … 0 + 2008 = 35000 … 2008. Thus, the required answer = 3 + 5 + 2 + 8 = 18. (b) Let d be the common difference of the A. P., then 4 = abc = (b – d) b(b + d) = b(b2 – d2) Þ b3 = 4 + bd2 ³ 4 [Q b > 0, d2 ³ 0]] Þ b ³ 22/3 Thus, minimum possible value of b is 22/3, that is the case when d = 0.

From (i)

3 and

x3 = 2 2 + 6 x - 2 + 3 2 x2 + 2 - 2 x 2 + 3,

r1

\ AB =

)

3

10.

E b

2b

a a C B D Since, ÐB = 2ÐC, it follows that ÐEBC = ÐECB. We obtain EB = EC. Consider the triangles BEA and CED. We observe that BA = CD, BE = CE and ÐEBA = ÐECD. Hence, BEA @ CED giving EA = ED. If ÐDAC = b, then we obtain ÐADE = b. Let I be the point of intersection of AD and BE. Now, consider the triangles AIB and DIE. They are similar since ÐBAI = b = ÐIDE and ÐAIB = ÐDIE. It follows that ÐDEI = ÐABI = ÐDBI. Thus, BDE is isosceles and DB = DE = EA. We also observe that ÐCED = ÐEAD + ÐEDA = 2b = ÐA. This implies that ED is parallel to AB. Since, BD = AE, we conclude that BC = AC. In particular ÐA = 2ÐC. Thus, the total angle of ABC is 5ÐC giving ÐC = 36°. We obtain ÐA = 72°. (d) For these conditions to be met, we must have

a2 + b ³ b2 - a

b2 + a ³ a2 - b

( a - b + 1)( a + b) ³ 0 ( b - a + 1)( a + b) ³ 0

SOLUTIONS – MOCK TEST-1

99

a ³ b -1

b ³ a -1 or b + 1 ³ a Therefore, a = b, b – 1, b + 1. Notice that we can only account for a = b – 1 and then reverse the solutions.

a2 + a We must then have 2 is an integer.. a -a a2 + a a + 1 2 = = 1+ a -1 a2 - a a - 1 This gives the solution pairs (a, b) = (2, 2), (3, 3).

Notice that

Case II : a = b – 1 We notice that in this case a2 + b = b2 – a, therefore, we only have to consider b2 + a ( a + 1) + a a 2 + 3a + 1 = 2 = 2 a2 - b a - a -1 a - a -1 2

4a + 2 a - a -1 Notice that for a ³ 6, however, then we have a2 – a – 1 ³ 4a + 2, contradiction. Therefore, we consider a Î {1, 2, 3, 4, 5}. Testing these, we see that only a = 1, 2 give solutions. 2

We, therefore, get the solution pair (a, b) = (1, 2), (2, 3) and permutations (for a = b + 1). In conclusion, all solutions are of the form (a, b) = (1, 2), (2, 1), (2, 3), (3, 2), (2, 2), (3, 3). 11.

= 2( PA2 + PE 2 )

(QPA = PC)

Similarly, EB + ED = 2(QD + QE2). 2

2

2

Therefore, EA2 + EB2 + EC2 + ED2

Case I : a = b

= 1+

= EP 2 + PA2 + 2 EP × PA + PC 2 + PE 2 - 2PC × PE

(a) Let O be the centre of the circle and P, Q the feet of perpendiculars from O to AC and BD. Clearly, OPEQ is a rectangle.

D

= 2(PA2 + PE2) + 2(QD2 + QE2)

= 2( PA2 + OQ 2 ) + 2(QD 2 + OP 2 ) = 2( PA2 + OP 2 ) + 2(QD 2 + OQ 2 ) = 2(OA2 + OD 2 ) = 4 R 2 Þn=4 12. (d) Given equation is 1/m +4/n =1/12 or 1/m = 1/12 – 4/n or m =12n/(n – 48) since m is positive hence n > 48 but n < 60 hence possible values of n is 49, 51, 53, 55, 57 and 59, out of which only 49, 51 and 57 gives integral values of m.

13. (d) We need to maximize the median in each group to maximize the average of all median. Highest possible median is 18 at there should be 3 numbers higher than it in a group of 7. Se we have 1 2

3

18

19

20

21

Similarly, 4 5

6

14

15

16

17

7 8

9

10 11 Medians are, 18, 14 and 10.

12

13

Mean = 14. (c)

18 + 14 + 10 42 = = 14 3 3 B A

E

Q

O A

P

C

C

E B

Now,

EA2 + EC 2 = ( EP + PA)2 + ( PC - PE )2

N

(i)

D The diagonals of a square are perpendicular and bisect one another. Therefore, AE = EC and AE ^ BE. AN · NC = AN (NE + EC) = AN (AN + 2 NE) = AN2 + 2 AN · NE.

EBD_7839 100

KVPY-SA hw = 10 + 15 = 25 cm., hs = 12.5 + 15 = 27.5 cm. Pressure due to water column, P1 = 0.25 × 103 × g Nm–2. Pressure due to mercury column, P2 = 0.275 × 0.8 × 103 × g Nm–2 Since P1 > P2 the mercury will rise in the spirit arm. If h is the difference in the levels of mercury in the two arms then P1 – P2 = hrg 0.25 × 103 × g – 0.275 × 0.8 × 103 × g = h × 13.6 × 103 g or (0.25 – 0.275 × 0.8) = h × 13.6

In DABN, AB2 = BN2 + AN2 + 2 AN · NE (Since, in an obtuse angled triangle, the square on the side opposite the obtuse angle is greater than the sum of the squares on the sides containing the obtuse angle, by twice the rectangle contained by either of these sides and the projection, on this side produced, of the other side adjacent to the obtuse angle). Hence, AB2 – BN2 = AN · NC. (ii) AN2 = (AE – NE)2 and

or h =

NC = (CE + NE) = (AE + NE) . 2

2

2

17.

Adding gives AN + NC = 2 (AE + NE ) 2

2

2

2

= 2 (BE2 + NE2) = 2 BN2. 15.

P0 + rgh1 +

v2 = h1 + h2 ...(i) Equation of continuity yields A1 v1 = A2 v2 ...(ii) Eliminating v1 from equations (i) and (ii), we obtain

PR = l2 Given l1 + l2 = m also perimeter = 2p \ PQ = p/2 In DPOQ

2

v2 =

2

2

æ l1 ö æ l2 ö æ pö (PO) + (OQ) = (PQ) Þ ç ÷ + ç ÷ = ç ÷ è 2ø è 2ø è 2ø 2

v2 = h

Þ m2 – 2l1 . l2 = p2 Þ 2l1 . l2 = – p2 + m2

2

1 m -p .l .l = 2 1 2 4 PHYSICS

Q Area of rhombus = 16.

2

(b) When we pour 15 cm. of water and spirit in the respective arms, then

2 gh 2

æA ö 1- ç 2 ÷ è A1 ø Putting A1/A2 = h,we obtain

2

1 m2 - p2 m2 - p 2 Þ l1 . l2 = 2 2 4

æ A2 ö v2 ÷ + 2 gh ç è A1 ø

Þ v2 =

2

Þ l12 + l22 = p2 Þ (l1 + l2)2 – 2l1 l2 = p2

Þ l1 . l2 =

(c) Applying Bernoulli’s theorem at reservoir and power plant for the flowing water, we get, 1 1 rv12 = P0 + rgh2 + rv22 2 2 Þ v22 = v12 + 2g (h1 – h2). Putting (h1 – h2) = h1 we obtain

(a) Let SQ = l1

2

0.25 - 0.2200 0.03 = m = 0.22 cm 31.6 13.6

18.

2 gh

h2 - 1 (c) Newton’s law of cooling can be written as:

é T1 + T2 ù T1 - T2 - T0 ú = kê 2 t ë û In first case ; T1 = 60°C, T2 = 40°C, T0 = 10°C and t = 7 minute

\

é 60 + 40 ù 60 - 40 - 10ú = kê 7 ë 2 û

SOLUTIONS – MOCK TEST-1

101

1 14 In second case; T1 = 40°C and T2 = ?, T = 7 minute

or

\

k=

1 é 40 + T2 ù 40 - T2 - 10ú = ê 14 ë 2 7 û

or

80 - 2T2 = 20 +

or

T2 = 28°C

T2 - 10 2

19. (a) Here, fall in the temperature of child = 101 – 98 = 3°F 5 5 ´ 3 = °C 9 3 m = 30 kg, S = 1000 cal kg–1 °C–1 Heat lost by child 5 DQ = ms DT = 30 ´ ´ 1000 3 = 5 × 104 J Let m is the mass of water evaporated in 20 min. Then

=

DQ 5 ´ 104 = = 86.2 g L 580 Average rate of evaporation

22. (d) When a steady current flows in a metallic conductor of non uniform cross section then drift speed I I and Electric field E = neA sA Drift velocity (vd) and electric field (E) vary because cross section of the conductor is non-uniform. Hence only current remains constant along the length of the conductor. 23. (a) Since hollow prism contains only air, refraction and dispersion do not occur in the same medium. There will neither be angular deviation nor dispersion. 24. (a) g1 = g - w 2 R cos 2 0 = g - w 2 R

vd =

g2 = g - w 2 R cos 2 60° =g-

w2 R 4

g1 g - w2 R = f = g2 w2 R g4

DQ = m'L or m ¢ = \

=

86.2 = 4.31 g min -1 20

I1 20. (c) 40 = 10log I , 0 I2 and 20 = 10log I , 0

I1 \ I = 104 0 I2 \ I = 102 0

I1 \ I = 100 2

r22 I1 r22 Also I = 100 = 2 = 2 r1 2 1 or r2 = 10 m. 21. (b) IR = x q = 1 × 20° =1×

20° ´ p ´ 100 = 36.4 cm 180°

Also g =

…(i)

æ 4 3ö çè pR ÷ø 3

GM

…(ii) = Gr R2 R2 On solving above equation, we get g w1 æ 4 - f ö 3p = 1 =ç w 2 g 2 è 1 - f ø÷ 4GT 2

25. (c) Given that F(t) = F0 e - bt dv = F0 e - bt dt F0 - bt dv e = dt m

Þ m

v

ò dv 0

t

F0 -bt = m ò e dt 0

t

F0 v= m

é e - bt ù F ê ú = 0 é - e - bt - e -0 ù û êë -b úû 0 mb ë

Þ v=

F0 é 1 - e - bt ù û mb ë

(

)

EBD_7839 102 26.

KVPY-SA (b)

lsunTsun = lstar Tstar

29.

(b)

30.

Tsun lstar 350 35 = = \T = l 510 51 star sun = 0.69 (a) mg = Fb

( rFe ) V0 g

= (k1V0 )(rHg )0 g

… (i)

and (rFe )60Vt g = (k2Vt )(rHg )60 g The forces acting on the bead as seen by the observer in the accelerated frame are : (a) N ; (b) mg ; (c) ma (pseudo force). Let q is the angle which the tangent at P makes with the X- axis. As the bead is in equilibrium with respect to the wire, therefore N sin q = ma and N cos q = mg

a g But y = k x2. Therefore,

\

tan q =

dy = 2kx = tan q dx

or

é (r Hg )0 ù = k2 ê úg ëê1 + g Hg ´ 60 ûú From above equations, we get

CHEMISTRY 31.

… (ii)

From (i) & (ii) 2kx =

27.

28.

4.29 ´ 365.25 ´ 864 ´ 3 ´ 1010

parsec 3.08 ´ 1016 [Q1 parsec = 3.08 ´ 1016 m] = 1318656.9 × 10–6 parsec = 1.319 parsec. Required parallax = 2q, where q is the annual parallax. Since parsec is the distance corresponding to an annual parallax of one second of arc, therefore q is 1.319 second of arc. \ Required parallax = 2 × 1.319 second of arc = 2.638 second of arc.

=

… (ii)

k1 é 1 + 60 g Fe ù =ê ú k2 ë1 + 60 g Hg û

… (i)

a a Þ x= g 2kg (a) Angular momentum of system remains constant I w 1 20 I µ Þ 2 = 1 = Þ I 2 = 2 I1 = 2 I w I1 w2 10 (a) Distance = 4.29 light years = (4.29 × 365.25 × 86400) × 3 × 108m

é (r Fe )0 ù ê1 + g ´ 60 ú g ë û Fe

32.

Na+ Mg2+ Al3+ Si4+ Protons 11 12 13 14 Electrons 10 10 10 10 Size of isoelectronic cations decreases with increase in magnitude of nuclear charge \ Order of decreasing size is Na+ > Mg2+ > Al3+ > Si4+ (a) C + 2S ¾¾® CS2 DH = SDHP – SDHR = –1108.76 – [–393.3 – 2 ´ (293.7)] = –1108.76 + 393.3 + 587.4 = –128.06 kJ (a)

33.

(c) Correct IUPAC name of the given compound is 3-ethyl- 4-methyl pentan-2one 34. (b) Higher the basic character of a compound, lesser will be its pH value Rock Salt (NaCl) < Baking Soda (NaHCO3) < Washing Soda (Na2CO3) < Slaked lime (CaCO3) 35. (a) Number of Eq. of NaOH = Number of Eq. of oxalic acid

36.

Then 100 ´ 1 = wt.of oxalic acid 1000 63 \ W= 6.3 g (b) e/m for He2+ = 2/4 e/m for H+ = 1/1 + e/m for He = 1/4 e/m for D+ = 1/2 \ Value of e/m is highest for H+.

SOLUTIONS – MOCK TEST-1

103 c hc or l = n l 3 ´108 = 3.75 ´10 -8 m Þl= 8 ´1015 In nanometer l = 3.75× 10 » 4 × 101 42. (c) P µ n PH 2 nH 2 22 = = PCO 2 n CO 2 1

41. (d)

37. (c) (a)

E = hn =

PH2 = 22 × PCO2

(b)

PCO = 1 atm (given) 2

\ PH2 = 22 atm 43. (b) Alkyl groups with at least one hydrogen atom on the a-carbon atom, attached to an unsaturated carbon atom, are able to release electrons in the following way. H H+ s| _ p –C–C=C –C=C–C

(c)

I

F C F

(d)

F

F

Non - polar (m = 0) Non - planar (Tetrahedral)

38. (b)

II

The delocalisation involves s and p bond orbitals (or p orbitals in case of free radicals) ; thus it is also known as s – p conjugation. This type of conjugation is called hyperconjugation. 44. (d) Reacting of aluminium with oxygen is an exothermic reaction i.e evolution of heat occurs. 4Al + 3O 2 ¾¾ ® 2Al 2 O 3 , DH = - ve

45. (c)

+

N 2 = 7 + 7 – 1 = 13 electrons Electronic Configuration is : s1s 2 s *1s 2 s 2s 2 s * 2 s 2 2 1 p2 px 2 = p2 p y s2 pz

1 [No. of bonding electrons 2 – No. of antibonding electrons]

Bond order = 39. (c) CO2 has the maximum oxidation state of +4. So it cannot go to a higher oxidation state. 40. (c) Given : V = 2 L, Molarity = 0.5M, Moles = ? Molarity = No. of moles of solute Vol. of solution in L No.of Moles 2 \ No. of Moles = 2 × 0.5 = 1.0 0.5 =

1 1 (9 - 4) = ´ 5 = 2.5 2 2 BIOLOGY 46. (b) Cholecystokinin (CCK or CCK-PZ) is a peptide hormone of the gastrointestinal system responsible for stimulating the digestion of fat and protein. Its presence causes the release of digestive enzymes and

=

EBD_7839 104

47.

48.

49.

50.

51.

52.

KVPY-SA

(a)

(c)

(c)

(a)

(a)

(a)

bile from the pancreas and gallbladder, respectively, and also acts as a hunger suppressant. Factor VII or proconvertin is not used up during clotting. During clotting, it is changed into convertin which accelerates tissue or extrinsic thromboplastin formation. Factor XI is activated to factor XIa by factor XIIa. Calcium is used for both intrinsic and extrinsic thromboplastin formation and also in the conversion of prothrombin into thrombin. Heparin is present as anticoagulant in the normal blood circulation. The discovery of Emerson effect stated that one group of pigments absorbs light of both shorter and longer wavelengths (more than 680 nm) and another group of pigment absorbs light of only shorter wavelengths (less than 680 nm). These two groups of pigments are known as pigment systems or photosystems. Right cerebrum controls the left side of the body so that left arm weakens and never the right arm. Deep tendon reflexes like knee-jerk operate at the level of spinal cord (not brain) and hence loss of neither left nor right kneejerk reflex can be affected by this stroke. Right side of the cerebrum sees the left visual field and vice versa. Each eye sends information to each side of the brain so that blindness in neither eye is expected as the direct result of a stroke on only one side of the brain. Euphotic zone (where sufficient light penetrates) is responsible for main bulk of primary production. Below this zone, there is disphotic zone where no effective plant production occurs. The lightless region below the disphotic zone is termed as aphotic zone. Neurotransmitters cause rapid responses of the nervous system. Changes in heart rate are mediated by the nervous system, and alterations have to be made rapidly. Meristem is a tissue, which is primarily concerned with protoplasmic synthesis and formation of new cells by division. Since the meristems have capacity to synthesise new cells, that is why it is called totipotent.

53.

54.

55. 56.

57.

58.

59. 60.

(d) The medulla (a sub-region of the brainstem) is a major control centre for the autonomic nervous system. The hypothalamus acts to integrate autonomic functions and receives autonomic regulatory feedback from the limbic system to do so. (c) The lysosome is filled with 40 types of acid hydrolases (digestive enzymes). The acidic pH £ 5 (4.6 to 5) is due to the action of ATP fueled proton pump in the membrane of the lysosome. All the enzymes do not occur in the same lysosome. These digestive enzymes are synthesised on RER and packed into lysosomes. (a) In muscle contraction, a calcium ion facilitates the binding of a cocked myosin head to a site on an actin molecule. (b) Chronic alcoholic patients are frequently deficient in one or more vitamins. The deficiencies commonly involve folate, vitamin B6, thiamine, and vitamin A. (d) Rabies, influenza and AIDS are viral diseases. Amoebiasis, ascar iasis and trypanosomiasis are caused by protozoans. Taeniasis, ascariasis and elephantiasis are the diseases caused by helminths but cancer, tuberculosis and tetanus are not related diseases. Tuberculosis and tetanus are bacterial diseases while cancer is caused by abnormal cell division. (a) During the transport of CO2 through the blood, bicarbonate ions diffuse out of RBCs while chloride ions from plasma, enter the RBCs to maintain ionic equilibrium. This is called chloride shift. (b) Shorter the food chain, more is the availability of food to consumer. (d) The thoracic duct, which carries lymph from most of the body, drains into the venous system. It ends on the left side of the body behind the left clavicle, usually draining into the left subclavian on left brachio-cephalic vein. PART – II

61.

MATHEMATICS (b) The given equation is equivalent to (106 – 1)n = Þn=

10 k - 1 9

10 k - 1

9 (10 6 - 1)

with k = 6m.

SOLUTIONS – MOCK TEST-1

105 Þ 2pr = r¢a Þ 2p × 5 = 13 × a

1 + 106 + K + 10 ( ) . 9 The numerator becomes a multiple of 9 if m = 9. 6 m-1

Then n =

Thus, the smallest n is n =

1054 - 1

(

)

9 10 - 1 62. (a) In the following figure :

A

6

Þ

10 p 13

64. (b)

.

B

a=

C

There are 4 bus routes from A to B and 3 routes from B to C. Therefore, there are 4 × 3 = 12 ways to go from A to C. It is round trip so the man will travel back from C to A via B. It is restricted that man cannot use same bus routes from C to B and B to A more than once. Thus, there are 2 × 3 = 6 routes for return journey. Therefore, the required number of ways = 12 × 6 = 72. 63. (a) l¢ = r¢a ...(1) r¢ = radius of sector l¢ = length of sector a = angle of sector

by mid point theorem, PS || AC || QR and, PQ || BD || SR. \ PQRS forms a parallelogram. Let area of DABC = a then ar DBPS =

a (direct result by mid point 4

theorem) also, let area of DACD = b. b . 4 now, let ar DABD = c

\ ar DQRD =

c 4 also, let ar DBCD = d

\ ar DAPQ =

\ ar DSCR =

d 4

a+b+c+d 2 \ area of region PQRS

now, area of ABCD = Now, cone is made by this sector = Slant height, l = radius of sector

and circumference of cone = l¢ \

l = r 2 + h2 = 52 + 122 Þ l = 13 cm = r¢ = radius of sector also, circumference = l¢ = r¢a

a + b + c + d æ a b c dö -ç + + + ÷ è 4 4 4 4ø 2

ar (PQRS) =

a b c d + + + 4 4 4 4

ar (PQRS) =

1 æa +b + c+ dö ç ÷ø 2è 2

1 ar (ABCD) 2 65. (c) APB P CQD and transversal pass through PQ such that PH, PO, QH and QO bisectors of ÐAPQ, ÐBPQ, ÐCQP and ÐPQD respectively.

ar (PQRS) =

EBD_7839 106

KVPY-SA B

P

A

O

O M

H D

C

45

Q

R o

q °-q

P

Since AB P CD Þ ÐBPQ + ÐDQP = 180º (sum of interior angles on the same side of transversal) 1 1 ÐBPQ + ÐDQP = 90º 2 2 Q PO and QO bisects ÐBPQand ÐPQD Þ

Þ ÐOPQ + ÐPQO = 90º ...(i) In D OPQ, by angle sum properly Þ ÐOPQ + ÐPQO + ÐPOQ = 180º Þ 90º + ÐPQO = 180 o

Þ ÐPOQ = 90º (from (i)) Since AB P CD and PQ is transversal Þ ÐAPQ = ÐPQD (alternate interior angles) 1 1 ÐAPQ = ÐPQO Þ ÐHPQ = ÐPQO 2 2 But these are alternate angles Þ HP P QO Þ

Similarly PO P HQ Now HP P QO and PO P HQ ÞPOQH is a parallelogram with one angle as right angle, ÐPOQ = 90º Þ POQH is a rectangle PHYSICS 66.

(a) At equilibrium, taking torque of liquids about O æ Torque due to ö æ Torque due to ö çè liquid of density r÷ø = çè liquid of density1.5r÷ø

m2g × QM = m1g × PN \ m2g R sin (45° + q ) = m1gR sin (45° – q ) VrgR sin (45° + q ) = 1.5VrgR sin (45° – q ) ... (i)

N

45

Q K

m2g

m1g

Þ

sin (45 + q) = 1.5 sin (45 - q)

Þ

sin 45° cos q + cos 45° sin q 3 = sin 45° cos q - cos 45° sin q 2

1 5 Let us now displace the liquids in anticlockwise direction along the circumference of tube through an angle a. The net torque Þ tan q =

t = m2gR sin (45° + q + a) – m1gR sin (45° – q – a) = VrgR sin (45° + q + a) – 1.5VrgR (45° – q – a) = VrgR sin ( q + 45°) cos a + VrgR cos (45° + q ) sin a – 1.5 VrgR sin (45° – q) cos a + 1.5 VrgR cos(45° – q) sin a Using eq. (i) we get t = V rgR éëcos ( 45° + q ) sin a +1.5 cos ( 45° - q ) sin aùû t = VrgR [cos (45° + q ) + 1.5 cos (45° – q )] sin a when a is small (given) sin a » a t = VrgR [cos (45° + q ) + 1.5 cos (45° – q )]a Since, t and a are proportional and directed towards mean position. \ The motion is simple harmonic. Moment of inertia about O is, I = VrR2 + 1.5 VrR2

T= 2p = 2p

I C (V r ´ 2.5R 2 ) [cos(45 + q ) + 1.5 cos(45 - q)]V rgR

or, T = 2p

1.803R g

1 (using the value tan q = ) 5

SOLUTIONS – MOCK TEST-1

107 70. (d) As we know,

67. (b) As we know, dU = F.dr r

ar 3 U = ò ar dr = 3 2

0

Acceleration, a = ...(i)

mv 2 = ar 2 r m2v2 = mar3

For cylinder, a c =

2 1 McR 2 R2 2 M .g. sin qc = c or, ac = g sin qc 2 3 M R Mc + c 2 2R For sphere, M g sin qs Ms g sin qs as = s = I 2 Ms R 2 M s + s2 Ms + r 5 R2 5 or, a s = g sin qs 7 given, ac = as i.e., 2 g sin qc = 5 g sin qs 3 7 5 g sin qc 7 15 \ = = 2 sin qs 14 g 3 CHEMISTRY

As,

Mc +

1 3 ar ...(ii) 2 Total energy = Potential energy + kinetic energy Now, from eqn (i) and (ii) Total energy = K.E. + P.E. ar 3 ar 3 5 3 + = ar = 3 2 6 68. (c) The distance of bottom of the beaker from mirror

or, 2m(KE) =

æ 1ö = h - d ç1 - ÷ è mø æ 1ö So it will be at a distance = h - d ç1 - ÷ è mø from mirror. Now distance between bottom of beaker and image æ 1ö æ m - 1ö = h + h - d ç1 - ÷ = 2h - d ç . è mø è m ÷ø 69. (a)

mg sin q I m+ 2 r M c .g. sin qc

+

(a) -CCl 3 ,- NO 2 an d –NH3 are electron withdrawing group, therefore, they are m - directing in nature. 72. (b) For the equation 71.

B 2 H 6 (g) + 3O 2 (g) ¾¾ ® B 2 O 3 (g) + 3H 2 O (g)

As uˆ1 and uˆ2 and the unit vectors and so a2 + b2 = c 2 + d 2 Now using Snell's law, we have

m2 sin q = m1 sin q '

or

a / a2 + b2 2 4 a = = . or 2 2 1.5 3 c c/ c +d

Eqs. (i) + 3 (ii) + 3 (iii) – (iv) DH = – 1273 + 3(–286) + 3(44) – 36 = – 1273 – 858 + 132 – 36 = – 2035 kJ/mol 73. (b) 1 lb = 454g (given) Mass of copper in the statue = 2.0 × 105 × 454g (in grams) = 908 × 103 No. of moles of copper in the statue Given mass 908 ´ 105 = At.mass 63.5 Now, No. of atom of copper on the statue = No. of moles × NA =

908 ´105 ´ 6.022 ´1023 63.5

EBD_7839 108

74.

75.

KVPY-SA = 8.6 × 1029 Therefore, 8.6 × 1029 no. of atoms of copper are on the statue. (d) v = 2.188 × 106 Z/n Following the above relation, option (d) represents the incorrect graph between v and n. (d)

CH3 CH3 | | + D+ CH3 – CH – CH = CH2 ¾ ¾ ® CH3 – CH – CH – CH2 – D [1, 2-Hydride Shift]

CH3 |+ CH3 – C – CH2 – CH2D H 2O

CH3 | CH3 – C – CH2 – CH2–D | OH

76.

BIOLOGY (c) Net primary productivity is the amount of dissolved oxygen gained in the sample exposed to light. This is calculated by subtracting the amount of oxygen in the control from the amount of oxygen in bottle kept in light. From bottle 2, Net primary productivity (NPP) = Oxygen in bottle 2 – Oxygen in bottle 1 = 10 – 9 = 1 mg/L/h Respiration rate is determined by the amount of oxygen lost in the sample kept in the dark. This is calculated by subtracting the amount of oxygen in the bottle kept in dark from the amount of oxygen in control bottle. From bottle 3 Rate of respiration (R) = Oxygen in bottle 1 – Oxygen in bottle 3 =9–4 = 5 mg/L/h Gross primary productivity is calculated by adding the values of net primary productivity and the respiration rate. Gross primary productivity = NPP + R =1+5 = 6 mg/L/h

77.

(a) Molecular weight of average amino acid = 110 Molecular weight of polypeptide chain = 11,000 Number of amino acid residues in polypeptide chain = 11,000/ 110 = 100 The number of nucleotides in DNA strand coding polypeptide chain with 100 amino acids = 100 × 3 = 300 nucleotides 78. (d) Turgor pressure is the force within the cell that pushes the plasma membrane against the cell wall. All the cells except lignified cells are living cells. So, it is not required for maintaining the shape of lignified cell. So, it is not required for maintaining the shape of lignified cells. 79. (a) Solute potential (ys) and pressure potential (yp) are the two main components that determine water potential. y = yp + ys From the given options, Water potential of Water potential of root parenchyma root xylem (a) 10 –195 (b) 20 60 (c) –20 60 (d) –20 –60 By convention, the water potential of pure water at standard temperatures, which is not under any pressure, is taken to be zero. Hence, the values of water potential in options (a), (b) and (c) cannot be considered. Water will move from the system having higher water potential to the one having lower water potential. So, according to the water potential of root parenchyma and root xylem in option (d), transpiration pull will cause water to move from root parenchyma to the root xylem. 80. (a) Allele for green body colour = XG Allele for rosy body colour = Xg Rosy body gcolour female × Green body colour male g XG Y

X X

X

XG Xg Green body colour (Daughter)

XG X g Xg Y Xg Y Green body Rosy body Rosy body colour colour colour (Son) (Daughter) (Son)

MOCK TEST-2 PART – I MATHEMATICS 1.

(a) Draw CN ^ AB and join CD and CE. 4.

C

Þ K2 – K1 = 3 ...(2) From (a) & (b) K2 = 1; K1 = –2 \ K1 × K2 = –2 (d) Since, the mean of a group of eleven consecutive natural numbers is m, then

G A

2.

B D N E Since, BC = BD, therefore ÐBCD = ÐBDC. But ÐBCD = ÐBCN + ÐNCD and ÐBDC = ÐDCA + ÐA (Since, the exterior angle is equal to the sum of the interior nonadjacent angle). Hence, ÐBCN + ÐNCD = ÐDCA + ÐA ...(1) and since CN ^ AB, therefore, ÐBCN = ÐA (since both complement ÐNCA) ...(2) From eq. (a) and eq. (b), ÐNCD = ÐDCA But ÐCND = ÐCFD = right angle Therefore, ÐCND and ÐCFD are congruent. Therefore, DN = DF Similarly, EN = EG and therefore by adding, DE = DF + EG. (b) 3x = 5y = 75z = k Þ 3 = k1/x, 5 = k1/y, 75 = k1/z \ 75 = 52 × 3 Þ k1/z = k2/y. k1/x Þ k1/z = k2/y + 1/x Þ

3.

x + x + 1 + ... + x + 10 =m 11 11x + 55 = 11 m Þ x + 5 = m Þ x = m – 5 Let n be the mean when next six consecutive natural numbers are included in the group then

F

2 1 z (2 x + y ) 1 = 1. = + Þ y x xy z

(c) P(x) = x2010 + K1x2011 + K2 Q P(a) = 0 = (a)2010 + K1(a)2011 + K2 1 + K1 + K2 = 0 ...(1) also P(–1) = 4 = (–1)2010 + K1 (–1)2011 + K2 Þ 4 = 1 + K2 – K1

x + x + 1 + ..... + x + 16 =n 17 16 ´ 17 = 17 n 17x + 8 × 17 = 17 n 2 m – 5 + 8 = n Þ n = m + 3 (\ x = m – 5) Hence, required percentage change in the mean 17 x +

= 5. 6.

n-m m+ 3-m 300 ´ 100 = ´ 100 = % m m m

(a) (c) x2 + y2 = 25 and xy = 12 2

æ 12 ö Þ x + ç ÷ = 25 Þ x4 + 144 – 25x2 = 0 èxø Þ (x2 – 16) (x2 – 9) = 0 Þ x2 = 16 and x2 = 9 Þ x = ± 4 and x = ± 3. 2

7.

(a)

1 1 1 1 - = a b b c æ 1 1 1 öæ 1 1 1 ö \ ç + - ÷ç + - ÷ è a b c øè b c a ø æ 2 1 öæ 2 1 ö 4 1 æ 2 2 ö 1 - ç + ÷+ = ç - ÷ç - ÷ = è a b ø è c b ø ac b è a c ø b 2 =

4 2æ2ö 1 4 3 = - ç ÷+ 2 ac b è b ø b ac b 2

EBD_7839 110

KVPY-SA

8.

(a) Q(a – b)2 + (b – c)2 + (c – a)2 > 0 Þ 2(a2 + b2 + c2 – ab – bc – ca) >0 Þ 2 > 2(ab + bc + ca) Þ ab + bc + ca < 1

9.

(a) D

F

G

C

11. (c)

360°

C

E

A

I 1

H x

B

J 2–x

q

r

3

1

A

B

D

2

2

2

q ´ pr 2 Area of sector ADB ° 360 = Area of sector ACD 360° – q ´ pr 2 360°

Draw EH, FI, GJ parallel to BC. AJ = GJ = 3 Therefore, ΧGAJ = 45° Now, in DAEH, EH = 1 + x Now, in DBHE and DBIF,

Þ q = 120°

EH BH = FI BI

¼ = q ´ 2pr = 2pr \ ADB 360° 3

x +1 4 - x 8 = or x = 3 4 7

¼ = 4pr Þ ACB 3

Þ

1 q = 2 360° – q

Therefore, ar ( DAEB ) =

1 æ 8ö ´ 5 ´ ç1 + ÷ è 7ø 2

Þ a = 75 10. (b) Since, x is the product of four consecutive integers, it is always divisible by 4, i.e., it is always even. So, 1 + x is always odd. n=1+x and x = (y –1)(y)(y + 1)(y + 2) = y (y 2 – 1)(y + 2) = (y3 – y)(y + 2) = y4 +2y3 – y2 – 2y Therefore, 1 + x = y4 + 2y3 – y2 – 2y + 1 = y4 + y2 + 1 + 2y3 – 2y2 – 2y = (y2 + y –1)2 So, 1+ x is a perfect square. Hence, n is odd as well as a perfect square.

h1

1

r

h2

2

r1

r

r2

¼ = circumference of base = 2pr1 ADB 2pr r = 2 pr1 Þ r1 = 3 3

Similarly, r2 =

2r 3

h1 = r 2 – r12 = r 2 –

r 2 2 2r = 9 3

SOLUTIONS – MOCK TEST-2

111 13. (c) Const: Join B to O, C to O

5r Similarly, h 2 = 3

» = CD » as » AB = BC

1 2 2 pr h V1 3 1 1 æ r1 ö æ h1 ö 1 2 2 = =ç ÷ ç ÷ = ´ V2 1 2 è r2 ø è h 2 ø 4 5 pr h 3 2 2

\ Angle made by them will be equal. \ ÐAOB = 30° = ÐBOC = ÐCOD

1

=

10 A

12. (c)

B q

a

P

Qa q

D

C Let AP = x

AP 1 = AD n

AD = nx PD = nx - x

In DAQP & DCQB ÐPAQ = ÐBCQ = q ÐAQP = ÐBQC = a \ DAQP ~ DCQB

(Alternate) (V.O.A) (By AA)

In DAOC, AO = OC (Radius) ÐAOC = 60° \ ÐOAC = ÐOCA = 60° In DCOR, ÐOCR = 60° ÐROC = 30° \ ÐCRO = 90° similarly, we get ÐBPO = 90° In quadrilateral PQRO, ÐP + ÐR = 180° \ PQRO is a cyclic quadrilateral \ x + ÐO = 180° Þ x + 30° = 180° Þ x = 150° 14. (c) Draw AF ^ BC. A

AQ AP = CQ BC

Þ

AQ x = CQ nx

æQ BC = ADö ç = AP + PD÷ ç ÷ BC = nxø è

Þ

AQ 1 CQ n = Þ = CQ n AQ 1

Þ

CQ CQ + AQ +1 = n +1Þ = n +1 AQ AQ

Þ

AC AQ 1 = n +1 Þ = AQ AC n + 1

C

FO

D

B

E (i) Since, D is the mid-point of BC. Therefore, AD = DB = DC (Since, in a right triangle, the median from the right vertex to the hypotenuse is equal to half the hypotenuse.) Therefore, ÐDAB = ÐDBA But ÐDBA or ÐFBA = ÐFAC (Q they are complementary to ÐFAB)

EBD_7839 112

KVPY-SA

Therefore, ÐDAB = ÐFAC But, since AO bisects ÐA. Therefore ÐOAB = ÐOAC and subtraction gives ÐOAD = ÐOAF Since, DE || AF (both perpendicular to BC), Therefore, ÐOAF = ÐOED, and hence ÐOAD = ÐOED. Therefore, AD = DE. (ii) As shown above, ÐDAB = ÐDBA or ÐB. Since, ÐFAB = ÐC (both complement ÐFAC), therefore, (ÐFAB – ÐDAB) = (ÐC – ÐB) = 2 ÐDAO or ÐDAE = (ÐC – ÐB). 15. (d) a > b > c = c + 10 = a – 15 = k b = 5 c = k – 10 a = k + 15 a + b + c = 3k k + 15 + 5 + k – 10 = 3k 10 = k a = 25 b= 5 c=0 mean =

252 + 52 + 02 650 2 = = 216 3 3 3

PHYSICS 16. (d) Quantity of gas in these bulbs is constant i.e., initial no. of moles in both the bulbs = final number of moles µ1 + µ2 = µ1 + µ2 PV PV 1.5PV 1.5PV + = + R ( 273 ) R ( 273 ) R ( 273) R ( t ) 2 1.5 1.5 = + Þ T = 819 K = 546ºC 273 273 T (c) From figure, kx0 + FB = Mg

Þ

17.

kx0 + s

L Ag = Mg 2

Þ x0 =

Mg -

sLAg 2 k

kx0 FB

=

Mg æ LAs ö 1k çè 2M ÷ø

Hence, extension of the spring when it is in equilibrium is, x0 = 18.

Mg æ LAs ö ç1 ÷ 2M ø k è

(c) As the length of the string of simple pendulum is exactly l m (given), therefore the error in length Dl = 0. Further the possibility of error in measuring time is 1s in 40s. \

Dt DT 1 = = t T 40

The time period T = \

40 = 2 seconds 20

DT 1 DT 1 = Þ = Þ DT = 0.05sec T 40 2 40

We know that T = 2p

l l Þ T 2 = 4 p2 g g

l

\

g = 4p 2

\

Dg Dl DT ´ 100 = ´100 + 2 ´100 g l T

\

Dg æ 1 ö ´100 = 0 + 2 ç ÷ ´100 = 5% g è 40 ø

T2

19. (b) The vertical component of the tension in the two ends of the wire, 2Tsin(15°), must equal the total loaded weight W. The volume of the wire is pRw2 L, where Rw is the radius of the wire, and the total volume of ice is p(Ri2 – Rw2) L, where Ri is the radius of the ice-covered wir e. Multiplyin g by the respective densities, the total mass of the 500m-long ice-covered wire is 694.5 + 3531.5 = 4226 kg, having a total weight of 41415N. The tension in the wire is thus 41415/ (2sin15°) = 80000 N, and dividing by the cross-sectional area of 1.77 × 10–4 m3, the stress is 4.52 × 108 N/m2

SOLUTIONS – MOCK TEST-2 20. (c)

113

s = h a k Bb c g or MT–3K–4

[ML2T–1]a [ML2T–2k–1]b[LT–1]g

= = Ma + b L2a + 2b + g T–a – 2b – g × k–b \ b = 4, a + b = 1, 2a + 2b + g = 0 and – a –2b – g = – 3. after solving, we get a = – 3, b = 4 and g = – 2 21. (c)

v 8 = x 100

25. (a)

Ptop V¢ = Pbottom V¢

Ex = –

26. (b)

L ´ 23. (b) We know that Y = 2 l D p 4 2

pD l

=

(

p 0.4 ´10

-3

) ´ ( 0.8 ´10 ) 2

-3

= 2.0 × 1011 N/m2

DY 2DD Dl = + Now Y D l [\ the value of m, g and L are exact] 0.01 0.05 + = 2 × 0.025 + 0.0625 0.4 0.8 = 0.05 + 0.0625 = 0.1125 Þ DY = 2 × 1011 × 0.1125 = 0.225 × 1011 = 2.0 × 1011 N/m2 Note : We can also take the value of Y from options given without calculating it as it is same in all options. \ Y = (2 ± 0.2) × 1011 N/m2

= 2´

27. (a) White light is broken up into several component colours due to refraction phenomenon. m P RT ; \ = Cm M T

m 3m slope of B =3 = B= mA m slope of A

or

4 ´ 1´ 9.8 ´ 2

¶V -¶ = ( - x 2 + y 2 ) = -2 y ¶y ¶y

r Now E = E x iˆ + E y ˆj = 2 xiˆ - 2 yjˆ

28. (a) PV =

mg

¶V ¶ = – (– x 2 + y 2 ) = 2 x ¶x ¶x

and E y = -

dv 2 æ dx ö 2 æ 2 ö 4 = ç ÷ = ´ç x÷ = x dt 25 è dt ø 25 è 25 ø 625

4mgL

d = d , u =15 m/s (u - 10)

As Pbottom > Ptop; \ V > V¢.

x = 100 m, a = 0.64 m/s2. r Thus, a - x graph is a straight line with maximum value 0.64 m/s2. 22. (b) Circular part in the centre of retina is called yellow spot.

ÞY=

(10 - 5) ´

Pbottom = (Ptop + rgh)

2 x or v= 25 Acceleration,

a=

d u - 10 Separation by this time has increased by ‘d’ between A and C, hence

24. (b) B catches C in time t then t =

l A If wire is bent in the middle then

29. (c) Resistance of wire (R) = r

l¢ =

l , A¢ = 2 A 2

l r l¢ 2 \ New resistance, R¢ = r = A¢ 2A

rl R = . 4A 4

=

30. (a) F = 20t – 5t2 \

Þ

a=

FR = 4t - t 2 I w

t

0

0

(

Þ

)

dw = 4t - t 2 dt

2 ò d w = ò 4t - t dt

EBD_7839 114

KVPY-SA \ 0.75 g of sample contains

x w

= u

37.

w = 2t 2 -

q

6æ

2

-

t3 3

38.

t3 ö ÷ dt 3 ÷ø

0è

Þ

q = 36 rad Þ n =

36 I– > C6H5O– > OH– > Br– > Cl– (b) Higher the K sp, more soluble is that compound in H2O. \ Ksp of FeS (11 × 10–20) is highest. So it is more soluble and has maximum solubility in H2O. (a) 10 mL, 1 M H2SO4 = 20 mL, 1 M NH3 Q Wt of N in one mole NH3 = 14 \ Wt. of nitrogen in 20 × 10– 3 mol NH3 = 20 × 10– 3 × 14

39.

(d) Relation between wavelength and energy hc (E) of a photon is, E = l Graph (d) correctly represents the above relation. (d)

H CH2 = C = C ­ sp

Cl

H – C º C – H CH3 – C º N ­ ­ sp sp

(v)

(x)

(y)

40.

(a) Quantum numbers n = 3, l = 2, m = +2 represent an orbital with 1 s=± (3d xy or 3d 2 2 ) x -y 2 which is possible only for one orbital.

41.

(a) KO2 reacts with CO2 to give oxygen, therefore, it is used in oxygen cylinders in space and submarine. 4KO 2 + 2CO2 ® 2K 2 CO3 + 3O2

42.

(b) It has chain of 4C atoms and double bond is given preference over Cl atom.

43.

(a) In trigonal bipyramidal geometry, lone pair electrons cannot occupy axial positions. Hence, trigonal planar geometry is not possible. (c) Cyclohexene + KMnO4 (cold & alkaline) ¾ ¾® cis-Hexanediol

44. 45.

(b)

O -2 (17) = KK s 2s 2s*2 s 2 s2 px2 p2 py2 = p 2 pz2

p*2 py2 = p*2 p1z

O2- has one unpaired electron.

SOLUTIONS – MOCK TEST-2 BIOLOGY 46. (b) Despite the fact that the atmosphere has 79% nitrogen gas, plants cannot use the element in that form and frequently suffer from nitrogen deficiency. 47. (d) Phospholipids are composed of phosphate group and one or more fatty acids. They have hydrophilic (polar) phosphate group and long hydrophobic (non-polar) hydrocarbon ‘tails’. The phospholipids readily form membrane like structure in water. 48. (b) Allergens are non-infectious foreign substances that causes allergic reaction. The common allergens are dust, pollen, mould spores, fabrics, lipsticks, nail paints, fur, heat, bacteria, etc. Allergy involves mainly the secretion of IgE antibodies and histamines. Allergy causes marked dilation of all the peripheral blood vessels. 49. (b) In birds, forelimbs are modified as wings for flying. Therefore, the forelimbs are not found in birds. 50. (c) Blood serum does not coagulate because it does not contain white or red blood cells or a clotting factor. It is the blood plasma without the fibrinogens. It includes all proteins not used in blood clotting (coagulation) and all the electrolytes, antibodies, antigens, hormones, and any extra substances (such as drugs and microorganisms). 51. (a) Cancer cells are HeLa cells. HeLa cells have been used to test the effects of radiation, cosmetics, toxins, and other chemicals on human cells. They have been instrumental in gene mapping and studying human diseases, especially cancer. 52 (a) Fermentation is the incomplete oxidation of glucose under anaerobic conditions by sets of reactions where pyruvic acid is converted to CO2 and ethanol (alcoholic fermentation) or lactic acid (lactic acid fermentation). In fermentation, there is a net gain of only 2 ATP molecules for each molecule of glucose degraded to pyruvic acid.

115 53. (b) Phloem parenchyma are unspecialised cells, found in the phloem. They are absent in the monocots as they have abundant companion cells. 54. (c) Distilled water has the highest water potential. The value of water potential is always negative or less than zero. Water potential of a solution is determined by using pure water as the standard of reference, which has zero water potential at normal temperature and pressure. The presence of solute particles reduces free energy of the water. Hence, it decreases the water potential in negative value. So, water potential of a solution is always less than zero. 55. (a) The number of essential amino acids in humans is 9. They are lysine, isoleucine, leucine, methionine, phenylalanine, tryptophan, threonine, histidine and valine. They must be ingested in food for the survival, as they are not synthesised in the body. 56. (c) Fumaric acid is not a product of anaerobic respiration. It is an intermediate compound of aerobic respiration (Kreb's cycle). 57. (b) CT-cells are cytotoxic T-cells which are also known as killer T-cells. These cells are involved in killing of the cancerous cells. 58. (b) Kwashiorkar is caused due to the deficiency of proteins. Meat, lentils, milk and eggs are rich sources of proteins. Scurvy is caused due to deficiency of vitamin C whose sources are citrus fruit, tomatoes, peppers, leafy green vegetables. Deficiency of vitamin D (sources - milk, egg yolk and liver) leads to rickets. Anaemia is caused due to deficiency of folic acid (sources - yeast, liver, fish, green vegetables) or vitamin B12 (sources - liver, eggs, meat, fish) or iron (egg yolk, spinach). 59. (d) Ti plasmid found in Agrobacterium tumefaciens, produces crown gall (tumour) in a large number of dicot species. A. tumefaciens is a gram negative soil bacterium that infects a wide range of plants and causes crown galls.

EBD_7839 116 60.

KVPY-SA ABCD ® || gram DABK ~ DMDK DABK = DMDK = g DAKB = DMKD = b

(a) The correct sequence is : 1. Phosphorus input from sewage and agricultural runoff increases. 2. Algal blooms occur. 3. Respiratory demand from decomposers increases. 4. Oxygen levels drop in deeper water.

\ Þ

MATHEMATICS

63. (c)

AK AB = MK MD

6 AB AB 3 = Þ = 4 MD MD 2 Let AB = 3x, MD = 2x \ AB = CD = MD + CM Þ CM = x DAMD ~ DLMC

PART – II

61. (c) We have (i) 1010 < n < 1011 (ii) Sum of the digits of n = 2 Clearly, nmin = 10000000001 (1 followed by 9 zeros and finally 1) Obviously, we can form 10 such numbers by shifting 1 by one place from right to left again and again. Again, there is another possibility for n, i.e., n = 20000000000 So, finally number of different values of n = 10 + 1 = 11 62. (c) I. Number of 3 letter words (repetition not allowed) = 6 × 5 × 4 = 120 (as first place can be filled in 6 different ways, second place can be filled in 5 different ways and third place can be filled in 4 different ways) II. Number of 3 letter words (repetition is allowed) = 6 × 6 × 6 = 216 (as each of the place can be filled in 6 different ways)

(By AA)

Þ

AM MD 10 2x = = Þ Þ LM = 5 LM MC LM x

64. (d)

h1 ® height of cone h2 ® depth of water r1 ® radius of cone r2 ® radius of water Q DABC ~ DADE (AA) h1 r2 = h r1

Volume of water =

p 2 V r 2 h2 = 3 27

p 2 1 p ´ ´ r12 h1 r 2 h2 = 3 27 3 2

h1 æ r2 ö çè r ÷ø h2 = 27 1

SOLUTIONS – MOCK TEST-2

117 The weight of system

2

h1 æ h2 ö çè h ÷ø h2 = 27 1

=

3

1 æ h2 ö çè h ÷ø = 27 1

This is to be balanced by the buoyant force. This can be possible only when the light sphere is completely submerged. In this way the buoyant force

h2 1 = h1 3

65. (c)

P

Q

D

A

éæ 4 ö ù B = êç pR3 ÷ ´ 2ú ´ (2r) ´ g = ø û ëè 3

l

B

m

given l || M \ area of rectangle ABCD = area of parallelogram ABQP as they both lie on same base AB and between two parallel lines. In DPDA and DQCB, PA > DA and QB > CB as PA and QB are hypotenuse, also AB = PQ = DC. now, Perimeter of rectangle Perimeter of parallelogram =

é4 3 ù 4 3 4 pR (3r) g + pR 3rg = 4 ê pR rg ú 3 3 ë3 û

AB + CB + CD + AD 2 AB + 2 DA AB + BQ + PQ + PA = 2 AB + 2PA

Q 2PA > 2DA Perimeter of rectangle \ Perimeter of parallelogram < 1

PHYSICS 66. (a) Consider the equilibrium of the system of both spheres and the spring.

é4 ù 4 ê pR 3rg ú ë3 û

Now considering the equilibrium of the heavy sphere Fs + B = W \ Fs = W – B 4 4 \ Kx = pR3 (3r) g - pR 3 (2r) g 3 3

4 R3rg \ x= p 3 K 67. (d) When the ball is thrown upwards, at the point of throw (O) the linear momentum is in upwards direction (and has a maximum value) and the position is zero. As the time passes, the ball moves upwards and its momentum goes on decreasing and the position becomes positive. The momentum becomes zero at the topmost point (A). Momentum

0

A

Position

As the time increases, the ball starts moving down with an increasing linear momentum in the downward direction (negative) and reaches back to its original position. These characteristics are represented by graph (d).

EBD_7839 118

KVPY-SA

68. (b) Molar mass of N2 gas in M = 28g

a1

Number of moles of N2, n0 = (m/M) = 1 One-third molecules are dissociated into atoms. No. of moles of diatomic N2 = 2/3 For monatomic gas Cv1 = (3/2) R For diatomic gas Cv2 = (5/2) R

2

– v2 = 2a1

…(2) …(3)

Equating (1) and (2) and solving, we get

70.

(c) Intensity of the source at the cross section A I=

=

Increase in internal energy, DU = (n1 + n2 ) Cn(T2 – T0)

Let a1 and a2 be the retardations offered to the bullet by wood and iron respectively.

1.25 ´ 103 4 p (50 r ) 2

80 æ 1.25 ´ 103 ö 2 ´ç ÷ (pr ) = 0.1 J/sec 100 è p ´ 104 r 2 ø

Using, Wien’s displacement law, to determine the temperature of the end B lm TB = 0.003

Þ T2 = T0 2

8 4 ´ 0.2 R (T0 2 - T0 ) = RT0 ( 2 - 1) 3 3

4 p (50 r ) 2

=

A = 80% of {I× pr2}

V1 = V0; g-1

P

Power absorbed by the end

For adiabatic compression, T1V1g -1 = T2V2g-1

69. (b)

…(1)

Adding, we get, –u2 – 2(2a1 + 2a2)

Piston is displaced back to its initial position, and during the adiabatic compression, volume of the gas mixture is halved.

=

2cm

4a1 + a2 – 2a1 + 2a2 Þ a2 = 2a1

= 1.5

æV ö T2 ç 0 ÷ è2ø

4cm

02

and

Þ Cp = 3R

T0V0g-1 =

Iron

u

For A¢ ® B¢ ® C¢, v22 – u2 = 2a2

æ2öæ3ö æ2öæ5ö çè ÷ø çè ÷ø R + çè ÷ø çè ÷ø R 3 2 3 2 = 2R Cp – Cn = R = 2 2 + 3 3

Where T1 = T0 ; T2 = ?; V2 = V0/2

Wood

A¢

Adding, we get –u2 = 2(4a1 + a2)

n1 + n2

Cv

C¢ v2

02 – v12 = 2a2

and

n1Cv1 + n2Cv2

Cp

0

For A ® B ® C, v12 – u2 = 2a1

Average molar specific heat at constant volume

g=

v2 C¢ 0

No. of moles of monatomic nitrogen = 2 × (1/3)

Cn =

B¢

a2 C¢

TB =

0.003 100000 ´ 10-10

= 300 K

æT ö æ dQ ö 2 dT çè ÷ø = - ç ÷ (pr ) dt dx è TA ø 1

0.1TA dx

ò 4 ´ 10-4 0

TB

= - ò TdT \ T = 500 K A TA

SOLUTIONS – MOCK TEST-2

119

CHEMISTRY 71. (a) In NH3 the atomic dipole (orbital dipole due to lone pair) and bond dipole are in the same direction, whereas in NF3 these are in opposite directions so in the former case, they are added up whereas in the latter case net result is reduction of dipole moment. It has been shown in the following figure :

75. (d) Order of Acidic strength µ stability of conjugate base.

NH2 < I

OH < II

–

NH


>

OCH3 OCH3 II Stabilised by –I effect

I Destabilised by +R’ stabilised by –I

III Destabilised by +R ’ stabilised by –I effect which is minimum in para isomer

73. (d) Molarity,

=

wt. of solute per litre of solution Mol. wt. of solute 1000 mol/L 18 ƒ H+ + OH–

Molarity of H2O = H2O c (1 – a)

ca

ca

¾¾®

¾¾®

¾¾®

BIOLOGY 76. (b) This is an example of non-competitive inhibition. Binding of inhibitor alters the enzyme conformation thus the reaction catalysed by the enzyme is affected. Since inhibitor binds at the site other than active site, the affinity for the substrate is not affected and such inhibition cannot be overcome by increasing the concentration of substrate. Vmax decreases and Km stays constant. 77. (c) Lyases enzyme catalyses breakdown without addition of water. Isomerase enzyme catalyses the conversion of an aldose sugar to a ketose sugar. Oxidoreductase enzyme catalyses the transfer of electrons from one molecule to another molecule. Ligases enzyme catalyses the bonding of two RNA molecules.

EBD_7839 132 78.

KVPY-SA (a) Dead space represents the volume of ventilated air that does not participate in gas exchange. The two types of dead space are anatomical dead space and physiological dead space. Anatomical dead space is represented by the volume of air that fills the conducting zone of respiration made up by the nose, trachea, and bronchi. This volume is considered to be 30% of normal tidal volume (500 mL); therefore, the value of anatomical dead space is 150 mL.

79.

OR Alveolar ventilation = Respiratory rate (Tidal volume – Dead space) 4200 = 12 (500 – Dead space) Dead space = 150 mL

80.

Anatomical dead space ventilation = Respiratory rate × Dead space = 150 mL ×12 = 1800 mL/min (a) The total number of progeny = 800 In F2 population, the ratio (9 : 3 : 3 : 1) represent 9 – Yellow and round 3 – Yellow and wrinkled 3 – Green and round 1 – Green and Wrinkled So, Total number of yellow and wrinkled 3 ´ 800 = 150 seeds = 16 (b) Water surrounding the body in Hydra and blood in prawn are the media of circulation and transport.

MOCK TEST-4 When number is not divisible by 11 then smallest such number is 8 × 9 × 5 × 7 × 13 When number is not divisible by 13 then smallest such number is 8 × 9 × 5 × 7 × 11 And smallest of all these numbers is 8 × 9 × 5 × 7 × 11 = 27720

PART-I MATHEMATICS 1.

(a) Since, there is no restriction with participants from China and Russia hence these 10 members can be arranged around circle in (9!) ways, this will create 10 spaces, so now 5 participants from USA can be arranged in these 10 places in

3.

(d) Since that number is divisible by 14 numbers from 1 to 15, so let us start eliminating the number. The number has to be divisible by 2 because if it is not divisible by 2 then that number can not be divisible by 4, 6, 8, 10, 12, and 14 but the given number is not divisible by only one number from 1 to 15 hence 2 cant be that number, on the same logic, 3, 4, 5, 6, 7, are eliminated (e.g if number is not divisible by 7 then number can not be divisible by 14 also, hence it has to be divisible by 7) the given number may not be divisible by 8 and 9, now consider 10, since that number is divisible by 2 and 5 both hence that number must be divisible by 10, consider 11, that number may not be divisible by 11, since number is divisible by 3 and 4 hence it has to be divisible by 12, number may not be divisible by 13, again since number is divisible by 2 and 7 hence number has to be divisible by 14, similarly number has to be divisible by 15. Hence number may not be divisible by 8, 9, 11 and 13. When number is not divisible by 8 then smallest such number is 4 × 9 × 5 × 7 × 11 × 13 When number is not divisible by 9 then smallest such number is 8 × 3 × 5 × 7 × 11 × 13

b
HF (Any two sides of a triangle are together greater than the third) Therefore, AE + DF > HF

...(1)

Similarly in DEGH, EH + EG > HG or AD + EG >HG Adding eq. (1) and eq. (2) gives AE + EG + AD + DF > HF + HG. Therefore, AF + AG > HF + HG.

...(2)

EBD_7839 134 5.

KVPY-SA (b)

A

B

10.

O D

C

Q

6.

AO2 + OB2 = AB2B ...(i) 2 2 2 AO + OD = AD ...(ii) 2 2 2 OC + OD = CD ...(iii) 2 2 2 OC + OB = BC ...(iv) After adding (i), (ii), (iii) and (iv) we get (AC)2 + (BD)2 = 4AB2 (b) Given two digit number is ab = 10a + b 10a + b has quotient = 6 & remainder = 7 a+b

11.

(a) Let the four digit number be ‘abcd’. a+ b= c+ d ...(1) [Since the sum of the first two digits is equal to that of the last two digits ] a+d=c ...(2) [The sum of the first and the last digit is equal to the third digit] b + d = 2 × (a + c) ...(3) [the sum of the second and fourth digits is twice the sum of the other two digits] After solving these equations together we get. b = 2d; d = 4a; c = 5a. Hence we can say that c has to be a multiple of 5 as a is an integer. The only single digit number multiple of 5 is 5. Hence c, the third digit has to be 5. (a) Let the vertices of a triangle be A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3). A(x1, y1, z1)

Hence 10a + b = 6a + 6b + 7 (0, 8, 5) E

4a - 7 =b 5

B(x2, y2, z2)

Now by trial & error method we will get a = 3, b = 1 or a = 8, b = 5

7.

8.

9.

But a = 3, b = 1, a + b = 4 so it can’t give us remainder 7, hence only 1 two digit number exist. (c) The three numbers are 14x, 14y and 14z then any two numbers in pairs are co-prime to each other. If all x, y and z are co-prime to each other then their LCM = 14xyz. But L can be less than 14xyz in all other cases hence option C best describes the L. (c) um + v m = wm, this one is similar to Pythagoras theorem, so we can take some Pythagoras triplets like 3, 4, 5 or, 5, 12, 13 i.e 32 + 42 = 52 than correct option is (c). (d) Without restriction they can be arranged in (27!) ways, and the number of ways in which at least one manager is next to director is 102(25!) [from question number 4]. So required number of ways is (27!) – 102(25!) = 26 × 27(25!) – 102(25!) = 600(25!)

D (5, 7, 11)

F(2, 3, – 1)

C(x3, y3, z3)

Since D, E and F are the mid-points of AC, BC and AB æ x + x 2 y1 + y2 z1 + z 2 ö \ ç 1 , , ÷ = ( 0,8, 5 ) 2 2 ø è 2 Þ x1 + x2 = 0, y1 + y2 = 16, z1 + z2 = 10 …(i)

Þ

æ x 2 + x 3 y 2 + y3 z 2 + z 3 ö , , = ( 2,3, -1) ç 2 2 2 ÷ø è x2 + x3 = 4, y2 + y3 = 6, z2 + z3 = – 2 …(ii)

æ x1 + x3 y1 + y3 z1 + z3 ö = ( 5,7,11) , , and ç 2 2 ÷ø è 2 Þ x1 + x3 = 10, y1 + y3 = 14, z1 + z3 = 22. …(iii) On adding eqs. (i), (ii) and (iii), we get 2(x1 + x2 + x3) = 14, 2(y1 + y2 + y3) = 36. 2(z1 + z2 + z3) = 30, Þ x1 + x2 + x3 = 7, y1 + y2 + y3 = 18, z1 + z2 + z3 = 15. …(iv) On solving eqs. (i), (ii), (iii) and (iv), we get x3 = 7, x1 = 3, x2 = – 3; y3 = 2, y1 = 12, y2 = 4 and z3 = 5, z1 = 17, z2 = – 7. Hence, vertices of a triangle are (7, 2, 5), (3, 12, 17) and (– 3, 4, – 7).

SOLUTIONS – MOCK TEST-4

135

12. (a) Let the cost price of one chair be ` x and that of one table be ` y. Profit on a chair = 25% \ selling price of one chair = x+

25 125 x= x 100 100

Profit on a table = 10% \ selling price of one table = y+

10y 110 = y 100y 100

By the given condition, we have 125 110 x+ y = 1520 100 100

Þ 125x + 110y = 152000 ...(i) Þ 25x + 22y = 30400 If profit on a chair is 10% and on a table is 25%, then total selling price is `1535. \ Þ

10 ö æ 25 ö æ x÷ + ç y + y÷ = 1535 çè x + ø è 100 100 ø 110 125 x+ y = 1535 100 100

Þ 110x + 125y = 153500 Þ 22x + 25y = 30700 ...(ii) Eqn. (ii) – (i) gives 3x – 3y = – 300 Þ x – y = – 100 ...(iii) (i) + (ii) gives 47x + 47y = 61100 Þ x + y = 1300 ...(iv) (iii) + (iv) gives 2x = 1200 Þ x = 600 and (iv) – (iii) gives 2y = 1400 Þ y = 700 ...(iv) Hence, the cost price of a chair is ` 600 and that of a table is ` 700. 13. (b) PQ = PS = 1 = QR In DPQR PQ = QR = 1 \ ÐQRP = ÐQPR = 2° \ ÐQ + 2° + 2° = 180° ÐQ = 176°

In DPQS ÐPQS = ÐPSQ = a 2a + 2° = 180° Þ a = 89° \ ÐRQS = ÐQ – ÐPQS Þ ÐRQS = 176° – 89° Þ ÐRQS = 87° 14. (c) No. of ways getting one correct

1ö æ 1 1 1 = 7 C1 6!ç1 - + - + ... + ÷ = 7 C × 265 1 6! ø è 1! 2! 3! No. of ways getting two correct 1ö æ 1 1 1 = 7C2 × 5!ç1 - + - + ... - ÷ = 7 C × 44 2 1! 2! 3! 5! è ø No. of ways getting three correct æ 1 1 1 1ö = 7 C3 × 4!ç1 - + - + ÷ = 7 C3 × 9 è 1! 2! 3! 4! ø Required no. of ways = 7 C3 × 9 + 7 C2 × 44 + 7 C1 × 265 . 15. (c) Suppose the number is x + 10y + 100z Now, on reversing the new number = z + 10y + 100x According to question x + 10y + 100z – z – 10y – 100x = 297 Þz=x+3 Also, x + y + z = 8 So, possible cases are 512, 431 but hundred possible value is 5 (maximum) So, tens digit is 1.

PHYSICS 16. (a) Man and wedge will displace as one unit of mass 3 M. If its displacement is Dx1, towards left, then

0 = M ( 2 - Dx1 ) + 3M ( -Dx1 ) \

Dx1 =

2 = 0.5 m 4

EBD_7839 136 17.

KVPY-SA

=

(c) As a-particles are doubly ionised helium He++ i.e. Positively charged and nucleus is also positively charged and we know that like charges repel each other. 21. (c) From Snell’s law, 20.

g

(b) g¢ =

2

hö æ çè 1 + ÷ø R gR 2

( R + h)

\

2

=

gR 2

( 3 R / 2)

2

4g 9

=

60°

m =

4g W¢ = mg¢ = 200× 9

4 ´ 10 = 889 N. 9 18. (b) The electric field due to the sheet,

22.

= 200 ×

E=

s Î0

+ +q

Now for the equilibrium of the ball T sin q = Eq and T cos q = mg

+

19.

q

3

30° 90° 30°

(d) Let c be the centre of the cross-section considered at any section of the wire. If r be the radius of a circular strip concentric with the cross-section considered and, dr be its thickness then (fig.) current through the circular strip of thickness dr will be r r di = j.dS = (a + br) (2prdr) cos 0

Eq

+ +

Eq sq \ tan q = = . mg Î0 mg

Thus

T

sin 60° = sin 30°

A

30°

or

di = 2p (ar + br 2 ) dr

mg

+

C dr

s µ tan q

(b)

r R

The total current i can be obtained by summing up the currents flowing through individual circular strips, R

i.e.,

0

p(nd )2 pd 2 x´ = y 4 4

or

\ x = n2y From Pascal’s law rw gh = r Hg g ( x + y) or \

1 × gh = sg (n2 y + y ) y=

i = ò di = ò 2p (ar + br 2 ) dr

h s(1 + n 2 )

é i = 2p êa ê ë

R R é r2 ù é r3 ù ù + b ê ú ê ú ú ëê 2 ûú 0 ëê 3 úû 0 úû

2pR 2 (3a + 2bR) 6 23. (c) Torque working on the bob of mass m is, =

t = mg ´ l sinq . (Direction parallel to plane of rotation of particle)

SOLUTIONS – MOCK TEST-4

q

l

137 æ 1ö 25. (d) We know that C = sin -1 ç ÷ . è mø

m

mg

ur As t is perpendicular to L , direction of L changes but magnitude remains same. 24. (b) When body rolls dawn on inclined plane with velocity V0 at bottom then body has both rotational and translational kinetic energy. Therefore, by law of conservation of energy,

26.

(b)

If R is the resistance of each lamp, then current (initially) in x and y is same, i =

e . 2R

2

same, when switch is closed,

V2 1 mk 2 0 2 R02

…(i)

é V ù 2 êQ I = mk , w = ú R0 û ë

When body is sliding down then body has only translatory motion. 1 æ5 ö \ P.E. = K.Etrans = m ç v0 ÷ 2 è4 ø

Dividing (i) by (ii) we get é K2 ù 1 mV02 ê1+ 2 ú P.E. 2 êë R0 úû = = 1 25 P.E. ´ ´ mV02 2 16 K2 25 K2 9 = 16 = 1 + 2 Þ 2 = 16 R0 R0 3 or, K = R0 . 4

-1 æ m ö -1 æ sin i A ö = sin ç B ÷ = sin ç è mA ø è sin iB ÷ø

e2 æ e ö Px = i 2 R = ç ÷ R = and Py also the è 2R ø 4R

1 1 mV02 + I w 2 2 2

1 2 = 2 mV0 +

æ 1 ö C AB = sin -1 ç = sin -1 ( A m B ) è B m A ÷ø

So power consumed,

P.E. = K.Etrans + K.Erotational =

Given , iB > i A \ m B < m A . So B is rarer and A is denser. Light will be totally reflected, when it passes from A to B. Now critical angle for A to B

l

i=

e 2e = . R ö 3R æ + R çè ÷ 2ø 2

2

4e 2 æ 2e ö ´R = , and ÷ è 3R ø 9R

Now P 'x = i ' R = ç 2

e2 æ e ö Py = ç ÷ R = . è 3R ø 9R Clearly x-increases, y-decreases.

2

...(ii)

27. (d) The least count (L.C.) of a screw guage is the smallest length which can be measured accurately with it. 1 As least count is 0.001 cm = cm 1000 Hence measured value should be recorded upto 3 decimal places i.e., 5.320 cm 28. (c) According to law of sines or Lami’s Theorem vB

75°

v B / A = vB - vA 60°

45° vA

EBD_7839 138

KVPY-SA Þ

v vA vB = = B/ A sin 75° sin 60° sin 45°

Þ vB = 717 kmh–1

35.

(d)

H

H ¾¾® sp3 Carbon

29. (c) W = m (V2 – V1) é GM1 GM 2 ù when, V1 = - ê + , 2a úû ë a é GM 2 GM1 ù + V2 = - ê 2a úû ë a

\ W= 30.

31.

32.

Cyclopentadiene is non aromatic, as it has sp3 carbon in the ring. 36. (c) Na 2SO 3 + 2H 2SO 4 (dil) ¾¾ ® Na 2SO 4 + H 2O + SO2

Gm( M 2 - M1) ( 2 - 1) a 2

(c) When ball fall from great height, its initial velocity before entering into liquid is quite enough. So viscous force together with buoyant force becomes greater than weight of the cone. So first ball retarted and thereafter will move with constant velocity. CHEMISTRY (c) This is Avogadro’s hypothesis. According to this, equal volume of all gases contain equal no. of molecules under similar condition of temperature and pressure. (a) For isothermal expansion, relation between P and Vm, and, U and Vm are as:

K 2 Cr2 O 7 + H 2SO 4 + 3SO 2 ¾¾ ® K 2SO 4 + Cr2 (SO4 )3 + H 2O Green

37.

38.

39. 40.

33.

Therefore (B) and (D) are not correct representation. (a) H 2O + C ¾¾ ® CO2 + H 2 Steam

34.

Water oxidises C to CO2, hence water act as an oxidising agent, whereas, carbon act as reducing agent since it reduces water to hydrogen gas. (d) According to Hund’s rule electron pairing in p, d and f orbitals cannot occur until each orbital of a given subshell contains one electron each or is singly occupied.

41.

The above reactions indicates the presence of SO2– 3 ion in the compound. (b) Higher the positive charge, greater is the EN and higher the negative charge lesser is the EN. So, EN order is M– > M2– > M3– > M4– (d) Gypsum is CaSO4 . 2H2O, whereas Plaster 1 of Paris is CaSO 4 × H 2 O. 2 Therefore, difference in number of water 1 molecules in them is 1 . 2 (b) IUPAC name of the given compound is 2-hydroxy benzoic acid. (c) More effective axial and sideways overlapping between atomic orbitals of carbon and oxygen atom is due to smaller size of oxygen atom. Oxygen is more electronegative than carbon atom, due to these factors CO has highest bond energy. (d) The pOH of a buffer consisting of NH3 (i.e. NH4OH) and salt NH4Cl (salt) is given by the equation pOH = pKb + log = 5.0 + log

[1.0] [ 0.1]

[salt] [ammonia]

= 5.0 + log 10 = 5 + 1 = 6 \ pH = 14 – pOH = 14 – 6 = 8

SOLUTIONS – MOCK TEST-4

139

42. (d) In the reactivity series, as we move down the series the reactivity of the element decrease. 43. (b) Methyl carbanion is sp3 hybridised, with three bond pairs and one lone pair, same is the case with NH3.

44. (a) H2 (g) + O2 (g) ¾¾® H2O2 (g)

DH reaction = B.E.reactants - B.E.products = [B.E.(H - H) + B.E.(O = O)]

45.

-[2B.E.(O - H) + B.E.(O - O)] = [438 + 498] – [2 × (464) + 138] = 936 – 1066 = –130 kJ mol–1 (d) Bond structure of H4P2O7 is following O

H

s

O

O

s p s s

s p s s

s

s

P O

O

s

P

H O

s

O

s

H

H

12s, 2dp – pp bonds. 46. (d)

47. (c)

48. (a)

49. (b)

BIOLOGY Debris normally gets trapped in mucus lining of the respiratory passage ways. This mucus is then swept up and away from the lungs by the action of cilia. Cholesterol is the precursor of the five major classes of steroid hormones: progestogens, glucocorticoids, mineralocorticoids, androgens, and oestrogens. These hormones are powerful signal molecules that regulate a host of organismal functions. The route of water and solutes through the nephron is as follows: From the glomerulus ® Bowman’s capsule ® proximal tubule ® loop of Henle ® distal tubule ® collecting ducts. In gram negative bacteria, peptidoglycan layer is thin and does not retain crystal violet stain like gram positive bacteria do.

50. (a) Substrate level phosphorylation means synthesis of ATP by utilising energy released directly by the substrates. In glycolysis, there are two steps in which ATP is synthesised. In Kreb’s cycle, there is only one step where GTP is synthesised when succinyl CoA is converted into succinic acid. 51. (a) Heat-killed S-type bacteria caused a transformation of the R-type bacteria into S-type bacteria but he was not able to understand the cause of this bacterial transformation. The occurrence of living S-type virulent bacteria is possible only by their formation from R-type non-virulent bacteria, which pick up the trait of virulence from dead bacteria. This phenomenon is called Griffith effect or transformation. Griffith proposed that a chemical substance released by heat- killed bacteria changed the R-bacteria into Sbacteria which was a permanent genetic change. 52. (c) Human embryo have gills. This shows ontogeny repeats phylogeny. Ontogeny is the life history of an organism while phylogeny is the evolutionary history of the race of that organism. Modern theory of origin of life was propounded by Oparin and Haldane which is based on chemical evolution. Chemical evolution, also called chemogeny, involves the synthesis of simple organic molecules. Miller and Urey experimentally supported Oparin and Haldane theory with the help of stimulation experiment. Analogous organs are those organs which are similar in shape and function but their origin, basic plan and development are dissimilar. Example, wings of butterfly, bird and bat. Such similarities are because of convergent evolution for adaptation to a common condition. 53. (c) Baculoviruses are suitable for speciesspecific, narrow spectrum insecticidal applications. 54. (a) After cellulose, chitin is the second largest structural polysaccharide and principal component of exoskeleton of insects and crustaceans. It is a polymer of N-acetyl glucosamine.

EBD_7839 140 55.

56.

57.

58.

59.

60.

KVPY-SA (c) Oogenesis is initiated during the embryonic development stage when some million oogonia (A) are formed within each foetal ovary. No more oogonia are formed and added after birth. These cells start division and enter into prophase I (B) of meiotic division and get arrested (C) at a stage called primary oocyte (D). (c) A geological time scale is a diagram that details the history of earth’s geology, noting major events like the formation of the earth, the first life forms and mass extinctions. The first geological time scale was proposed in 1913 by the British geologist Arthur Holmes (1890–1965). The history of the earth has been subdivided into eras, eras into periods and periods into major divisions. (d) PS II is located within the appressed granal region, whereas PS I and ATP synthase complex is located within the nonappressed thylakoid membrane regions and stroma lamellae. (c) Tropic hormones are those hormones that have other endocrine glands as their target to control their function. Most of the tropic hormones are produced and secreted by anterior pituitary. (c) A nucleotide is made up of a sugar molecule, phosphate group and a heterocyclic base while a nucleoside is made up of a sugar molecule and a heterocyclic base. (b) During prophase, each chromosome consists of two chromatids (46 chromosomes, each with 2 chromatids = 92 chromatids).

Let area of DPTU = R, and ar DTUV = a 1 QS (by mid point theorem) 2 and also, TU || SQ. \ DTUV ~ DQSV

Q TU =

TU 2 ar DTUV = \ SQ 2 ar D SQV QS 2 a = Þ Þ ar DSQV = 4a. 2 ar D SQV (2 QS ) Q TU is median for DPTQ, \ ar DPTU = ar DTUQ Þ ar DTUQ = R \ ar DUQV = R – a In D PSU, as UT is median, \ ar DUTS = R \ ar DPSU = 2R, In DPSQ, as SU is median, \ ar DSUQ = ar DPSU = 2R. Q SQ is diagonal, \ ar (PQRS) = 8R Q ar DSUQ = ar DSVQ + ar DUQV Þ 2R = 4a + R – a Þ R = 3a now, ar QRSV ar DSQR + ar DSVQ = ar DPQT ar DPQT Þ

Þ ar QRSV = ar DPQT 62.

PART-II MATHEMATICS 61.

ar QRSV 4 R + 4a = ar DPQT 2R

(d) 63.

4R 3 =8 2R 3

4R +

(d) Let a two digit number 10a + b if we reverse the digits then we have new number 10b + a. Summation of these two numbers are 11(a + b) hence this summation or all the elements of the 1st must be divisible by 11, and so the required HCF would be 11. (b) Consider the figure, Area of sector OACB = pr 2

q 60 1 p p = p ´1´ = . = 360° 360 2 3 6

SOLUTIONS – MOCK TEST-4

141 Putting b = B – 15°, we get a relation 1 + 2 cos(b + 30) = cosb We write this in the form

(1 - 3)cos b + sin b = 1. Since, sin b £ 1, it follows that

(1 - 3)cos b ³ 0. p Area of shaded region = – area of DOAB 6

We conclude that cos b £ 0 and hence that b is

p 3 = 6 4 Hence, area of lune = Area of semi-circle – area of shaded region

So, is angle B and hence ÐADC.

=

=

2 æ 1 æ 1ö p 3ö pç ÷ -ç ÷ è ø è 2 2 6 4 ø

3 p p 3 p + - = 4 8 6 4 24

64. (a)

obtuse. We have the relation (1 - 3)cos b + sin b = 1.

æ bö If we set x = tan çè ÷ø , then we get, using 2 cos b =

(1 - x 2 ) 2x , sin b = , (1 + x 2 ) (1 + x 2 )

( 3 , 2) x2 ∗ 2 x , 3 < 0. Solving for x, we obtain x = 1 or x < 3(2 ∗ 3). If x < 3(2 ∗ 3), then

æ bö tan ç ÷ > 2 + 3 = tan 75° giving us b > 150°. è 2ø

O A

This forces that B > 165° and hence, B + A > 165° + 15° = 180°, a contradiction.

a 15 °

B

D

a

F E C

Let a denote the equal angles ÐBAD = ÐDCA. Using sine rule in triangles DAB and DAC, we get AD BD CD AD < , < . sin B sin  sin15↓ sin  Eliminating a (using BD = DC and

p . 2 This gives B = 105° and hence, a = 30°.

Thus, x = 1 giving us b =

Thus, ÐDAO = 60°. Since, OA = OD, the result follows. 7+4 3 - 7-4 3

65. (d)

2 2 + ( 3) 2 + 2(2)( 3)

=

- 2 2 + ( 3) 2 - 2(2)( 3)

2a + B + 15° = p),

(2 + 3)2 - (2 - 3) 2

we obtain 1 + cos(B + 15°) = 2 sin B sin 15°

=

But, we know that

= 2+ 3 -2+ 3= 2 3

2sin B sin 15° = cos(B – 15°) – cos(B + 15°)

EBD_7839 142

66.

KVPY-SA PHYSICS (a) If v1 and v2 are the velocity of ball and the wedge after collision, then v1

J

v2

60° 60°

z2 = 0.273 +

68. (c) As we know, time period, T = 2p

2v2 = v1 + v0 3

2

TM = T

2

...(i)

v1 + 3v2 ...(ii) v0 After simplifying above equations, we get

=

3 (1 + e)v0 5

67. (d) Let zmin be the minimum reference level of the dotted parabola and z1 and z2 the liquid levels above the base. z1 =

r12w 2 0.12 ´ 102 + zmin = + zmin , 2g 2 ´ 9.81

r22 w 2 0.22 ´ 102 + zmin = + zmin 2g 2 ´ 9.81 But z1 + z2 = 2 × 0.4 = 0.8m

z2 =

Hence, 0.8 =

\

69.

0.12 ´ 102 = 0.324m 2 ´ 9.81

Mgl ù é êëQ Dl = AY úû

2 ù A 1 éæ TM ö = êç 1 ú ÷ Y êè T ø úû Mg ë

(a) From the figure it is clear that (a) s2 > s1 (b) r2 > s2 [As the string is taut] (c) r1 < s1 [As the string is taut] \ r1 < s1 < s2 < r2 When P alone is in L2 2pr 2 (r1 – s 2 ) g is negative as r < s 1 2 9h2 Where r is radius of sphere. When Q alone is in L1

VP =

2 pr 2 (r2 – s1 ) g is positive as r > s 2 1 9 h1 ur ur Therefore V P . V Q < O VQ =

0.22 ´ 102 + 0.1 ´ 10 2 + 2 zmin 2 ´ 9.81

whence, zmin = 0.273m Consequently, z1 = 0.273 +

l + Dl l + Dl æT ö or ç M ÷ = è T ø l l

Mg æT ö or, ç M ÷ = 1 + AY è T ø

[v cos60° - ( -v2 cos30°)] e= 1 (v0 cos 60° - 0)

v2 =

l + Dl g

TM = 2p

60°

For ball: 2 J cos 60° = mv1 – (–mv0) or J = mv1 + mv0 And J sin 60° = mv2 – 0 From above equations, we get

Also

l g

When additional mass M is added then

J

v2

60°

0.22 ´ 102 = 0.477m 2 ´ 9.81

s1 T T s2

SOLUTIONS – MOCK TEST-4

143 CHEMISTRY

VP r1 – s 2 h1 Also V = r – s ´ h 2 1 2 Q

...(i)

...(ii)

...(iv) VP

\

VQ

=

h1 h2

70. (b) The time period of a simple pendulum is given by

Þ

l 2 2 l 2 l \ T = 4p g Þ g = 4 p 2 g T

Dg Dl DT ´ 100 = ´100 + 2 ´ 100 g T l

Case (i) Dl = 0.1 cm, l = 64cm, DT = 0.1s, T = 128s \

Dg ´ 100 = 0.3125 g

Case (ii) Dl = 0.1 cm, l = 64cm, DT = 0.1s, T = 64s \

Dg ´ 100 = 0.46875 g

Case (iii) Dl = 0.1 cm, l = 20cm, DT = 0.1s, T = 36s \

Dg ´ 100 = 1.055 g

Clearly, the value of in case (i).

OH -

1-Propanol

...(iii)

(iii) – (ii) gives r1 – s2 = s1 – r2 From (i) and (iv)

T = 2p

2 2 2(CH3CH 2 CH 2 )3 B ¾¾¾®

6CH3CH 2 CH 2 OH+2H3BO3

4 3 4 pr r1g = pr 3 s1g 3 3

h VP =– 1 h2 VQ

ether, 0 °C H O

4 3 4 pr s 2 g = pr 3 r 2 g 3 3

For equilibrium of P T+

B H

2 6 6CH3 - CH = CH 2 ¾¾¾®

1-Propene

For equilibrium of Q T+

71. (d)

Dg ´ 100 will be least g

72. (a) H2SO4 can be formed by adding all the given equations. The obtained reaction is, H2 + S + 2O2 ¾¾® H2SO4 Now, enthalpy of formation of H2SO4 will be obtained by adding DH of the given reactions, i.e., (i) + (ii) + (iii) + (iv) So, –298.2 kJ + (–98 kJ) + (–130.2 kJ) + (–287.3 kJ) = –814.4 kJ 73. (d) The reaction may be given as ¾® 2Z + 3H2O Z2 O3 + 3H2 ¾ 0.1596 g of Z2O3 react with H2 = 6 mg = 0.006 g \ 1 g of H2 react with = 0.1596 0.006 = 26.6 g of Z2O3 \ Eq. wt. of Z2O3 = 26.6 g (from the definition of eq. wt.) Eq. wt. of Z + Eq. wt. of O = E + 8 = 26.6 g Þ Eq. wt. of Z = 26.6 – 8 = 18.6 g Valency of metal in Z2O3 = 3

Eq. wt.of metal
Li2 > B2. (b) The more substituted alkenes have greater number of hyperconjugated forms. Hence they are more stable. H3C H

C == C

H CH3

2-Butene

H3C H3C

C == C

78. CH3 CH3

2,3-Dimethyl But-2-ene

BIOLOGY 76.

77.

(d) A frameshift mutation occurs due to insertion and/or deletion of a number of nucleotides in a DNA sequence that is not divisible by three. This changes the reading frame resulting in a completely different translation from the original. A missense mutation is a point mutation in which change in a single nucleotide results in a codon that specifies a different amino acid. A nonsense mutation is also a point mutation that changes a sense codon into a nonsense codon. As a result, the altered DNA sequence prematurely stops the translation of mRNA leading to the formation of a truncated protein that may be non-functional. Silent mutations are mutations in DNA that do not have an observable effect on the organism's phenotype. (c) Bundle sheath is a layer of cells that forms a sheath surrounding the vascular bundles in plant leaves and stems. Some C3 plants

79.

80.

do not have bundle sheath. If they have bundle sheath, its cells do not contain chloroplasts. In contrast, C4 plants have bundle sheath cells with chloroplasts. Dicot plants have reticulate venation and monocots have parallel venation. Hence, plant A with reticulate venation but no bundle sheath is a C3 dicot, plant B with parallel venation and bundle sheath containing chloroplasts is a C4 monocot, plant C with reticulate venation and bundle sheath containing chloroplasts is a C4 dicot and plant D with parallel venation and bundle sheath without chloroplasts is a C3 monocot. (b) Air at sea level with an atmospheric pressure of 700 mm Hg has a partial pressure of oxygen of 160 mm Hg (760 × 21%). Therefore, the partial pressure of O2 at Antonito, Colorado, is 600 × 21% = 126 mm Hg. (d) Analogous organs have similar functions but different origins. Options (a), (b), and (c) are the examples of analogous organs whereas tendrils of pea and spines of barberry are examples of homologous organs. (b) Tidal volume is the normal volume of air displaced between normal inhalation and exhalation. It is approximately 500 mL. The inspiratory reserve volume (IRV) is the additional amount of air that can be inhaled after a normal inspiration (tidal volume). It is between 2500-3000 mL in humans. The expiratory reserve volume (ERV) is the amount of extra air above a normal breath, exhaled during a forceful breath out. The average ERV volume is about 1100 mL in males and 800 mL in females. The residual volume is the amount of air remaining in the lungs at the end of a normal exhalation. This averages 1100 mL to 1200 mL.

MOCK TEST-5 For P to be minimum (b + c)/(a + d) should be maximum that means b + c should be maximum when b = c = 9, and a + d should minimum when a = 6 and d = 0. Then N = 6990 and then P = 110 + 891/4 = 332.75

PART-I MATHEMATICS 1.

(b) As A1B1 and C1D1 are parallel and half of one of the diagonals of ABCD, the quadrilateral A1B1C1D1 is a parallelogram. Similarly, A2B2C2D2 is a parallelogram. As it is also given to be a rectangle, the diagonals of A1B1C1D1 are perpendicular to each other with lengths 8 and 12. Thus, it is a rhombus of side length

5. (b)

D

C 1 3

52 Thus, the lengths of diagonals of

4

é n ù é 208 ù Therefore, ê ú =ê ú = 2. ë 100 û ë 100 û 2. (d) Since unit digit of all the factorials more than 4 is 0 hence unit digit 9 is given by either 0! + 2! + 3! Or 1! + 2! + 3! In both the cases the value of {(a!)(b!)(c!)} = 1 × 2 × 6 = 12 3. (c) Adding both equations x2 + xy + x + y2 + xy + y = 42, Þ (x + y)2 + (x + y) = 42 Þ (x + y) (x + y + 1) = 42 Possible solutions, x + y = 6, obtained by factorizing 42. Where (x + y) = 6 and (x + y + 1) = 7 or x + y + 1 = –6 and x + y = –7 Sum of all possible solutions is (–7) + (6) = –1 4. (c) Let N = abcd = 1000a + 100b + 10c + d Then S = (1000a + 100b + 10c + d) + (1000d + 100c + 10b + a) = 110(a + b + c + d) + 891(a + d) From the given condition

2 A

R Join DP and produce DP to meet AB in R.

Ð2 [Q AC cuts || lines AB and DC] Ð3 =Ð4

[Vert. opp. Ðs]

CP = AP

[Given ]

\ DDPC @ DRPA, and in particular DP = PR and DC = AR In D DRB, Q PQ joins P and Q the mid-points of DR and BD respectively. \ PQ || RB And, PQ =

6.

1 RB 2

Þ PQ = (AB – AR)

1 (AB – CD) [QDC = AR] 2 (a) Unit digit of x is prime hence it can be 2, 3, 5, or 7 (We have only 4 prime digits) As the product of all digits of x is prime hence rest all digits must be 1, then possible values of X are 1112, 1113, 1115 or 1117.

=

a+b+c+d 891( a + d ) a+b+c+d 891 = 110 + ì b+c ü í1 + ý î a+d þ

B

In DDPC and DRPA, we have Ð1 = alt.

{110 ( a + b + c + d ) + 891( a + d )} = 110 +

Q

P

ABCD is 2 52 each, giving the required product as 208 i.e., n = 208.

EBD_7839 146

KVPY-SA

7. (b) If a number is divisible by 36 then it has to be divisible by 4 and 9. Given number is 368A37982B. If it is divisible by 4 then last two digits should be divisible by 4 hence B must be either 0, or 4 or 8 (as 20, 24 and 28 the last two digit is divisible by 4). Now since the number is divisible by 9 hence sum of the digits should be divisible by 9 or 46 + A + B should be divisible by 9. Hence 46 + A + B = 54 or 63 If 46 + A + B = 54 then A + B = 8 hence possible pairs of (A, B) = (8, 0), (4, 4) and (0, 8) If 46 + A + B = 63 then A + B = 17 hence possible pairs of (A, B) = (9, 8) Hence possible values of A are 0, 4, 8 and 9, and sum of all possible values of A is 21 8. (a) Given that S = (2, 3, ….2n + 1), hence total numbers of element in S is 2n. Odd elements are (3, 5, 7, ….2n + 1) total n elements And even elements are (2, 4, ….2n) total n elements Hence X = [3 + 5 + 7 + ….(2n + 1)]/n And Y = [2 + 4 + 6 + ….. + 2n]/2 So X – Y = [(3–2) + (5–4) + ….(2n + 1–2n)]/n = [1 + 1 + 1….n times]/n = n/n = 1 Hence X–Y = 1 (Point to note that it is independent from the value of n) 9.

(d) In DOAB, OA = OB = 10 2 cm ( radius ) and AB = 20 cm (Given as the side of the rectangle) Þ

AB2 = OA 2 + OB2 (Each = 400)

Þ

Ð AOB = 90°

We take a point P on the arc AB of the semi-circle. Now, area of the segment APB = The area of the sector OAPB – The area of the DOAB

2 1 ì 90 ü ´ p ´ (10 2 ) - ´ 10 2 ´ 10 2 ý cm 2 =í 2 î 360 þ

1 æ1 ö æp ö = ç ´ p ´ 200 - ´ 200 ÷ cm 2 = ç - 1÷ ´ 100cm 2 è4 ø è2 ø 2

The area of the shaded region = The area of the rectangle ABCD – The area of the segment APB æp ö = 20 ´ 10cm 2 - ç - 1 ÷ ´ 100cm 2 è2 ø

p ì ü = í400 - ´ 100 + 100ý cm 2 2 î þ pö æ = ç 5 - ÷ ´ 100 cm 2 . è 2ø

10. (a) P(x) = x2018 + x2017 +....+ x + 1 remainder = P(–1) Þ P(–1) = (–1)2018 + (–1)2017 +....+ (–1) + 1 Þ

P ( -1) = 1144 - 1 + 14 -2444 1.... + 1 31 + 1 equal +ve and -ve terms

\ P(–1) = 1 11. (d) Since sum of digits is minimum hence number of 4’s should be maximum. Let us assume that number of 4’s is x and that of 8 is y then sum of digits is 4x + 8y. Since N is divisible by 72 hence it must be divisible by 8 and 9 both hence sum of digits should be divisible by 9. So now we have two equations 4x + 8y = multiple of 9 or 4(x + 2y) = multiple of 9 or x + 2y = 9k (i.e multiple of 9) and x + y = 99 (total number of digits) from these two questions we will get y = 9(k – 11) that means y should be multiple of 9. Minimum value of y = 0 but then we have only 4’s and last three digits is 444 which is not divisible by 8 hence number cannot be divisible by 72. Next minimum value of y is 9 then x = 90 and sum of digits = 4 × 90 + 9 × 8 = 432 Here in this case last 3 digits can be 448, 848, 488 and 888 all these are divisible by 8.

SOLUTIONS – MOCK TEST-5

147 Since, a, b are the roots of this equation,

12. (c) As per to question, figure isA

therefore, a + b =

E

F X

But given

Y

B

Þ

C

Q EB || AC and EY || BC

\ BXFC is a parallelogram. Q BXFC and EYCB lies on same base BC and between 2 parallel lines,

( l - 1) l2

\ ar (EYCB) = ar (BXFC) ...(1)

now, consider DAEB, as they both lie on same base BE and between parallel lines BE and AC. Q EC is diagonal of EYCB

-

5 l

( l - 1)

1 ar (EYCB) 2

2

...(2)

l 2 - 16l + 1 = 0

It is quadratic in l. So, let the roots be l1 and l2 then l1 + l 2 = 16 and l1l 2 = 1

now consider DAFC, ar DAFC = ar DBFC,

Therefore,

as they both lie on same base FC and between parallel lines AB and FL.

=

1 \ ar DAFC = 2

ar (BXFC) ...(3)

from (1), (2) and (3) we get ar DEBC = ar DAFC or ar DABE = ar DAFC. (b) The given equation can be written as lx 2 - ( l - 1) x + 5 = 0

10 l =4 5

- 10l 4 = 5l 5

Q BF is diagonal of BXFC,

13.

2

l2 - 12l + 1 = 4l

ar DAEB = ar DEBC,

ar DEBC =

2

- 2ab 4 = ab 5

also, XF || BC and CF || AB.

1 1 ar (EYCB) = ar (BXFC) 2 2

a b 4 + = b a 5

a 2 + b2 4 = ab 5

( a + b)

\ EYCB is a parallelogram,

\

l -1 5 and ab = l l

=

( l1 + l 2 )

l1 l 2 l12 + l 22 + = l 2 l1 l1l 2 2

- 2l1l2 l1l2

(16)

2

-2

1

= 256 – 2 = 254 é l1 l 2 êl + l 1 Now, ê 2 ë 10

ù ú é 254 ù ú=ê = [ 25.4] = 25 û ë 10 ûú

14. (b) Time taken by bucket to ascend = 1 min, 28 sec. = 88 sec. speed = 1.1 m/sec.

EBD_7839 148

KVPY-SA Length of the rope = Distance covered by bucket to ascend = (1.1 × 88) m = (1.1 × 88 × 100) cm = 9680 cm. Radius of the wheel =

x From eqn (1) a = cos wt 1 and from eqn (2) y = cos 2 wt =[2cos2 wt –1] a2

77 cm. 2

2

17.

æ xö y = 2a 2 ç ÷ – a 2 è a1 ø (d) The acceleration of the particle at equilibrium position is zero and is maximum at the extreme position.

18.

(b) WABC = WDEF =

19. Circumference of the wheel

22 77 ö æ = 2pr = ç 2 ´ ´ ÷ cm = 242cm 7 2 ø è \ =

Number of revolutions

Length of the rope æ 9680 ö = = 40 Circumference of the wheel çè 242 ÷ø Hence, the wheel makes 40 revolutions to raise the bucket.

15. (c)

21.

xn + ax + b = ( x - x1 )( x - x2 ) ¼( x - xn ) Þ ( x - x2 )( x - x3 ) ¼( x - xn ) =

x n + ax + b ( x - x1 )

Þ ( x1 - x2 )( x1 - x3 )¼¼( x1 - xn ) x n + ax + b = nx1n-1 + a x® x1 x - x1

PHYSICS (a) Two perpendicular S.H.Ms are x = a1 cos wt and y = a2 cos 2wt

22.

p ´ 32 p ´ 32 + + (15 - 12) ´ 18 2 4

= 6.75p + 54 WDEF > WABC (a) The motion of the train will affect only the horizontal component of the velocity of the ball. Since, vertical component is same for both observers, the h m will be same, but R will be different. (d) We know that, to cross the river by shortest u path; sin a = v where u = river flow speed, v = speed of person. But as given in question u > v Þ sin a > 1, which is not possible. (a) KE and PE completes two vibration in a time during which SHM completes one vibration. Thus frequency of PE or KE is double than that of SHM. (b) As block B falls, L will decrease. Block B will displace, volume of liquid (V1) Equal to its own volume when it is in the liquid. V1 = VB =

= lim

16.

20.

pr 2 p (6) 2 = = 18p 2 2

....(1) ....(2)

MB rB

...(i)

When block B is on block A, it will displace the volume of liquid (V2) whose weight is equal to the weight of the block B. V2rLg = MBg = (V1rB)g or V2rL = V1rB. Since H increases, V1 > V2 So, rL > rB.

SOLUTIONS – MOCK TEST-5

149

23. (d) As maximum height attained by each one is same, so uy is also same. As 2u y T= , g So T1 = T2 = T3. 24. (d)

From eqs (ii) and (iii) 1 2m 2 - 1 = m 2m 2 Squaring both sides and then solving, we get

1 1 sin C = = m 2

\ C = sin

1 ö ÷÷ = 45 º ç è 2ø

26. (d)

sin 45º 1 sin C 1 = Now = or sin r sin r m 2 sin r = 1 or r = 90º 25. (d) At point A by Snell’s law m=

Here, T =

… (i)

1 m

=

45°

i2

A i1 90° m

-

1 \ (sin 90° - r ) = m Þ cos r =

1 m

Now cos r = 1 - sin 2 r = 1 -

=

2m 2 - 1 2m 2

0.1 æ 1ö + 2 ç ÷ = 0.027 è 90 ø 20.0

Thus, the percentage error in g is 100 (Dg/g) = 100 (DL/L) + 2 × 100 (DT/T ) = 3%. 27. (a) Let the speed of the instrument package is v when it grazes the surface of the planet. Conserving angular momentum of the package about the centre of the planet. mv0 × 5R sin(p – q) = mvRsin 90° Þ v = 5v0sin q .....(i) Conserving mechanical energy

From figure, i1 = 90° - r

B

Dt t and DT = . n n

DT Dt = . T t The errors in both L and t are the least count errors. Therefore, (Dg/g) = (DL/L) + 2 (DT/T )

At point B, for total internal reflection,

Air

g = 4p 2 L / T 2 ;

Therefore,

sin 45° 1 Þ sin r = sin r m 2

sin i1 =

3 2

m=

-1 æç

GMm 1 2 + mv = R 2

… (ii)

2m 2

… (iii)

(

)

v0 m

q

5R

R

1 4GMm m v 2 - v02 = 2 5R 8 GM 2 2 Þ v - v0 = .....(ii) 5R Substituting the value of v from eq. (i) in equation (ii) 8GM 25v02 sin 2 q - v02 = 5R

Þ

1

GMm 1 2 + mv0 5R 2

EBD_7839 150

KVPY-SA The angle, package just graze the surface of the planet, q = sin

28.

æ1 8GM ç 1+ ç5 5v02 R è

(a) Since, F = GM R

m

ò

GM nR

smaller bubble.

ö ÷ ÷ ø

mdv =av2 dt

m R

(

)

n -1

a GM (b) Let wA and wB be the absolute angular speeds of A and B. Since they are in the same orbit, their time periods must be the same, i.e., wA = wB. Considering the dynamics of circlar motion in the cases, 2 mwA R =

GMm 2

R (Q GM = gR2)

Similarly wB = And wA = wB =

Þ wA =

CHEMISTRY 31.

(b) – I group destablises carbocation, and since inductive effect decreases with increasing length of carbon chain. Therefore (b) is the correct option.

32.

(a) Suppose there are x-atoms of N. then % of N = 28.9 =

33.

g R

g R

x ´ 14 ´ 100 Þ x = 4 194

(b) The main gas responsible for the global warming is carbon dioxide. Other contributors include methane released from landfills and agriculture, nitrous oxide from fertilizers etc.

34. (a)

is aromatic + 2pe–

9.8

(4 ´ 0 + 2) pe– in conjugation

6.37 ´106 = 124 × 10–5 rad s–1 Now wAE = wA – wE = 124 × 10–5 – 7.3 × 10–5 = 116.7 × 10–5 rad s–1 wBE = (velocity of B relative to E) = wB – (–wE) = wB + wE = 131.1 × 10–5 rad s–1 1 mω 2ΒΕ r 2 K 131.32 Therefore, B = 2 = 1.27 = 2 KA 1 mω 2ΑΕ r 2 116.7 2 30.

4T , is more for the r

Therefore, when valve T is opened, then air rush from smaller to the bigger bubble. The smaller bubble shrinks and the bigger bubble expands.

é R nR ù + êúm t GM GM dv ë û =t = a ò dt Þ a v2 0

or, t =

29.

-1

Therefore Pexcess =

(a) When the stop cock T is opened, then air will rush from higher pressure region. For the smaller bubble, radius is less i.e. r < R.

and

are antiaromatic

–

+

4pe–

8pe–

(4 × 1) pe– (4 × 2) pe– is non aromatic due to the presence of sp3 carbon. 35.

(d) The subshell, whose n value is 3 and l value is 2, is 3d. 3d-subshell contains maximum 10 electrons.

SOLUTIONS – MOCK TEST-5

151

36. (d) Higher the negative value of reduction potential, more will be the reducing power. According to the given information, reduction potential of Fe2+, Al, and Br – are –0.77V, + 1.66V and –1.09V respectively. Hence, The correct order is Al < Fe2+ < Br– 37. (d) Tautomerism is exhibited by the oscillation of hydrogen atom between two polyvalent atoms present in the molecule. As (d) has a-hydrogen atom ther efore it shows tautomerism whereas other structures do not. O R CH2 N Nitro form

OH R CH

O

N

aci-Nitro form

O

38. (a) The beans are cooked earlier in pressure cooker because boiling point increases with increasing pressure. 39. (d) Order of acidic strength is CHºCH > CH3—CºCH > CH2=CH2 40. (b) Milliequivalents of NaOH = 10 × 0.1 = 1 Milliequivalents of H2SO4 = 10 × 0.05 = 0.5 2NaOH + H 2SO 4 ¾¾ ® Na 2SO4 + H 2O 2 equ.

41.

42.

1 equ.

\ 1 equivalent of NaOH reacts with 0.5 eq. of H2SO4 to give neutral (pH = 7) solution. (d) (i) dsp2 hybridisation, number of 90° angle between bonds = 4 (ii) sp3d or dsp3 hybridisation, number of 90° angle between bonds = 6 (iii) sp3d2 hybridisation, number of 90° angle between bonds = 12 (c) The given reaction is exothermic in nature, due to release of heat. An exothermic reaction is favoured by low temperature to proceed in forward direction.

43. (d)

44. (d) Dipole moment cancels each other in BeF2 because of linear shape and thus it is nonpolar, whereas H2O is V-shape and is polar. F — Be — F BeF2

O H

H H 2O

45. (a) The stability of dihalides (MX2) increases down the group. Except C and Si, the other members form dihalides. BIOLOGY 46. (c) In a given DNA segment, 200 guanine bases will pair with 200 cytosine bases. Similarly, 200 thymine bases will pair with 200 adenine bases. Thus, total number of nucleotides in the given segment of DNA is 800. 47. (a) During meiosis I, DNA content gets reduced by half and after meiosis II further it reduced by half. 48. (d) The organs which are present in reduced form and do not perform any function in the body but correspond to the fully developed functional organs of related animals are called vestigial organs. They are believed to be remnants of organs which were complete and functional in their ancestors. Some of these are nictitating membrane, muscles of pinna (part of external ear), vermiform appendix, caudal vertebrae (also called coccyx or tail bone), third molars (wisdom teeth), hair on body, and nipples in male. 49. (c) Oxidative phosphorylation refers to the synthesis of ATP from ADP and inorganic phosphate by chemiosmosis. It occurs with the help of energy obtained from oxidation of reduced enzymes formed in cellular respiration.

EBD_7839 152 50.

51.

52.

53.

54.

KVPY-SA (a) A histone octamer is a complex of eight positively charged histone proteins (two of each H2A, H2B, H3 and H4) that aid in the packaging of DNA. Negatively charged DNA wraps around the histone octamer to form the nucleosome. The DNA is held there by ionic bonds. Linker histone H1 binds to each nucleosome where the DNA enters and exits and this draws a string of nucleosomes closer together to form the 10 nm fibre. The nucleosomes in chromatin are seen as beads-on-string structures when viewed under electron microscope. (b) Due to direct chemical control on respiratory centres, CO2 stimulates respiratory centres in CNS. (d) Cytochromes are small proteins (intrinsic membrane proteins) that contain a cofactor, haem which holds an iron atom. The iron carries electrons between + 2 and + 3 oxidation states. These form a part of electron transport chain in mitochondria and chloroplast and act as an electron transporter or electron acceptor in respiration and photosynthesis. (a) Anaphase I with its feature is correctly matched. Metaphase I: Spindle apparatus appears and the chromosomes are arranged on equatorial plate, with the centromeres towards the pole. Spindle fibres become attached to the centromeres of the two homologous chromosomes. Interphase: It is the time during which cell is preparing for division by undergoing both cell growth and DNA replication. Prophase I: It is the lengthy phase when compared with mitotic phase. It is subdivided into 5 sub-phases: leptotene, zygotene, pachytene, diplotene and diakinesis. (c) Sexually transmitted diseases (STDs) are caused by infections that are passed from one person to another during sexual contact. Most STDs initially do not cause

55.

(a)

56.

(b)

57.

(b)

58.

(a)

59.

(a)

60.

(d)

symptoms. Symptoms and signs of disease may include vaginal discharge, penile discharge, ulcers on or around the genitals and pelvic pain. B-lymphocytes are involved in humoral immunity by producing antibodies. Depending on the shifting of plane of polarised light, the sugars are denoted as ‘L’ or ‘D’ forms. The D and L forms are mirror images based on the penultimate carbon atom. Splitting of water in photosynthesis is called photolysis. Sucrose is composed of glucose and fructose. In 1953, Stanley Miller and Harold Urey created, in the laboratory, conditions comparable to those of early Earth, with water vapour, hydrogen, methane, and ammonia. The Miller-Urey apparatus produced a variety of amino acids and other organ ic compounds found in living organisms today. Silicon, like carbon, needs four electrons to fill its outer shell and could, therefore, form similar compounds. PART-II MATHEMATICS

61. (a) DABC is rt D, as AB is diameter \ ÐABC = 90° – 24° = 66° Also ÐADE = ÐEDC = a by Alternate segment theorem ÐADE = ÐDCE = a Q ÐACB = 90° \ a + q = 90° ...(1) also In DDBC, ÐD + q + ÐB = 180° Þ ÐD + q = 114° Þ ÐD = 114° – q Q AB is a straight line \ 2a + 114° – q = 180° Þ 2a – q = 66° ...(2)

SOLUTIONS – MOCK TEST-5

153 Also, ar DADE = ar DAQD, as they both lie on same base. AD and between the two parallel lines.

from (1) and (2) 3a =156° Þ a = 52° Þ q = 24° 62. (c)

A

\

ar (ABCDE) = ar (DAPC) + ar (DACD) + ar (DAQD)

\

ar (ABCDE) = ar (DAPQ).

65. (c) Highest power of 10 in N! is 28 hence 120 £ N < 124

P

If N = 120, 121, 122, or 123 then (N + 1)! Will have Highest power of 10 in N! is 28 and if N = 124 then (N + 1)! Will have highest power

Q E

H

of 10 as 31.

B D R Q Q, R are mid points of AB & BC.

\ We can say QR || AC (By converse of

PHYSICS

C

66. (b) For the deflection of spring by x, the energy stored in the spring,

Mid-pt. theorem)

x

\ÐBHR = ÐBEC = 90º

ò Fdx

U=

0

Q QR is a line

\

ÐBHQ + ÐBHR = 180º

Þ

ÐBHQ = 90º

x

= ò 300 x 2 dx = 300 0

also ÐPQR = ÐBHQ = 90º(alternate angles)

Thus by conservation of mechanical energy, we have

as Q, P are mid points of AB & AH 63. (b) Let A = {q : sin q = tan q} and B = {q : cos q = 1} sin q ü ì \ A = íq : sin q = ý cos q þ î = {q : sin q (cos q – 1) = 0}

or

ar DABC = ar DAPC, as they both lie on same base AC and between two parallel lines.

1 mv2 2

100 x3 =

1 × 200 × (1.0)2 2

x=1m

The maximum restoring force acts on the car

For B : cos q = 1 Þ q = p, 2p, 4p,......

64. (a) ar (ABCDE) = ar DABC + ar DACD + ar DADE

100 x3 =

or

= {q = 0, p, 2p, 3p,.....} This shows that A is not contained in B. i.e. A Ë B. but B Ì A.

x3 = 100 x3 3

F= \

300 x2 = 300 (1)2 = 300 N

Deceleration of the car =

67. (c) Here, V1 = 234 ´ V2 = 180 ´

F 300 = = 1.5 m/s2 m 200

5 = 65 ms -1 18

5 = 50 ms - 1 , A = 25 m2 18

EBD_7839 154

KVPY-SA \

Applying Bernoulli’s theorem above and below the wings 1 1 P1 + rv12 = P2 + rv22 2 2

=

69.

(b) Using Snell’s law (at origin),

v2 = 180 km/h

m=

1 1 ( P2 - P1 ) = r (v12 - v22 ) = ´ (652 - 502 ) 2 2 1 -2 = (65 - 50) (65 + 50) = 862.5 Nm 2 \ Upward thrust on the two wings = 862.5 × (25 × 2) This force supports the weight of the aeroplane i.e., m × 9.8 = 862.5 × 50

68.

m=

\

(d) Potential at point A,

\

70.

(potential due to each q =

kq and a

potential due to each – q =

-kq ) a 5

q

2a

2a A

(a)

q

B

1 m2 -1

1

=

1 + e x/ d - 1

d

A

x

0

0

x 2d

x = d ln 4.

Let A be the area of cross-section of the tube. Since temperature is the same, applying Boyle's law on the side AB P × (x × A) = P2 × (x2 × A)... (i) Applying Boyle's law in section CD P × (x × A) = P1 × (x1 × A)... (ii) From (i) and (ii) P1 × (x1 × A) = P2 × (x2 × A) Þ P1x1 = P2x2 where P2 = P1 + Pressure due to mercury column

x

x

P

P B

-

- x /2 d ò dy = ò e dx

5c

–q –q Potential at point B, VB = 0 (\ Point B is equidistant from all the four charges)

=e

dy - x /2 d =e dx

or

or

2kq 2kq a a 5

1 sin 90° or sin q = m sin q

\ tan q =

862.5 ´ 50 = 4400 kg 9.8

VA =

2kqQ é 1 ù 1ú a êë 5û

æ 1 ö 2Qq é 1 ù = ç 4pe ÷ a ê1 ú è ø 5û ë 0

v1 = 234km/h

or

Using work energy theorem, (WAB)electric= Q(VA – VB)

5cm

C

m

x2

P1

x1

P2 D

A

B 30O

Pressure due to mercury column

C

D

SOLUTIONS – MOCK TEST-5 F mg sin 30° Vdg sin 30° = = A A A

P=

=

155

(A ´ 5) ´ dg sin 30° =5 sin 30° cm of Hg A

P2 = P1 + 5 sin 30° = P1 + 2.5

V=

RT 0.082 ´ 373 = = 30.58 L ; 30.6 L P 1

>

74. (b)

>

>

Substituting this value in (iii) P1 × x1 = [P1 + 2.5] × x2 P1 × 46 = [P1 + 2.5] × 44.5 \

P1 =

44.5 ´ 2.5 1.5

Substituting this value in (ii) P×x =

44.5 ´ 2.5 ´ 46 1.5

m-Xylene is more reactive than p-xylene towards electrophilic aromatic substitution, because of stable s-complex. CH3 CH3

é 46 + 44.5 ù 44.5 ´ 2.5 ´ 46 Þ P× ê úû = 2 1.5 ë x + x2 Þ P = 75.4 cm [ Q x = 1 ] 2

CHEMISTRY gg

71. (b) In NH3 central atom is nitrogen which is sp3 hybridized hence, it will be at the centre of tetrahedron with H-atoms at three vertices.

+

+

CH3

CH3

E Stable (One methyl reinforces the other)

E

(One methyl counteracts the other)

75. (d) Since molarity of solution is 3.60 M. It means 3.6 moles of H2SO4 is present in its 1 litre solution. Mass of 3.6 moles of H2SO4 = Moles × Molecular mass = 3.6 × 98 g = 352.8 g

\ 1000 mL solution has 352.8 g of H2SO4 N H

H

H

(Ammonia, NH3)

72. (a) S2O62– : 2x – 12 = –2 Þ x = 5 SO32– : x – 6 = –2 Þ x = 4 S2O42– : 2x – 8 = –2 Þ x = 3 where x represents oxidation state. 73. (c) Volume of 1 mole of an ideal gas at 273 K and 1 atm pressure is 22.4 L and that at 373 K and 1 atm pressure is calculated as ;

29% H2SO4 by mass means 29 g of H2SO4 is present in 100 g of solution. \ 352.8 g of H2SO4 is present in

=

100 ´ 352.8 g of solution. 29

= 1216 g of solution Mass 1216 = = 1.216 g/mL Volume 1000 = 1.22 g/mL

Density =

EBD_7839 156

KVPY-SA BIOLOGY

76.

77.

78.

(d) According to Michaelis-Menten equation, the rate of an enzyme-catalysed reaction depends on the substrate concentration. When the solution is diluted by adding water, the reaction rate is slowed down because it becomes difficult for the enzyme and substrate to come in contact with each other. (b) Parents: AABBCC × aabbcc F1 genotype: AaBbCc If n is the number of heterozygous alleles, Number of F1 gametes = 2n = 23 = 8 Number of F2 genotypes = 3n = 33 = 27 Number of F2 phenotypes = 2n = 23 = 8 (a) Tidal volume (TV) = 500 mL Expiratory residual volume (ERV) = 1000 mL Expiratory capacity (EC) = TV + ERV = 500 + 1000 = 1500 mL

79.

80.

(a) During protein synthesis, the sequence of nucleotides in a messenger RNA (mRNA) molecule specifies the sequence of amino acids in a polypeptide. Ribosomes provide the site for the protein synthesis and tRNA reads the triplet code on mRNA and brings corresponding amino acids to the site of protein synthesis. So, in the given situation, mRNA of human is added so it will specify the sequence of amino acids so the nature of polypeptide synthesised, will be of the nature of body cells of human. (c) Recessive alleles are not lost from the population because they occur in heterozygous individuals who are carriers. Heterozygous and homozygous dominant both individuals will be able to survive and reproduce. In heterozygous individuals, both dominant and recessive alleles segregate during gamete formation and are passed on to offsprin g with equal probabilities.

MOCK TEST-6 3.

PART – I MATHEMATICS 1.

(c) We use the facts that the line joining the midpoints of the sides of a triangle is parallel to the third side and any median of a triangle bisects its area and two triangles having equal bases and bounded by same parallel lines have equal area. (Note: [·] denotes area) C R D

X

S

O

4.

Q

Y

B A P (i) Now BD is parallel to PS as well as OX. So OX is parallel to PS. Hence, [PXS] = [POS]. Adding [PAS] to both sides we get [APXS] = [APOS]. (ii) Now, [APXS] = [APX] + [ASX]

5.

(a) Since p is product of 5 consecutive number. Hence p must be divisible by 5! So p = 5!n = 120n Then q = 120n + 2 Hence q must be even When divided by 4 & 3 it leaves remainder 2 So it is not divisible by 3 Since it is even and it can’t be prime (b) Set is in the form of S = { 20, 21/2, 2, 23/2, ...... 210} Since P has 2 elements – If 1st element is 210 then next element can be any one of 1st 20 elements number of such sets = 20. If 1st element is 219/2 then next element can be any one from 21/2 to 29 number of such sets = 18 If 1st element is 29 then next element can be any one from 21 to 217/2 number of such sets = 16 And so on hence total number of elements = 20 + 18 + 16 + …. + 0 = 110 P

(b)

1 1 = [ ABX ] + [ ADX ] 2 2 1 1 = [ ABC ] + [ ADC ] 4 4

h

1 = [ ABCD ]. 4

1 Hence, by (i), [ APOS ] = [ ABCD]. 2 Similarly, by symmetry each of the areas [AQOP], [CROQ] and [DSOR] is equal to

2.

1 [ ABCD ]. 4 Thus, the four given areas are equal. (b) Since 45! Contains 510 if a = 5 then we can get 5(45!) is completely divisible by 511 (50! Contains 512)

B 17

10 A a Q

c

a

S 21

b

R

Let ABCD side = a – AB = BC = CD = AD QD = c, CR = b

DPQS ~ DAQD DPRS ~ DBRC PQ PS PR PS Þ = = ... (1) Þ ...( 2) AQ AD BR BC

EBD_7839 158

KVPY-SA s=

10 + 17 + 21 = 24 2 24 ´ ( 24 - 10 )

area ( DPQR ) =

( 24 - 17 )( 24 - 21)

D = 84 1 Þ ´ h ´ 21 = 84 Þ h = 8 2 From (1)

10 2

a +c

2

=

8 aÞ

Þ a2 + c2 =

a 2 + c2 =

7. 10 a 8

100a 2 64

6a 36a 2 Þ c= 8 64

Þ c2 = From (2)

17 a 2 + b2

=

8 a Þ

17 a a +b = 8 2

8.

2

15a 289a 2 Þ b= 8 64 \ a + b + c = 21

Þ a2 + b2 =

Þ a+ Þ

6a 15a + = 21 8 8

29a 168 = 21 Þ a = 8 29

Perimeter = 4a =

6.

4 ´ 168 = 23.2 29

(d) It is given that x is odd, divisible by 3 but not divisible by 7. Consider the number of multiples of 3 between 100 and 200 (both included) is [200/3] – [100/3] = 33. Out of these 33, 16 are odd and 17 are even [because the starting multiple 102 and last multiple 198 both are even]

9.

Now we need to find the number of multiples of 21 between 100 and 200, 200/21 quotient is 9 and 100/21 quotients is 4, hence number of multiples of 21 between 100 and 200 is 9–4=5 Of these 5, 3 are odd and 2 are even. Here we are looking for odd multiples of 3, not divisible by 7, And it is 16 – 3 = 13 (a) Since abcd is divisible by 4 hence cd must be divisible by 4 say 4k. Now consider acdb it is divisible by 8 that means cdb is divisible by 8 or (10cd) + b is divisible by 8 but we know that cd = 4k hence cdb = 40k + b it is divisible by 8 that means b = 0 or 8. Since we need maximum value of a + b + c + d hence b = 8, and cd = 88 (8 + 8 = 16 is the maximum value). Then required maximum value will be 9 + 8 + 8 + 8 = 33 (d) The easiest way to solve this question is that take some values of n such th at seed(n) =9 these are 18, 27, 36 etc, that means multiples of 9, so what actually we need to find is the number of multiples of 9 less than 500, So 500/9 and quotient is 55 hence we will have 55 such numbers. (c) Q All 3 segments are congruent Area of 1 segment (PDC) is Area segment (PDC)

=

1 2 æ 60º ´p ö - sin 60º÷ ( 2) çè ø 2 180º

=

æp 1 3ö ( 4) ç - ÷ 2 è3 2 ø

Q 3 segments are there

æp 3ö \ Total area = 3 ´ ( 2) ç ÷ 3 2 è ø

(

= 2p - 3 3

)

SOLUTIONS – MOCK TEST-6 10. (d) f(1) = 1; f(2) = 3; f(3) = 5; f(4) = 7; f(5) = 9 f(1) – 1 = 0 f(2) – 3 = 0 f(3) – 5 = 0 f(4) – 7 = 0 f(5) – 9 = 0 Let there is a new polynomial p(x) = f(x) – (2x – 1) p(x) has x = 1, x = 2, x = 3, x = 4, x = 5 as its factors. \ P(x) = (x – 1)(x – 2)(x – 3)(x – 4)(x – 5) = f(x) – (2x – 1) f(x) leading coefficient is 2009, p(x) leading coefficient will also be 2009. Þ p(x) = 2009(x – 1)(x – 2) (x – 3)(x – 4)(x – 5) = f(x) – (2x – 1) for f(6), substituting x = 6 Þ f(6) = 2009 × 5! + (12 – 1) = 241091 11. (a) If ab is a two digit number then as per the given condition a > b When a = 1 then possible number 10 When a = 2 then possible numbers are 20, 21 (i.e. 2 such numbers) When a = 3 then possible numbers are 30, 31, 32 (i.e. 3 such number) Similarly when a = 9 then possible numbers are 90, 91,…. 98 (i.e 9 such numbers) Hence total number of such two digit numbers are 1 + 2 + 3 + … + 9 = 45. And 45 two digit numbers will take 90 digits, but we have to consider only 75 digits. So let us consider 1 + 2 + 3 + … + 8 = 36 and 36 two digit number will use 72 digits. Hence if N has 75 digits then after 72 digits numbers that starts with 9 will appear so next numbers will be 90, 91 etc, Hence N = 102021303132 …… 87909 and last four digit is 7909.and hence remainder when 7909 divided by 16 remainder is 5. D

12. (c)

Y

N

A

C

X

M

B

159 let the area of parallelogram ABCD = 16 sq. units. Construction 1. from Y, draw a line parallel to AD. which cuts AB at M We have YMAD as a parallelogram 1 (ar of ABCD) 2 \ ar YMAD parallelogram = 8 sq. units. Q AY is the diagonal of parallelogram YMAD \ ar DADY = 4 sq. units. Construction 2. draw a line parallel to AB from X. which meets AD at N Similarly, we get ar DAXB = 4 sq. units. Construction 3. Join D to B, as DB is diagonal, it will divide parallelogram in equal area. \ ar DBDC = 8 sq. units Q X and Y are mid points

Whose area is

\ ar DXYC =

1 (ar DBDC) = 2 sq. units 4

now ar (DAXY) = ar (ABCD) – ar (DADY) – ar (DXYC) – ar (DAXB) ar D(AXY) = 16 – (4 + 4 + 2) = 6 sq. units, 3 ar (ABCD) 8 13. (c) If x e – 0, then the given equation assumes the form, x2 + (a – 5) x + 1 = 0 …..(1) If x < 0, then it takes the form, x2 + (a + 1) x + 1 = 0 …..(2) For these two equations to have exactly three distinct real solutions we should have (I) either (a – 5)2 > 4 and (a + 1)2 = 4; (II) or (a – 5)2 = 4 and (a + 1)2 > 4. Case I : From (a + 1)2 = 4, we have a = 1 or –3 But only a = 1 satisfies (a – 5)2 > 4. Thus, a = 1 Also, when a = 1, eq. (1) has solutions x = 2 + ; and eq. (2) has solutions x = –1,–1. As 2 ± > 0 and –1 < 0, we see that a = 1 is indeed a solution.

\ ar DAXY =

EBD_7839 160

KVPY-SA

Case II: From (a – 5)2 = 4, we have a = 3 or 7. Both these values of a satisfy the inequality (a + 1)2 > 4. When a = 3, eq. (1) has solutions x = 1, 1 and eq. (2) has the solutions x = –2 ± As 1 > 0 and –2 ± < 0, we see that a = 3 is in fact a solution. When a = 7, eq. (1) has solutions x = –1, –1, which are negative contradicting x e–0. Thus, a = 1, a = 3 are the two desired values. 14. (a) Area of outer circle = p.OA2 and area of inner circle = p×OD2 As given p×OA2 = 4×pOD2 Þ

OD 2 OA2

Hence, side AB = 2 ´

sin ÐOAD =

3 3 36 9 3 ´ side2 = . ´ = 4 4 p p 15. (c) The given expression can be written as 3sec

2

x -1

y2 -

2y 2 + £1 3 9 2

Þ 3sec

2

x

1ö 1 æ çè y - ÷ø + £ 1 3 9

where sec2 x ³ 1 2

OD 1 = \ÐOAD = 30° OA 2

So, we should have and sec 2 x = 1 and y -

1 3 \ Total number of solutions is four.

O

PHYSICS

C 16.

[Since, ÐBAC = 2ÐOAD ] Hence, DABC is equilateral. The next step is to find side AB and OD.

12 1 12 and OD = 2 p p

Using Pythagoras theorem, we get, DA2 = OA2 – OD2 ; =

12 1 12 9 - ´ = p 4 p p

(a) Net power given to N2 gas = 100 – 30 = 70 cal/sec The nitrogen gas expands isobarically. \ Q = nCP

17.

dT dt

or 70 = 5 ´ 7 R dT 2

dT = 2K / sec dt

\

Since pr 2 = 12, we get,

3 p

1 =0 3

Þ x = 0, p, 2p, 3p and y =

A

Þ DA =

1ö 1 1 æ çè y - ÷ø + ³ 3 9 3

Þ 3sec 2 x ³ 3 and

B

r = OA =

p

=

\ ÐOAD = 30°

D

p

6

=

Area of DABC

1 OD 1 = Þ 4 OA 2

=

3

(d) Given, F = cs -1/3 Acceleration, a = or v

F c -1/3 = s m m

dv c -1/3 = s ds m

v

s

0

0

c -1/3 ds or ò vdv = m ò s

dt

SOLUTIONS – MOCK TEST-6

161 2 2 2 2 at x = 0, F = r Aw l and f = F = rw l

v 2 c s2/3 = 2 m 2/3

8

3c 2 /3 s or v = m

or

v = ks

Now power = Fv = cs -1/3 ´ ks1/3 = s0

8

Rupture of rod will occur when f = s

2

1/3

A

\

rw 2 l 2 =s Þw = 8

=

8s rl 2

or n

1 8s 2p rl2

19. (b) Angular momentum of M2 about O is

18. (a) O

The stresses are zero at the free end and maximum at the axis. Therefore the rod will ruptures at the middle. Let us consider an element of rod at a distance x from the axis, the mass of element dm = rAdx Applying Newton’s second law

M2vR

L2 = M2vR clockwise Angular momentum of M1 about O is L1 = M1vR clockwise Ldisc = Iw clockwise \ Ltotal = L1 + L2 + Ldisc 1 MR2w 2 Now, v = Rw as rope does not slip

= M2vR + M1vR +

1 ù é \ L = ê M 2 + M1 + M ú vR 2 û ë F - ( F + dF ) = (dm)an

or - dF = (rAdx)w2 x where w is the rotation speed. \

ò

ò

= -rAw2

x2 +C 2

2 F = - dF = - (rAdx)w x

at x = l/2, F = 0 \

C=

rAw 2l 2 8

Now F = r

Aw2 æ l 2 2ö ç -x ÷ 2 è 4 ø

Therefore, k = 2 20. (c) The ball will stop after a long time. The final displacement of the ball will be equal to the height. The motion is first accelerated, then retarded, then accelerated and so on. Hence the correct graph is (c). 21. (c) For isotopes Z is same and A is different. Therefore the number of neutrons A-Z will also be different. 22. (a) In ohm's law, we check V = IR where I is the current flowing through a resistor and V is the potential difference across that resistor. Only option (a) fits the above criteria. Remember that ammeter is connected in series with resistance and voltmeter parallel with the net resistance.

EBD_7839 162

KVPY-SA A

A V

V

23. (d)

I1 D12 t1 = I 2 D22 t2 [D = diameter of aperture]

Here, D is constant and I = So,

Þ 24.

25. 26.

L r2

(c) i = neAvd and vd µ E (Given) or, i µ E i2 µ E i2 µ v Hence graph (c) correctly depicts the V-I graph for a wire made of such type of material. 29. (b) x = h cot q 28.

L1 L ´ t1 = 22 ´ t2 2 r1 r2

60 120 ´ 10 = ´ t Þ 20sec = t (2) 2 (4) 4

(b) In the space, the external gravity is absent, but there will be a very small gravitational force between the astronauts, due to which both will move toward each other with a very small acceleration so, the best correct answer should be (b). (d) This is the defect of hypermetropia. (b) Using Newton’s second law,

\

dx = h d (cot q) dt dt

= h (–cosec2q)

æ - dq ö or v = h cosec2q ç è dt ÷ø

30.

= (8 × 103) cosec260° × 0.025 = 266.67 m/s = 960 km/h (a) Here, Mw = 28, T = 300 K, R = 8.31 × 107 erg mol–1 K–1 Kinetic energy of 2g of nitrogen, = 2´

27.

3 RT 3RT 3 ´ 8.31 ´ 10 7 ´ 300 = = 28 2 Mw Mw

= 267.1 × 107 erg

kx = mw 2 ( l + x )

or

kx = mw 2 l + mw 2 x

\

x=

mw 2 l

dq dt

( k - mw ) 2

(a) Fractional error =

DM as DM A = DM B M

DM A DM B as M B > M A Þ M > M A B

CHEMISTRY 31. 32.

(b) Both He and Li+ contain 2 electrons each, therefore their spectrum will be similar. (a) Compounds (b), (c) and (d) are cyclic planar and have 6p electron (4×1+2pe–) in complete conjugation and thus are aromatic.

gg N H

:

:

SOLUTIONS – MOCK TEST-6

163 O

HOCH2

Compound (a) is anti-aromatic as

H

OH H

H

it has 4p electrons in complete conjugation and doesn’t obey Hückel rule. 33. (b) For compounds containing cations of same charge, lattice energy increases as the size of the cation decreases. Thus, NaF has highest lattice energy. The size of cations is in the order : Na+ < K+ < Rb+ < Cs+ 34. (c) Hydrogen is a much lighter element than alkali metals or halogen. 2Ca + O2 ¾¾ ® CaO (ionic compound)

35. (a)

CaO + H 2 O ¾¾ ® Ca(OH) 2 ˆˆ† 2NO; Dn = 0 36. (c) N 2 + O 2 ‡ˆˆ

37. (d) For p sub-shell, m = – 1, 0, + 1. Therefore, m = – 2 orbital will not be present in p sub-shell. 38. (d) Order of screening effect is s > p > d > ¦. 39. (b) BH3 has sp2 hybridization and hence does not have tetrahedral structure while all others have tetrahedral structures. 40. (d) Isothermal expansion PVm = K (graph - C) K P = V (graph - A) m

U

P Vm

Vm

Therefore (d) is not correct representation. 41. (c) Diamond is a crystalline solid which contains only carbon-atom. 42. (a) M =

Wt ´ 1000 5 ´ 1000 = = 0.5 M.Wt ´ V 40 ´ 250

43. (b) RNA has D (–) – Ribose and the DNA has 2–Deoxy D (–) – Ribose as the carbohydrate unit.

H OH OH Ribose

5

O

HOCH2 4

H

H

3

OH

OH H 2

1

H

H

2-Deoxy ribose

From the structures it is clear that 2nd carbon in DNA do not have OH group. 44. (c) KHSO3 furnish H+ ions in solution, therefore, it is acidic. 45. (c) It has four O atoms as peroxide with oxidation number –1 and one O atom with oxidation number –2. Hence, x + 4(–1) + 1 (–2) = 0 or x = +6 BIOLOGY 46. (c) The normal range of pH of arterial blood is 7.35 – 7.45. As the pH decreases (< 7.35), it implies acidosis, while if the pH increases (> 7.45), it implies alkalosis. 47. (c) Plants require a periodic exposure to light to induce flowering. This light period is called photoperiod. On the basis of photoperiod, some plants are long day plants, some are short day plants and some are day neutral plants. 48. (c) Yellowing and blackening of Taj Mahal at Agra is due to SO2 and other pollutants released by Mathura refinery. 49. (b) Magnesium is an essential element. It is present in enzymes and every cell type. These en zymes eg. hexokinase, phosphofructokinase etc., require magnesium ion for their catalytic action. 50. (d) Adrenal gland, located on top of each kidney, does not involve in the process of digestion. It is an endocrine gland which secretes hormones like adrenaline and the steroids aldosterone and cortisol.

EBD_7839 164

51.

KVPY-SA

(b)

52.

(a)

53.

(c)

54.

(c)

55.

(d)

56.

57.

58.

(d)

(c)

(a)

Liver is involved in the process of digestion by secreting bile which helps in the emulsification of fats. Pancreas secretes digestive juices containing many digestive enzymes. It plays an essential role in converting the food we eat into fuel for the body’s cells. Although contraction of smooth muscle is controlled differently from that of skeletal muscle but these muscle contains actin and myosin. Unlike cardiac and skeletal muscles, smooth muscles do not contain the calciumbinding protein troponin. Myoglobin is the main oxygen-carrying molecule present in skeletal muscle. Endospores are dormant, tough, nonreproductive structures produced by small number of bacteria. The primary function of these is to ensure survival of bacterium through extreme environmental stress. Food (carbohydrates) is translocated from leaves to roots along the phloem in form of sucrose. Glucose and galactose are stereoisomers (have atoms bonded together in the same order, but differently arranged in space). They differ in their stereochemistry at carbon 4. Bamboo and grasses elongate by the activity of intercalary meristems. These meristems are found at the base of leaves of many monocotyledons. Intercalary meristems are mainly responsible for the vertical growth of plant organs. Amylopsin is pancreatic amylase which hydrolyses polysacchar ides (except cellulose) into disaccharides (mostly maltose) in alkaline medium. Micronutrients are the essential elements needed by organisms in small quantities. Micronutrients include boron (B), chlorine (Cl), molybdenum (Mo), sodium (Na), manganese (Mn), copper (Cu) and Zinc (Zn).

59.

60.

(a) Hormones are chemical messengers which are secreted from endocrine glands. They are secreted directly into the blood which carries them to organs and tissues of the body. These hormones stimulate the glands to do their functions. (c) Plants are autotrophs and synthesise their food by the process of photosynthesis with the help of chloroplast (plastid). Plastids are not found in animals. PART – II MATHEMATICS

61. (a) ÐA = ÐB = ÐC = 60° MN = QR = 2r in quadrilateral BTQM ÐB + ÐBMQ + ÐMQT + ÐQTB = 360° 60° + 90° + 90° + ÐQTB = 360° Þ ÐQTB = 120° DBTQ @ DBQM A

P

a

r

Q r 60° r

r

T

B

a

r r

x

M

R

r 90° 2r N C a

Þ ÐTQB = ÐMQB = 60° In DBQM tan 60° = 3=

BM x = QM r

x Þx=r 3 r

Similarly CN = r 3 BC = M + MN + CN = a 2x + 2r = a Þ 2(x + r) = a

Þ 2(r 3 + r) = a Þ a = 2r( 3 + 1)

SOLUTIONS – MOCK TEST-6

165

62. (a) ÐDAB + ÐABC = 180° QConsecutive interior angles. ÐABC = 180 – ÐDAB ..... (1) Q SRDA is a square ÐQAB = 90° ..... (2) ÐSAD = 90° ..... (3) Q ABPQ is a square ÐSAD + ÐDAB + ÐQAB + ÐSAQ = 360° 90° + ÐDAB + 90° + ÐSAQ = 360° ÐDAB + ÐSAQ = 180° ..... (4) ÐSAQ = 180° – ÐDAB .... (5) By (1) & (5) ÐSAQ = ÐABC 63. (d) P = {q : sin q - cos q = 2 cos q} sin q = ( 2 + 1) cos q , tanq =

Þ

Þ

Þ

Þ

1 1 1 × m (a + x) = × × (m + n) (a + b) 2 2 2 2 ( a + x) a+b 2 ( a + x) a+b

=

m+n m

=1+

n m

2 ( a + x) a + x =1 a+b b+ x

(from (1))

Þ 2 (a + x) (b + x) – (a + x) (a + b)

2 +1

= (a + b) (b + x) Þ 2x2 = a2 + b2

Q = {q : sin q + cos q = 2 sin q} cos q = ( 2 - 1)sin q or tan q =

1 ar (ABCD) 2

Q ar (ABNM) =

2 +1

Þ 2MN2 = AB2 + CD2 65. (b) We know that only perfect squares can have odd number of factors.

\ P=Q 64. (c) Let AB = a, CD = b, MN = x

Hence set S can be defined as = {100, 121, 144, …., 961} Product of all the elements is 100 × 121 × 144 × …. × 961 = 102 × 112 × 122 × …. × 312

Let distance between AB and MN be “m” and let distance between CD and MN be “n”. 1 given: ar (ABNM) = ar (ABCD) 2

Þ 2ar (ABNM) = ar (ABNM) + ar (CDMN) Þ ar (ABNM) = ar (CDMN) Þ

Since 12 = 22 × 31 Highest power of 2 in 31! is 26 and that in 9! Is 7 hence highest power of 2 in (31!)/(9!) is 26 – 7 = 19 and so highest power of 22 in (31!)/(9!) is 19. Similarly highest power of 3 in 31! Is 14 and that in 9! Is 4 hence highest power of 3 in (31!)/(9!) is 14 – 4 = 10 and so highest power of 3 in [(31!)/(9!)]2is 20.

1 1 × m (a + x) = × n (b + x) 2 2

m b+ x Þ n =a+x

(10 × 11 × 12 × .. × 31)2 = [(31!)/(9!)]2

...(1)

Therefore, highest power of 12 is 19

EBD_7839 166

KVPY-SA 68.

PHYSICS 2 66. (d) Terminal velocity, v = 2r (r – s) g ...(i) h 9

But we know, v = \

T = 2p

s t

\ T 2 = 4p 2

l g

Dg Dl DT ´ 100 = ´ 100 + 2 ´ 100 g T l Case (i) Dl = 0.1 cm, l = 64cm, DT = 0.1s, T = 128s Þ

Given s = 2 × 10–2 m, t = 1h = 3600 s \ h = 1 ´ 10 –2 poise = 1× 10–3 kg m–1s–1

Substituting given values, we get

\

9 ´ 2 ´ 10 –2 1 ´ 10 –3 ´ 2 ´ 3600 1.8 ´ 103 – 1 ´ 103 ´ 10

(

l g

l Þ g = 4p 2 2 T

s 2 2 (r – s) g h 9s = r Þ r2 = t 9 h 2t (r – s) g

r2 =

(b) The time period of a simple pendulum is given by

)

Dg ´ 100 = 0.3125 g

=

9 1 ´ ´ 10 –10 36 8

Case (ii) Dl = 0.1 cm, l = 64cm, DT = 0.1s, T = 64s Dg \ ´ 100 = 0.46875 g

=

1 ´ 10 –10 32

Case (iii) Dl = 0.1 cm, l = 20cm, DT = 0.1s, T = 36s

\ r=

\

100 ´10 –6 m = 1.77 ´10 –6 m and 32

Diameter D = 2r = 2 × 1.77 mm = 3.54 mm

Clearly, the value of

67. (d) Let m and d be the mass and diameter of the sphere, then the density r of the sphere is given by r=

mass m 6m = = volume 4 æ d ö 3 pd3 pç ÷ 3 è 2ø

Taking log and differentiating partially we get dr dm 3d (d) = ´ r m d æ dr ö dm 3d (d) = + for ç ÷ or m d è r ø max

æ dr ö dm 3d (d) = ´ 100 + ´ 100 çè r ´ 100ø÷ m d max

= 2 + 3 (3) = 11%

Dg ´ 100 = 1.055 g

69.

Dg ´ 100 will be least g

in case (i). (c) Total heat given to the system is Q – Q'. So from first law of thermodynamics, Q – Q' = total work done by the gas in both the chambers (W) + change in internal energies of both the gases (DU) ... (i) Here W = sum of potential energies stored in the springs é 1 æ l ö 2 ù kl 2 W = 2 ê kç ÷ ú = 4 êë 2 è 2 ø úû

... (ii)

Since the temperature of left part remains constant (piston does not conduct heat), internal energy of left part does not change. DU of right part can be given as DU = nC V DT =

3 nRDT 2

... (iii)

SOLUTIONS – MOCK TEST-6

167

DT can be found from the condition of equilibrium at the end of the process. Pressure on right side = pressure on left side or

nR(T + DT) nRT 2K(l / 2) = + A(l + l / 2) A(l - l / 2) A

3kl 2 + 2T Simplifying this we get, DT = 2nR ...(iv) From equations (i), (ii), (iii) and (iv), we get Q¢ = Q -

æ 3kl ö kl 3 - nR ç + 2T ÷ 4 2 è 2nR ø 2

2

5 2 or Q ¢ = Q - kl - 3nRT 2

70. (a)

i + r = 90°, and Ði = Ðr \ i = 45° Also i + q = 90° \

q = 90° – i = 90° – 45° = 45° Given y2 = 2 x or 2 y

dy =2 dx

\

dy 1 = dx y

or

tan 45° =

\

1 y

y =1 Now x =

y2 12 1 = = 2 2 2

CHEMISTRY 71. (b) Alkynes having terminal hydrogen react with Na in liquid ammonia to yield H2 gas. CH3CH2CºCH can react with Na in liquid NH3. Na in liquid NH3

CH3CH 2 C º CH ¾¾¾¾¾ ® 1 CH3CH 2 C º C – Na + + H 2 ( g ) 2

72. (b) Ba(OH)2 + CO2 ® BaCO3 + H2O Atomic wt. of BaCO3 = 137 + 12 + 16 × 3 = 197 No. of mole =

Wt. of substance Mol Wt.

Q 1 mole of Ba(OH)2 gives 1 mole of BaCO3 \ 0.205 mole of Ba(OH)2 will give 0.205 mole of BaCO3 \ Wt. of 0.205 mole of BaCO3 will be 0.205 × 197 = 40.385 g » 40.5 g. 73. (b) Enthalpy of reaction = B.E(Reactants) – B.E(Products) = éë B.E (C= C) + 4 B.E.(C–H) + B.E.(H - H) ùû - éë B.E.(C - C) + 6 B.E.(C - H) ùû

= [606.1 + (4 × 410.5) + 431.37)] – [336.49 + (6 × 410.5)] = –120.0 kJ mol–1 74. (c) Easily liquefiable gases have greater intermolecular forces which is represented by high value of 'a'. The greater the value of 'a' more will be liquefiability. So, the order is Q < S < P < R. 75. (d) + CH2Cl2 +

anhydrous AlCl3

¾¾¾¾¾¾® CH2 Diphenylmethane

+ 2HCl

EBD_7839 168

KVPY-SA BIOLOGY

76.

77.

(a) Semi-circular canal is interconnected tubes located inside each ear. They are lined with cilia and filled with a fluid known as endolymph. With every movement of head, the endolymph moves the cilia. This works as a type of motion sensor, as the movements of the cilia are communicated to the brain. The vestibule is the central part of the osseous labyrinth and responds to gravity and movements of head. Cochlea is the auditory portion of the inner ear. It is a spiralshaped cavity in the bony labyrinth. Perilymph is an extracellular fluid located within the cochlea (part of the inner ear) in two of its three compartments: the scala tympani and scala vestibuli. The major cation (positively charged ion) in perilymph is sodium. Endolymph is the fluid contained in the membranous labyrinth of the inner ear. (b) A recessive allele is not weaker than the dominant allele. It (recessive allele) does not show its effect (in the presence of dominant allele) because of modified or different enzymes. A recessive allele makes its gene product even when paired with the dominant allele. It is not necessary that dominant allele is always better (in the case of dominant disease).

78.

79.

80.

(d) The periderm consists of three different layers: inner phelloderm or secondary cortex which is inside of cork cambium and composed of living parenchyma cells, middle phellogen or cork cambium which is a meristem that gives rise to periderm and outer phellem or cork which is dead at maturity and composed of suberised cells. (d) The mother would have genotype IAIB and the father would have genotype ii. Thus, the two genotypes possible for their offspring would be IAi and IBi, with equal probability. Hence, the expected phenotypes would be 1 type A : 1 type B. (a) Diagram given in option (a) correctly represents the transformation of P570 into P412 on absorbing a quantum of light along with the release of H+ and transformation of P412 into P570 with the gain of H+. Also the formation of ATP from ADP and Pi on movement of H+ from outside to inside of the cell from F0-F1 system is correctly depicted. Option (b) correctly represents the formation of ATP from F0-F1 system but cyclical changes in bacteriorhodopsin are not correctly shown. Option (c) correctly represents the cyclical changes in bacteriorhodopsin but not the formation of ATP from F0-F1 system. Option (d) neither correctly depicts the cyclical changes in bacteriorhodopsin nor the formation of ATP from F0-F1 system.

MOCK TEST-7 PART-I

6.

MATHEMATICS 1.

æ1 ö æ1 ö ç 2 ED ´ AM ÷ ´ ç 2 BE ´ CN ÷ è ø è ø

(b) For real roots q2 – 4 p r ³ 0 2

æ p+rö ç ÷ – 4 p r ³ 0 (\ p, q, r are in A. P. ) è 2 ø Þ p2 + r2 – 14 p r ³ 0

Þ

=

p p2 – 14 + 1 ³ 0 r r2 2

p æp ö Þ ç - 7÷ - 48 ³ 0 Þ - 7 ³ 4 3. èr ø r

2.

3.

4.

5.

(b) Since unit digit of two digit numbers is same as that of its square hence b = 1, 5 or 6 and square is a four digit number hence ab ³ 32 So we have possible options 41, 35, 45, 36 & 46 By checking the squares of these 5 numbers we will find only 45 Satisfy the given condition with 452 = 2025

7.

3( 3n - 1)

since 3n – 1 is 2 not divisible by 3 hence A has only single power of 3 and for any value of n it cannot be a perfect square. (d) 2x + y = 10 (x + y) + x = 10 x + y is max when x = 0 \ (x + y)max = 10 2x + y = 10 2(x + y) – y = 10 \ x + y is min. y = 0 (x + y)min. = 10 = 5 2 \ (x + y)max. + (x + y)min. = 10 + 5 = 15 (a) Since xi will be either +1 or – 1 hence each term must be either +1/–1 Since the sum of all the terms is zero, out of the n products that are being added, half should be +1 and the other half should be –1, so n must be an even number. (d) The value of A =

(a) Draw AM ^ BD and CN ^ BD area DAED × area DBEC =

8.

1 ED × AM × BE × CN 4

æ1 ö æ1 ö = ç BE ´ AM ÷ ´ ç DE ´ CN ÷ è2 ø è2 ø = area DABE × area DCDE Hence, area DAED × area DBEC = area DABE × area DCDE (d) AE + EB = CE + ED Þ AB = CD D also, OP ^ AB & OQ ^ CD \ AB = 2AP CD = 2DQ Q O Þ 2AP = 2DQ B Þ AP = DQ E P OD = OA = radius A In DOAP C In DODQ OD2 = OQ2 + DQ2...(2) OP2 = OA2 – AP2 2 2 2 OA = OP + AP ...(1) OA2 = OD2 Þ OP2 + AP2 = OQ2 + DQ2 (Q AP = DQ, \AP2 = DQ2) Þ OP2 = OQ2 Þ OP = OQ (b) In right angled DABC, AB 2 + AC 2 = BC 2 [By Pythagoras theorem] \BC = 5 units Area of shaded region = area of semicircle on side AB + area of semicircle on side AC – area of semicircle on side BC + area of DABC Area of semicircle on side AB = =

pr 2 2

22 9 1 99 ´ ´ = sq. units 7 4 2 28

EBD_7839 170

KVPY-SA Area of semicircle on side AC = =

From equation (1) & (2) b + 2(3b) = 180

22 1 ´ 4´ 7 2

44 sq. units 7

b=

Area of semicircle on side 22 25 1 275 BC = ´ ´ = sq. units 7 4 2 28

1 Area of DABC = ´ AB ´ AC 2

æ 180 ö 5 ÐAQP = 180° – 2 ç ÷= p è 7 ø 7 10.

(b) We have, area of square metal plate = 40 × 40 cm 2 = 1600 cm2 Area of each hole

1 = ´ 3 ´ 4 = 6 sq. units 2

Now, Area of shaded region =

= pr 2 =

= 441 ´

= (1600 – 346.5) cm2 = 1253.5 cm2

b

2b

B

2b a

11. b

Q

11 2 cm = 346.5 cm 2 14

Hence, area of the remaining square plate

A

(d)

2

22 æ 1 ö 11 ´ ç ÷ cm2 = cm2 7 è2ø 14

\Area of 441 holes

99 44 275 + +6 28 7 28

Area of shaded region = 6 sq. units 9.

180 7

a

P

(c)

Let A had a rupee and b paisa, and B had c rupee and d paisa Initial amount with A = 100a + b paisa and that with B = 100c + d paisa

a

C

In DABC AB = AC Þ ÐC = ÐB Þ ÐB = ÐC = a By angle sum properly in DABC, b + a + a = 180 Þ b + 2a = 180° ...(1) In DQPB Þ ÐQPB = 180 – 4b Since ‘APC’ is a straight line Þ 180 – 4b + a + b = 180 Þ a = 3b ...(2)

After interchanging amount with A = 100b + a paisa and that with B = 100d + c paisa. Increase in amount of A = 100b+a–100a–b = 99(b–a) Decrease in amount of B = 100c + d–100d–c = 99(c–d) these two has to be equal hence 99(b–a) = 99(c–d), so b + d = a + c, the maximum value that a, b, c, and d can take is 99 as there is no paisa more than 99. So maximum value with A and B is 99.99 but in that case B is giving zero amount to A and that is not the case here so for maximum sum A had 98.99 and B had 99.98 and their sum is 198.97.

SOLUTIONS – MOCK TEST-7

171

12. (b) Truncated cone ® Frustum

13. (c) If elements are not repeated, then number of elements in A1 È A2 È A3 ... È A30 is 30 × 5. But each element is used 10 times, so 30 ´ 5 = 15 S= 10 If elements in B1, B2 , ....., Bn are not repeated, then total number of elements is 3n but each element is repeated 9 times, so S=

Volume, V =

(

ph 2 2 r1 + r2 + r1r 2 3

New diameter, d2¢ = d 2 +

= 100 +

)

3n Þ n = 45 9 14. (b) If n = 10a + b then the ratio (10a + b)/a + b is maximum. Maximum ratio is 10 & that comes when b = 0 e.g 10,20.30…….90. So 9 such number exists.

Þ 15 =

21 ´ d2 100

21 ´ 100 = 121 mm 100

15. (b)

121 mm 2 r1, h ® same r2¢ =

New volume, V ¢ = V + Þ

(

ph 2 r 1 + r ¢22 + r1r2¢ 3

Þ =

121r 12

+ 100r ¢22

+ 121r 22

Þ 100 ´

–1

f (B)

Þ f ( f -1 ( B )) Ì B

)

(

121 2 2 r 1 + r2 + r1r2 100

+ 100r1r2¢

+ 121r1r2

121 121 12100 ´ + r1 2 2 2 2

= 21r 12 + 121 ´ ( 50 ) + 121 ´ 50r1

Þ (121)2 × 25 – 121 × (50)2 = 21r12 Þ 63525 = 21r12 Þ r1 = 55 mm

Y B f (A)

)

(

100r 12

f ( f -1 ( B )) = { f ( x ) : x Î f -1 ( B )}

= { f ( x) : f ( x ) Î B} X A

21 121V ´V = 100 100

121 ph 2 r 1 + r 22 + r1r2 = 100 3 2 2 Þ r 1 + r ¢2 + r1r2¢ =

3n 9

)

Now if x Î B Ì Y Þ x Î Y . It may happen that f–1(x) does not Exist in x as function is not given to be subjective. f ( f -1 ( B )) ¹ B .

Also, f -1 ( f ( A)) = {x Î X : f ( x ) Î f ( A)} but f ( A) = { f ( x ) : x Î A} From f

-1

above,

we can't

conclude

( f ( A) Ì A.

If the function is non-injective, then it may happen that x Ï A and f ( x ) Î f ( A). Þ f -1 ( f ( A) Ë A Þ f -1 ( f ( A)) ¹ A.

EBD_7839 172

KVPY-SA PHYSICS

16. (b)

17. (a)

19. (b)

In portion OA, slope of x – t graph is decreasing. \ Vel. is decreasing, acc. is negative. In portion AB, displacement is constant v = 0, a = 0. In portion BC, slope of x – t graph is increasing. \ vel. is increasing, acc. is positive. In portion CD, slope is constant, v = const., a =0 The maximum possible error in Y due to l

20. (a)

r r r Here, FAB + FBCDA = 0 r r r r Þ FBCDA = - FAB = - F (Q FAB = F ) A stretched catapult has elastic potential energy stored in it æ 1 YAl 2 ö ç U = ç 2 L ÷÷ ´ 2 è ø This energy, when imparted to the stone, it flies off a height 20 m. Energy possessed by the stone = mg(h + l). Now,

DY Dl 2Dd = + Y l d The least count

and d are

=

U = mgh Þ

Pitch Number of divisions on circular scale

0.5 mm = 0.005 mm 100 Error contribution of l

\

=

mghL Al 2

=

1´ 10-1 ´ 9.8 ´ 25 ´ 10-1

(

10 ´ 10-6 ´ 5 ´ 10-2

3R R – =R 2 2

Error contribution of d

Q = DU + W or 0 = nCv(DT) – P0V0

dq = 2 \ q = 2t dt Let BP = a, \ x = OM = a sin q = a sin (2t)

Þ

d2x dx =a ´ 2cos (2t) Þ 2 = – 4a sin 2t dt dt

or 22. (d)

23. (d)

y B

24. (d)

a P

q

l A O

x M

x

P0

R

0 = n g - 1 (T – T0 ) - nRT0

\ T = gT0 Initially centre of gravity is at the centre. When sand is poured it will fall and again after a limit, centre of gravity will rise. Adding detergents to water helps in removing dirty greasy stains. This is because dirt is held suspended surrounded by detergent molecules. Reynold’s number = Coefficient of friction = [M0L0T0] Curie is the unit of radioactivity (number of atoms decaying per second) and frequency also has the unit per second. Latent heat =

Hence M executes SHM within the given time period and its acceleration is opposite to x that means towards left.

2

V0

=

2Δd 2 × 0.005 mm 1 = = d 0.5 mm 50

)

Young's modulus, Y = 9.8 × 107 N/m2 The work done of thermodynamics,

Dl 0.005 mm 1 = = = l 0.25 mm 50

= 18. (b)

21. (a)

Y=

YAl 2 = mg(h + l) ; mgh L

Q and m

Gravitation potential =

W . m

SOLUTIONS – MOCK TEST-7 25. (c)

173

The distance travelled by each car is equal to the area under the graph. These areas are equal at 10.0 s.

26. (c)

æ RDisc ö 4 3 2 Þ pRDisc çè ÷ = pR 6 ø 3 Sphere

RDisc é ù êt = 6 , given ú ë û

N q

3 3 Þ RDisc = 8RSphere Þ RSphere =

mg

RDisc 2

Moment of inertia of disc

q

1 2 MRDisc = I (given) 2 \ M (RDisc)2 = 2I Moment of inertia of sphere 2 2 ISphere = MRSphere 5 I Disc =

N cos q = mg and N sin q = mw2r \

tan q =

w 2r g

… (i)

Given y = x2 \

2

=

dy = 2x dx

30. (c)

or tan q = 2 × 1 = 2 … (ii) From above equations, we get w= 27. (d)

h is, g h =

28. (c)

[m = 1]

d (i)

29. (c)

+x

I

O d +A

gh =

r1 = r2

9.8 ´ (6400 ´ 103 ) 2 (6400 ´ 103 + 520 ´ 103

= 8.4 m/sec2

M2 Þ r1 = r2 if M1 = M2 M1

Hence, compounds are N2O and CO2 as both have same molar mass.

(ii)

4 3 pRSphere 3

; where Re is the

31. (a) Rate of diffusion depends upon molecular weight.

d +A

2 VDisc = VSphere Þ pRDisc t=

(R e + h) 2

CHEMISTRY

+x

From Fig. (i) and (ii), it is clear that if the mirror moves distance 'A', then the image moves a distance '2A'. According to problem disc is melted and recasted into a solid sphere so their volume will be same.

g R 2e

radius of the earth

S2 is correct because whatever be the g, the same force is acting on both the pans. Using a spring balance, the value of g is greater at the poll. Therefore mg at the pole is greater S4 is correct. S2 & S4 are correct.

d

Acceleration due to gravity at are altitude

(r = 1 m)

2g

I

2 æ RDisc ö M 2I I ( RDisc ) 2 = Mç = = ÷ è ø 5 2 10 10 5

32. (a)

® SO 42– + H 2 O. HSO-4 + OH - ¾¾ base

Conjugate acid

33. (d)

+1

-3

+3

® PH3 + H3 PO3 H3PO2 ¾¾

n1 = | – 3 – (1) | = |– 4| = 4 n2 = | 3 – (1) | = | 2 | = 2

EBD_7839 174

KVPY-SA n1 ´ n 2 4 ´ 2 4 n-factor = n + n = 4 + 2 = 3 1 2

Eq. mass =

mol mass M ´ 3 = n - factor 4

34.

(b) The first ionization enthalpy will be minimum if an atom after losing an electron acquires stable configuration. Hence, option (b) is correct.

35.

(a)

H SO

2 4 CH3 – CH2 – CH2 – CH3 ¾¾¾¾ ® 475 K

39.

40.

(b) For any given element, size of its negative ion, netural atom and positive ion is in the order: negative ion > neutral atom > positive ion. (c) Hybridisation Shape

CH3 – CH = CH – CH 3

More symmetrical (major product)

CH3 – CH2 – CH2 –

H 2SO4 CH3 ¾¾¾¾ ® 475 K

CH 2 = CH – CH 2 – CH3 Less symmetrical or unsymmetrical (minor product)

36.

It is based on Saytzeff’s rule. According to this more symmetrical or more alkylated alkene predominates. (c) Proton donors are acids. Thus conjugate base of an acid has one H+ less than the acid.

41.

HBr ¾¾ ® Br - + H + Acid

37.

Conjugate base

(a) Hydrogen bonding is possible only in compounds having hydrogen attached with F, O or N.

42.

C2H5 - OH (H-bonding possible) CH3 - O - CH3 (H-bonding not possible)

43.

O ÷÷ CH3 - C - CH 3 (H-bonding not possible) O ÷÷ CH3 - C - H (H-bonding not possible)

38.

(c) Dipole moment is maximum for p–nitro aniline, since the dipole moment of nitro group and amide groups adds up because they are in same direction.

44.

45.

NO +2

sp

Linear

NO -2

sp2

Angular

PCl5

sp3 d

BrF5

sp3 d 2

XeF4

sp3 d 2

Trigonal bipyramidal Square pyramidal Square planar

ICl -4

sp3 d 2

Square planar

TeCl4

sp3 d

Trigonal bipyramidal 3 XeO4 sp Tetrahedral (d) Sulphate of alkaline earth metal are sparingly soluble or almost not soluble in water, whereas, BeSO4 is soluble in water due to high degree of solvation. Be(OH)2 is insoluble in water but soluble in NaOH. (d) DH = DE + PDV For isochoric process, DV = 0 \ DH = DE (a)

Correct IUPAC name of above compound is trans-2-chloro-3-iodo-2-pentene. (d) 2I- ® I2 is oxidation (loss of electrons) ; Cr (+6) changes to Cr (+3) by gain of electrons. Hence Cr is reduced. (c) hn = hno + KE For the above relation, the correct graph is option (c).

SOLUTIONS – MOCK TEST-7 BIOLOGY 46. (a) According to principle of dominance, out of two factors or alleles representing different traits of a character, only one expresses itself. But when both express themselves, it is co-dominance. 47. (d) Bilirubin and biliverdin are the pigments present in the bile juice secreted from liver. They provide yellowish brown colour to the stool. So, malfunction of liver leads to appearance of whitish grey stool. 48. (a) Secondary air pollutants are O3, PAN and aldehydes which are produced from oxides of nitrogen and hydrocarbon by the reaction with UV rays. 49. (a) Digestion of food begins in mouth. Both chemical and physical digestion takes place in mouth. The teeth cut the food into small pieces, chew and grind it. So, teeth help in physical digestion. Salivary gland produces saliva that contains amylase enzyme. The amylase enzyme digests the starch present in food into sugar, thereby helping in chemical digestion. 50. (d) An arthropod has a segmented body covered by an exoskeleton made from chitin and other chemicals. This exoskeleton serves as protection and provides places for muscle attachment. Arthropods must moult because their exoskeletons do not grow with them. The body feature from which the phylum takes its name is the join ted appendages, which in clude antennae and mouth parts as well as walking legs. 51. (a) Copper-T is an intrauterine device which prevents the fertilised egg to becoming implanted in the wall of the womb. 52. (c) A functional strand of mRNA must have a start and a stop codon. The start codon often also codes for the amino acid methionine, which may or may not end up being a part of the final protein. However, the stop codon would not code for an amino acid. Thus, with 66 codons in the mRNA, there could be as many as 65 amino acids in the protein product.

175 53. (a) Process of passing out of urine is called micturition. During discharge of urine, urethral sphincter relaxes and smooth muscles of bladder wall contract gradually. 54. (b) If 30 percent of DNA is adenine then by Chargaff’s rule, 30 percent will be thymine. The remaining 40 percent of the DNA will be cytosine and guanine. Since the ratio of cytosine to guanine must be equal then each accounts for 20 percent of the bases. 55. (a) Ribozymes (ribonucleic acid enzymes) are also called catalytic RNA. They are RNA molecules capable of catalysing specific biochemical reactions, similar to the act of protein enzymes. Ribozyme, discovered in 1982, demonstrated that RNA can be both genetic material (like DNA) and a biological catalyst (like protein enzymes). Examples of ribozymes include the hammerhead ribozyme, the VS ribozyme, Leadzyme and the hairpin ribozyme. 56. (d) Hepatitis B is a viral disease, transmitted through both blood transfusion and sexual intercourse. 57. (c) The second heart sound indicates the beginning of diastole. The second heart sound, "Dup" is due to the closure of the semilunar valves. The heart is in systole between the first and second heart sounds. During systole, the heart pumps blood from the ventricles, and the atrioventricular valves are closed to prevent regurgitation into the atria. The closure of the semilunar valves is the start of a brief period of “isovolumetric relaxation”. All valves are closed during the second heart sound. 58. (c) Hugo de Vries, a Dutch botanist, one of the independent rediscoverers of Mendelism, put forward his views regarding the formation of new species in 1901. According to him, new species are not formed by continuous variations but by sudden appearance of variations, which he named as mutations. Hugo de Vries stated that mutations are heritable and persist in successive generations. He conducted his experiments on Oenothera lamarckiana (evening primrose).

EBD_7839 176 59.

60.

KVPY-SA (a) Somatostatin from the hypothalamus inhibits the pituitary gland's secretion of growth hormone and thyroid stimulating hormone. In addition, somatostatin is produced in the pancreas and inhibits the secretion of other pancreatic hormones such as insulin and glucagon. (c) DNA replication is the process in which a double-stranded DNA molecule is copied to produce two identical DNA molecules. In DNA replication, each new DNA molecule has half the original parent strand hence called semiconservative and one strand is synthesised continuously and other discontinuously hence called discontinuous.

Area of a hexagon occupied by circles =2×

Area of a shaded region = 54 3 – 24p = 6 (9 3 – 4p) sq. cm. 64.

PART-II MATHEMATICS 61.

62.

(d) Since N = 6 × 12 × 18…..600 = (6100)(100!) which is divisible by all the prime numbers till 100, so smallest prime number must be more than 100. (b) Here, a + b = – p, ab = q Let t = xn Þ when x = a, t = an when x = b, t = bn \ t2 + pnt + qn = 0 \ an + bn = – pn, an bn = qn Again

a b and are roots of xn + 1 + (x + 1)n = 0 b a n

n

æaö æa ö Þ ç ÷ + 1 + ç + 1÷ = 0 èbø èb ø Þ an + bn + (a + b) n = 0 Þ – pn + (– p)n = 0 which is true only if n is even integer. 63.

(a) Interior angle of hexagon =

(2 ´ 6 - 4) × 90 = 120º 6

Area of hexagon =

1 3 ×6×6× × 6 = 54 3 sq. cm 2 2

120 × p × 62 = 24p sq. cm. 360

65.

(a) BP || AC and AD || EQ, Since triangles on the same base between the same parallels are equal in area ar (DABC) = ar (DAPC) .... (1) and ar (DADE) = ar (DADQ) .... (2) adding (1) and (2), we get ar (DABC) + ar (DADE) = ar (DAPC) + ar (DADQ) Adding ar (DACD) to both sides, we get ar (DABC) + ar (DADE) + ar (DACD) = ar (DAPC) + ar (DADQ) + ar (DACD) Hence ar (ABCDE) = ar (DAPQ) (d) Let the four-digit number be xxyy Then xxyy = 1000x + 100x + 10y + y = 1100x + 11y = 11(100x + y) Since the number is a perfect square i.e 11(100x + y) is a perfect square, it is possible only when 100x + y be in the form of 11k2, so that the number becomes 11(11k2) or 112 × k2 a perfect square. Now we have 11k2 =100x + y = a three digit number in the form of x0y (middle digit is zero) K = 3 will give us 99 a two digit number hence ruled out, so we will start from k=4 K = 4, 11k2 = 176, k = 5, 11k2 = 275, if k = 6, 11k2 = 396, if k = 7 11k2 = 539, if k = 8, 11k2 = 704, if k = 9, 11k2 = 891. Out of all these three digit number only 704 has middle digit 0 or is in the form of 100 x + y hence only one number exist with x = 7, y = 4, or 7744 = 882

SOLUTIONS – MOCK TEST-7

177

PHYSICS

66. (a)

H qA

q1

q2

I K1

II K2

d1

d2 q A > qB

H III q K3 B d3

At steady state

d 2 d ( x + y 2 ) = (l2 ). dt dt

K1 A(q A - q1 ) ... (i) First material : H = d1

Second material: H =

K2 A(q1 - q 2 ) ... (ii) d2

K Third material : H = 3 A(q2 - q B ) ... (iii) d3

From (i) q A - q1 =

H d1 A K1

... (iv)

From (ii) q1 - q 2 =

H d2 A K2

... (v)

From (iii) q 2 - q B =

H d3 A K3

2y

dy dl = 2l dt dt

or

l dl dy u = = dt y dt cosq

Similar relation can be obtained from DBCD . That is dy v = dt cosq dy (u + v ) = dt 2 cos q

... (vi) Thus velocity of block,

H æ d1 d 2 d3 ö + + A çè K1 K 2 K 3 ÷ø

rection is 68. (b)

(q A - q B ) A Þ H= d1 d 2 d 3 + + K1 K 2 K 3

67. (c)

(u + v ) . 2 cos q

If the temperature of surrounding increases by DT , the new length of rod becomes Dueto changein length, moment of inertia of rod also changes and it is about an end P and is given as

(27 - 0)1 = 41.54 J/s 0.01 0.05 0.01 + + 0.8 0.8 0.8

l P¢ =

Let at any instant the block is at a distance y from line AB and length of string between A and C is l. In DACD, we have x 2 + y 2 = l2

dy in upward didt

l ¢ = l (1 + aDT )

Substituting the values H=

....(iii)

Adding equations (ii) and (iii), we get

Adding the above three equations, we get q A - qB =

....(ii)

.... (i)

M l ¢2 3

As no e xternal force or torque is acting on rod thus its angula rmomentum remains constant and during heating, thus we have

I P w = I P¢ w '

EBD_7839 178

KVPY-SA [If w¢ is the final angular velocity of rod after heating]. or

M l2 M l 2 (1 + aDT ) 2 w= w¢ 3 3

For t =

v ; the slope is zero. a

For t >

v ; the slope is positive. a

or w¢ = w(1 - 2aDT ) [using binomial expansion for small a]. Thus percentage change in angular velocity of rod due to heating can be given as, Dw =

69. (c)

w - w¢ ´ 100% = 2aDT ´100% w

71.

(c)

CH3

or sin r µ = sin i On differentiating partially, we have sin r (dµ) + µcos r (dr) = 0 dr =

These characteristics are represented by graph (b). CHEMISTRY

sin i m= sin r

or

CH3 CH3

… (i)

dµ =

CH3 CH3

-2bdl

CH3

… (ii)

l3

From above equations, we have dr = 70. (b)

2b tan r al 3 + bl

Methyl shift

¾¾¾¾¾

CH3 CH3

As µ = a + bl–2 or

CH3 ¾ ¾ ®

CH3

- sin r (dm ) m cos r

v ; the slope is negative. a

For t
15y and 16y > 40x. Since 40x>15y hence 8x > 3y or 8x > = 3y + 1 …(i) 16y > 40x or 2y > 5x or 5x < = 2y – 1 …(ii) Multiply eq (i) with 5 and eq(ii) with 8 and then compare the result we will get 16y – 8> = 40x> = 15y + 5 or 16y – 8> = 15y + 5 hence y> = 13 so minimum possible value of y is 13. Now 8x>3y or x>3y/8 or x>39/8 so minimum possible value of y is 5 so minimum possible value of x + y = 13 + 5 = 18.

12. (a)

V1 = Vol. of air between cube and sphere 4 3 4 3 3 a3 pr1 - a3 = p - a3 3 3 8

Þ

V1 =

Þ

æ 3p ö 3 V1 = ç - 1÷ a è 2 ø

Ratio =

V1 Vol. of air of cube

æ 3p - 2 ö a3 3p - 2 =ç ÷ø 3 = è 2 2 a 13. (b) Case-I: x Î[0,9] ; 2(3 - x ) + x - 6 x + 6 = 0

Þ x - 8 x + 12 = 0 Þ x = 4, 2 Þ x = 16, 4 Since x Î[0,9] \ x = 4 Case-II: x Î[9, ¥] ; 2( x - 3) + x - 6 x + 6 = 0

Þ x - 4 x = 0 Þ x = 16,0 Let r1 be radius of hollow cylinder A and C are opposite corners of cube such that So AC is longest diagonal So,

3a = 2r1 r1 =

3a 2

Since x Î[9, ¥] \ x = 16 Hence, x = 4 & 16 14. (b) Unit digit of product of two digit number is 5. Hence unit digit of these two digit must be either odd or 5. Hence numbers are 15, 51, 35, 53,75, 57, 95, 59 & 55. So total 9 such 2 digit number exist.

EBD_7839 184 15.

KVPY-SA (a) Here, 4

2

4

2

p -1 x + x +1 x + 2x + 1 - x = 2 = 2 p + 1 ( x + x + 1) ( x 2 + x + 1)2 =

( x 2 + 1) 2 - x 2 2

( x + x + 1)

2

=

2

18.

( x 2 + x + 1)2

ve =

\

Þ

2 p 2( x 2 + 1) = 2 2x

As, f ( x) =

Þ p = x+

1 x

...(1)

1- x 1+ x

æ æ 1öö 1 Þ f ( f ( x )) + f ç f ç ÷ ÷ = x + è ø è x ø x

...(2)

1cm = 0.125cm 8

\ 5VSD = 4 ×

L.C

1 cm = 0.5cm 8

= 1MSD – 1VSD = 0.125cm – 0.1cm = 0.025cm

Screw gauge 3L 4

of wire 3W 4

force area (c) Smooth and polished plates are poor radiators of heat. Hence, heat coming out from A1 is small, even though B1 being a black and rough plate is a good absorber. Effectively the heat coming to the left of pellet P is small.

stress =

17.

1 MSD =

æ 1 öö f ç ÷÷ = p è x øø

(c) Force = weight suspended + weight of

= W1 +

(b) Vernier callipers

\ 1 VSD = 0.1cm

PHYSICS 16.

19.

6 3 Gm . a

5 VSD = 4MSD

From Eqs. (1) and (2), we get æ f ( f ( x )) + f ç è

-G mm ' 1 ´ 3 + m ' ve2 = 0 r 2 -3Gm 1 + ve2 = 0 or a ( / cos30°) 2 2

( x 2 + x + 1)( x 2 - x + 1)

p - 1 x2 - x + 1 = , using componendo p + 1 x2 + x + 1 and dividendo.

(c)

Black and rough plates are good radiators of heat. Hence, plate B2 radiates heat to a satisfactory level; however, plate A2, being smooth and polisher, is a bad absorber. Effectively, the heat coming to the right of P is also small.

One complete revolution = 2M.S.D If the pitch of screw gauge is twice the L.C of vernier callipers then pitch = 2 × 0.025 = 0.05cm. L.C of screw Gauge pitch = Total no. of divisions of circular scale

=

0.05 cm = 0.0005cm = 0.005 mm. 100

(II) is a correct statement Now if the least count of the linear scale of the screw gauge is twice the least count of venier callipers then. L.C of linear scale of screw gauge = 2 × 0.025 = 0.05cm. Then pitch = 2 × 0.05 = 0.1cm.

SOLUTIONS – MOCK TEST-8 Then L.C of screw gauge =

185 0.1 cm 100

23. (d) Initial position of cm =

= 0.001cm = 0.01mm. (III) is a correct statement.

Also Dcm =

20. (d) Efficiency of a transformer, h=

Power output Power input

\

for an ideal transformer, h = 1 \ power output = power input = 60 W 21. (a) Consider an element part of solid at a distance x from left end of width dx. dx

x

m2l m1v0t final position = m + m + m + m 1 2 1 2

24. (a) The direction of current in the loop will be such as to oppose the increase in the magnetic field. r 25. (a) Since torque t is rotational analogue of r r force F and F = mass × acceleration, r therefore, torque t = (moment of inertia) × (angular acceleration) = (I a) as moment of inertia is rotational analogue of mass.

Resistance of this elemental part is, r xdx = 0 pa pa 2

rdx

t1 =

l l 2l , t2 = = v v/2 v

t3 =

(n - 1)l 3l , ..... , tn–1 = v v

2

L

R = ò dR = ò

0

r0 xdx pa 2

=

m1Dx1 + m2 Dx2 m1v0t + 0 = m1 + m2 m1 + m2

26. (d)

V

dR =

m2l m1 + m2

r0 L2 2 pa 2

l 2l 3l (n - 1)l + + + ..... + v v v v

So T =

Current through cylinder is,

V V ´ 2pa 2 = I= R r0 L2

=

l [1 + 2 + 3 + ..... + (n – 1)] v

Potential drop across element is,

=

n(n - 1)l 2v

vcm =

mv + 0 + 0 + .... + 0 v = . nm n

dV = I dR =

E(x) =

2V L2

´ dx

dV 2V = x dx L2

22. (b) If v ^ B, then path is circular and if v has a component along B, then path will be helical.

27. (a) Mass of the removed sphere M '=

2

4 æRö M ´ pç ÷ = 4 3 3 è2ø 8 pR 3 M

EBD_7839 186

KVPY-SA Thus, F =

28.

29.

30.

GMm d

2

-

36.

GM ' m (d - R / 2)

2

(c) The three curves AB, CD and EF meet at point P which is called the triple point of water. It is the point where all three states solid, liquid and gas of water co-exists. (b) Temperature of B will be higher as due to expansion centre of mass of B will come down same heat is supplied but in B, Potential energy is decreased therefore internal energy gain will be more and internal energy is directly proportional to temperature. (c) Since horizontal component of the velocity of the bomb will be the same as the velocity of the aeroplane, therefore horizontal displacements remain the same at any instant of time.

37.

31.

(b) In KCN, K+ and CN– have ionic bond and C º N has covalent bond.

32.

(b) Geometrical isomerism is shown by compounds which have C=C and two groups attached to same C atoms are different. Choice (b) fulfills both conditions.

(b) Moles of C12 H 22 O11 =

= 0.0747 × 22 × 6.023 × 1023 = 9.91 × 1023 34. 35.

(b) Dng = 2 – 4 = – 2, DH = DE – 2RT. (a)

+2 +2

+4

FeC2 O4 ® Fe3+ + 2CO 2 + 3e – ; In acidic medium MnO 4– changes to Mn 2+ and consumes 5e–

\ 3e– will be consumed by 3/5 moles of KMnO4.

n=5

Ti(22)

[Ar]3d2 4s2

n=2

V (23)

[Ar]3d3 4s2

n=3

Al (13)

[Ne]3s2 3p1

n=1 –

(c) H2SO4 is a strong acid and HSO4 is its weak conjugate base. +

38.

(c)

+1 0 H 2O ® H2

Oxidation number of H decreases from +1 to 0. Hence, H2O is reduced to H2. 39.

(b) Chain propagating step is Clg + CH 4 ¾¾ ® g CH3 + HCl

40.

(a) Due to the inert pair effect, thallium exists in more than one oxidation state. Also, for thallium + 1 oxidation state is more stable than +3 oxidation state.

41.

(b) O < S < F < Cl Electron gain enthalpy increases along the period and decreases along the group but in between III and II period elements, III period elements have greater electron gain enthalpy because of very high electron density of II period elements.

42.

(c) Metal + dil. HCl ® Metallic salt + Hydrogen Ag is below hydrogen in reactivity series. So it fail to displace hydrogen from dil.HCl. Thus there is no product formed and no gas evolved.

43.

(a)

25.6 = 0.0747 342.3

Number of H-atoms

[Ar]3d5 4s2

NH3 is a strong base and NH 4 is its weak conjugate acid of NH3.

CHEMISTRY

33.

(a) Mn(25)

Kc =

[Z] [X][Y]

1 1 Þ [Y] = [X] = [Z] = a(say) 2 2

\ [Z] = 2a, [Y] = 2a, [X] = a 104 =

2a Þ a = 10–4 a.2a

[Z] = 2a = 2 ×10–4

SOLUTIONS – MOCK TEST-8 44. (a) M1 = 64 ; r2 = 2r1 2

é r1 ù 1 M2 = M1 ê ú = 64 × = 16 4 ë r2 û

45. (a) NaOH is a strong alkali. It combines with acidic and amphoteric oxides to form salts. Since CaO is a basic oxide, hence, it does not reacts with NaOH. BIOLOGY 46. (d) Maltose, also known as maltobiose or malt sugar, is a disaccharide formed from two units of glucose joined with an a–1, 4–glycosidic bond. 47. (b) The sarcomere is the functional unit of muscle. As actin myofilaments slide over myosin, the sarcomere shortens, but not the individual components that make up the sarcomere. 48. (a) Amphitrichous bacteria have a single flagellum at each of the two ends or poles. e.g., Spirillum volutans. 49. (d) Pancreatic lipase (formely called steapsin) hydrolyses fats into glycerol and fatty acids. 50. (a) It is the long distance movement of organic substances from the source or supply end (region of manufacture or storage) to the region of utilisation or sink end. 51. (d) Pectin, a structural heteropolysaccharides, are abundant in fruits particularly in the citrus fruits like oranges and lemons. They are present in the cell wall and in the intracellular substance. They contain arabinose, galactose and galacturonic acid. 52. (a) E. coli resides in the large intestine of human. If they are present in water supply, it represents that water supply is contaminated.

187 53. (b) Monosaccharides are the class of sugars that cannot be hydrolysed to give a simpler sugar. Pentoses and hexoses are the examples of monosaccharides. 54. (a) Ca 2+ and Mg2+ are involved in muscle contraction. Ca 2+ ion promotes the formation of actomyosin which is a combination of myosin and actin. Ca2+ and Mg2+ ions are also helpful in relasing energy by the oxidation of food in the form of ATP which is broken down in ADP, phosphorous and energy. Released energy is used up in the contraction of muscle fibres. 55. (a) Lateral meristems are present along the lateral sides of stem and roots. Intrastelar cambium ring formed by intrafascicular and interfascicular cambia and extrastelar cork cambium are the examples of lateral meristem. 56. (d) Relaxin is a protein hormone, secreted by the placenta that causes the cervix to dilate and prepares the uterus for the action of oxytocin during labour. 57. (b) Gibberellin overcomes vernalisation requirement (low temperature) for flowering. 58. (d) Blood clotting involves none of the given options. In blood clotting pathways, prothrombin is converted into thrombin and fibrinogen is converted into fibrin. 59. (b) Contraction of gall bladder is induced by cholecystokinin. It is secreted by I-cells in the small intestine. 60. (a) Blood platelets are 2–3 µm in diameter. They are considered to be the cell fragments instead of being cell themselves.

EBD_7839 188

KVPY-SA PART-II

Sum of areas of two circles = 153p px2 + py2 = 153p

MATHEMATICS 61.

Þ x2 + y2 = 153

(b) Since highest power of 2 in 50! Is 47 hence highest power of 8 in 50! Is [47/3] = 15

x + y = 15 (x +

If we assume a = 15 then 158 or 38 and 58

225 –153 72 = = 36 2 2 Þ x (15 – x) = 36 Þ 15x – x2 – 36 = 0

Þ xy =

(c) x2 – 2x sec q + 1 = 0 Þ x = sec q ± tan q

Þ (x – 12)(x – 3) = 0 Þ x = 12 or 3 and y = 3 or 12

p p y Þ x = 12 and y = 3 Ratio of circumferences smaller to larger circle

p p Þ sec > sec q > sec 6 12

and - tan

also tan

p tan p < tan q < 6 12

p p < - tan q < tan 12 6

= 64.

\ a1 = sec q – tan q and b1 = sec q + tan q

2p(3) = 3 : 12 = 1 : 4 2p(12)

(a) DABF and ||gm ABCD are on the same base AB and between the same parallels AB and DF. \

a1,b1 are roots of x2 – 2x sec q + 1 = 0 and a1> b1

x2 – 15x + 36 = 0

or

and x2 + 2x tan q – 1 = 0 Þ x = –tan q ± sec q Q -

.... (ii) (Given)

y)2 = x2 + y2 + 2xy

Þ 152 = 153 + 2xy

So we have to check that whether 38 and 58 are divisible by 50! Or not, Since highest power of 3 in 50! Is 22 and that of 5 is 12, hence a = 15 will satisfy the condition. 62.

.... (i)

ar (DABF) =

D

1 ar (||gm ABCD) .... (1) 2

C

a2, b2 are roots of x2 + 2x tan q – 1 = 0 and a2 > b2

F

E

\ a2 = –tan q + sec q, b2 = – tan q – sec q \ a1 + b2 = sec q – tan q – tan q – sec q = – 2tan q 63.

(b) Let the radius of two circles with centres O and O¢ be x and y respectively

A

B

sum of radii = 15

Q Diagonal BD of a ||gm ABCD, divides it into two congruent triangles.

Þ

\ ar (DBCD) = ar (||gm ABCD)

x + y = 15

.... (2)

ar (DABF) = ar (DBCD) .... (3) [From (1) and (2)]

O x

y O'

Also, DABE and DDBE are on the same base BE and between the same parallels BE and AD.

SOLUTIONS – MOCK TEST-8 \ ar (DABE) = ar (DDBE)

189 ... (4)

5V0

Subtracting (4) from (3), we get ar (DABF) – ar (DABE)

P1

= ar (DBCD) – ar (DDBE)

T0

Piston

Hence, ar (DBEF) = ar (DCDE). 65. (d) Since the number is between 10 to 1000, that means n may be a two digit or a three digit number, so we have two different cases-

P2

3V0

24 5 V0 P'1

Case (i) if the n is a two digit number, ab = 10a +b, here none of the digit is zero. From given information a + b + ab = 10a + b or ab = 9a or b = 9, and the two digit numbers that satisfy this condition is 19, 29, 39, 49, 59, 69, 79, 89 and 99 (Total 9 numbers)

2T0

Piston P'2

16 5 V0

Case (ii) if the number is a 3digit number then Pn = abc and Sn = a + b + c

where n1 and n2 are the number of moles in the two portions.

a + b + c + abc = 100a + 10b + c

Finally, from the F.B.D. of the piston,

or abc = 9(11a + b)

P1A + W = P2A

Here LHS is always less than the RHS, because c £ 9 and ab < 11a, no number satisfy this condition.

or

Thus there are only 9 values (all are two digit number) of n that satisfy the condition. PHYSICS 66. (a) Let the weight of the piston W and crosssectional area of the cylinder be A. Let the initial and the final pressures of the upper and the lower regions be P1, P¢1 and P2, P¢2 respectively. Further, if initial temperature be taken to be T0, then final temperature will be 2T0. From the F.B.D. of the piston, initially,

n1R(2T0 ) W n 2 R(2T0 ) + = 3(8V0 / 5) A 2(8V0 / 5)

Subtracting eq. (ii) from eq. (i), we have æ1 5 ö æ 1 5ö n1 ç - ÷ = n 2 ç - ÷ or n : n = 35 : 26 1 2 è 5 12 ø è 3 8ø

67. (c) Shearing strain is created along the side surface of the punched disk. Note that the forces exerted on the disk are exerted along the circumference of the disk, and the total force exerted on its center only. D

F

P1A + W = P2A or

n1RT0 W n 2 RT0 + = 5V0 A 3V0

... (i)

... (ii)

h

EBD_7839 190

KVPY-SA The equilibrium extended length of wire

Let us assume that the shearing stress along the side surface of the disk is uniform, then

ò

F=

= L + DL = L+

dFmax

surface

=

ò

s max dA = s max

surface

ò

69.

dA

surface

æ Dö = s max .2p çè ÷ø h 2

2

(b) As no external force acts in z-direction, hence z-coordinate of the centre of mass of the ball should be zero. To make z-coordinate zero other ball should fall symmetrically with respect to z-axis. Hence z-coordinate of other ball = –5 m.

8 æ1 -2 ö -2 = 3.5 ´10 ´ ç ´10 ÷ ´ 0.3 ´10 ´ 2p 2 è ø 4 4 = 3.297 ´ 10 Þ h = 3.3 ´ 10 N

68.

æ 1 Mg ö = L ç1 + ÷ 3pR Y è 3 pYR 2 ø MgL

(c) Consider a small element dx of radius r,

2R r= x+R L

The balls do not have any external force in x-direction. Hence in x-direction the centre of mass should move with constant velocity. x-coordinate of centre of mass at t = 1.5 s, = 200 × 2 = 400 m xCM =

Hence,

m1 x1 + m2 x2 m1 + m2

R

y x

200 m/s r

dx

30 m L

250 m

z

400 =

3R Mg

At equilibrium change in length of the wire 1

ò dL = ò 0

Mg dx 2

é 2R ù pê x + Rú y ë L û

1 Mg é L ù MgL ´ = ê L py 2 R ú 3pR 2 y ê é 2 Rx + R ù ú úû êë êë L úû 0

x

20 ´ 50 + 20 x2 20 + 20

x2 = 800 – 250 = 550 m Position of centre of mass at t = 2 s, h=

1 ´ 10 ´ 2 2 = 20 m 2

Hence y-coordinate of centre of mass = 30 – 20 = 10 m

Taking limit from 0 to L DL =

5m

Hence yCM = Þ 10 =

m1 y1 + m2 y2 m1 + m2

20 ´ 0 + 20 ´ y2 20 + 20

\ y2 = 20 m

SOLUTIONS – MOCK TEST-8

191

70. (a) If x1 is the overhang length on second blocks, then,

æl ö M M ç - x1 ÷ = ( x1) 2 è2 ø

\ x1 =

l 3

73. (d) If two chirality centres are created as a result of an addition reaction four stereoisomers can be obtained as products.

Now of x2 is the overhang table then

æl ö 3M M ç - x2 ÷ = x2 2 2 è ø

\ x2 =

l 5

Br CH3 | | CH3 – CH – CH – CH2 – CH3

l l 8l Nowx = x1 + x2 = + = . 3 5 15

* * 2-Bromo-3-methyl pentane (2 chiral centre)

CHEMISTRY 71. (d)

No. of stereoisomers = 2n = 22 = 4 where n = chiral centre CH3

CH3

H

Br

H

CH3

Br

H

H3C

H

C2H5

C2H5

(I)

(II)

CH3

72. (d) At NTP 22400 c.c. of N2O = 6.02 × 1023 molecules 6.02 ´ 10 23 molecules \ 1 cc N2O = 22400 23

3 ´ 6.02 ´ 10 atoms 22400 1. 8 = ´ 1022 atoms 224 No. of electrons in a molecule of N2O = 7 + 7 + 8 = 22 Hence no. of electrons =

=

6.02 ´ 1023 ´ 22 22400

=

1.32´1023 electrons 224

CH3

Br

H

H

CH3 C2H5 (III)

H

Br

H3C

H C2H5 (IV)

74. (c) PbCl2 is most ionic because on going down the group the metallic character increases and also the inert pair effect predominates. Between PbCl2 and PbCl4; PbCl2 is more ionic according to Fajan’s rule. 75. (c) The b.p. of p-nitrophenol is higher than that of o-nitrophenol because in p-nitrophenol there is intermolecular H-bonding but in o-nitrophenol there is intramolecular H-bonding.

EBD_7839 192

KVPY-SA BIOLOGY

76.

77.

(a) In oxidative decarboxylation, the transition reaction converts the two molecules of the 3-carbon pyruvate into two molecules of 2-carbon molecule of acetyl CoA and two molecules of carbon dioxide. (b) In grasshoppers, human and Drosophila, the sex determination is similar because all the male individuals consist of one X chromosome and the female individuals consist of two X chromosomes. In grasshopper and Drosophila, no Y chromosome is foun d. So, sex determination is done on the basis of presence or absence of X-chromosome.

78.

(c) It is the part of brain where the optic nerves partially cross. The optic chiasma is located at the bottom of the brain immediately inferior to the hypothalamus.

79.

(b) The inhibitory gene ratio is 13:3 Complementary gene ratio is 9:7 Recessive epistasis ratio is 9:3:4 Dihybrid test cross ratio is 1:1:1 Dominant epistasis ratio is 12:3:1

80.

(d) [H+] [OH–] = 10–14 1.3 × 10–4 × [OH–] = 10–14 [OH–] =

10 –14

1.3 ´ 10 –4 = 0.769 ´ 10–10 M

= 7.7 ´ 10–11 M

M

MOCK TEST-9 \ pq = 2005, – 2004

PART-I MATHEMATICS 1.

4.

(a)

(a) Q P(a) = P(b) = P(c) = 0 a, b, c are roots x3 – 3x2 + 2x + 5 = (x – a)(x – b)(x – c) Replacing x by 2 23 – 3(2)2 + 2(2) + 5 = (2 – a)(2 – b)(2 – c)

10 3

(2 – a)(2 – b)(2 – c) = 5 2. (d) Let the two digit number 10a + b Then (10a + b)2 + (10b + a)2 ends with 3. Or 100(a2 + b2) + 40ab + (a2 + b2) has unit digit 3. But here a2 + b2 will give us unit digit Square of any numbers ends with either 0, 1, 4, 5, 6 or 9 and unit digit after adding any two of them is 3 then these two must be 4 & 9 Hence unit digit of two squares must be end with 4 & 9. If square has unit digit 4 then unit digit of that no must be 2 or 8. And if square of a no has unit digit 9 then the nos must has unit digit 3 or 7. Hence numbers are 23, 27, 83, 87 or reverse of these 32, 72, 38 or 78 Hence total 8 such numbers exists. 3. (c) 2005 + p = q2 .....(1) 2 2005 + q = p .....(2) From equation (1), (2), (p – q) = q2 – p2 Þ (p + q) = – 1

.....(3)

also from eqn. (1) × (2), Þ (2005 + p) (2005 + q) = p2.q2 Þ (2005)2 + 2005 (p + q) + pq = p2q2

10 3

10 3

Let the side of eq D is a. \

3 2 æ1 ö . a = ç ´ (4) ´ a ÷ + 4 è2 ø

æ1 ö + ç ´ (6) ´ a ÷ è2 ø Þ

3 2 1 . a = × a × (15) Þ a = 10 3 4 2 for height of D;

(10 3 ) = (5 3 ) 2

Þ p2q2 – pq – 2005 (– 1) – (2005)2 = 0 Þ (pq)2 – (pq) – 2004 × 2005 = 0 Þ (pq)2 – (2005pq) + (2004pq) – 2004 × 2005 = 0 Þ (pq – 2005) (pq + 2004) = 0

2

+ h2

Þ 300 = 75 + h2 Þ 225 = h2 Þ h = 15 5.

(d) (2!)x = 2x (5!)y = 23y, 3y, 5y (7!)2 = 24z, 32z, 5z, 7z Hence (2!)x × (5!)y × (7!)z +z

= 2x + 3y + 4z × 3 y+ × 5 y × 7 z 2z

from (3),

æ1 ö ç 2 ´ (5) ´ a ÷ è ø

Since 100! = 297 × 348 × 524 × 716 Maximum value of z = 16 Maximum value of y + z = 24 Maximum value of z = x + 3y + 4z = 97; hence for maximum value of x + y + z, y = 0, z = 0 and x = 97, then x + y + z = 97

EBD_7839 194 6.

(b)

KVPY-SA

s=

Therefore, DFAC ~ DEBC (AA similarity)

4+6+8 =9 2

Þ

1 D = 9.1.3.5 = 3 15 = .4.h 2

M 6

B

Þ

FA AC = DB BC

Therefore, from (1) and (2) we have :

h1

AC OA = BC OB N

8

i.e,

h

OC - OA OA = or OB - OC OB

OB.OC – OA.OB = OA.OB – OA.OC or OB.OC + OA.OC = 2OA.OB or

C

4

(OB + OA).OC = 2OA.OB or

D 15 r= = s 3 h1 = h – 2r =

1 1 2 + = OA OB OC

[Dividing both the sides by OA.OB.OC] 8.

3 2 5 15 - 15 = 15 2 3 6

(d) Area of circle = p Area of square = p

DAMN ~ DABC 5 4. 15 h1 MN 4h1 20 = Þ MN = = 6 = 3 h 4 h 9 15 2

7.

(d) In DAOF and DBOD ÐO = ÐO (same angle) and ÐA = ÐB (each 90°) Therefore, DAOF ~ DBOD (AA similarity) So,

....(2)

But EB = DB (B is mid-points of DE)

3 Þh = 15 2 A

FA AC = EB BC

OA FA = OB DB

....(1)

Also, in DFAC and ÐEBC, ÐA = ÐB (each = 90°) and ÐFCA = ÐECB (Vertically opposite angles).

\

Side of square = p

Þ

OM =

p 2

Now: MQ2 + OM2 = 12 Þ

MQ2 +

p =1 4

SOLUTIONS – MOCK TEST-9 Þ

MQ 2 =

195 25 = 6y – 7

4-p 4

y=

PQ = 2MQ = 4 - p 9.

(b) Here, area of square ABCD = 4cm2 According to question, Area of D ADE = Area of DBEC = 1 Area of D DEC =

32 16 = 6 3 2

20 æ 16 ö Þ x 2 = ç ÷ + 42 = 3 è 3ø 11. (d) Let the number of deer = x According to question

4 =2 2

x 3æ x ö + +9= x 2 4 çè 2 ÷ø

x 3x x– - =9 2 8

Area of D DFC =

a+ b+ c

Area of D DGC = 1/2 Area of D DHC = 1/4

Then 5 + 10 + 15 + 6 15 3 æ5ö = 5+2 ç ÷ +2 +2 4 2 è2ø

1 1 = cm2 4´2 8

=

10. (c) Join OT which intersect PQ at M

(

a+ b+ c

)

2

= a + b + c + 2 ab + 2 ac + 2 bc

DPQT is an isosceles Triangle and OT is angle bisector of PQ. Þ OT ^ PQ Þ PM = 4cm and OM = 3 cm Let PT = x, MT = y In DPMT, by Pythagoras theorem x2 = y2 + 42

... (i)

In DOPT, by Pythagoras theorem (y + 3) = 5 x 2

x = 9 Þ x = 72 8

Difference between number of deer who are grazing and those who are playing = 36 – 27 = 9 So, it is a multiple of 9. 12. (a) Let the required square root be

2 =1 2

\ Area of D KDC =

Þ

2+

2

Subtracting (i) from (ii), we get

... (ii)

Now equating the rational and irrational part we will get a + b + c = 5, ab = 5/2, bc = 15/4 and ac = 3/2 Since in denominator of bc is 4 while that of other two terms is only 2 hence we can assume that b = 3/2 and c = 5/2 then a 1 Hence required square roots is æ æ 3ö æ5öö ç 1+ ç ÷ + ç ÷÷ ç è 2ø è 2 ø ÷ø è

EBD_7839 196

KVPY-SA

13. (c) Since N = (xyz – x)/99. Hence 396N will always assume an integral value. 14. (d) Since highest power of 7 in N! is k and that in (N + 3)! is K + 2 it is possible only when N + 1, N + 2 or N + 3 is a multiple of 7^2 = 49. There are 2 such numbers exist these are 49 and 98, two values of N is 48 and 97. Highest power of 7 in 46!, 47!, and 48! is 6 and that in 49!, 50!, and 51! is 8. Highest power of 7 in 95!, 96!, and 97! is 14 and that in 98!, 99! And 100! is 16. Hence total 6 numbers exist. 15. (d) Let number of girls is ‘g’ then Number of group having 4 boys and 1 girl = (4C4) (gC1) = g And number 4 of groups having 3 boys and 2 girls = (4C3) (gC2) = 2g (g – 1) Thus, total number of tests is g + 2g(g – 1) = 66 Or 2g2 – g – 66 = 0 only integral value of g

is 6

17.

As there is no slipping between any point of contact hence distance moved by the man is 2L. (b) According to the question, t = (90 ± 1) or,

As we know, t = 2p

Þg=

or,

Since time taken by the axis to move a distance L is equal to t = L/v0. In the same interval of time distance moved by the topmost point is s = 2v0 ´

L = 2L v0

\

l g

4 p 2l t2

Dg Dt ö æ 0.1 1ö æ Dl = ±ç +2 ÷ = ç +2´ ÷ è l g t ø è 20 90 ø

Dg % = 2.7% g

18. (a) From 0 to 6 second the acceleration varies linearly with time, therefore we have,

da 5 = + dt 6

or,

v=v0+wR

or v=v0–wR

Dg %=? g

= 0.027

PHYSICS

v = v0 + wR = 2v0

Dl 0.1 = 20 l

l = (20 ± 0.1) or,

Hence total number of students is 5 + 6 = 11 16. (a) Let v0 be the linear speed of the axis of the cylinder and w be its angular speed about the axis. As it does not slip on the ground v0 hence w = . Where R is the radius of R the cylinder. Speed of the topmost point is

Dt 1 = t 90

d é d 2s ù 5 ê ú=+ dt ëê dt 2 ûú 6

d 3s dt

3

= +

5 6

SOLUTIONS – MOCK TEST-9

197

Integrating equation (i) w.r.t. time, we get

d 2s dt

2

=

Now,v0 sin q t =

5 t + c1 6

or \

and

ds 5t 2 = dt 12

s=

1 é 2v0 cos q ù v0 sin q = a ê ú g 2 ë û

\

5 3 .t 36

tan q =

a g

tan q=

4 = 0.4 10

…(iii)

20. (d) For path ab : (DU)ab = 7000 m

5 ´ (6)3 = 30 m s1 = 36

By using DU = mCV DT

The velocity at t = 6 s,

7000 = m ´

v=

5 2 5 t = ´ 62 = 15 m/s 12 12

Now from, 6 s to 12 s u = 15 m/s, a = 5m/s2 1 2 1 2 \ s2 = ut + at = 15 ´ 6 + ´ 5 ´ 6 2 2

= 180 m. Therefore total distance travelled on runway = 30 + 180 = 210 m. (d) If t is the time taken by ball to return the boy’s hand, then v0 cos q

5 R ´ 700 Þ m = 0.48 2

For path ca : ( DQ)ca = ( DU )ca + ( DW )ca

...(i)

Q ( DU )ab + (DU )bc + ( DU )ca = 0 Q 7000 + 0 + ( DU )ca = 0 Þ ( DU )ca

= –7000 m

...(ii)

Also ( DW )ca = P1 (V1 - V2 ) = mR (T1 - T2 ) = 0.48 × 8.31 × (300 – 1000) = – 2792.16 m ... (iii) On solving equations (i), (ii) and (iii)

( DQ )ca = -7000 - 2792.16 = –9792.16 J » –9800 m 21. (b) From geometry

v0

q v0 sin q

1 2 0 = v0 cos qt - gt 2

or

t=

2

…(ii)

The distance travelled from 0 to 6s, from equation (ii), we get

19.

1 2 at 2

2v0 cos q g

a r = f tan a \pr2 = pf 2 tan 2 a or pr2 µ f 2

f r

EBD_7839 198

KVPY-SA

22.

(a) In fig. A force due to 5M, 2M cancel and similarly in fig. B force due to 2M, 2M cancel, hence FA = FB. In fig. C also force due to 5M cancel but net force will be more in comparison to fig. A or fig. B as man at the centre is double similarly think for fig D. 23. (b) We choose ball and cart as our system, No external force acts on the system in x-direction; therefore momentum along xaxis is conserved.

From conservation of momentum, Pi = Pf m B v 0 = (m A + m B )v or v =

24.

B

Initial position

.......(i)

Differentiating both sides, we get, 2v

GM 2GM dv dv =– Þv = 2 ......(ii) 2 dR dR R R 1 1 dv =– 2R v dR

Dividing (ii) by (i),

(vA)i = v0

(vB)i= v0

2GM R

2GM R 2 v = (2GM)R–1

When the ball reaches maximum height, the cart and ball move horizontally with same velocity at the extreme position.

Reference level

(a) Escape velocity = v =

Þ v2 =

The ball will continue to move upwards until its velocity relative to the cart is zero. r r r i.e., v BA = v B - v A = 0 r r or vB = vA

A

m Bv0 mA +mB

Þ

25.

1 dv × 100 = × 4% = 2% 2 v

\ If the radius decreases by 4%, escape velocity will increase by 2%. (c) This graph suggest that when

u = – f, v = +¥

v (cm)

(vA)f = v

A

(vBA)i = 0 h

B

(vB)f =(vA)f =V

f –f

u (cm)

Final position Pi = m B v B = m B v 0 Pf = m B (v B )f + m A (v A )f = (m B + m A )v

When the object is moved further away from the lens, v decreases but remains positive. When u is at –¥, v = f.

SOLUTIONS – MOCK TEST-9

199

This is how image formation takes place for different positions of the object in case of a convex lens. 26. (c) Here, s = (13.8 ± 0.2) m Velocity, v =

v = (2pr)n

(where n is frequency)

= (2p ´ 9)(6) = 1.8p m / s

t = (4.0 ± 0.3) s

s 13.8 = = 3.45 ms–1 t 4.0

= 3.5 ms–1 (rounding off to two significant figures)

N1

Dv æ Ds Dt ö = ±ç + ÷ è s v t ø

N2 mg

( 0.8 + 4.14) æ 0.2 0.3 ö = ±ç + =± ÷ è 13.8 4.0 ø 13.8 ´ 4.0 Dv 4.94 =± = ±0.0895 Þ v 13.8 ´ 4.0

N1 N2 X

D v = ± 0.0895 × v = ± 0.0895 × 3.45 = ± 0.3087 = ± 0.31 (rounding off to two significant figures) Hence, v = (3.5 ± 0.31) ms–1 % age error in velocity =

Dv ´ 100 v

= ± 0.0895 × 100 = ± 8.95 % = ± 9% 27. (a) In x = A cos wt, the particle starts oscillating from extreme position. So at t = 0, its potential energy is maximum. 28. (a) The woman experiences three forces : mg, her weight acting vertically downwards; N1, reaction due to her weight; N2, horizontal reaction which provides the centripetal acceleration.

mg

Therefore, N2 =

(50)(1.8p ) 2 = 178 N 9

N1 = mg = 490 N The magnitude of her weight is the magnitude of the resultant force exerted on her by the chair. N = N12 + N 22 = 4902 + 1782 = 521 N

29. (a) Angle of incidence is given by

From Newton’s second law, cos (p–i) = SFx = N 2 =

mv2 r

SFy = N1 - mg = 0

Y

(6

– cos i = –

)

3iˆ + 8 3 ˆj - 10kˆ .kˆ 20

1 2

Ð i = 60 °

EBD_7839 200

KVPY-SA

Þm³

z

sin q cos q 2 - cos 2 q

\ Minimum coefficient of friction, i

2

æ sin q cos q ö 1 m min = ç = 2 ÷ è 2 - cos q ømax 2 2

x 3

CHEMISTRY

r

31.

(d)

SnCl2

SnCl4

119 : 2 × 35.5

From Snell’s law, 2 sin i = 3 sin r 30.

Ð r = 45° (a) Consider a general position of rod as shown in figure. Let applied force be F. As the end is moved slowly so the rod has no acceleration. Applying Newton’s law in x-direction f - F sin q = 0 Þ f = F sin q

32.

Unhybridised orbital = 1 [1p-bond] 33.

(a)

ˆˆ† M 2 + + 2X MX 2 ‡ˆˆ s

2s

Ksp = s × (2s)2 = 4s3 34. 35.

F y

Chlorine ratio in both compounds is = 2 × 35.5 : 4 × 35.5 = 1 : 2 (c) In sp2-hybridisation, Hybridised orbital = 3 [3s-bonds]

... (i)

In y-direction, F cos q + N = mg ... (ii) l Making t A = 0 Þ F l - mg cos q = 0 2

119 : 4 × 35.5

B

= 4 × (0.5 × 10–4)3 = 5 × 10–13 (d) The electron goes to 4p after filling up to 3d. (a) Due to hydrogen bonding between the two OH groups, gauche conformation of ethylene glycol (a) is the most stable conformation.

N

O

mg A

Þ F=

H O

H x

f

mg cos q 2

H

H

... (iii)

H H

For no slipping

f £ mN Þ m ³

f F sin q Þm³ N mg - F cos q

36.

(b) In SO2 the oxidation state of S varies from –2 to + 6. Hence, it can behave as reducing as well as oxidising agent both.

SOLUTIONS – MOCK TEST-9

201

37. (d) In CCl4, dipole moment of each C–Cl bond is cancelled by dipole moment of the other C–Cl bond.

Gram equivalent of H2SO4 100 ´ 0.1 = 0.01 1000 Hence solution will be neutral.

=

45. (b) Chiral conformation will not have plane of symmetry. Since twist boat does not have plane of symmetry it is chiral. 38. (a) Carbon atom which is connected with four different groups is chiral. 39. (a)

CO32 - + H 2 O ¾¾ ® HCO3- + OH Base

Acid

Acid

Base

40. (a) As the basicity of metal hydroxides increases down the group from Be to Ba, the thermal stability of their carbonates also increases in the same order. Further, group 1 compounds are more thermally stable than group 2 because their hydroxide are much basic than group 2 hydroxides, therefore, the order of thermal stability is BeCO3 < MgCO3< CaCO3< K2CO3. 41. (c) Sodium reacts vigorously with cold water while Mg reacts slowly with hot water. Under normal conditions Zn and Fe do not react with water. Hence, the reactivity order is Na > Mg > Zn > Fe. 42. (c) When double and triple bonds, both are present. Then bond nearest to terminal carbon will be preferred. But, if both are equidistant from terminal carbon then double bond will be preferred over triple bond. CH 3 - CH = CH - C º CH 5

4 3 2 Pent -3- en -1- yne

1

43. (c) Glucose is considered as a typical carbohydrate which contains –CHO and –OH group. 44. (b) Gram equivalent of Na2CO3.H2O =

0.62 = 62

0.01(Mol. wt. of Na2CO3.H2O = 124)

BIOLOGY 46. (d) Fats make more electrons available to the electron transport system and more hydrogen ions available for chemiosmosis. 47. (a) A - Stele: It is the central core of the stem and root of a vascular plant, consisting of the vascular tissues (xylem and phloem) and associated supporting tissues. B - Endodermis: It is an inner layer of cells in the cortex of a root and of some stems, surrounding the stele. C - Casparian strip: It is a band of cell wall material deposited on the radial and transverse walls of the endodermis, and is chemically different from the rest of the cell wall, which is made of lignin and without suberin whereas the casparian strip is made of suberin and sometimes lignin. D - Bark: It is the outermost layer of stems and roots of woody plants. Plants with bark include trees, woody vines and shrubs. 48. (a) As in heavy rainfall, minerals are also drained along with water, hence it will not make minerals more available to plants. 49. (d) Glutamic acid is an amino acid formed by hydrolysis of proteins. It is the only amino acid, which is metabolised by the brain. 50. (b) Epinephrine is secreted by medulla of adrenal gland along with closely related noradrenaline. It is used in medicine as a heart stimulant and to constrict blood vessels. Chemically, epinephrine is a derivative of amino acid (aminohydroxy phenyl-propionic acid).

EBD_7839 202 51.

52.

53.

54.

55.

KVPY-SA (c) Insulin is secreted by the Islets of Langerhans in the pancreas when the blood sugar level is high. This increases the rate of glucose uptake from the blood into the muscle cells. (c) The animal found by the boy belongs to phylum mollusca. Mollusca is the second largest phylum after arthropoda and include predominantly marine animals. They are triploblastic, bilaterally symmetrical, schizocoelic and unsegmented protostomes. They have moist skin, a complete digestive tract, a ventral nerve cord, and had gone through torsion. (d) Plants can absorb water through their entire surface. However, water is found in the soil and only positively geotropic part, i.e., root system is specialised to absorb water. In root system, the most efficient region of water absorption is the root hair zone or zone of cell differentiation.

56.

(c) ELISA test is a technique used to detect and quantitate extremely small amount of a protein, antibody or antigen with the help of enzyme. The commonly used enzymes are peroxidase and alkaline phosphatase. Southern blotting and DNA probes are used in molecular analysis of DNA. Catalase is not involved in ELISA. (d) Primary succession on rocks starts with lichens like Rhizocarpon, Rinodina and Lecanora. They produce some acids which bring about weathering of rocks. These lichens are then replaced by foliose type of lichens. Due to deep depressions and retention of water by them, they form a fine thin soil layer on rock surface and thus there is a change in the habitat.

60.

57.

58.

(a) AIDS is characterised by reduction in the number of helper T- lymphocytes because of HIV infection. It suppresses human immune system due to which any secondary infection may lead to death. T-lymphocytes are the main cells of immune system. (c) Primary respiratory substrate is carbohydrate and secondary respiratory substrate is fat. (a) When blood is shed, th e platelets disintegrate and liberate thromboplastin which activates prothrombin to thrombin.

59.

(c) Lipoproteins are conjugated proteins having polypeptides in association with lipids. Immunoglobulins are the constituent of antibodies. Interferons (INFs) are a group of three vertebrate glycoproteins (i.e. a, b, g,). Out of these three a and b are produced within virally infected cells. (c) Chitinase is an enzyme that cleaves the glycosidic bonds in chitin, thereby breaking down the structural polysaccharide component of the hard outer covering of many animals and of the cell wall of fungi. PART-II MATHEMATICS

61. (c) CD || AE and CY || BA. Since triangle on the same base and between the same parallels are equal in area, so we have ar (DABC) = ar (DABY) Þ ar (DCBX) + ar (DABX) = ar (DABX) + ar (DAXY) Hence, ar (DCBX) = ar (DAXY) [Cancelling ar (DABX) from both sides]

SOLUTIONS – MOCK TEST-9

203

log x 62. (a) 4096. ( f ( x, x )) 2 = x13

Þ

h 5 – r PC = = 12 5 AC

Þ

h 5–r 2r 5 – r = Þ = 12 5 12 5

2

Þ 4096. x log 2 x = x13 Take log on base 2 Þ 12 + Þ log2x = 1, 3, – 4 \

Sum =

log32 x

= 13log2x

1 \ x = 2, 8, 16

Þ 10r = 60 – 12r

161 16

63. (a)

Þp=

Volume of cylinder = pr2h

Q BE is median, \a+ b= c+d

...(1)

=

also, CF is median, \a+ c= b+d ...(2) From eqn. (1) – (2), (a + b) – (a + c) = (c + d) – (b + d) Þ b – c = c – b Þ 2b = 2c Þ b = c \ ar DGBC = ar (AFGE) 64. (c) Let h be the height and r be the radius of the cylinder. Given h = 2r Since, DPSC ~ DAOC

22 æ 900 ö 60 » 127.50 7 çè 121 ÷ø 11

65. (b) Sum of 16 observations = 16 × 16 = 256 Sum of resultant 18 observations = 256 – 16 + (3 + 4+5) = 252 252 = 14 Mean of observations = 18 PHYSICS 66. (c) If x is the object distance, then image distance v = D – x. Thus

\ By similarity of two triangles, we have

PS SC PC = = AO OC AC

30 60 and h = 11 11

m=

v D-x = u x

or D – x = m x

EBD_7839 204

KVPY-SA Ray comes out from CD, when it get reflected from AD.

D \ x = (1 + m )

and D – x =

n1 sin a max = n2 sin r1

mD 1+ m

ù -1 é m Þ amax = sin ê 1 sin r1 ú ë m2 û

1 1 1 Now using lens formula, - = , we f v u

Also r1 = 90° – q = 90° – C

have

æ 1 ö or r1 = 90° – sin -1 ç è 2 m1 ÷ø

1 1 1 Þ = f mD D æ ö æ ö +ç ÷ -ç ÷ è1+ m ø è 1+ m ø

æm ö or r1 = 90° – sin -1 ç 2 ÷ è m1 ø

After solving, we get f=

mD

(1+ m) 2

é ù ì -1 m1 -1 m 2 ü \amax= sin ê sin í90° – sin ýú m1 þúû êë m 2 î

.

67. (b) In figure, "C" reaches the position where "A' already reaches after wt =

p and "A" 2

reaches the position where "B" already p reaches after wt = . 2 68.

69.

(d) In case of destructive interference Phase difference f = 180° or p So wave pair (i) and (ii) will produce destructive interference. Stationary or standing waves will produce by equations (iii) and (iv) as two waves travelling along the same line but in opposite direction.

(a)

µ2 D

A r

a max B

é ìï -1 æ m 2 ö üï ù -1 m1 = sin ê m cos ísin çè m ÷ø ý ú ï ûú 1 þ îï ëê 2

70.

(c) Let q0 be the temperature of surroundings. According to Newton's law of cooling, The rate of cooling µ excess temperature Body cools 80°C to 64°C : Average rate of cooling =

80 - 64 °C / min 5

é 80 + 64 ù °C Mean temperature = ê ë 2 úû é 80 + 64 ù - q0 \ Excess temperature = ê ë 2 úû 80 - 64 é 80 + 64 ù =K ê - q0 ú 5 ë 2 û

µ1 C

... (i)

where K is a constant of proportionality.

SOLUTIONS – MOCK TEST-9

205

Body cools 64°C to 52°C : Average rate of cooling =

64 - 52 5

Mean temperature =

72. (a) OH Å H

+

[1, 2]-methyl shift

¾®

64 + 52 2

\ 64 - 52 = K é 64 + 52 - q0 ù ê 2 ú 5 ë û

...(ii)

Dividing eqn.(i) by (ii), we get 16 72 - q 0 = 12 58 - q0

By solvin g, we get temperature of surrounding, q0 = 16°C Let q0 be the temp. at the end of next 5 minutes (or after 15 minutes from start)

73. (a) Given: T = 27°C = 27 + 273 = 300 K V = 10.0 L Mass of He = 0.4 g Mass of oxygen = 1.6 g Mass of nitrogen = 1.4 g nHe = 0.4/4 = 0.1

\Average rate of cooling

nO = 1.6/32 = 0.05

52 - q °C / min = 5

nN = 1.4/28 = 0.05

52 + q 2 52 - q é 52 + q ù =K ê - q0 ú \ 5 ë 2 û

2

2

ntotal = nHe + nO2 + nN = 0.1 + 0.05 + 0.05 2

Mean temperature =

ntotal = 0.2 ...(iii)

Dividing eqn. (iii) by (i) and putting q0 = 16°C

P=

n RT 0.2 ´ 0.082 ´ 300 = = 0.492 atm V 10

74. (a) I 2 (s) + Cl 2 (g) ¾¾ ® 2ICl(g) DrH = [DH(I2(s) ® I2(g)) + DHI–I + DHCl–Cl] – [DHI – Cl]

52 + q - 16 52 - q = 2 16 72 - 16

= 62.76 +151.0 + 242.3 –2 × 211.3 = 33.46

By solving, we get q = 43°C

Χf H°(ICl)
T2 > T1

The temperature of Sun is higher than that of welding arc which in turn is greater than tungsten filament. 19. (b) The given rod system is equivalent to as shown in figure.

)

= v x $i + v y - 3 $j Given; tan 45° =

vy - 3 vx

Þ vy = vx + 3 = 3 + 3 = 6 m/s

17. (a) Suppose the liquid in left side limb is displaced slightly by y, the liquid in right limb will increase by y/2. The restoring force

trest = -2mg ´ é l sin qù / 2 ê 2 ú ë û

1 2 mg l(-q) trest 2 ; and a = 2ù I é (2m) ( l/ 2) ú ê 3 ë û

æ 3y ö F = –PA = –rg ç ÷ ´ 2 A = 3rgA( - y) . è 2ø a=

F = 3rgA( - y ) / m m

=

3 g (-q) 2 2 l

Comparing with , a = -w2q we get w=

3g 2 2l and T = 2p . 3g 2 2l

EBD_7839 212

KVPY-SA

20. (d)

R=

V V ± DV Þ R ± DR = I I ± DI

æ DR ö V ç 1 ± DV / V æ R ç1 ± = ç R ÷ø è I ç 1 ± DI è I

Þ dt = ö ÷ ÷ ÷ ø

22.

(d) The real gas cannot be liquefied above critical temperature by applying pressure. (d) As, g = 4p

2

Time taken for the sphere to cool down temperature 2T0,

=

t=

R=

T2

\

23. (a) R = (2 p r) (1/p) = 4 p

Specific heat of the material of the sphere varies as, C = aT3 per unit mass (a = a constant) Applying formula,

(

dT sA 4 4 T - Tsurr = dt McJ

Þ

)

T0 s 4 pR 2 é = (3T )4 - ( T0 ) 4 ùúû dt M a ( 3T )3 J êë 0 0

B

Heater R

\ Reff = (3/4) p Þ V/Reff = 8 A. 24. (c) In the given problem, fall in temperature of sphere,

Initial temperature of sphere, Tinitial = 3T0

V 2 100 ´ 100 = = 10 W P 100

V = R ´ P ' = 10 ´ 62.5 = 625 = 25 Since the voltage drop across the heater is 25m hence voltage drop across 10W resistor is (100 – 25) = 75 m. A

Here 1 ohm and 3 ohm will now be in parallel

Temperature of surrounding, Tsurr = T0

æ 16 ö ln ç ÷ 16pR s è 3 ø 2

The power on which it operates is 62.5 m

0.1 1 ´100 + 2 ´ ´ 100 = 2.72 ; 3% 20 90

dT = (3T0 – 2T0) = T0

Ma

25. (c) The resistance of the heater is

l

Dg Dl DT ´ 100 = ´ 100 + 2 ´ 100 So, g l T

s 4pR 2 ´ 80T04

Solving we get,

æ DR ö æ DV ö æ DI ö ç R ÷ =ç ÷ + ç ÷ = (3 + 3)% = 6% è ø è V ø è I ø 21.

M a 27T04 J

100 V

\

The current in AB = I =

V 75 = = 7.5 A R 10

This current divides into two parts. Let I1 be the current that passes through the heater. Therefore 25 = I1 × 10 Þ I1 = 2.5 A Thus current through R is 5A. Applying Ohm's law across R, we get 25 = 5 × R Þ R = 5W 26. (b) In fig, C1 and C2 are IC (instantaneous centre of rotation) of the two cylinders. The cylinder can be considered as rotating about C1 and C2. In the absence of slipping between the plank and the cylinders, points A1 and A2 have the same velocity.

SOLUTIONS – MOCK TEST-10

213

Angular velocity of the larger cylinder is 2v v = 4R 2R

A2

A1 R

2R

2v v

CHEMISTRY 31. (a) No. of molecules in different cases (a) Q 22.4 litre at STP contains = 6.023 × 1023 molecules of H2 \ 15 litre at STP contains =

C1

= 4.03×1023 molecules of H2 (b) Q 22.4 litre at STP contains = 6.023×1023 molecules of N2 \ 5 litre at STP contains

C2 æ v ö v CM = ( 2R ) ç =v è 2R ÷ø

27. (a) According to given conditions TIR must take place at both the surfaces AB and AC. Hence only option (a) is correct. 28. (d)

v2 =

w2(A2 – x2)

and a2 = (w2x)2 = w4x2 From above equations, we have v2 = -

a2 2

w

… (i) … (ii)

5 ´ 6.023 ´ 10 23 22.4

= 1.344 × 1023 molecules of N2 (c) Q 2 g of H2= 6.023×1023 molecules of H2 0.5 ´ 6.023 ´10 23 2

= 1.505 × 1023 molecules of H2

It represents straight line with negative slope. 29. (b) Sound waves in air are longitudinal while light waves are transverse. Sound waves require material medium to propagate but transverse waves do not require any material medium. 30. (a) For whole system ... (i)

m

////////////////////////////////////// f1 f2

For rear cylinder 100 – f1 = 0.5 (a) For front cylinder f2 = 0.5 (a) From (i), (ii) and (iii) 100 = 4a a = 20 m/s2

=

\ 0.5 g of H2=

+ w2 A2 Þ Y = mX + c

f1 - f 2 = 3 (1)a

15 ´ 6.023 ´ 10 23 22.4

... (ii) ... (iii)

(d) Similarly 10 g of O2 gas =

10 ´ 6.023 ´ 10 23 molecules of O2 32

= 1.88 × 1023 molecules of O2 Thus (a) will have maximum number of molecules. 32. (b) Due to mesomeric effect (+M) of –OH group the electron density on benzene ring increases. So the electrophile easily attack on these electron rich center. 33. (b) During the dissolution of alkali metal hydrides energy is released in large amount, i.e., it is exothermic in nature. 34. (b) PbO2 is a powerful oxidizing agent and liberate O2 when treated with acids. 2PbO 2 + 4HNO 3 ¾¾ ®

2Pb(NO3 ) 2 + 2H 2 O + O 2 ­

35. (d) Copper is more reactive than silver hence displaces silver from silver nitrate solution.

EBD_7839 214 36.

37.

38.

39.

40.

KVPY-SA (d) A compound is said to be aromatic if it is planar and there is complete delocalization of p-electrons, which is only possible if it is a conjugated cyclic system and number of p electrons used in delocalization is (4n + 2). 1, 3, 5-heptatriene is not an aromatic c om pou n d because comp l et e delocalization of p-electrons, is not possible in it. (b) The sub-shells are 3d, 4d, 4p and 5s; among these 4d has highest energy as n + l value is maximum for it. (c) Draw a line at constant P parallel to volume axis. Take volume corresponding to each temperature. From volume axis, V1 > V2 > V3 Hence, T1 > T2 > T3. (a) Higher the value of reduction potential higher will be the oxidising power whereas lower the value of reduction potential higher will be the reducing power. (b) Eq. of acid = Eq of base,

\

BIOLOGY 46.

0.45 20 ´ 0.5 = , = E.Wt = 45 E.Wt 1000

M.Wt 90 = =2 E.Wt 45 41. (a) CaO is basic as it forms strong base Ca(OH)2 on reaction with water. CaO + H2O ¾® Ca(OH)2 CO2 is acidic as it dissolves in water forming unstable carbonic acid. H2O + CO2 ¾® H2CO3 Silica (SiO2) is insoluble in water and acts as a veryweak acid. SnO2 is amphoteric as it reacts with both acid and base. SnO2 + 2H2SO4 ¾® Sn(SO4)2 + 2H2O SnO2 + 2KOH ¾® K2SnO3 + H2O 42. (b) For isomeric alkanes, the one having longest straight chain has highest b.p. because of larger surface area. 43. (a) Total moles Basicity =

=

Þ P = 1.125 ´ .0821 ´ 273 P = 25.22 atm 44. (b) Diacidic base has two replaceable hydroxyl groups. –OH groups present in compounds (a) and (c) represent alcoholic group, not base. 45. (c) When gas is compressed its entropy decreases, so, DS is negative.

4 2 + = 1.125; PV = nRT 32 2

47. 48.

49.

50.

(c) SA (sinoatrial) node lies in the wall of right atrium near the opening of superior vena cava. It initiates the electrical impulse to stimulate contraction. From SA node, cardiac impulse travels to AV node (atrioventricular node, which lies between right atrium and ventricle), then passes to AV bundle (also called bundle of His) and its branches reaches to the Purkinje fibres in ventricles. Purkinje fibres conducts the impulses five times more rapidly than surrounding cells. It forms a pathway for conduction of impulse that ensures that the heart muscle contracts in the most efficient manner. (d) Peroxisomes contain oxidative enzymes such as catalase. (d) For an action potential to occur in an axon, the membrane must be depolarised above a certain level. This level is known as the threshold. (c) In monocot stem, vascular bundles are scattered. Each bundle is surrounded by a thick-walled bundle sheath. Due to disintegration of protoxylem elements, a lysigenous water cavity is formed. (a) Excessive nutrients run off from the land into water body. These nutrients such as nitrates, phosphates and urea act as nutrients and accelerate the growth of algae that may form a mat on water surface. This process of enrichment of water body with nutrients is called entrophication.

SOLUTIONS – MOCK TEST-10 51. (b) Only 10 percent of the energy is transferred between trophic levels. 52. (a) Vaccine contains dead, attenuated form or antigen of a pathogen which can be injected to provide immunity towards that pathogen. Monoclonal antibodies are homogeneous immunological reagents of defined specificity, so that these can be utilised for diagnosis and screening with certainty. 53. (b) Photosynthesis occurs most efficiently with red or blue light because the chlorophyll pigments can absorb light maximally at these wavelengths and the least at green light. 54. (c) Atropine is a substance that blocks the inhibitory effects of muscarinic receptors in muscle tissue, especially the heart. 55. (c) Purkinje fibres are a vital component of the system of blood circulation in humans. These fibres are present at both ventricular myocardium for the proper contraction of ventricles. They receive conductive signals originating at the atrioventricular node (AVN), and simultaneously activate the left and right ventricles by directly stimulating the ventricular myocardium. 56. (c) Oxytocin stimulates muscle contractions during childbirth and milk letdown. 57. (b) The enzyme salivary amylase present in saliva, converts starch into maltose. It is the maltose that tastes sweet. Maltase secreted by the intestine converts maltose into glucose. 58. (a) Resting membrane potential is greater in a large size nerve fibre than in a small size nerve fibre. In larger nerve fibres, it can go up to –80mV and in small size fibre it is –60mV. 59. (d) Lactic acid is a three-carbon organic acid formed in muscle resulting in fatigue. 60. (a) Inhaled air passes from bronchi into bronchioles that lead to alveolar ducts, which terminate at alveoli.

215 PART – II MATHEMATICS 61. (a) Here in quadrilateral APCB, AB || CP Since triangles on the same base and between same parallel lines are equal in area. Þ Area (DAPC) = Area (DBPC) .....(i) Also in quadrilateral ADQC, AD || CQ \ Area (DDCQ) = Area (DACQ) or Area (DDPQ) + Area (DPCQ) = Area (DAPC) + Area (DPCQ) Þ Area (DDPQ) = Area (DAPC) .....(ii) From (i) & (ii) Þ Area (DBPC) = Area (DDPQ)

62. (c) Let t = 5x + 2, then A = {t : 0 £ t £ p} \ f (t) = cos t which is bijective in [0, p]. Hence, f (x) is bijective. 63. (d)

Construction: Join D to C. Q DP || QC \ ar DDPC = ar DQDP

EBD_7839 216

KVPY-SA (same base DP, and between two parallel lines DP and QC) Q D is mid point, \ DC is median, \ ar (DBDC) =

1 ar (DABC) 2

Þ ar (DBDP) + ar (DDPC) = \ ar (DBDP) + ar (DQDP) =

Hence, the required volume of sphere 3

=

65.

(a) There are 2n observations x1, x2, ..., x2n. 2n

1 ar (DABC) 2

So, mean =

part is

n

å xi + 5n .

In second part 3 is subtracted from each

P

So, total of second part is

h

n

2n

i =1

i =n +1

å xi + 5n + å R

2h 3 l Þl= 3 2

3

=

66. (b) Remote half from the sun is the arc B'A'B. ...(1)

OS PS

OS PS Þ OS = = PS 3

h 3

B

A

h 3 3

Therefore, the radius of sphere is r = OS =

2n x xi + 2 n = å i +1 2n 2n i =1 i =1

å

PHYSICS

l h \ In DOPS, PS = = 2 3

1

i =1

So, it increase by 1.

Q ÐOPS = 30°

Þ

2n

xi - 3n = å xi + 2n

2n

Mean =

In DPOS, tan 30° =

xi - 3n.

Total of 2n terms are

l 60° S r

Q

h=

2n

å

i = n +1

6 0° O

i =1

i =1

64. (d) Since vertical angle of the cone = 60º

l

x

å 2ni .

Let these observations be divided into two parts x1, x2, ..., xn and xn+1, ..., x2n Each in 1st part 5 is added, so total of first

1 ar (DABC) 2

1 ar (DABC) 2

\ ar (DBQP) =

4 æ hö 4ph3 4 3 pr = p ç ÷ = 3 è 3ø 81 3

=

h 3

S

C

A'

B'

Let t be the time of description of the arc B'A'B, therefore rate of description of the sectorial area SB'A'BS =

area SB ¢A¢BS t

SOLUTIONS – MOCK TEST-10

217

Since the whole area of the ellipse is described in a year, therefore rate of description of the sectorial area is area of the ellipse = a year

¢ ¢ Hence, area SB A BS = area of the ellipse t a year

or

t area SB¢A¢BS = year area of the ellipse

=

e

=

PAVA n = A PBVB nB

b

1 year + 2 days nearly 2

4 + 2 cot q 2 / sin q or v = 8 v

... (i) dv =0 dq

A

Q æ S Dq ö æ 1 ö = ha ç è h ÷ø çè rg ø÷ A

or, [MT–3] = [Ma–cL–a + b + 2c T –a – 2b + 2c] Equating powers and solving we get, a = 1, b = 1, c = 0 Q S Dq =h A h

70. (a) Let P be the common point of the two parabolas where the two particles collide. Let v1, v2 be the velocities of m1 and m2 at the time of collision.

C

v0

Truck

v q B

4m

c

[MT–3] = [ML–1T–1]a [LT–2]b [M–1L2T2]c

P 90°

2m

nB = 6. nA

Using dimensional method

\

8 2sin q + cos q

For minimum v,

8 8 = 3.57 m/s = æ 2 ö 1 5 2ç ÷ + è 5ø 5 68. (d) From, PV = nRT, we have

vmin =

69. (d) Let

Solving e = 1/60 67. (c) Let the man starts crossing the road at an angle q as shown in figure. For safe crossing the condition is that the man must cross the road by the time the truck describes the distance 4 + AC or 4 + 2cotq. \

= 0 or 2 cos q – sin q = 0

After substituting values, we get

æ ö \ t = ç + ÷ year = year + year p 2 è2 pø 1

(2 sin q + cos q) 2 tan q = 2 From equation (i),

æ 4T ö 4 3 çè 8 + r ÷ø ´ 3 prA nA A Þ = n B æ 4T ö 4 3 çè 8 + r ÷ø ´ 3 prB B

1 1 pab + .2bae 1 e 2 2 = + = 2 p pab e

or

-8(2 cos q - sin q)

Þ

1 area of the ellipse + D SBB¢ = 2 pab

1

or

S

EBD_7839 218

KVPY-SA Now substituting for v12 from (1), we have

Since the two paths are parabolic, therefore v12 =

2m = v22 R

(at the point P)

m12 + m22 2m æ2 1ö =mç - ÷ ( m1 + m2 )2 R è R a¢ ø

... (i)

é ù m ê acc. = ú (distance) 2 ûú ëê

\

After collision the two masses combine into one, let V be the velocity after impact of the single mass (m1 + m2 ). By th e principle of conservation of momentum the resultant momentum must be the same after and before the impact : 2

2

\ ( m1 + m2 ) V =

m12 v12

+

m22 v22

... (ii)

Because v1 and v2 are at right angles. From (i) it is seen that v2 = v1 and then from (ii), we have

i.e. 2a¢ =

i.e. V 2 =

+ m22

( m1 + m2 ) 2

v12

71.

(a) Total energy (En) = K.E + P.E in first excited state é Ze 2 ù 1 1 Ze 2 Ze2 E = mv 2 + ê ú =+ 2 2 r r êë r úû -3.4 eV = -

... (iii)

2m , which is the condition for R

\ K.E =

1 Ze2 2 r

1 Ze2 = +3.4 eV 2 r

72. (c)

describing an ellipse. So after the collision the path will be an ellipse. Let 2a' be its major axis. For the path to be an ellipse, æ2 1 ö v2 = m ç - ÷ è r a¢ ø

At the point P, just after the collision v = V and r = R. So, V 2 = m æç

2

èR

-

1ö ÷ a¢ ø

Substituting for V2 from (3), we have m12 + m22

( m1 + m2 ) 2

æ2 1ö v12 = m ç - ÷ è R a¢ ø

( m1 + m2 ) 2 ×R 2m1m2

CHEMISTRY

It is seen from (iii) that after the collision the (velocity)2 becomes less than v12, that is less than

(m1 + m2 )2 .R 2 m1m2

That is, major axis =

(m1 + m2 )2 V 2 = (m12 + m22 )v12 m12

1 ìï m2 + m22 üï 2 2m1m2 2 = í1 - 1 × ý× = 2 2 a¢ ïî ( m1 + m2 ) ïþ R (m1 + m2 ) R

C H3

+

SOLUTIONS – MOCK TEST-10 73. (d) Second ionization energy is amount of energy required to take out an electron from the monopositive cation. M (g) ¾¾ ® M 2 + (g) + 2e- ... (5)

M (g) ¾¾ ® M + (g) + e - ... (3) Hence I.E. from (5) – I.E. from (3) will be I.E.2 of M. 74. (c) (i) [Cr(H 2 O)6 ]Cl3 : x + 6 × 0 + (–1) × 3 =0 x =+3 (ii) [Cr(C6H6)2]: y+ 2× 0 = 0 y=0 (iii) K2[Cr(CN)2(O)2 (O2)(NH3)] : 2 × 1 + z + 2 × (–1) + 2 × (–2) + (–2) + 0 =0 z = +6 The oxidation states of Cr in given compounds are +3, 0 an d + 6 respectively. 75. (c) Because of high electronegativity of the halogen atom, the carbon halogen (C – X) bond is highly polarised covalent bond. Thus, the carbon atom of the C – X bond becomes a good site for attack by nucleophiles (electron rich species). Nucleophilic substitution reactions are the most common reactions of alkyl halides. BIOLOGY 76. (b) Water moves from lower osmotic pressure to higher osmotic pressure. So, when a plant cell ‘A’ is kept in solution of 10 atm osmotic pressure, endosmosis will occur till equilibrium is achieved that is, both the solution and plant cell has 11 atm osmotic pressure. When a plant cell ‘B’ is kept in solution of 8 atm osmotic pressure, endosmosis will occur till both solution and plant cell ‘B’ has 9 atm osmotic pressure.

219 When plant cell ‘A’ with 11 atm osmotic pressure is kept in intimate contact with plant cell ‘B’ with 9 atm osmotic pressure, water will move from plant cell ‘B’ to ‘A’. 77. (d) Human beings have 44 autosomes and 2 sex chromosomes in each cell. Females have two X chromosomes (46, XX) while males have one X chromosome and one Y chromosome (46, XY). Females receive one X chromosome from their mother and the other X chromosome from their father. Males receive X chromosome from their mother and Y chromosome from father. So, the female has received her X chromosome having TDF gene from her father. Though TDF gene is found on the Y chromosome but this gene is moved to X chromosome due to pairing (synapsis) of the X and Y ch romosomes and cr ossing over (recombination) between them. 78. (b) Allele for black coat colour = XB Allele for orange coat colour = Xb Since males have only one X chromosome, they can either have allele XB or Xb for black or orange coat colour, respectively. In cont ra st, fema les h ave t wo X ch romosomes so thr ee di fferent genotypes - XBXB, XBXb, XbXb with black, calico and orange coat colour, respectively are possible. But, due to X inactivation, females heterozygous for alleles can have mosaic pattern i.e., patches of both black and orange coat colour. 79. (d) Plant B is a long day plant because it needs a minimum of 15.5 hrs of light to flower. Plant A with critical day length of 15.5 hrs, requires 8.5 or more hours of darkness to flower. So, it is a short day plant. Plant C and D are also short day plants because they flower when they receive light less than critical day length of 10 hrs and 9.5 hrs, respectively.

EBD_7839 220 80.

KVPY-SA (c) The steps in glycolysis are as follows: Glucose first ATP priming 1 reaction ADP Glucose 6-phosphate (a) 2 Fructose 6-phosphate (b) second ATP priming 3 reaction ADP Fructose 1, 6-bisphosphate (c) cleavage of 6-carbon sugar phosphate to two 3-carbon sugar phosphates

4

Glyceraldehyde 3-phosphate (d) + Dihydroxyacetone phosphate (e) 5

Glyceraldehyde 3-phosphate (2) (f) 2P 2NAD+ oxidation and 6 + phosphorylation 2 NADH + H 1, 3-Bisphosphoglycerate (2) (g) 2ADP 7 2 ATP

first ATPforming reaction (substrate-level phosphorylation)

3-Phosphoglycerate (2) 8 2-Phosphoglycerate (2) 9

2H2O

Phosphoenol pyruvate (2) (h) 2ADP 10 2 ATP

second ATPforming reaction (substrate-level phosphorylation)

Pyruvate (2)