Super 10 Mock Tests for NTA IIT JEE Main 2021 with Optional Questions 4th Edition Disha Experts [4 ed.]

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EBD_7308

Contents 1. Mock Test - 1

MT-1–10

2. Mock Test - 2

MT-11–22

3. Mock Test - 3

MT-23–36

4. Mock Test - 4

MT-37–48

5. Mock Test - 5

MT-49–60

6. Mock Test - 6

MT-61–72

7. Mock Test - 7

MT-73–82

8. Mock Test - 8

MT-83–94

9. Mock Test - 9

MT-95–106

10. Mock Test - 10

Mock Test Solutions (1-10)

MT-107–118

MT-119-244

EBD_7308

1 Time : 3 hrs.

Max. Marks : 300 INSTRUCTIONS

1. 2. 3. 4.

5.

6. 7.

This test will be a 3 hours Test. This test consists of Physics, Chemistry and Mathematics questions with equal weightage of 100 marks. Each question is of 4 marks. There are three parts in the question paper consisting of Physics (Q.no.1 to 30), Chemistry (Q.no.31 to 60) and Mathematics (Q. no.61 to 90). Each part is divided into two sections, Section A consists of 20 multiple choice questions & Section B consists of 10 Numerical value answer Questions. In Section B, candidates have to attempt only 5 questions out of 10. There will be only one correct choice in the given four choices in Section A. For each question 4 marks will be awarded for correct choice, 1 mark will be deducted for incorrect choice and zero mark will be awarded for unattempted question. For Section B 4 marks will be awarded for correct answer and zero for marked for each review / unattempted/incorrect answer. Any textual, printed or written material, mobile phones, calculator etc. is not allowed for the students appearing for the test. All calculations / written work should be done in the rough sheet provided.

PHYSICS 2.

Section - A 1.

Two balls of same mass and carrying equal charge are hung by threads of length l from a fixed support. At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, x between the balls is proportional to : (1) l

(2) l 2

(3) l 2/3

(4)

l 1/3

A 20 kg block B is suspended from a cord attached to a 40 kg cart A. Find the ratio of the acceleration of block in cases (i) and (ii) shown in the figure immediately after the system is released from rest. (neglect friction)

B

A

Case (i)

A Case (ii)

B

EBD_7308 JEE MAIN

MT-2

(1) 3.

2 3

(2)

3 2

(3)

3 3 (d) 2 2 2

7.

The diagram showing the variation of gravitational potential of earth with distance from the centre of earth is (1) V (b) V O

R

O

r

A certain amount of gas is taken through a cyclic process (ABCDA) that has two isobars, one isochore and one isothermal. The cycle can be represented on a P-V indicator diagram as : P

(1)

R

B

B P

A

r

C

(2)

D

A

C

D

V

(3)

O

4.

5.

R

V

O

r

P

R

d1K1 + d 2 K 2 d1 + d 2

Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the centre of the ring is m0 q m 0 qf (2) (1) 2f R 2R (3)

m 0q 2pf R

8.

(4)

m 0 qf 2 pR

10.

D

B

C

(4) A

D

V

V A goods train accelerating uniformly on a straight railway track, approaches an electric pole standing on the side of track. Its engine passes the pole with velocity u and the guard’s room passes with velocity v. The middle wagon of the train passes the pole with a velocity. 1 2 u+v u + v2 (1) (2) 2 2

(3) 9.

V

C

P

A

d1 + d 2

(d) (d / K + d / K ) 1 1 2 2

B

(3)

r

In “Al” and “Si”, if temperature is changed from normal temperature to 70 K then (1) The resistance of Al will increase and that of Si will decrease (2) The resistance of Al will decrease and that of Si will increase (3) Resistance of both decrease (4) Resistance of both increase Two rods of length d1 and d2 and coefficients of thermal conductivities K1 and K2 are kept touching each other. Both have the same area of cross-section, the equivalent thermal conductivity is (1) K1 + K2 (2) K1d1 + K2d2 (3)

6.

(d)

V

uv

(4)

æ u 2 + v2 ö ç ÷ 2 ø è

A wheel is rotating at 900 r.p.m. about its axis. When power is cut off it comes to rest in 1 minute. The angular retardation in rad/s2 is (1) p/2 (2) p/4 (3) p/6 (4) p/8 Two springs of force constants 300 N/m (Spring A) and 400 N/m (Spring B) are joined together in series. The combination is compressed by 8.75 cm. The ratio of energy E E stored in A and B is A . Then A is equal EB EB to : 4 16 (1) (2) 3 9 9 3 (3) (4) 16 4

Mock Test -1 11.

12.

13.

A particle of mass m is acted upon by a force F R given by the empirical law F = 2 v(t). If this t law is to be tested experimentally by observing the motion starting from rest, the best way is to plot : 1 (1) log v(t) against (2) v(t) against t2 t 1 (3) log v(t) against 2 (4) log v(t) against t t In case of a p-n junction diode at high value of reverse bias, the current rises sharply. The value of reverse bias is known as (1) cut off voltage (2) zener voltage (3) inverse voltage (4) critical voltage Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choice(s) given below. B/A

MT-3

14.

15.

16.

8 6 4 2 0

(1) (2) (3)

(4)

100 200 A Fusion of two nuclei with mass numbers lying in the range of 1 < A < 50 will release energy Fusion of two nuclei with mass numbers lying in the range of 51 < A < 100 will release energy Fission of a nucleus lying in the mass range of 100 < A < 200 will release energy when broken into two equal fragments Fission of a nucleus lying in the mass range of 120 < A < 180 will release energy when broken into two equal fragments

17.

18.

If 10% of a radioactive material decays in 5 days, then the amount of the original material left after 20 days is approximately (1) 60% (b) 66% (3) 70% (4) 75% When white light passes through a dispersive medium, it breaks up into various colours. Which of the following statements is true? (1) Velocity of light for violet is greater than the velocity of light for red colour. (2) Velocity of light for violet is less than the velocity of light for red. (3) Velocity of light is the same for all colours (4) Velocityof light for different colours has nothing to do with the phenomenon of dispersion A plate of mass (M) is placed on a horizontal frictionless surface and a body of mass (m) is placed on this plate. The coefficient of dynamic friction between this body and the plate is m. If a force 3mmg is applied to the body of mass (m) along the horizontal, the acceleration of the plate will be (1)

mm g M

(2)

mmg M+m

(3)

3mmg M

(4)

2mmg M+m

Lights of two different frequencies, whose photons have energies 1 eV and 2.5 eV respectively, successively illuminate a metal whose work function is 0.5 eV. The ratio of the maximum speeds of the emitted electrons will be (1) 1 : 5 (b) 1 : 4 (3) 1 : 2 (4) 1 : 1 Which of the plots shown in the figure represents speed (vn) of the electron in a hydrogen atom as a function of the principal quantum number (n)?

EBD_7308 JEE MAIN

MT-4

22.

In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance P = 4 W and the neutral point N is at 60 cm from A. Now an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A. The value of unknown resistance R is ________ohm. Q

P

G

19.

20.

(1) B (2) D (3) C (4) A An engine has an efficiency of 1/6. When the temperature of sink is reduced by 62°C, its efficiency is doubled. Temperatures of source and sink are (1) 99°C, 37°C (2) 124°C, 62°C (3) 37°C, 99°C (4) 62°C, 124°C A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCRcircuit. Given that R= 5 W, L= 25 mH and C = 1000 mF. The total impedance, and phase difference between the voltage across the source and the current will respectively be : (1) 10 W and tan–1 æç 5 ö÷ è 3ø (2) 7 W and 45° æ8ö (3) 10 W and tan –1 ç ÷ è3ø 5 æ (4) 7 W and tan–1 ç ö÷ è 3ø Section - B

21.

A toy–car, blowing its horn, is moving with a steady speed of 5 m/s, away from a wall. An observer, towards whom the toy car is moving, is able to hear 5 beats per second. If the velocity of sound in air is 340 m/s, the frequency of the horn of the toy car is close to ________ Hz.

A

E

23.

24.

25.

26.

B

N

Rh

( ) K

The circular head of a screw gauge is divided into 200 divisions and move 1 mm ahead in one revolution. If the same instrument has a zero error of –0.05 mm and the reading on the main scale in measuring diameter of a wire is 6 mm and that on circular scale is 45. The diameter of the wire is ________ mm. The radius of curvature of a thin plano-convex lens is 20 cm (of curved surface) and the refractive index is 1.5. If the plane surface is silvered, then it behaves like a concave mirror of focal length ________ cm. Three resistors of 4 W, 6 W and 12 W are connected in parallel and the combination is connected in series with a 1.5 V battery of 1 W internal resistance. The rate of Joule heating in the 4 W resistor is ________ watt. A bottle has an opening of radius a and length b. A cork of length b and radius (a + Da) where (Da a < < a) is compressed to fit into the opening completely (see figure). If b the bulk modulus of cork is B and frictional coefficient between the bottle and cork is m then the force needed to push the cork into the bottle is (x pmBb)DA. Find the value of x.

Mock Test -1 27.

28.

MT-5

A sinusoidal voltage of amplitude 25 volt and frequency 50Hz is applied to a half wave rectifier using P-n junction diode. No filter is used and the load r esistor is 1000W. The forward resistance Rf of ideal diode is 10W. The percentage rectifier efficiency is _______. An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 sec. What is the linear speed (in cm/s) of the motion?

29.

30.

The magnetic field of earth at the equator is approximately 4 × 10–5 T. The radius of earth is 6.4 × 106 m. Then the dipole moment of the earth of the order of 10x Am2. Find the value of x. A particle starts S.H.M. from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is 3E/4, its displacement a at this instant is y = . Find the value of x. x

CHEMISTRY 34.

Section - A

:

Which of the following resonance structure is lowest in energy? 35.

:

(3)

:

: :

: :

A.

H O – | | B. H – C – C + | O– H O– H B H ||

H O || | H–C –C | B H

: :

31.

H O – | | C. H – C – C | + O– H B H (1) A (2) B (3) C (4) All have same energy Which of the following pairs have identical bond order ? (1) N 2 , O 22+ (2) N 2 , O 2– – (3) N 2 , O 2 (4) O 2+ , N 2 In a compound AOH, electronegativity of ‘A’ is 2.1, the compound would be (1) Acidic (2) Neutral towards acid & base (3) Basic (4) Amphoteric

36.

: :

: :

33.

ClO3-

(4)

ClO-4

Ka Given, HF + H 2 O ¾¾¾ ® H 3O + + F – ; K

|

32.

Which of the following orders is wrong? (1) Electron affinity– N < O < F < Cl (2) Ist ionisation potential – Be < B < N < O (3) Basic property– MgO < CaO < FeO < Fe2O3 (4) Reactivity–Be < Li < K < Cs The following species will not exhibit disproportionation reaction (1) ClO– (2) ClO-2

b ® HF + OH – . F– + H 2 O ¾¾¾ Which relation is correct ?

(1) Kb = Kw

Kb =

1 Kw

Ka = Kw Kb The oxidation states of sulphur in the anions SO32- , S2 O24- and S2 O62- follow the order

(3) Ka × Kb = Kw

37.

(2) (4)

(1)

SO 32 - < S 2 O 24 - < S2 O 62 -

(2)

S 2 O 24- < S2 O 62 - < SO 32 -

(3)

S 2O 62- < S2 O 24- < SO32-

(4)

S 2 O 24- < SO 32 - < S 2 O 62 -

EBD_7308 JEE MAIN

MT-6

38.

39.

Among the electrolytes Na 2 SO 4 , CaCl 2 , Al 2 (SO 4 ) 3 and NH4 Cl, the most effective coagulating agent for Sb2S3 sol is (1) Na2SO4 (2) CaCl2 (3) Al2(SO4)3 (4) NH4Cl Which of the following will not be soluble in sodium carbonate solution? OH

O2N

COOH

NO2

(1)

44.

45.

(b) 46. NO2 OH

SO2OH

NO2

(3) 40.

41. 42.

43.

(4)

Although Al has a high oxidation potential it resists corrosion because of the formation of a tough, protective coat of (1) Al(NO3)2 (2) AlN (3) Al2O3 (4) Al2(CO3)2 Which is used as medicine? (1) PVC (2) Terylene (3) Glyptal (4) Urotropine In Lassaigne’s test, the organic compound is fused with a piece of sodium metal in order to (1) increase the ionisation of the compound (2) decrease the melting point of the compound (3) increase the reactivity of the compound (4) convert the covalent compound into a mixture of ionic compounds An aqueous solution of colourless metal sulphate M gives a white precipitate with NH4OH. This was soluble in excess of NH4OH. On passing H2S through this solution a white ppt. is formed. The metal M in the salt is (1) Ca (2) Ba (3) Al (4) Zn

47.

Which of the following oxidising reaction of KMnO4 occurs in acidic medium? (i) Fe2+ (green) is converted to Fe3+ (yellow). (ii) Iodide is converted to iodate. (iii) Thiosulphate oxidised to sulphate. (iv) Nitrite is oxidised to nitrate. (1) (i) and (iii) (2) (i) and (iv) (3) (iv) only (4) (ii) and (iv) Which of the following compound cannot be used in preparation of iodoform? (1) CH3CHO (2) CH3COCH3 (3) HCHO (4) 2-propanol Anhydrous AlCl3 cannot be obtained from which of the following reactions ? (1) Heating AlCl3.6H2O (2) By passing dry HCl over hot aluminium powder (3) By passing dry Cl2 over hot aluminium powder (4) By passing dry Cl2 over a hot mixture of alumina and coke Identify X in the sequence given : NH 2 CHCl3

KOH

(Y)

HCl

(300 K)

X + methanoic acid

Cl

(1)

NH2

(2)

C

N

Cl

(3)

N

C

Cl

(4)

CH3–NH

Cl

Cl

Mock Test -1 48.

Cl

Select the rate law that corresponds to the data shown for the following reaction A + B ¾ ¾® C Expt. No. (i) (ii) (iii) (iv)

49.

MT-7

[A] 0.012 0.024 0.024 0.012

[B] 0.035 0.070 0.035 0.070

Initial Rate 0.10 0.80 0.10 0.80

(1) Rate = K[B]3 (2) Rate = K [B]4 3 (3) Rate = K [A] [B] (4) Rate = K [A]2 [B]2 An alkene upon ozonolysis yield CHO – CH2– CH2– CH2 – CHO only. The alkene is (1) CH2= CH—CH2— CH2 — CH2— CH2 — CH3

Cl Cl

53.

54.

55.

56.

(4)

50. An inorganic compound gives off O2 when heated, turns an acidic solution of KI violet and reduces acidified KMnO4. The compound is (1) SO3 (2) KNO3 (3) H2O2 (4) All of these

57.

Section - B 51.

Ionization energy of gaseous Na atoms is 495.5 kJ mol–1. Calculate the lowest possible frequency of light that ionizes a sodium atom in terms of x × 1015s–1 (h = 6.626 × 10–34 Js, NA = 6.022 × 1023 mol–1) Cl

52.

of H2SO4, then calculate the number of moles of H2SO4 left in terms of x × 10–3. The initial volume of a gas cylinder is 750.0 mL. If the pressure of gas inside the cylinder changes from 840.0 mm Hg to 360.0 mm Hg, calculate the final volume of the gas. In an amino acid, the carboxyl group ionises at pKa 1 = 2.34 and ammonium ion at pK a 2 = 9.60.

(2)

(3)

Cl

If 3.01 × 1020 molecules are removed from 98 mg

The dipole moment of chlorobenzene

58.

Fe +3 + e – ¾ ¾® Fe +2 ; E° = + 0.77V and ¾® Fe ; E° = – 0.44V. Calculate the Fe +2 + 2e - ¾

¾® Fe E° of the half cell Fe +3 + 3e – ¾

59. is 60.

1.5 D. Find the dipole moment of

Find the isoelectric point (pI) of the amino acid AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2, AB and B2 are in the ratio 1:1 :0.5 and enthalpy of formation of AB from A2 and B2 is –100 kJ mol–1. Calculate the bond energy of A2 A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol–1) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm–3, calculate molar mass of the substance If the following half cells have the E° values as

What is the oxidation number of Mn in the product of alkaline oxidative fusion of MnO2. A solid AB crystallises as NaCl structure and the radius of the cation is 0.100nm. Calculate the maximum radius of the anion

EBD_7308 JEE MAIN

MT-8

MATHEMATICS Section - A 61.

(3)

p (2) 2

p 4

(4) None 5

The value of

å r =1

n

r

n

Cr

C r -1

(1) 5 (n – 3) (3) 5n 63.

14

67. =

(2) 5 (n – 2) (4) 5 (2n – 9)

68.

å 17-i C6 = i =1

64.

(1)

16

C7

(2)

17

C7

(3)

17

C8

(4)

16

C8

If ey(x + 1) = 1, then, (1)

d2y dx 2

69.

2

3

æ dy ö (4) 1 ç dx ÷ è ø Let ABC be a triangle with vertices at points A (2, 3, 5), B (–1, 3, 2) and C (l, 5, m) in three dimensional space. If the median through A is equally inclined with the axes, then (l, m) is equal to : (1) (10, 7) (2) (7, 5) (3) (7, 10) (4) (5, 7) (3)

65.

æ4ö cos -1 ç ÷ è9ø

æ3ö (2) cos -1 ç ÷ è9ø

(3)

æ2ö cos -1 ç ÷ è9ø

æ1ö (4) cos -1 ç ÷ è9ø

The contrapositive of (p Ú q) Þ r is

ò (x - b)

(2) ~ r Þ (p Ú q)

dx

(x - a)(b - x )

is

(1)

2 x -α +C α -β β - x

(2)

2 (x - α) ( β - x) + C α -β

a -b (x - a ) b - x 2 (4) None of these. Consider the following planes P : x + y – 2z + 7 = 0 Q : x + y + 2z + 2 = 0 R : 3x + 3y – 6z – 11 = 0

(3)

70.

r Þ (p Ú q)

(3) ~ r Þ ~ p Ù ~ q (4) p Þ (q Ú r) If (2, 3, 5) are ends of the diameter of a sphere x2 + y2 + z2 – 6x – 12y – 2z + 20 = 0. Then coordinates of the other end are (1) (4, 9, –3) (2) (4, 3, 5) (3) (4, 3, –3) (4) (4, –3, 9)

is

æ dy ö (2) ç ÷ è dx ø

dy dx

(1)

(1)

3

C7 +

The angle between the two lines x +1 y + 3 z - 4 x - 4 y + 4 z +1 = = = = & is -1 2 2 1 2 2

If 2 sec 2a = tan b + cot b then one of the values of (a + b) = (1) p

62.

66.

Mock Test -1 (1) (2) (3) (4) 71.

MT-9

75.

P and R are perpendicular Q and R are perpendicular P and Q are parallel P and R are parallel

2 é cos q sin qù and A = ê cos q ú sin 2 q úû êëcos q sin q

é cos 2 f cos f sin fù B= ê ú , q – f is sin 2 f ûú ëêcos f sin f

p4 , then the value If 4 + 4 + 4 + ..... + ¥ = 90 1 2 3 1

1

1

p 2 (2) an odd multiple of p

1

1 1 of 4 + 4 + 4 + ......¥ is 1 3 5

(1)

72.

p4 96

(1) an odd multiple of

(2)

p4 45

(3) an even multiple of

89 4 p (3) (4) None of these 90 The domain of the function

76.

f (x) = exp( 5x - 3 - 2x 2 ) is

73.

(1) [3/2, ¥) (2) [1, 3/2] (3) (–¥, 1] (4) (1, 3/2) The value of the determinant

77.

1 a a2 cos(n –1) x cosnx cos (n + 1) x is zero, if a ¹ 1 sin (n –1) x sinnx sin (n + 1) x

74.

If AB = 0, then for the matrices

(1) sin x = 0

(2) cos x = 0

(3) a = 0

(4) cos x =

1 + a2 2a

æ 2p ö æ 2p ö If a = cos ç ÷ + i sin ç ÷ , then the quadratic è 7ø è 7ø equation whose roots are a = a + a2 + a4 and b = a3 + a5 + a6, is (1) x2 – x + 2 = 0 (2) x2 + x – 2 = 0 2 (3) x – x – 2 = 0 (4) x2 + x + 2 = 0

78.

p 2

(4) 0 If f(x) = xex(1 – x), x Î R , then f(x) is (1) decreasing on [–1/2, 1] (2) decreasing on R (3) increasing on [–1/2, 1] (4) increasing on R The area bounded by the curves x = y2 and 2 x= is 1 + y2 2 2 (1) p (2) p + 3 3 2 (4) None of these 3 An inverted cone is 10 cm in diameter and 10 cm deep. Water is poured into it at the rate of 4cm3/min. When the depth of water level is 6 cm, then it is rising at the rate

(3)

-p -

(1)

9 cm 3 / min . 4p

(2)

1 cm 3 / min . 4p

(3)

1 cm 3 / min . 9p

(4)

4 cm 3 / min . 9p

EBD_7308 JEE MAIN

MT-10

79.

The equation of tangent to 4x2 – 9y2 = 36 which is perpendicular to straight line 5x + 2y – 10 = 0 is æ

(1) 5 (y–3) = 2çç x è

117 ö÷ 2 ÷ø

84.

(2) 2y – 5x + 10– 2 18 = 0 (3) 2y – 5x – 10– 2 18 = 0 (4) None of these 80.

log p

òlog (1) (3)

p2

ö æ1 e 2 x sec 2 ç e 2 x ÷ dx is equal to : ø è3

(2)

3 3 3 2

(4)

1 2 3

82.

log an+1

log an +2

log an+1

log an +2

log an +3

log an+ 2

log an +3

log an +4

86. 87. 88.

If the foci of the ellipse

x2 y 2 + = 1 coincide 16 b 2

then value of b2 is ________. If f(x) = | x – 2| and g(x) = f (f(x)) then for x > 10, g'(x) is. If ab2c3, a2 b3c4, a3b4c5 are in A.P. (a, b, c > 0), then the minimum value of a + b + c is. The sum of the coefficients in the expansion of æ 2 1ö çè x - ÷ø 3

199

1ö æ ´ ç x3 + ÷ è 2ø

200

is ________.

89.

If the median and the range of four numbers {x, y, 2x + y, x – y}, where 0 < y < x < 2y, are 10 and 28 respectively, then the mean of the numbers is _________.

90.

If

is _____.

The probability that in the random arrangement of the letters of the word ‘UNIVERSITY’, the two I’s does not come together is ______.

A point is selected at random from the interior of a circle. The probability that the point is close to the centre, than the boundary of the circle, is ________. Three persons A, B, C throw a die in succession. The one getting 'six' wins. If A starts then the probability of B winning is __________.

2 2 with the foci of the hyperbola x - y = 1 , 144 81 25

3

If a1, a2, a3, ................. are positive numbers in G.P. then the value of log an

85.

1

Section - B 81.

83.

ò cos3 x

dx

= (tan x)A + C(tan x)B + k ,

2sin 2x where k is a constant of integration, then value of A + B + C is _________.

2 Time : 3 hrs.

Max. Marks : 300

INSTRUCTIONS 1. 2. 3. 4.

5.

6. 7.

This test will be a 3 hours Test. This test consists of Physics, Chemistry and Mathematics questions with equal weightage of 100 marks. Each question is of 4 marks. There are three parts in the question paper consisting of Physics (Q.no.1 to 30), Chemistry (Q.no.31 to 60) and Mathematics (Q. no.61 to 90). Each part is divided into two sections, Section A consists of 20 multiple choice questions & Section B consists of 10 Numerical value answer Questions. In Section B, candidates have to attempt only 5 questions out of 10. There will be only one correct choice in the given four choices in Section A. For each question 4 marks will be awarded for correct choice, 1 mark will be deducted for incorrect choice and zero mark will be awarded for unattempted question. For Section B 4 marks will be awarded for correct answer and zero for marked for each review / unattempted/incorrect answer. Any textual, printed or written material, mobile phones, calculator etc. is not allowed for the students appearing for the test. All calculations / written work should be done in the rough sheet provided.

PHYSICS Section - A 1.

2.

N divisions on the main scale of a vernier callipers coincide with N + 1 divisions on the vernier scale. If each division on the main scale is of ‘a’ units, the least count of the instrument is a a (1) (2) N +1 N -1 2a 2a (3) (4) N -1 N +1 The gravitational field, due to the 'left over part' of a uniform sphere (from which a part as shown,

has been 'removed out'), at a very far off point, P, located as shown, would be (nearly) : Mass of complete sphere = M

Removed Part R

P

R

x

(1) (3)

5 GM 6 x2 7 GM 8 x2

(2) (4)

8 GM 9 x2 6 GM 7 x2

EBD_7308 JEE MAIN

MT-12

3.

n moles of an ideal gas undergo a process A ® B as shown in the figure. Maximum temperature of the gas during the process is (1)

(3) 4.

nR

9P0V0 2nR

(2)

(4)

3P0V0 2 nR

9P0V0 4nR

(2) (3) (4)

B

P0 P

V

V0 2V 0

8.

1 20 23 ® 10 Ne + 24 He 11 Na + 1H ¾¾ 4 1 10 ® 13 5 B + 2 He ¾¾ 7 N + 1H 1 10 ® 11 5 B + 0 n ¾¾ 5 B+b + n 1 14 ® 12 7 N + 1H ¾¾ 6 C+b + n

White light is incident on the interface of glass and air as shown in the figure. If green light is just totally internally reflected then the emerging ray in air contains (1) yellow, orange, red (2) violet, indigo, blue (3) All colours (4) All colours

6.

A

2P0

Which of these nuclear reactions is possible – (1)

5.

9 P0V0

7.

Air Glass

A body is executing simple harmonic motion. At a displacement x from mean position, its potential energy is E1 = 2J and at a displacement y from mean position, its potential energy is E2 = 8J. The potential energy E at a displacement (x + y) from mean position is (1) 10J (2) 14J (3) 18J (4) 4J A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width ‘d’. If a be the angle of deviation of proton from initial direction of motion (see figure), the value of sin a will be : Bd 2m

(1)

qV

(2)

B qd 2 mV

(3)

B q d 2mV

B a

Green

d

White

except green What is the conductivity of a semiconductor sample having electron concentration of 5 × 1018 m–3, hole concentration of 5 × 1019 m–3, electron mobility of 2.0 m2 V–1 s–1 and hole mobility of 0.01 m2 V–1 s–1 ? (Take charge of electron as 1.6 × 10–19 C) (1) 1.68 (W – m)–1 (2) 1.83 (W – m)–1 –1 (3) 0.59 (W – m) (4) 1.20 (W – m)–1

(4) Bd 9.

q 2mV

Two wires are made of the same material and have the same volume. However wire 1 has cross-sectional area A and wire 2 has crosssectional area 3A. If the length of wire 1 increases by Dx on applying force F, how much force is needed to stretch wire 2 by the same amount? (1) 4 F

(2) 6 F

(3) 9 F

(4) F

Mock Test -2 10.

A homogeneous solid cylinder of length L (L < H/2). Cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquidliquid interface with length L/4 in the denser liquid as shown in the fig. The lower density liquid is open to atmosphere having pressure P0. Then density of solid is given by (1)

5 d 4

(2)

4 d 5

(3) d

11.

d

H/2

L

(3) is p ml 3qE

b ( m - 1) æ t1 ö l çè t ÷ø 2

ml but ¹ p 3qE E A thin convex lens of focal length ‘f’ is put on a plane mirror as shown in the figure. When an object is kept at a distance ‘a’ from the lens mirror combination, its image is formed at a a distance in front of the combination. The 3 value of ‘a’ is : (4) is proportional to

13.

1

(1) 3f

2d

H/2

(2)

mb t1 l t2

l b ( m - 1) t1 - t2 ) (4) ( m - 1) ( t1 + t2 ) ( b l A dipole consisting of two charges +q and –q joined by a massless rod of length l, is seen oscillating with a small amplitude in a uniform electric field of magnitude E. The period of oscillation (1) is proportional to E (3)

(2) is proportional to 1/E

3L/4

d (4) 5 In a Young’s double slit experiment with light of wavelength l, fringe pattern on the screen has fringe width b. When two thin transparent glass (refractive index m) plates of thickness t1 and t2 (t1 > t2) are placed in the path of the two beams respectively, the fringe pattern will shift by a distance

(1)

12.

MT-13

(2)

3 f 2

(3) f 14.

(4) 2f A sinusoidal wave is propagating in negative xdirection in a string stretched along x-axis. A particle of string at x = 2m is found at its mean position and it is moving in positive y-direction at t = 1 sec. The amplitude of the wave, the wavelength and the angular frequency of the wave are 0.1 meter, p/2 meter and 2p rad/sec respectively. The equation of the wave is (1) y = 0.1 sin (4p (t – 1) + 8 (x – 2)) (2) y = 0.1 sin ((t – 1) – (x – 2)) (3) y = 0.1 sin (2p (t – 1) + 4 (x – 2)) (4) None of these

EBD_7308 JEE MAIN

MT-14

15.

v

Velocity–time graph for a body of mass 10 kg is shown in figure. Work–done on the body in first two seconds of the motion is : v (m/s) -1

x

50 ms

a

(1)

a x

(2)

a

(0,0)

16.

17.

18.

10s

t(s)

(2) 12000 J (1) – 9300 J (3) –4500 J (4) –12000 J A charged particle of mass m and having a charge Q is placed in an electric field E which varies with time as E = E0 sin wt. What is the amplitude of the S.H.M. executed by the particle? QE 0 1 QE 0 (2) (1) 2 2 mw 2 mw 2QE 0 (3) (4) None of these mw 2 In a photoelectric experiment, with light of wavelength l, the fastest electron has speed v. If the exciting wavelength is changed to 3l/4, the speed of the fastest emitted electron will become – 3 4 (2) v (1) v 4 3 4 4 (3) less than v (4) greater than v 3 3 A particle moves along x-axis with initial position x = 0. Its velocity varies with x-coordinate as shown in graph. The acceleration ‘a’ of this particle varies with x as –

(3)

19.

20.

x

a x (4)

x

A ring of mass M and radius R lies in x-y plane with its centre at origin as shown. The mass distribution of ring is non-uniform such that at any point P on the ring, the mass per unit length is given by l = l0 cos2q (where l0 is a positive constant). Then the moment of inertia of the ring about z-axis is – y (1) MR2 1 P M MR 2 (2) R 2 q x 1M (3) 2 l R 0 1 M (4) p l R 0 The electric potential V is given as a function of distance x by V = (5x2 + 10x – 9) volt. Value of electric field at x = 1m is (1) –20 V/m (2) 6 V/m (3) 11 V/m (4) 20 V/m

Mock Test -2

MT-15

26.

Section - B 21.

The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle q should be _____ degree.

A series LR circuit is connected to an ac source of frequency w and the inductive reactance is equal to 2R. A capacitance of capacitive reactance equal to R is added in series with L and R. The ratio of the new power factor to the old one is

27. q 2m m

22.

23.

24.

25.

m

A current of 4A produces a deflection of 30° in the galvanometer. The figure of merit is _____ A/rad. A uniform thin rod AB of length L has linear bx mass density m (x) = a + , where x is L measured from A. If the CM of the rod lies at æ 7ö a distance of ç ÷ L from A, then a and b are è 12 ø related as b = xa. Find the value of x. A transmitting antenna at the top of a tower has height 32 m and height of the receiving antenna is 50 m. What is the maximum distance (in km) between them for satisfactory communication in line of sight (LOS) mode? A ball of mass 160 g is thrown up at an angle of 60° to the horizontal at a speed of 10 ms–1. The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is _____ kg m2/s. (g = 10 ms–2)

28.

29.

30.

x . Find the value of x. 2

A mass of 50g of water in a closed vessel, with surroundings at a constant temperature takes 2 minutes to cool from 30°C to 25°C. A mass of 100g of another liquid in an identical vessel with identical surroundings takes the same time to cool from 30° C to 25° C. The specific heat of the liquid is _____ kcal/kg. (The water equivalent of the vessel is 30g.) The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is _____ V/m. An ideal monatomic gas with pressure P, volume V and temperature T is expanded isothermally to a volume 2V and a final pressure Pi. If the same gas is expanded adiabatically to a volume 2V, P the final pressure is Pa. The ratio a is _____. Pi Six resistors of 10W each are connected as shown. The equivalent resistance between points X and Y is _____. X

Y

EBD_7308 JEE MAIN

MT-16

CHEMISTRY Section - A 31.

Consider the following reactions (i) Cd 2+ (aq ) + 2e - ¾¾® Cd (s) , E° = – 0.40 V (ii) Ag + (aq) + e- ¾¾® Ag(s), E° = 0.80 V For the galvanic cell involving the above reactions. Which of the following is not correct? (1) Molar concentration of the cation in the cathodic compartment changes faster than that of the cation in the anodic compartment. E cell increase when Cd2+ solution is diluted. (3) Twice as many electrons pass through the cadmium electrode as through silver electrode. (4) E cell decreases when Ag + solution is diluted. A container contains certain gas of mass ‘m’ of high pressure. Some of the gas has been allowed to escape from the container and after some time the pressure of the gas becomes half and its absolute temperature 2/3 rd. The amount of the gas escaped is (1) 2/3 m (2) 1/2 m (3) 1/4 m (4) 1/6 m The hybrid state of S in SO3 is similar to that of (1) C in C2H2 (2) C in C2H4 (3) C in CH4 (4) C in CO2 Choose the correct options – (1) Tranquillizers are called psychotherapeutic drugs. (2) Vitamin C, Vitamin E and b-carotene are antioxidants. (2)

32.

33.

34.

35.

36.

(3) Sodium or potassium salt of a long chain fatty acid is called soap. (4) All of these. Electrode potential of the half cell Pt (s) | Hg (l) | Hg2Cl2(s) | Cl– (aq) can be increased by : (1) Increasing [Cl–] (2) Decreasing [Cl–] (3) Increasing Hg2Cl2 (s) (4) Decreasing Hg (l) What is the name of the following reaction :

COOEt EtONa

¾¾® COOEt

37.

(1) Knoevenagel reaction (2) Perkin reaction (3) Reformatsky reaction (4) Dieckmann’s condensation. Choose the correct options – (1) The change of atmospheric temperature with altitude is called the lapse rate. (2) The gases responsible for greenhouse effects are CO2, water vapour, CH4 and ozone. (3) The order of contribution to the acid rain is H2SO4 < HNO3 < HCl (4) Both (1) and (2)

Mock Test -2 38.

MT-17

Optically active isomer (A) of (C 5H9Cl) on treatment with one mole of H2 gives an optically inactive compound (B) compound (A) will be: (1) CH3 CH CH CH2

43.

CH2Cl

(2)

Cl

CH

CH

CH

CH 3

CH 3

39.

40.

41.

42.

(3)

CH3

CH

CH2

CH

CH2

(4)

CH3

Cl CH2

CH

CH

CH2

Cl Which statement is/are correct about ICl 3 molecule ? (1) All I – Cl bonds are equivalent. (2) Molecule is polar and non-planar. (3) All adjacent bond angles are equal. (4) All hybrid orbitals of central atom having equal s-character. Isopropyl alcohol is obtained by reacting which of the following alkenes with concentrated H2SO4 followed by boiling with H2O? (1) Ethylene (2) Propylene (3) 2-Methylpropene (4) Isoprene In electro-refining of metal the impure metal is used to make the anode and a strip of pure metal as the cathode, during the electrolysis of an aqueous solution of a complex metal salt. This method cannot be used for refining of (1) silver (2) copper (3) aluminium (4) sodium Which of the following compounds does not give nitrogen gas on heating?

44.

(1) NH4NO2 (2) (NH4)2SO4 (3) NH4ClO4 (4) (NH4)2Cr2O7 Flocculation value of BaCl2 is much less than that of KCl for sol A and flocculation value of Na2SO4 is much less than that of NaBr for sol B. The correct statement among the following is : (1) Both the sols A and B are negatively charged. (2) Sol A is positively charged and Sol B is negatively charged. (3) Both the sols A and B are positively charged. (4) Sol A is negatively charged and sol B is positively charged. dil. OH –

HgSO

4 ® (A) ¾¾¾¾® (B) HC º CH ¾¾¾¾ H SO 5 ºC 2

4

Give the IUPAC name of “B” is – (1) 2-butanal (2) 3-hydroxy butanal (3) 3-formyl-2-propanol (4) 4-oxo-2-propanol Br

Alcoholic KOH/ D

2/ hn ® Major (X) ¾¾¾¾¾ ¾¾¾¾ ®

45.

H - Br Peroxide

Major (Y) ¾¾¾¾® Major (Z)

Major final product (Z) is – Br

(1)

(2) Br

(3)

Br

(4)

Br

EBD_7308 JEE MAIN

MT-18

46.

Choose the incorrect option – (1) When a dilute solution of an acid is added to a dilute solution of a base, neutralization reaction takes place. (2) In acid-base titrations, at the end point, the amount of acid becomes chemically equivalent to the amount of base present. (3) In the case of a strong acid and a strong base titration, at the end point the solution becomes acidic. (4) In acid-base titrations the end point is determined by the hydrogen ion concentration of the solution. NO2 Br Fe, D

O

NO2

O

(4)

(X)

49. OCH3 NO2

(3)

48.

SOCl

C H NH

6 5 2® 2 ® C ¾¾¾¾¾ A ¾¾¾¾ ® B ¾¾¾¾ (ii) HÅ

Conc. HNO

NaOH

3 ® E (Major) D ¾¾¾¾¾ E can be – Conc. H 2SO4 (1 equivalent)

¾¾¾¾ ® Y+ Z (Y and Z both give

the Iodoform test). The compound X is – (1) CH3 – CH = CH – O – CH2 – CH3

OCH3

OCH3

H 3O +

(C5 H10O)

(2)

(4)

(i) NaOI

NH2

O2N Cl

Br

NO2

NO2 H

(2)

OCH3

NH

(3)

Cl Product (Y) of this reaction is – NO2 NO2

Cl

NH

(2)

CH ONa CH3OH, D

(1)

NO2

O

3 2 ® X ¾¾¾¾¾ ¾¾¾ ®Y

47.

NH

(1)

H | CH 3 - C - O - CH 2 - CH 3 | CH 3

CH3 - C - O - CH 2 - CH3 || CH 2 (4) Both (1) and (3) (3)

Mock Test -2 50.

MT-19

If the elevation in boiling point of a solution of non-volatile, non-electrolytic and nonassociating solute in a solvent (Kb = x K kg mol–1) is y K, then the depression in freezing point of solution of same concentration would be (Kf of the solvent = z K kg mol–1) (1)

2xz K y

(2)

yz K x

(3)

xz K y

(4)

yz K 2x

55.

Rate constant k = 2.303 min–1 for a particular reaction. The initial concentration of the reactant is 1 mol/litre then calculatle the rate of reaction after 1 minute.

56.

How many free radicals can be produced during following reaction (ignoring resonating structure) ? NBS, hn CCl4

¾¾¾¾®

Section - B 51.

The density of CaF2 (fluorite structure) is 3.18 g/cm3. Find the length of the side of the unit cell is _____.

57.

52.

When the following aldohexose exists in its Dconfiguration, the total number of stereoisomers in its pyranose form is _____.

An electron has a speed of 30,000 cm sec–1 accurate upto 0.001%. What is the uncertainty (in cm) in locating it’s position?

58.

The standard enthalpy of formation of NH3 is – 46.0 kJ/mol. If the enthalpy of formation of H2 from its atoms is – 436 kJ/mol and that of N2 is – 712 kJ/mol, calculate the average bond enthalpy of N - H bond in NH3.

59.

The rate coefficient (k) for a particular reaction is 1.3 × 10–4 M–1 s–1 at 100 °C and 1.3 × 10–3 M–1 s–1 at 150 °C. What is the energy of activation (EA) (in kJ/mol) for this reaction? (R = molar gas constant = 8.314 JK–1 mol–1)

60.

The e.m.f. of the cell Zn | Zn 2+ (0.01M) | | Fe2+ (0.001M) | Fe at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction 10x. Find x.

CHO — CH2 — CHOH — CHOH — CHOH — CH2OH 53.

A 25.0 mm × 40.0 mm piece of gold foil is 0.25 mm thick. The density of gold is 19.32 g/cm3. How many gold atoms are in the sheet in terms of x × 1022? (Atomic weight : Au = 197.0)

54.

In a chemical reaction A is converted into B. The rates of reaction, starting with initial con centra tions of A a s 2 × 10 –3 M and 1 × 10–3 M, are equal to 2.40 × 10–4 Ms–1 and 0.60 × 10–4 Ms–1 respectively. Find the order of reaction with respect to reactant A.

EBD_7308 JEE MAIN

MT-20

MATHEMATICS Section - A 66. 61.

62.

63.

64.

65.

2 ì ï 10 - x if - 3 < x < 3 Given f (x) = í ï2 - e x -3 if x ³ 3 î

The graph of f (x) is – (1) continuous and differentiable at x = 3 (2) continuous but not differentiable at x = 3 (3) differentiable but not continuous at x = 3 (4) neither differentiable nor continuous at x = 3 The greatest and the least absolute value of z + 1,where | z + 4 | £ 3 are respectively (1) 6 and 0 (2) 10 and 6 (3) 4 and 3 (4) None of these The equation (5x – 1)2 + (5y – 2)2 = (l2 – 4l + 4) (3x + 4y – 1)2 represents an ellipse if l Î (1) (0, 1] (2) (–1, 2) (3) (2, 3) (4) (–1, 0) If (– 4, 5) is one vertex and 7x – y + 8 = 0 is one diagonal of a square, then the equation of second diagonal is (1) x + 3y = 21 (2) 2x – 3y = 7 (3) x + 7y = 31 (4) 2x + 3y = 21 Which of the following is always true? (1)

(~ p Þ q) = ~ q Þ ~ p

(2) (3)

(~ p Ú q) º Ú p Ú ~ q ~ ( p Þ q) º p Ù ~ q

(4)

~ ( p Ú q) º ~ p Ù ~ q

æ x cos3 x - sin x ö Find ò esin x ç ÷ dx cos 2 x è ø (1) (2) (3) (4)

67.

x esin x – esin x sec x + C x ecos x – esin x sec x + C x2 esin x + esin x sec x + C 2x esin x – esin x tan x + C

The function f : [2, ¥) ® (0, ¥) defined by f (x) = x2 – 4x + a, then the set of values of ‘a’ for which f(x) becomes onto is (1) (4,¥) (2) [4, ¥) (3) {4} (4) f

68.

If a and b are the real roots of the equation x2 – (k – 2) x + (k2 + 3k + 5) = 0 (k Î R) . Find the maximum and minimum values of (a 2 + b 2 ) .

69.

(2) 18, 25/9 (1) 18, 50/9 (3) 27, 50/9 (4) None of these The sum of the coefficient of all the terms in the expansion of (2x – y + z)20 in which y do not appear at all while x appears in even powers and z appears in odd powers is – (1) 0

(2)

2 20 - 1 2

(3) 219

(4)

320 - 1 2

Mock Test -2 70.

71.

All the five digit numbers in which each successive digit exceeds is predecessor are arranged in the increasing order. The (105)th number does not contain the digit (1) 1 (2) 2 (3) 6 (4) All of these ˆ cˆ are non-coplanar unit vectors such that ˆ b, If a,

ˆ = aˆ ´ (bˆ ´ c)

72.

73.

74.

MT-21

75.

76.

(1)

3 x+ y =0 2

(2) x + 3 y = 0

(3)

3x + y = 0

(4) x +

3 y = 0. 2

2 1 + sin xdx = -4 cos (ax + b) + C then

the value of (a, b) is : 1 p , 2 4 (3) 1, 1

(1)

(2) a –1/2

(3) a 1/2

(4) None

The area bounded by the curve y2 (2a – x) = x3 and the line x = 2a is (1) 3pa2 sq. unit

(3) 77.

p 2 (4) None of these

(2) 1,

3pa 2 sq. unit 4

(2)

3pa 2 sq. unit 2

(4)

6pa 2 sq. unit 5

The expression satisfying the differential

(

+ 2 xy = 1 is ) dy dx

2 equation x - 1

Let P = (–1, 0), Q = (0, 0) and R = (3, 3 3 ) be three points. The equation of the bisector of the angle PQR is

ò

(1) a –4/3

1

ˆ then the angle between (bˆ + c) 2 the vectors aˆ , bˆ is (1) 3p/4 (2) p/4 (3) p/8 (4) p/2 The greatest and the least values of |z1 + z2| if z1 = 24 + 7i and |z2| = 6 respectively are (1) 31, 19 (2) 19, 31 (3) –19, –31 (4) –31, –19

If

If the tangent at any point on the curve x4 + y4 = a4 cuts off intercepts p and q on the co-ordinate axes then the value of p–4/3+q–4/3 is

(1)

x 2 y - xy 2 = c

(2)

( y 2 - 1) x = y + c

(3)

( x 2 - 1) y = x + c

(4) None of these 78.

Let f : R ® R and fn (x) = f (fn–1 (x))

n ³ 2, n Î N,

the roots of equation f3(x) f2(x) f (x) – 25f2(x) f (x) + 175 f (x) = 375, which also satisfy equation f (x) = x will be (1) 5

(2) 15

(3) 10

(4) Both (1) and (2)

EBD_7308 JEE MAIN

MT-22

79.

sin 2 x

ò0

80.

84.

The value of

sin

-1

t dt + ò

cos 2 x 0

cos

(1)

p

p (2) 2

(3)

p 4

(4) 1

If a is real and

-1

q2

q1

(1) [1, 2]

(2) [–1, 1]

(3) [1, ¥]

(4) [2–1/2, 21/2]

85.

86.

87.

Section - B

82.

83.

If two vertical poles 20 m and 80 m high stand apart on a horizontal plane, then the height (in m) of the point of intersection of the lines joining the top of each pole to the foot of other is _____. 3 then the value of cos q + i sin q + 2 4x – x2 – y2 is _____.

If x + iy =

Box contains 2 one rupee, 2 five rupee, 2 ten rupee and 2 twenty rupee coin. Two coins are drawn at random simultaneously. The probability that their sum is Rs. 20 or more, is _____.

501p where K

1003p p and q1 = . The value of K is 2008 2008 _____. The expansion of (1 + x)n has 3 consecutive terms with coefficients in the ratio 1 : 2 : 3 and can be written in the form nCk : nCk+1 : nCk+2. The sum of all possible values of (n + k) is _____. The mean and standard deviation of 6 observations are 8 and 4 respectively. If each observation is multiplied by 3, the new standard deviation of the resulting observations is _____.

q2 =

2ax + sin By + cos Bz = 0 ,

then the set of all values of a for which the system of linear equations has a non-trivial solution, is–

dq

ò 1 + tan q =

t dt is

x + cos By + sin Bz = 0 , – x + sin By – cos Bz = 0,

81.

The value of the definite integral,

88.

89.

90.

The value of lim

x ®0

tan x tan x - sin x sin x x3. x

is

_____. Three people each flip two fair coins. The probability that exactly two of the people flipped one head and one tail, is _____. If P(S) denotes the set of all subsets of a given set S, then the number of one-to-one functions from the set S = {1, 2, 3} to the set P(S) is _____. A triangle ABC satisfies the relation 2 sec 4C + sin 2 2A + sin B = 0 and a point P is taken on the longest side of the triangle such that it divides the side in the ratio 1 : 3. Let Q and R be the circumcentre and orthocentre of D ABC. If PQ : QR : RP = 1 : a : b, then the value of a2 + b2 is _____.

3 Time : 3 hrs.

Max. Marks : 300

INSTRUCTIONS 1. 2. 3. 4.

5.

6. 7.

This test will be a 3 hours Test. This test consists of Physics, Chemistry and Mathematics questions with equal weightage of 100 marks. Each question is of 4 marks. There are three parts in the question paper consisting of Physics (Q.no.1 to 30), Chemistry (Q.no.31 to 60) and Mathematics (Q. no.61 to 90). Each part is divided into two sections, Section A consists of 20 multiple choice questions & Section B consists of 10 Numerical value answer Questions. In Section B, candidates have to attempt only 5 questions out of 10. There will be only one correct choice in the given four choices in Section A. For each question 4 marks will be awarded for correct choice, 1 mark will be deducted for incorrect choice and zero mark will be awarded for unattempted question. For Section B 4 marks will be awarded for correct answer and zero for marked for each review / unattempted/incorrect answer. Any textual, printed or written material, mobile phones, calculator etc. is not allowed for the students appearing for the test. All calculations / written work should be done in the rough sheet provided.

PHYSICS 2.

Section - A 1.

A glass prism of refractive index 1.5 is immersed 4 in water (refractive index ) as shown in figure. 3 A light beam incident normally on the face AB is totally reflected to reach the face BC, if (1)

sin q ³

(2)

sin q ³

(3) (4)

5 9

2 3 8 sin q ³ 9 1 sin q ³ 3

B

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R? (1)

5GmM 6R

(2)

2GmM 3R

(3)

GmM 2R

(4)

3GmM 2R

A q

C

EBD_7308 JEE MAIN

MT-24

3.

What is the maximum value of the force F such that the block shown in the arrangement, does not move?

60º

5.

m=

m = 3kg

1 2 3

(3) 12 N (4) 15 N (1) 20 N (2) 10 N A particle falls freely near the surface of the earth. Consider a fixed point O (not vertically below the particle) on the ground. Then pickup the incorrect alternative (1) The magnitude of angular momentum of the particle about O is increasing (2) The magnitude of torque of the gravitational force on the particle about O is decreasing (3) The moment of inertia of the particle about O is decreasing (4) The magnitude of angular velocity of the particle about O is increasing Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons U with internal energy per unit volume u = µ T4 V 1æUö and pressure p = ç ÷ . If the shell now 3è V ø undergoes an adiabatic expansion the relation between T and R is :

Tµ

1 R

(2) T µ

(3) T µ e–R 6.

F

4.

(1)

1 R3

(4) T µ e–3R

Two particles A and B of equal mass M are moving with the same speed v as shown in the figure. They collide completely inelastically and move as a single particle C. The angle q that the path of C makes with the X-axis is given by: Y

C q A

(1) tanq = (2) tanq = (3) tanq = (4) tanq =

30°

X

45°

3+ 2 1- 2

3- 2 1- 2

1- 2 2(1 + 3) 1- 3 1+ 2

B

Mock Test -3 7.

8.

MT-25

The fundamental frequency of a sonometer wire of length l is n 0. A bridge is now introduced at a distance of Dl ( T1 ) . The rate of heat transfer through the slab, in a steady state is

27.

28.

by I = ( e1000V T - 1) mA, where the applied

æ A(T2 - T1 ) K ö ç ÷ f . Find the value of F. x è ø x

K

4x

2K

T1

f to come to rest. If the total distance traversed 2 ft 2 is 15 S , then S = . Find the value of x. x Consider an optical communication system operating at a wavelength of 800 nm. Suppose, only 1% of the optical source frequency is the available channel bandwidth for optical communication. How many channels can be accommodated for transmitting audio signals requiring a bandwidth of 8 kHz ? The current voltage relation of a diode is given

29.

voltage V is in volts and the temperature T is in degree kelvin. If a student makes an error measuring ±0.01 V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in ________ mA? Currents of a 10 ampere and 2 ampere are passed through two parallel thin wires A and B respectively in opposite directions. Wire A is infinitely long and the length of the wire B is 2 m.

Mock Test -3

MT-29

The force acting on the conductor B, which is situated at 10 cm distance from A will be x × 10–5 N. Find the value of x.

30.

Hot water cools from 60°C to 50°C in the first 10 minutes and to 42°C in the next 10 minutes. The temperature of the surroundings is ______ °C.

CHEMISTRY Section - A 31.

Let n1 be the frequency of the series limit of the Lyman series, n2 be the frequency of the first line of the Lyman series, and n3 be the frequency of the series limit of the Balmer series, then – (1) n3 =

1 (n – n ) 2 1 3

(2) n2 – n1 = n3

(3) n1 – n2 = n3 32.

33.

(4) n1 + n2 = n3

Match List I with List II and select the correct answer : List I (Ions) List II (Shapes) A. ICl -2

1.

Linear

B.

BrF2+

2.

Pyramidal

C.

ClF4-

3.

Tetrahedral

D.

AlCl4

4.

Square planar

(1) (2)

5. D 5 3

Angular

A 1 4

B 2 5

C 4 2

34.

(3) 1 2 4 3 (4) 5 1 3 4 Which one of the following is the correct statement? (1) Boric acid is a protonic acid. (2) Beryllium exhibits coordination number of six. (3) Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase. (4) B2 H 6 .2NH 3 is known as ‘inorganic benzene’. Choose the correct option for Anhydrous

O

¾¾¾¾® I HI | Conc. ¾¾¾¾® II HI

(1) I and II are identical (2) I and II are different (3) Mechanism of formation of I and II are not known (4) None of these

EBD_7308 JEE MAIN

MT-30

35.

36.

Reducing the pressure from 1.0 atm to 0.5 atm would change the number of molecules in one mole of ammonia to (1) 25% of its initial value (2) 50% of its initial value (3) 75% of its initial value (4) None of the above Structure of some important polymers are given. Which one represents Buna-S?

(1) (2)

(–CH2 – CH = CH – CH2 – CH – CH2 –)n | C6H5

(3)

(–CH2 – CH = CH – CH2 – CH – CH2 –)n | CN

(4) 37.

CH3 | (–C H2 – C = CH – C H2 –)n

38.

39.

40.

Cl | (–CH2 – C = CH – CH2 –)n

In electrolysis of NaCl when Pt electrode is taken, then H2 is liberated at cathode while with Hg cathode it forms sodium amalgam. This is because (1) Hg is more inert than Pt (2) more voltage is required to reduce H+ at Hg than at Pt (3) Na is dissolved in Hg while it does not dissolve in Pt (4) conc. of H+ ions is larger when Pt electrode is taken

41.

Which of the following option is having maximum number of unpaired electrons – (1) A tetrahedral d6 ion (2) [Co(NH3)6]3+ (3) A square planar d7 ion (4) A co-ordination compound with magnetic moment of 5.92 B.M. Which of the following carbide does not release any hydrocarbon on reaction with water? (1) SiC (2) Be2C (3) CaC2 (4) Mg2C3 An ether (A), C5H12O, when heated with excess of hot concentrated HI produced two alkyl halides which when treated with NaOH yielded compounds (B) and (C). Oxidation of (B) and (C) gave a propanone and an ethanoic acid respectively. The IUPAC name of the ether (A) is: (1) 2-ethoxypropane (2) ethoxypropane (3) methoxybutane (4) 2-methoxybutane Aqueous solution of (M) + (NH4)2S ® yellow (NH ) S

42.

4 2 2 ® insoluble. ppt (B) ¾¾¾¾¾ The cation present in (M) is – (1) CdS (2) SnS2 (3) Cd2+ (4) Sn2+ Both geometrical and optical isomerisms are shown by (1) [Co(en)2Cl2]+ (2) [Co(NH3)5Cl]2+ (3) [Co(NH3)4Cl2]+ (4) [Cr(ox)3]3–

Mock Test -3 43.

MT-31

aq. KOH Me – CH = CH2 + CHCl3 ¾¾¾¾ ® A (Major excess, D products) is –

(1)

2.303 500

(2)

2.303 100 log 500 90

(3)

2.303 90 log 500 100

(4)

100 10 ´ 500

O

(1)

OK OK

(2)

45.

O

The basic character of the transition metal monoxides follows the order (Atomic Nos.,Ti = 22, V = 23, Cr = 24, Fe = 26) (1) TiO > VO > CrO > FeO (2) VO > CrO > TiO > FeO (3) CrO > VO > FeO > TiO

(3)

(4) TiO > FeO > VO > CrO

O 46.

NH4ClO4 + HNO3 (dilute) ¾® X + HClO4 heat

X ¾¾¾ ® Y (gas) Gas (Y) is –

(4)

44.

OH OH

The decomposition of N 2 O 5 in carbon tetrachloride was followed by measuring the volume of O2 gas evolved :

2N 2 O5 ( CCl4 ) ® 2N2 O4 ( CCl 4 ) + O 2 ( g ) . The maximum volume of O2 gas obtained was 100 cm3. In 500 minutes, 90 cm3 of O2 were evolved. The first order rate constant (in min –1) for the disappearance of N2O5 is :

47.

(1) O2

(2) N2

(3) NO2

(4) N2O

Which of the following is/are formed when ozone reacts with the unburnt hydrocarbons in polluted air ? (i)

Formaldehyde

(ii) Acrolein

(iii) Peroxyacetyl nitrate

(iv) Formic acid

(1) (i) and (iv)

(2) (ii) only

(3) (iii) only

(4) (i), (ii) and (iii)

EBD_7308 JEE MAIN

MT-32

CO2 K O3

Conc. KOH

(4)

® (B) ¾¾® (A) ¾¾¾¾¾

48.

Zn (1 mole)

End product (B) of above reaction is –

CO2 K

49.

CHO

(1)

CH2OH

CO2 K

(2) CH2OH

50.

Precautions to be taken in the study of reaction rate for the reaction between potassium iodate (KIO3) and sodium sulphite (Na2SO3) using starch solution as indicator at different concentrations and temperature – (1) The concentration of sodium thiosulphate solution should always be less than the concentration of the potassium iodide solution. (2) Freshly prepared starch solution should be used. (3) Experiments should be performed with the fresh solutions of H2O2 and KI. (4) All of these Predict the product (A) of the following reaction

H+ D

CH2OH

(3) CH2OH

(1)

(2)

(3)

(4)

A

Mock Test -3

MT-33

55.

Section - B

56. 51.

52.

53.

54.

N

OH ¾¾® D

57.

Find the number of a-H in alkene which is major product in this reaction? At 675K, H2(g) and CO2(g) react to form CO(g) and H2O (g), Kp for the reaction is 0.16. If a mixture of 0.25 mole of H2(g) and 0.25 mol of CO2 is heated at 675K, calculate the mole % of CO(g) in equilibrium mixture In the spinel structure, oxides ions are cubicalclosest packed whereas 1/8th of tetrahedral voids are occupied by A2+ cation and 1/2 of octahedral voids are occupied by B3+ cations. The general formula of the compound having spinel structure is ABnO2n. Find the value of n. The value of P° for benzene is 640 mm of Hg. The vapour pressure of solution containing 2.5g substance in 39g benzene is 600mm of Hg. Find the molecular mass of X .

58. 59.

60.

Find the pH of a 2 litre solution which is 0.1 M each with respect to CH 3COOH and (CH3OO)2Ba. (Ka = 1.8 × 10–5) H3PO2 is the molecular formula of an acid of phosphorus. What is its basicity? A gas present in a cylinder fitted with a frictionless piston expands against a constant pressure of 1 atm from a volume of 2 litre to a volume of 6 litre. In doing so, it absorbs 800 J heat from surroundings. Determine increase in internal energy of process. Calculate the number of metamers represented by molecular formula C4H10O. A current of 2.0 A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At wt. = 177). Find the oxidation state of the metal in the metal salt. The instantaneous rate of disappearance of MnO4– ion in the following reaction is 4.56 × 10–3 Ms–1 2MnO4– + 10I– + 16H+

®

2Mn2+ + 5I2 + 8H2O

The rate of appearance I2 is x × 10–2 Ms–1

MATHEMATICS Section - A 61.

Let L1 be a straight line passing through the origin and L2 be the straight line x + y = 1. If the intercepts made by the circle x2 + y2 – x + 3y = 0

on L1 and L2 are equal, then which of the following equation can represent L1 ? (1) x + 7y = 0 (2) x – y = 0 (3) x – 7y = 0 (4) Both (1) and (2)

EBD_7308 JEE MAIN

MT-34

62.

Let a1, a2 and b1, b2 be the roots of ax2 + bx + c = 0 and px2 + qx + r = 0 respectively. If the system of equations a1y + a2z = 0 and b1y + b2z = 0 has a non-trivial solution, then (1)

(3)

b2

ac = 2 pr q a2 p

2

=

bc qr

c2

ab (2) 2 = pq r

67.

(4) None of these

63.

If f ''( x ) < 0, "x Î (a, b), then f '( x) = 0 occurs

64.

(1) exactly once in (a, b) (2) atmost once in (a, b) (3) atleast once in (a, b) (4) None of these If the function f : [0, 16] ® R is differentiable. If 16

0 < a < 1 and 1 < b < 2, then

ò f (t) dt is equal to–

68.

69.

then possible values of [ | 10a | ] where [ . ] denotes the greatest integer function will be (1) 1 (2) 5 (3) 10 (4) Both (1) and (3) The image of the pair of lines represented by: ax2 + 2hxy + by2 = 0 by the line mirror y = 0 is (1) ax2 – 2hxy – by2 = 0 (2) bx2 – 2hxy + ay2 = 0 (3) bx2 + 2hxy + ay2 = 0 (4) ax2 – 2hxy + by2 = 0 If x2 – 2x cos q + 1 = 0, then the value of x2n – 2xn cos nq + 1, n Î N is equal to – (1) cos 2nq (2) sin 2nq (3) 0 (4) some real number greater than 0 1

65.

66.

æ 1+ x2 ö –1 cos (cos–1 x) + sin–1 sin ç ÷ = 2 sec (sec x) è 2 ø

2

3

0

0

0

(1) 4 [a3f (a4) – b3 f (b4)] (2) 4 [a3f (a4) + b3 f (b4)] (3) 4 [a4f (a3) + b4 f (b3)] (4) 4 [a2f (a2) + b2 f (b2)] Three distinct points P (3u2, 2u3), Q (3v2, 2v3) and R (3w2, 2w3) are collinear then – (1) uv + vw + wu = 0 (2) uv + vw + wu = 3 (3) uv + vw + wu = 2 (4) uv + ww + wu = 1 Let ‘a’ denote the roots of equation

1

2

2

1

3

x and I 4 = ò 2 dx then 1

(1) 70.

2

x x x If I1 = ò 2 dx , I 2 = ò 2 dx , I 3 = ò 2 dx

I 2 > I1

(2)

I1 > I 2

(3) I 3 = I 4 (4) I 3 > I 4 The value of the definite integral 3p / 4

ò [(1 + x)sin x + (1 - x) cos x ] dx

is –

0

(1)

2( 2 + 1)

(2)

(3)

2 2

(4) 0

2 +1

Mock Test -3 71.

72.

MT-35

The locus of the centres of the circles which cut the circles x2 + y2 + 4x – 6y + 9 = 0 and x2 + y2 – 5x + 4y – 2 = 0 orthogonally is – (1) 9x + 10y – 7 = 0 (2) x – y + 2 = 0 (3) 9x – 10y + 11 = 0 (4) 9x + 10y + 7 = 0 The sum of the first n terms of the series

76.

along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then (1)

3a 2 - 10 ab + 3b 2 = 0

12 + 2.2 2 + 3 2 + 2.4 2 + 5 2 + 2.6 2 + ...

(2)

3a 2 - 2ab + 3b 2 = 0

n(n + 1)2 when n is even. When n is odd the 2 sum is

(3)

3a 2 + 10 ab + 3b 2 = 0

is

(1) (3) 73.

74.

75.

If the pair of lines ax 2 + 2 (a + b)xy + by 2 = 0 lie

é n(n + 1) ù êë 2 úû n(n + 1) 4

2

(2)

2

(4)

77.

n2 (n + 1) 2

3n(n + 1) 2

The straight line joining any point P on the parabola y2 = 4ax to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equation of the locus of R is – (1) x2 + 2y2 – ax = 0 (2) 2x2 + y2 – 2ax = 0 (3) 2x2 + 2y2 – ay = 0 (4) 2x2 + y2 – 2ay = 0 If z + 1/z = 2 cos q, then the value of |(z2n – 1)/(z2n + 1)| (1) | tan n q | (2) tan n q (3) | cot n q | (4) cot n q The two curves x3 – 3xy2 + 2 = 0 and 3x2y – y3 = 2 (1) cuts at right angle (2) touch each other p p (3) cut at an angle (4) cut at an angle 3 4

(4) 3a 2 + 2ab + 3b 2 = 0 Which of the following is a contradiction? (1)

(p Ù q)Ù ~ (p Ú q)

(3) (p Þ q) Þ p 78.

(2) p Ú ( - p Ù q) (4) None of these

If 1, w, w2 ,........ wn-1 are the n roots of unity,, then (1 - w) (1 - w2 )........(1 - wn-1 ) equals (1) 0 (3) n

79.

(2) 2 (4) n 2

Set of values of m for which two points P and Q lie on the line y = mx + 8 so that Ð APB = Ð AQB p = where A º (– 4, 0), B º (4, 0) is – 2 (1)

( -¥, - 3) È ( 3, ¥ ) - {-2, 2}

(2)

[ - 3, - 3] - {-2, 2}

(3)

(-¥, -1) È (1, ¥) - {-2, 2}

(4)

{- 3, 3}

EBD_7308 JEE MAIN

MT-36

80.

The trace Tr(A) of a 3 × 3 matrix A = (aij ) is defined by the relation Tr(A) = a11 + a22 + a33 (i.e., Tr(A) is sum of the main diagonal elements). Which of the following statements cannot hold ?

86.

A box contains 6 red, 5 blue and 4 white marbles. Four marbles are chosen at random without replacement. The probability that there is atleast one marble of each colour among the four chosen, is ______.

87.

Period of

88.

æ ö 16 sin q ÷÷ If the tangent at the point çç 4 cos q, 11 è ø

(1) Tr(kA) = kTr(A) (k is a scalar) (2) Tr(A + B) = Tr(A) + Tr(B) (3) Tr(I3) = 3 (4) Tr(A2) = Tr(A)2 Section - B 81. 82.

83.

84.

85.

The area above the x–axis enclosed by the curves x2–y2 = 0 and x2 + y – 2 = 0 is_____. If number of permutations 1, 2, 3, 4, 5, 6, 7, 8 and 9 taken all at a time are such that the digit. 1 appearing somewhere to the left of 2, 3 appearing to the left of 4 and 5 somewhere to the left of 6, is equal to k.7!, then value of k is _____. (e.g., 815723946 would be one such permutation) Vertices of a parallelogram taken in order are A (2, –1, 4), B (1, 0, –1), C (1, 2, 3) and D. If distance of the point P (8, 2, –12) from the plane k 6 of the parallelogram is , then value of k is__. 9 Given r r r ˆ B = ˆi + ˆj - 2kˆ and C = ˆi + 2ˆj + kˆ. A = 2iˆ + 3jˆ + 6k, r r r r r The value of | A ´ [A ´ (A ´ B)].C | is _____. If area of triangle formed by common tangents to the circles x2 + y2 – 6x = 0 and x2 + y2 + 2x = 0 is l 3 , then value of l is ____.

89.

sin q + sin 2q Kp is ,then k is ____. cos q + cos 2q 3

to the ellipse 16x2 + 11y2 = 256 is also a tangent to the circle x2 + y2 – 2x = 15, then the value of q is p ± . The value of k is _____. K One percent of the population suffers from a certain disease. There is blood test for this disease, and it is 99% accurate, in other words, the probability that it gives the correct answer is 0.99, regardless of whether the person is sick or healthy. A person takes the blood test, and the result says that he has the disease. The probability that he actually has the disease, is _____. p /2

90.

Let a n =

ò

(1 - sin t)n sin 2t dt then

0

n

å n ®¥ lim

1

an n is ______.

4 Time : 3 hrs.

Max. Marks : 300

INSTRUCTIONS 1. 2. 3. 4.

5.

6. 7.

This test will be a 3 hours Test. This test consists of Physics, Chemistry and Mathematics questions with equal weightage of 100 marks. Each question is of 4 marks. There are three parts in the question paper consisting of Physics (Q.no.1 to 30), Chemistry (Q.no.31 to 60) and Mathematics (Q. no.61 to 90). Each part is divided into two sections, Section A consists of 20 multiple choice questions & Section B consists of 10 Numerical value answer Questions. In Section B, candidates have to attempt only 5 questions out of 10. There will be only one correct choice in the given four choices in Section A. For each question 4 marks will be awarded for correct choice, 1 mark will be deducted for incorrect choice and zero mark will be awarded for unattempted question. For Section B 4 marks will be awarded for correct answer and zero for marked for each review / unattempted/incorrect answer. Any textual, printed or written material, mobile phones, calculator etc. is not allowed for the students appearing for the test. All calculations / written work should be done in the rough sheet provided.

PHYSICS Section - A 1.

(3) -

Charges +q and –q are placed at points A and B respectively which are a distance 2L apart, C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is R

A

qQ (1) 2pe0 L

C

D

B

(2)

qQ 6pe0 L

2.

qQ 6pe0 L

(4)

qQ 4pe0 L

Two pulses in a stretched string whose centres are initially 8 cm apart are moving towards each other as shown in the figure. The speed of each pulse is 2 cm/s. After 2 seconds, the total energy of the pulses will be (1) Zero (2) Purely kinetic 8 cm (3) Purely potential (4) Partly kinetic and partly potential

EBD_7308 JEE MAIN

MT-38

3.

4.

additional mass M is added to its bob, the time

A metallic wire of density d is lying horizontal on the surface of water. The maximum length of wire so that it may not sink will be

period changes to TM. If the Young's modulus of 1 is equal to: Y (g = gravitational acceleration)

2pT dg

(1)

2Tg pd

(2)

(3)

2T pdg

(4) any length

the material of the wire is Y then

Which logic gate with inputs A and B performs the same operation as that performed by the following circuit? A

B

7.

V Lamp

5.

6.

(2) OR gate (1) NAND gate (3) NOR gate (4) AND gate The specific heat capacity of a metal at low temperature (T) is given as 3 T æ ö C p (kJK –1kg –1 ) = 32 ç ÷ . A 100 g vessel è 400 ø of this metal is to be cooled from 20 K to 4 K by a special refrigerator operating at r oom temperature (27°C). The amount of work required to cool in vessel is (1) equal to 0.002 kJ (2) greater than 0.148 kJ (3) between 0.148 kJ and 0.028 kJ (4) less than 0.028 kJ A pendulum made of a uniform wire of cross sectional area A has time period T. When an

8.

9.

é æ T ö2 ù A M ÷ ú (1) ê1 - ç ë è T ø û Mg

é æ T ö2 ù A (2) ê1 - ç T ÷ ú Mg ëê è M ø úû

éæ T ö2 ù A M (3) êç ÷ - 1ú ëè T ø û Mg

éæ T ö2 ù Mg (4) êç M ÷ - 1ú ëè T ø û A

A bucket full of hot water is kept in a room and it cools from 75oC to 70oC in T1 minutes, from 70oC to 65oC in T2 minutes and from 65oC to 60oC in T3 minutes. Then (1) T1= T2 = T3 (2) T1 < T2 < T3 (3) T1 > T2 > T3 (4) T1< T3< T2 The length of an elastic string is x when the tension is 5N. Its length is y when the tension is 7N. What will be its length, when the tension is 9N? (1) 2y + x (2) 2y – x (3) 7x – 5y (4) 7x + 5y The load versus elongation graphs for four wires of same length and made of the same material are shown in the figure. The thinnest wire is represented by the line Load

D C B A

O

(1) OA (3) OD

Elongation

(2) OC (4) OB

Mock Test -4 10.

MT-39

Six equal resistances are connected between points P, Q and R as shown in figure. Then net resistance will be maximum between :

13.

P

r r

r

A flat plate P of mass ‘M’ executes SHM in a horizontal plane by sliding over a frictionless surface with a frequency V. A block ‘B’ of mass ‘m’ rests on the plate as shown in figure. Coefficient of friction between the surface of B and P is m. What is the maximum amplitude of oscillation that the plate block system can have if the block B is not to slip on the plate :

r r Q

11.

R

r

(1) P and R (2) P and Q (3) Q and R (4) Any two points Electrons are accelerated through a potential difference V and protons are accelerated through a potential difference 4 V. The de-Broglie wavelengths are le and lp for electrons and

(1) (3) 14.

le is given lp by : (given me is mass of electron and mp is mass of proton).

protons respectively. The ratio of

12.

mp

(1)

le = lp

(3)

le 1 me = l p 2 mp

me

(2)

le = lp

(4)

mp le =2 lp me

me mp

A straight section PO of a circuit lies along the xaxis from x = –a/2 to x = +a/2, and carries a steady current ‘I’. The magnitude of magnetic field due to the section PO at a point to y = + a will be (1) proportional to a

(2) proportional to a2

(3) proportional to 1/a (4) equal to zero

15.

mg 4p 2 V 2 m 4p 2 V 2 g

(2) (4)

mg 4p 2 V mg

2p 2 V 2

A glass slab has the left half of refractive index n1, and the right half of n2=3n 1. The effective refractive index of the whole slab is n1 (1) (2) 2n 2 2n 1 3n 1 (3) (4) 3 2 In the arrangement shown L1, L2 are slits and S1 and S2 are two independent sources. On the screen, interference fringes screen (1) will not be there L1 (2) will not be there if the × × intensity of light S1 S2 reaching the screen L2 from S1 and S2 are equal. (3) will be there under all circumstances (4) we will have only the central fringe

EBD_7308 JEE MAIN

MT-40

16.

17.

18.

For good demodulation of AM signal of carrier frequency f, the value of RC should be 1 1 (1) RC = (2) RC < f f 1 1 (3) RC ³ (4) RC >> f f The radioactivity of a sample is R1 at a time T1 and R2 at a time T2. If the half life of the specimen is T, the number of atoms that have disintegrated in the time (T2–T1) is proportional to (1) (R1T1 – R2T2) (2) (R1 – R2) (3) (R1 – R2)/T (4) (R1– R2) × T P–V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to

mass of each charge is m, then the electrostatic potential at the centre of line joining them will be æ 1 ö =k÷ ç è 4p Î0 ø. (1) 2 k mg tan q

(3)

P

1

(3) O2 and N2

21.

22.

19.

600 . The torque needed to maintain the needle in this position will be

(1)

3W

23.

2

(4) O2 and He V A magnetic needle lying parallel to a magnetic field requiers W units of work to turn it through

24.

(2) W

3 (4) 2 W W 2 Two small equal point charges of magnitude q are suspended from a common point on the ceiling by insulating mass less strings of equal lengths. They come to equilibrium with each string making angle q from the vertical. If the (3)

20.

(4)

k mg / tan q

Section - B

(1) He and Ar (2) He and O2

4 k mg / tan q

(2) 4 k mg tanq

25.

A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2. The car will catch up the bus after a time of s. A transformer is used to light a 140 W, 24 V bulb from a 240 V a.c. mains. The current in the main cable is 0.7 A. The efficiency of the transformer is _____%. A rod of length L is placed on x – axis between x = 0 and x = L. The linear density i.e., mass per unit length denoted by r, of this rod, varies as, r = a + bx. The dimensions of b is ML–2Tx. Find the value of x. The surface charge density of a thin charged disc of radius R is s. The value of the electric field at s the centre of the disc is . With respect to 2 Î0 the field at the centre, the electric field along the axis at a distance R from the centre of the disc is reduces by _____%. A body is thrown vertically upwards from the surface of earth in such a way that it reaches upto a height equal to 10Re. The velocity imparted to the body will be km/s

Mock Test -4

~

X C = 20W

30.

Two points of a rod move with velocities 3v and v perpendicular to the rod and in the same direction, separated by a distance r. Then the angular velocity of the rod xv . Find the value r of x. In the given circuit, the current drawn from the source is _____A. X L = 10W

28.

29.

R = 20W

27.

What is the ratio of the circumference of the first Bohr orbit for the electron in the hydrogen atom to the de Brogile wavelength of electrons having the same velocity as the electron in the first Bohr orbit of the hydrogen atom? A projectile can have the same range ‘R’ for two angles of projection. If ‘T1’ and ‘T2’ to be time of flights in the two cases, then the product of the two time of flights is xR . Find the value of x. g The electric field associated with an e.m. wave r in vacuum is given by E = iˆ 40 cos (kz – 6 × 108t), where E, z and t are in volt/m, meter and seconds respectively. The value of wave vector k is m–1.

V = 100x sin(100pt )

26.

MT-41

CHEMISTRY Section - A 31.

32.

The product E would be:

Which of the following structures does not contain any chiral C atom but represent the chirality in the structure. (1) 2 – Ethyl – 3 – hexene (2) 2, 3-Pentadiene (3) 1,3 – Butadiene (4) Pent – 3 – en – 1 – yne In a set of reactions p-nitrotoluene yielded a product E. CH3 Br

Sn/HCl

FeBr3

CH3

Br

Br (2)

(1)

Br

Br

CH3

CH2 Br

Br (3)

2 ® B ¾¾¾¾ ¾¾¾ ®C

NO2

CH3 Br

(4) Br

NaNO 2

CuBr

HCl

HBr

¾¾¾¾ ® D ¾¾¾® E

Br

EBD_7308 JEE MAIN

MT-42

33.

Compound A is formed by the interaction of CH3

O

O

36.

CH2COOH CH3

formation of (1) RCHOHR

OH

(3) RCH2CH2OH 37.

(3) CH3COCH2COOH and CHO

(4) CH3CHO and HO

COOH

N2 and O2 are converted to mono cations N2+ and O2+ respectively, which of the following is wrong? (1) In N2+, the N – N bond weakens (2) In O2+, the O – O bond order increases (3) In O2+, paramagnetism decreases (4) N2+ becomes diamagnetic The reaction in which hydrogen peroxide acts as a reducing agent is (1)

PbS + 4H 2 O 2 ® PbSO 4 + 4 H 2 O

(2)

2 Kl + H 2 O2 ® 2KOH + I 2

39.

R

CHCH2OH

1 e2 2 r

(2) -

e2 r

1 e2 me2 (4) 2 r r When chlorine water is added to an aqueous solution of sodium iodide in the presence of chloroform, a violet colouration is obtained. On adding more of chlorine water and vigorous shaking, the violet colour disappears. This shows the conversion of ...... into ...... (1) I2, HIO3 (2) I2, HI (3) HI, HiO3 (4) I2, HOI When tert-butyl chloride is made to react with sodium ethoxide, the major product is (1) dimethyl ether (2) di-tert-butyl ether (3) tert-butylmethyl ether (4) isobutylene

(3)

38.

(4)

If m and e are the mass and charge of the revolving electron in the orbit of radius r for hydrogen atom, the total energy of the revolving electron will be: (1)

OH

35.

(2) RCHOHCH3 R

COOH

(2) CH3CHO and HO

34.

2

CHO

(1) CH3COOH and HO

OH

2

O

[A]

HO

2 FeSO 4 + H 2SO 4 + H 2 O 2 ® Fe 2 (SO 4 ) 3 + 2H 2 O (4) Ag 2 O + H 2 O 2 ® 2Ag + H 2 O + O 2 Reaction of CH — CH with RMgX leads to (3)

Mock Test -4 40.

MT-43

Which statement is true regarding following reactions HCO H

3 ®A trans - 2 - Butene ¾¾¾¾

HCO H

3 ®B cis - 2 - Butene ¾¾¾¾

(1) Compounds A and B are formed by syn addition and they are racemic mixture and meso respectively

43.

(2) Compounds A and B are formed by anti addition and are racemic mixture and meso respectively (3) Compounds A and B are formed by anti addition and are meso and racemic mixture respectively (4) Compounds A and B are formed by syn addition and are meso and racemic mixture respectively 41.

An organic compound is treated with NaNO 2 and dil. HCl at 0°C. The resulting solution is added to an alkaline solution of b -naphthol where by a brilliant red dye is produced. It shows the presence of (1) – NO 2 group (2) aromatic – NH 2 group (3) – CONH 2 group

42.

(4) aliphatic – NH 2 group Point out the incorrect statment among the following : (1) The oxidation state of oxygen is +2 in OF2.

44.

(2) Acidic character follows the order H2O < H2S < H2Se < H2Te. (3) The tendency to form multiple bonds increases in moving down the group from sulphur to tellurium (towards C and N). (4) Sulphur has a strong tendency to catenate while oxygen shows this tendency to a limited extent. Removal of Fe, Cu, W from Sn metal after smelting is by ............... because ............. (1) poling; of more affinity towards oxygen for impurities (2) selective oxidation; of more affinity towards oxygen for impurities (3) electrolytic refining; impurities undissolved in elec-trolyte (4) liquation; Sn having low melting point compared to impurities. Among KO 2 , AlO2- , BaO2 and NO +2 ,unpaired electron is present in

45.

(1) NO +2 and BaO 2 (2) KO 2 and AlO -2 (3) KO 2 only (4) BaO 2 only If a 0.1 M solution of glucose (Mol. wt 180) and 0.1 molar solution of urea (Mol. wt. 60) are placed on two sided semipermeable membrane to equal heights, then it will be correct to say that (1) there will be no net movement across the membrane. (2) glucose will flow across the membrane into urea solution. (3) urea will flow across the membrane into glucose solution. (4) water will flow from urea solution to glucose solution.

EBD_7308 JEE MAIN

MT-44

46.

47.

48.

49.

Melamine plastic crockery is a codensation polymer of (1) HCHO and melamine (2) HCHO and ethylene (3) melamine and ethylene (4) None of these When pink complex, [Co(H2O)6]2+ is dehydrated the colour changes to blue. The correct explanation for the change is : (1) The octahedral complex becomes square planar. (2) A tetrahedral complex is formed. (3) Distorted octahedral structure is obtained. (4) Dehydration results in the formation of polymeric species. In the form of dichromate, Cr (VI) is a strong oxidising agent in acidic medium but Mo (VI) in MoO3 and W (VI) in WO3 are not because ____________ . (i) Cr (VI) is more stable than Mo(VI) and W (VI). (ii) Mo (VI) and W(VI) are more stable than Cr(VI). (iii) Higher oxidation states of heavier members of group-6 of transition series are more stable. (iv) Lower oxidation states of heavier members of group-6 of transition series are more stable. (1) (i) and (ii) (2) (ii) and (iii) (3) (i) and (iv) (4) (ii) and (iv) The correct statement among the following is : (1) The alkali metals when strongly heated in oxygen form superoxides. (2) Caesium is used in photoelectric cells. (3) NaHCO3 is more soluble in water than KHCO3. (4) The size of hydrated ions of alkali metals increases from top to bottom.

50.

If the following half cells have E° values as A3+ + e– ¾® A2+, E° = y2V A2+ + 2e– ¾® A, E° = –y1V The E° of the half cell A3+ + 3e– ¾® A will be 2 y1 - y2 y2 - 2 y1 (1) (2) 3 3 (3) 2y1 – 3y2 (4) y2 – 2y1 Section - B

51.

52.

53.

54.

55.

56.

The radii of Na+ and Cl– ions are 100 pm and 200 pm respectively. Calculate the edge length of NaCl unit cell. When 1.8 g of steam at the normal boiling point of water is converted into water at the same temperature, calculate Ds for the above change in J/K. (DHvapfor water = 40.8 kJ mol–1) Given n = 5, ml = + 1 Find the maximum number of electron(s) in an atom that can have the quantum numbers as given. What amount of sugar (C12H22O11) (in g) is required to prepare 2 L of its 0.1 M aqueous solution? (i) H2O 2

¾¾¾® Product (ii) heat

CH3 N(CH3)2 Find the number of unsaturated carbon in the product. A tetrapeptide has —COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For this tetrapeptide, find the number of possible sequences (primary structures) with —NH2 group attached to a chiral center.

Mock Test -4 57.

58.

MT-45

4 g of H2 effused through a pinhole in 10 s at some constant temperature and pressure. Calculate the amount of oxygen effused in the same time interval and at the same conditions of temperature and pressure. The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non volatile and non electrolyte solid weighing 2.175 g is added to 39.08 g of benzene. If the vapour pressure of the solution is 600 mm of Hg, calculate the molecular weight of solid substance.

59.

60.

When CO2 dissolves in water, the following equilibrium is established CO2 + 2H2O H3O+ + HCO3–; for which the equilibrium constant is 3.8 ´ 10–6 and pH = 6.0. What would be the ratio of concentration of bicarbonate ion to carbon dioxide? For a first order reaction, calculate the ratio between the time taken to complete 3/4th of the reaction and time taken to complete half of the reaction.

MATHEMATICS Section - A 61.

62.

63.

(2) If

The roots of the given equation (p – q) x 2 + (q – r) x + (r – p) = 0 are :

p-q ,1 r-p

(2)

(3)

r-p ,1 p-q

(4) None of these

(3)

1 log xy xyz

+

1 log yz xyz

+

1 log zx xyz

=2

(4) All are correct p and cot 2 a, cot b, cot g are in arithmetic progression, then the value of cot a cot g is (1) 1 (2) 3 2 (3) cot b (4) cot a + cot g 65. The equation of the chord of the hyperbola

64.

If a, b, c, d and p are distinct non zero real numbers such that (a 2 + b2 + c 2 ) p 2 – 2(ab + bc + cd)p + (b2 + c2+ d2) £ 0 then a,b,c,d are in (1) A.P.

(2) G.P.

(3) H.P.

(4) satisfy ab = cd

Which of the following is correct? (1) If a 2 + 4b 2 = 12ab , then log(a + 2b) =

then x a .y b .z c = abc

q-r ,1 p-q

(1)

log x log y log z = = , b-c c-a a-b

1 (log a + log b) 2

If 0 < a, b, g < p/2 such that a + b + g =

25 x 2 - 16 y 2 = 400 that is bisected at point (5, 3) is:

(1) 135 x – 48 y = 481 (3) 125 x – 4 y = 48

(2) 125x–48y=481 (4) None of these

EBD_7308 JEE MAIN

MT-46

66.

The number of solutions of the equation 70.

æ px ö ÷÷ = x 2 – 2 3 x + 4 sin çç è2 3ø

67.

(1) forms an empty set (2) is only one (3) is only two (4) is more than 2 If 1, a1, a2, a3, a4 are the roots of z5 – 1 = 0, then the value of w – a4 w2 – a4

68.

69.

w – a1 2

w – a1

×

w – a2 2

w – a2

×

w – a3 w 2 – a3

unity) (1) 1 (2) w 2 (3) w (4) None of these Which of the following is correct? (1) If A and B are square matrices of order 3 such that | A | = –1, | B | = 3, then the determinant of 3 AB is equal to 27. (2) If A is an invertible matrix, then det (A–1) is equal to det (A) (3) If A and B are matrices of the same order, then (A + B)2 = A2 + 2AB + B2 is possible if AB = I (4) None of these Let f(x) = | x – 1 |. Then (1)

(2) f (x + y) = f(x) + f(y) (3) f (| x |) = | f(x) | (4) None of these

(1)

a -b 1 + ab

(2)

b 1 + ab

(3)

b 1 - ab

(4)

a+b 1 - ab

× 71.

is (where w is imaginary cube root of

f(x2) = (f(x))2

2a 2b + sin -1 = 2 tan -1 x, then 2 1+ a 1 + b2 x is equal to

If sin -1

Let f (x) =

2 , g(x) = cos x and h(x) = x + 3 x +1

then the range of the composite function fogoh, is

72.

73.

(1) R+

(2) R – {0}

(3) [1, ¥)

(4) R+ – {1}

The set of points where f(x) = (x – 1)2 (x + | x –1 | ) is thrice differentiable, is (1) R

(2) R – {0}

(3) R – {1}

(4) R–{0,1}

Let f (x) = 1/(x – 1) and g (x) = 1/(x2 + x – 2). Then the set of points where (gof)(x) is discontinuous, is (1) {1}

(2) {–2,1}

(3) {1/2, 1, 2}

(4) {1/2 , 1}

m

74.

å n + r Cn

is equal to :

r =0

(1) (3)

n + m +1

Cn + 1

(2)

n + m +3

C n -1

(4) None of these

n + m +2

Cn

Mock Test -4

MT-47

( an -1) / n

75.

The value of

ò

1n

(1)

a 2

na - 2 2n The value of p/4

ò

a-x + x

(2)

(3)

76.

x

dx is

80.

1 (na + 2) 2n

(1) (2)

(4) None of these.

(3) (x | x | + sin3 x + x tan 2 x + 1) dx is

Then

78.

79.

(4)

(1) 0 (2) 1 (3) p/4 (4) p/2 Let (1 – x – 2x2)6 = 1+ a1x + a2x2 + …. + a12 x12.

a2

+

a4

+

a6

+ ...... +

a12

is equal to 22 24 26 212 (1) –1 (2) –1/2 (3) 0 (4) 1/2 The equation of a common tangent to y2 = 4x and the curve x2 + 4y2 = 8 can be (1) x – 2y + 2 = 0 (2) x + 2y + 4 = 0 (3) x – 2y = 4 (4) x + 2y = 4 The function f (x) = (x – 3)2 satisfies all the conditions of mean value theorem in [3, 4]. A point on y = (x –3)2, where the tangent is parallel to the chord joining (3, 0) and (4,1) is (1)

æ 7 1ö çè , ÷ø 2 2

(3) (1, 4)

æ 7 1ö (2) ç , ÷ è 2 4ø

(4) (4, 1)

2x 1- x

2

+

2y 1- y

2

-

2z 1- z2

is equal to

-p / 4

77.

If x + y – z + xyz = 0, then

xyz 2

[(1 - x )(1 - y 2 )(1 - z 2 )] - xyz 2

[(1 - x )(1 - y 2 )(1 - z 2 )] 8xyz 2

[(1 - x )(1 - y 2 )(1 - z 2 )] -8 xyz 2

[(1 - x )(1 - y 2 )(1 - z 2 )]

Section - B 81.

If the number of 5-element subsets of the set A= {a1, a2, ...., a20} of 20 distinct elements is k times the number of 5-element subsets containing a4, then k is _____.

82.

The value of cos360 cos420 cos780 is _____. é 5 -1 5 + 1ù and cos 36° = êGiven : sin18° = ú 4 4 û ë

83.

If x =1/5, the value of |cos (cos–1x + 2 sin–1 x)| is _____.

84.

Let A be the centre of the circle x2 + y2 – 2x–4y – 20 = 0, and B( 1,7) and D(4,–2) are points on the circle then, if tangents be drawn at B and D, which meet at C, then area of quadrilateral ABCD is _____.

EBD_7308 JEE MAIN

MT-48

85.

86.

87.

If the system of linear equations : x1 + 2x2 + 3x3 = 6 x1 + 3x2 + 5x3 = 9 2x1 + 5x2 + ax3 = b is consistent and has infinite number of solutions, then value of a + b is _____. If Ak = 0 ( A is nilpotent with index k), (I – A)p = I + A + A2 + . . . + Ak – 1, then value of |p| is _____. Let f ( x) =

x - {x + 1 x - {x + 2

} }

; where {x} is the

fractional part of x, then value of lim f ( x ) is x®1 / 3

_____.

88.

Let y (x) be a solution of

( 2 + sin x ) dy = cos x . (1 + y ) dx

æ pö If y (0) = 2, then value of y ç ÷ is _____. è 2ø 89.

1

ò f (x) dx = log(f (x))

2

+ C. If f(0) = 5 then value

of constant term of f(x) is _____. 1 a b

90.

In a DABC, if 1 c a = 0, then value of 1 b c sin 2 A + sin 2 B + sin 2 C is _____.

5 Time : 3 hrs.

Max. Marks : 300 INSTRUCTIONS

1. 2. 3. 4.

5.

6. 7.

This test will be a 3 hours Test. This test consists of Physics, Chemistry and Mathematics questions with equal weightage of 100 marks. Each question is of 4 marks. There are three parts in the question paper consisting of Physics (Q.no.1 to 30), Chemistry (Q.no.31 to 60) and Mathematics (Q. no.61 to 90). Each part is divided into two sections, Section A consists of 20 multiple choice questions & Section B consists of 10 Numerical value answer Questions. In Section B, candidates have to attempt only 5 questions out of 10. There will be only one correct choice in the given four choices in Section A. For each question 4 marks will be awarded for correct choice, 1 mark will be deducted for incorrect choice and zero mark will be awarded for unattempted question. For Section B 4 marks will be awarded for correct answer and zero for marked for each review / unattempted/incorrect answer. Any textual, printed or written material, mobile phones, calculator etc. is not allowed for the students appearing for the test. All calculations / written work should be done in the rough sheet provided.

PHYSICS Section - A 1.

2.

A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively (1) g, g (2) g – a, g – a (3) g – a, g (4) a, g Relative permittivity and permeability of a material are er and mr, respectively. Which of the

3.

following values of these quantities are allowed for a diamagnetic material? (1) er = 0.5, mr = 1.5 (2) er = 1.5, mr = 0.5 (3) er = 0.5, mr = 0.5 (4) er = 1.5, mr = 1.5 Let there be a spherically symmetric charge distribution with charge density varying as æ5 rö r(r ) = r0 ç - ÷ upto r = R , and r(r ) = 0 è 4 Rø for r > R , where r is the distance from the origin. The electric field at a distance r(r < R) from the origin is given by

EBD_7308 JEE MAIN

MT-50

4.

5.

(1)

r0 r æ 5 r ö ç - ÷ 4e 0 è 3 R ø

(2)

4pr0 r æ 5 r ö ç - ÷ 3e 0 è 3 R ø

(3)

r0 r æ 5 r ö ç - ÷ 4ε0 è 4 R ø

(4)

r0 r æ 5 r ö ç - ÷ 3ε 0 è 4 R ø

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V0 and its pressure is P0. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency (1)

1 AgP0 2p V0 M

(3)

1 2p

A 2 gP0 MV0

(2)

1 V0 MP0 2p A 2 g

(4)

1 2p

MV0 AgP0

The correct graph between the gravitational potential (Vg) due to a hollow sphere and distance from its centre will be r

(1) Vg

(2) Vg r

+Vg

(3)

+Vg

r = Re r

(4) –Vg

–Vg

r = Re r

6.

In a cubical vessel are enclosed n molecules of a gas each having a mass m and an average speed v. If l is the length of each edge of the cube, the pressure exerted by the gas will be (1) (3)

7.

8.

n mv 2 l3

m nv 2

3l

3

(2) (4)

n m2 v

2 l3

nmv 2l

A sphere, a cube and a thin circular plate all made of the same material and having the same mass, are initially heated to a temperature of 200oC. Which of these objects will cool slowest when left in air at room temperature? (1) the sphere (2) the cube (3) the circular plate (4) all will cool at same rate An electromagnetic wave in vacuum has the r r electric and magnetic field E and B , which are always perpendicular to each other. The r direction of polarization is given by X and that r of wave propagation by k . Then r r r r r (1) X || B and k || B ´ E r r r r r (2) X || E and k || E ´ B r r r r r (3) X || B and k || E ´ B r r r r r (4) X || E and k || B ´ E

Mock Test -5 9.

10.

MT-51

A particle of mass m1 collides head-on with another stationary particle of mass m2 (m2 > m1). The collision is perfectly inelastic. The fraction of kinetic energy which is converted into heat in this collision is (1) m2/(m1 + m2)

(2) m1/(m1 + m2)

(3) m1/(m1 – m2)

(4) m2/(m1 – m2)

The instantaneous values of current and voltage in an A.C. circuit are I = 4 sin wt and E = 100 cos

m0 i1i2 dl tan q 2 pr m0 i1i2 dl sin q (2) 1 2 2pr m0 (3) i1i2 dl cos q i1 i2 r 2 pr m0 dl q (4) i1i2 dl sin q 4 pr In Young's double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit is double of that other slit. If Im be the maximum intensity, the resultant intensity I when they interfere at phase difference f is given by :

(1)

13.

pö æ ç wt + ÷ respectively. The phase difference 3ø è

between voltage and current is (1)

(3) 11.

12.

p 3

(2)

5p 6

(4)

(1)

2p 3

(1) Wb/m 2

(2) Henry (H)

(3) H/m2

(4) Weber (Wb)

Wires 1 and 2 carrying currents i 1 and i 2 respectively are inclined at an angle θ to each other. What is the force on a small element dl of wire 2 at a distance of r from wire 1 (as shown in figure) due to the magnetic field of wire 1?

Im æ 2 fö ç 1 + 2 cos ÷ 3 è 2ø Im æ f 2 ö (3) ç 1 + 4 cos ÷ 5 è 2ø Im æ f 2 ö (4) ç 1 + 8 cos ÷ 9 è 2ø A pendulum consists of a wooden bob of mass m and length ‘l’. A bullet of mass m1 is fired towards the pendulum with a speed v1. The bullet emerges out of the bob with a speed v1/3, and the bob just completes motion along a vertical circle. Then v1 is (2)

7p 6

Which of the following units denotes the ML2 dimension , where Q denotes the electric Q2 charge?

Im (4 + 5 cos f) 9

14.

(1)

æ mö çè m ÷ø 5g l

(2)

3æ m ö 5gl 2 çè m1 ÷ø

(3)

2 æ m1 ö ç ÷ 5gl 3è m ø

(4)

æ m1 ö çè ÷ø g l m

1

EBD_7308 JEE MAIN

MT-52

15.

(a) 1490, 1510

In a p-n junction diode, a square input signal of 10 V is applied as shown in figure. 18.

– 5V The output signal across RL will be

10V (2)

(1) (3) 16.

–5V

(4)

5V

–10V

A conducting square loop of side L and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A magnetic induction B constant in time and space, pointing perpendicular and into the plane at the loop exists everywhere with half the loop outside the field, as shown in figure. The induced emf is

1 1 1 1 (d) , , 1490 1510 1510 1490 A ray of light is incident at an angle of 60° on one face of a prism of angle 30°. The ray emerging out of the prism makes an angle of 30° with the incident ray. The emergent ray is (1) Normal to the face through which it emerges (2) Inclined at 30° to the face through which it emerges (3) Inclined at 60° to the face through which it emerges (4) Inclined at 90° to the normal at face through which it emerges The anode voltage of a photocell is kept fixed. The wavelength l of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows :

(c)

5V RL

(b) 1510, 1490

19.

I

I

(1) O

v

L

(2) l

I

(3) (1) zero 17.

(2) RvB

(3) vBL/R (4) vBL Sinusoidal carrier voltage of frequency 1.5 MHz and amplitude 50 V is amplitude modulated by sin usoidal voltage of frequency 10 kHz producing 50% modulation. The lower and upper side-band frequencies in kHz are

(4) O

20.

l

O

I

l

O

l

A proton, a deutron and an a particle accelerated through the same potential difference enter a region of uniform magnetic field, moving at right angles to B. What is the ratio of their K.E.? (1) 1 : 2 : 2 (2) 2 : 2 : 1 (3) 1 : 2 : 1 (4) 1 : 1 : 2

Mock Test -5

MT-53

Section - B 21.

25.

In the circuit shown, the total current supplied by the battery is _________ A. B

In the figure battery B supplies 12 V. Take C1 = 1 mF, C2 = 2 mF, C3 = 3 mF, C4 = 4 mF. Charge on capacitor C1 when only S1 is closed is _____ mC. C1 C3

6W 2W

C

6 volts

A

22.

C2

1.5W

Three identical spheres, each of mass 1 kg are kept as shown in figure, touching each other, with their centres on a straight line. If their centres are marked P, Q, R respectively, the distance of PQ + PR centre of mass of the system from P is . x y Find the value of x. R

x

The half life of a radioactive substance is 20 minutes. The approximate time interval (t2 – t1) between the time t2 when and time t1 when

2 of it had decayed 3

1 of it had decayed is _____ 3

min. 24.

C4 S1

+ – D

P Q

23.

S2

3W

A wooden block of volume 1000 cc is suspended from a spring balance. It weighs 12 N in air. It is then held suspended in water with half of it inside water. What would be the reading in spring balance (in N) now?

26.

B p 2p0 p0

A D v0

B C 2v0 v

The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is x p0v0. Find the value of x. 27. A hydrogen-like atom has one electron revolving around a stationary nucleus. The energy required to excite the electron from the second orbit to the third orbit is 47.2 eV. The atomic number of the atom is _________. 28. A boy playing on the roof of a 10 m high building throws a ball with a speed of 10m/s at an angle of 30º with the horizontal. How far (in m) from the throwing point will the ball be at the height of 10 m from the ground ? 1 2

[ g = 10m/s2 , sin 30o = , cos 30o =

3 ] 2

EBD_7308 JEE MAIN

MT-54

29.

Two periodic waves of intensities I1 and I2 pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is x (I1 + I2). Find the value of x.

30.

A copper wire of length 1.0 m and a steel wire of length 0.5 m having equal cross-sectional areas

are joined end to end. The composite wire is stretched by a certain load which stretches the copper wire by 1 mm. If the Young’s modulii of copper and steel are respectively 1.0 × 1011 Nm–2 and 2.0 × 1011 Nm–2, the total extension of the composite wire is _______ mm.

CHEMISTRY Section - A 31.

32.

33.

In which of the following arrangements, the sequence is not strictly according to the property written against it? (1) CO2 < SiO2 < SnO2 < PbO2: increasing oxidising power (2) NH3 < PH3 < AsH3 < SbH3: increasing basic strength (3) HF < HCl < HBr < HI: increasing acid strength (4) B < C < O < N: increasing first ionisation enthalpy. Aluminothermy used for on the spot welding of large iron structure is based on the fact that(1) As compared to iron, aluminium has greater affinity for oxygen. (2) As compared to aluminium, iron has greater affinity for oxygen. (3) Reaction between aluminium and oxygen is endothermic. (4) Reaction between iron and oxygen is endothermic. The charge/size ratio of a cation determines its polarizing power. Which one of the following

34.

sequences represents the increasing order of the polarizing power of the cationic species, K+, Ca2+, Mg2+, Be2+? (1) Ca2+ < Mg2+ < Be+ < K+ (2) Mg2+ < Be2+ < K+ < Ca2+ (3) Be2+ < K+ < Ca2+ < Mg2+ (4) K+ < Ca2+ < Mg2+ < Be2+. A 1.0 M solution with respect to each of the metal halides AX3, BX2, CX 3 and DX2 is electrolysed using platinum electrodes. If E° E°

35.

A3+ / A C3+ /C

= 1. 50 V,

E°

= – 0. 74 V,, E°

B2+ /B

D2+ /D

= 0. 3 V, = – 2.37 V..

The correct sequence in which the various metals are deposited at the cathode is : (1) A, B, C, D (2) A, B, C (3) D, C, B, A (4) C, B, A An ideal gas undergoes isothermal expansion at constant pressure. During the process : (1) enthalpy increases but entropy decreases. (2) enthalpy remains constant but entropy increases. (3) enthalpy decreases but entropy increases. (4) Both enthalpy and entropy remain constant.

Mock Test -5 36.

MT-55

Among the reactions given below for B2H6, the one which does not take place is ® B2H5Cl + H2 (1) B2H6 + HCl ¾¾

39.

D

(2) 2B2H6 + 6NH3 ¾¾® B3N3H6 (borazine) ® 2(CH3)3 NBH3 (3) B2H6 + 2N(CH3)3 ¾¾ (4) B2H6 + 6C2H4 37.

Cl

(3) HgCl2 and ZnCl2 (4) ZnCl2 and HgCl24

en

+

40.

(2) NH4OH ® NH4+ + OH– (3) N2O5 + H2O ® 2HNO3

Co

Cl

(4) 2NO2 + H2O ® HNO3 + HNO2

Cl

en

en

I

41. II

en

en

3+

Cl

+

(2) MnO2, MnO2 and Mn2+ (3) MnO2, MnO2+ and Mn3+

Cl III

38.

(4) MnO, MnO2 and Mn2+

en

en

IV

(1) I only

(2) II only

(3) II and III

(4) IV only

Potassium permanganate acts as an oxidant in neutral, alkaline as well as acidic media. The final products obtained from it in the three conditions are, respectively (1) MnO42–, Mn3+ and Mn2+

Co

Co

In which reaction, there is change in oxidation number of N (1) 2NO2 ® N2O4

+

Cl

Co

en

(1) HgCl2 and Hg2Cl2 (2) Hg2Cl2 and HgCl2

H 3O + ¾¾¾®

3C2H5OH + 2B(OH)3 Which of the following ions are optically active? en

A metal gives two chlorides A and B. A gives black precipitate with NH4OH and B gives white. With KI, B gives a red precipitate soluble in excess of KI. A and B are respectively :

The pair of compounds having metals in highest oxidation state is:

42.

An organic compound ‘X’ on ozonolysis followed by reduction with Zn/H2O gives 2 moles O ||

O ||

O ||

of H - C - H and H - C - CH 2 - C - H . ‘X’ is (1) CH2 = CH – CH2 – CH = CH2

(1)

MnO2 and CrO2Cl2

(2) CH2 = CH–CH2–CH2–CH = CH2

(2)

[NiCl4]2– and [CoCl4]2–

(3)

[Fe(CN)6]3– and [Cu(CN)2]2–

(3) H – C º C – C º C – H (4) CH2 = CH – CH = CH2

(4)

[FeCl4]– and Co2O3

EBD_7308 JEE MAIN

MT-56

43.

44.

45.

46.

Which one of the following compounds is resistant to nucleophilic attack by hydroxyl ions? (1) Methyl acetate (2) Acetonitrile (3) Diethyl ether (4) Acetamide To detect iodine in presence of bromine, the sodium extract is treated with NaNO2 + glacial acetic acid + CCl4. Iodine is detected by the appearance of (1) yellow colour of CCl4 layer. (2) purple colour of CCl4. (3) brown colour in the organic layer of CCl4. (4) deep blue colour in CCl4. An organic compound A (C4H10O) has two enantiomeric forms and on dehydration it gives B(major product) and C (minor product). B and C are treated with HBr/ Peroxide and the compounds so produced were subjected to alkaline hydrolysis then(1) B will give an isomer of A (2) C will give an isomer of A (3) Neither of them will give isomer of A (4) Both B and C will give isomer of A

The respective compounds A and B are (1)

47 .

2+

H2SO4, Hg

A,

O

CºCH

H 2SO4, Hg

2+

B

O

COCH3

(3)

O CH2CHO

and O

CH2CHO

and

CH2CHO

(4)

COCH3

and

CH2CHO

(2)

O COCH3

and

NH2 (i) NaNO + HCl

¾ ¾ ¾ ¾2¾ ¾ ® A.

48.

(ii) CuCl

Compound ‘A’ is

Cl

NO2

(1)

In which of the following polymers, empirical formula resembles with monomer? (1) Bakelite (2) Teflon (3) Nylon-6, 6 (4) Dacron CºCH

COCH3

(2)

Cl (3) 49.

NO2

(4)

Allyl phenyl ether can be prepared by heating: (1) C6H5Br + CH2 = CH – CH2 – ONa (2) CH2 = CH – CH2 – Br + C6H5ONa (3) C6H5 – CH = CH – Br + CH3 – ONa (4) CH2 = CH – Br + C6H5 – CH2 – ONa

Mock Test -5

MT-57

54.

Cº N + C6H5MgBr

50.

Ether

H O+

3 ¾ ¾¾¾® A ¾¾¾ ®B The final product (B) in above reaction is

O

(Given E°

C (1)

55. N–MgBr C

56.

(2)

N–H C (3) 57. N–OH C (4)

58.

Section - B 51.

52. 53.

If radiation corresponding to second line of "Balmer series" of Li2+ ion, knocked out electron from first excited state of H-atom, then find the kinetic energy of ejected electron. What is the sum of number of sigma ( s ) and pi ( p ) bonds present in sulphuric acid molecule ? Copper crystallises in fcc with a unit length of 361pm. What is the radius of copper atom ?

To find the standard potential of M 3+/M electrode, the following cell is constituted : Pt/ M/M3+(0.001 mol L–1)/Ag+(0.01 mol L–1)/Ag The emf of the cell is found to be 0.421 volt at 298 K calculate the standard potential of half reaction M3+ + 3e– ¾® M at 298 K

59. 60.

Ag + /Ag

at 298 K = 0.80 Volt)

The number of p electrons present in 6.4 g of calcium carbide is x × NA where (NA = Avagadro’s number). Find the value of x. Two elements A & B form compounds having molecular formulae AB2 and AB4 . When dissolved in 20.0 g of benzene 1.00 g of AB2 lowers f.p. by 2.3 °C whereas 1.00 g of AB4 lowers f.p. by 1.3 °C. The molal depression constant for benzene in 1000 g is 5.1. Calculate the sum of atomic masses of A and B. An element (atomic mass =100 g/mol) having bcc structure has unit cell edge 400 pm. Calculate the density (in g/cm3) of the element. A substance "A" decomposes in solution following the first order kinetics. Flask I contains 1 L of 1 M solution of A and flask II contains 100 mL of 0.6 M solution. After 8 hr, the concentration, of A in flask I becomes 0.25 M. What will be the time for concentration of A in flask II to become 0.3 M? Concentration of NH4Cl and NH4OH in a buffer solution is in the ratio of 1 : 1, Kb for NH4OH is 10–10. Calculate the pH of the buffer. At 5 × 105 bar pressure and density of diamond and graphite are 3 g/cc and 2 g/cc respectively, at certain temperature 'T'. Find the value of DU – DH for the conversion of 1 mole of graphite to 1 mole of diamond at temperature 'T'.

EBD_7308 JEE MAIN

MT-58

MATHEMATICS Section - A

y=

61. Which of the following statements is false? (1) The length of sub-tangent to the curve x2y2 = 16a4 at the point (–2a, 2a) is 2a. (2) x + y = 3 is a normal to the curve x2 = 4y (3) Curves y = –4x2 and y = e–x/2 are orthogonal. (4) If a Î (–1, 0), then tangent at each point of 2 3 x - 2ax 2 + 2x + 5 makes 3 an acute angle with the positive direction of x-axis. (x 2 - 1)dx If ò æ x2 +1ö (x 4 + 3x 2 + 1) tan -1 ç ÷ è x ø = log | tan -1 f (x) | + C, then

(1) f (x) = x2 + 1

(2)

65.

63.

ellipse

x

2

a

2

+

y

2

b2

(2)

x 2 +1 2x 1 2 f ( x) = (x + 1) 2

x2 a2

+

y2 b2

=4

(4) None of these

Area included between y =

x2 and 4a

(2)

a2 (4 p + 3) 3

(4) None of these r r Given that a is ^ to b and p is a non zero scalar r r r r r r if pr + (r .b) a = c then r equals r r r r (1) c / p - éë(b.c) a ùû / p 2 r r r r (2) a / p - éë (c.a) b ùû / p2 rr r r 2 (3) b / p - éë(a.b)c ùû / p (4) None of these The domain of the function 2 f (x) = x - 1 - x is 1 ù é 1 ù é ,1ú (1) ê -1, úÈê 2û ë 2 û ë (2) [–1, 1] 1ù é 1 ö æ , + ¥÷ (3) çè -¥, - 2 ú È ê ø û ë 2

f ( x) =

= 1 , then locus of the mid- point

of PQ is x2 y2 1 + = (1) a 2 b2 2 x2 y2 + =2 (3) a 2 b2 64.

66.

x2 +1 (4) x ö æp If P(q) and Qç + q ÷ are two points on the ø è2

(3) f (x) =

is

x 2 + 4a 2 a2 ( 6p - 4) (1) 3 a2 (8p + 3) (3) 3

the curve y =

62.

8a 3

é 1 ù ,1ú ê ë 2 û Which of the following is an empty set? (1) The set of prime numbers which are even (2) The solution set of the equation

(4)

67.

2 2 (2x + 3) – + 3 = 0; x Î R x +1 x +1 (3) (A × B) Ç (B × A), where A and B are disjoint. (4) The set of real which satisfy x2 + ix + i – 1 = 0

Mock Test -5 68.

69.

70.

71.

72.

f (x) = x2 [x] (1) increases in (0,1) (2) decreases in (0 , 1) (3) increases in (–1,0) (4) None of these If y = ex + sin x , then d2x/dy2 is equal to (1) ex – sin x (2) –(ex + cos x)–2 (3) – (ex – sin x ) (ex + cos x)–2 (4) (sin x – ex) (cos x + ex)–3 The sum to n terms of the series 1 3 7 15 + + + + .............. is 2 4 8 16 –n (1) n – 1 – 2 (2) 1 (3) n – 1 + 2–n (4) 1 + 2–n If f : R ® R be a function defined by f (x) = 4x3 –7. Then (1) f is one-one -into (2) f is many-one - into (3) f is many-one onto(4) f is bijective

74.

75.

76.

(1) (2)

æ x - 2ö p sin -1 ç + è 2 ÷ø 6

2p æ x - 2ö + cos -1 ç è 2 ÷ø 3 (4) None of these The roots a and b of the quadratic equation px2 + qx + r = 0 are real and of opposite signs. The roots of a(x – b)2 + b(x – a)2 = 0 are (3)

dy 2 = for | x |< 1 dx 1 + x 2 -2 dy (2) = for | x | > 1 dx 1 + x 2 dy = 2 for x = -1 (3) dx dy (4) does not exist at | x | = 1 dx If the slope of the tangent at (x, y) to a curve

æ p ö is given by ç 1, ÷ è 4ø y æ yö - cos 2 ç ÷ , then the equation of the curve is x x

passing thr ough

3 sin x - cos x + 2 . Then f –1(x) is given

æ x - 2ö p sin -1 ç è 2 ÷ø 6

(1) positive (2) negative (3) of opposite signs (4) non-real The equation k sin q + cos 2q = 2k – 7 possesses a solution if : (1) 2 £ k £ 6 (2) k > 2 (3) k > 6 (4) k < 2 2 x If y = sin -1 , then which of the following 1+ x2 is not correct? (1)

é p 2p ù Let f : ê- , ® [0, 4] be a function defined ë 3 3 úû as f (x) = by

73.

MT-59

(1) (2) (3) (4) 77.

è ø y = tan -1 log (e / x )

y = e1+ cot y = x tan -1 log(e / x ) y = e1+tan(y/x) ( y / x)

æ 2x ö -1 ÷, then dy at x = 0 is : If y = tan çç 2 x +1 ÷ dx è1+ 2 ø 2 3 log 2 (1) – log 2 (2) 5 5 3 (3) - log 2 (4) None of these 2

EBD_7308 JEE MAIN

MT-60

a

78.

a2 1+ a3

82.

2 3 If b b 1 + b = 0 and the vectors

c

c2

the conditions f (0) = -1, f ¢(log 2) = 31 and

1 + c3

log 4

A = (1,a,a2) ; B = (1, b, b2) ; C = (1,c, c2) are

79.

non-coplanar then the product abc = (1) 0 (2) 1 (3) –1 (4) None A and B are two independent witnesses (i.e. there is no collusion between them) in a case. The probability that A will speak the truth is x and the probability that B will speak the truth is y. A and B agree in a certain statement. The probability that the statement is true is (1)

(3) 80.

x–y x+y

(2)

x–y 1 – x – y + 2xy

(4)

xy 1 + x + y + xy

(3)

83. 84.

85.

4(x12 + y12)

(2) 16(x12 + y12) (4)

8

(

x12

+ y12

)

Let f be a composite function of x defined by

1

1 , u ( x) = f (u ) = 2 . x -1 u +u -2 Then the number of points x where f is discontinuous is ________.

(f (x) - Rx)dx =

39 , then valve of P + Q 2

+ R is ______. The sum of 10 items is 12 and sum of their squares is 18, then standard deviation will be _______. If the system of equations : p3x + (p + 1)3 y = (p + 2)3 px + (p + 1) y = p + 2 x+y= 1 is consistent then valve of |P| is ______. If f (x) = 3x10 – 7x8 + 5x6 – 21x3 + 3x2 – 7, then lim

f (1 - a ) - f (1)

is ________. a3 + 3a A survey of 500 television viewers produced the following information, 285 watch football, 195 watch hockey, 115 watch basket-ball, 45 watch football and basket ball, 70 watch football and hockey, 50 watch hockey and basket ball, 50 do not watch any of the three games. The number of viewers, who watch exactly one of the three games are ________. The value of | x | for which sin (cot –1 (1+ x)) = cos (tan –1 x) is _______. a®0

86.

xy 1 – x – y + 2xy

Section - B 81.

ò0

Let x1 and y1 be real numbers. If z1 and z2 are complex numbers such that |z1| = |z2| = 4, then |x1z1 – y1z2|2 + |y1z1 + x1z2|2 = (1) 32(x12 + y12)

If the function f ( x ) = Pe 2 x + Qe x + Rx satisfies

87.

88. 89. 90.

1 - cos 3 x is _______. x ® 0 x sin x cos x

The value of lim

1 , th e number of points of 1- x discontinuity of f{f [f(x)]} is ________. The number of roots of equation If f (x ) =

cos x + cos 2 x + cos 3x = 0 ______.

is

(0 £ x £ 2p)

6 Time : 3 hrs.

1. 2. 3. 4.

5.

6. 7.

Max. Marks : 300

INSTRUCTIONS This test will be a 3 hours Test. This test consists of Physics, Chemistry and Mathematics questions with equal weightage of 100 marks. Each question is of 4 marks. There are three parts in the question paper consisting of Physics (Q.no.1 to 30), Chemistry (Q.no.31 to 60) and Mathematics (Q. no.61 to 90). Each part is divided into two sections, Section A consists of 20 multiple choice questions & Section B consists of 10 Numerical value answer Questions. In Section B, candidates have to attempt only 5 questions out of 10. There will be only one correct choice in the given four choices in Section A. For each question 4 marks will be awarded for correct choice, 1 mark will be deducted for incorrect choice and zero mark will be awarded for unattempted question. For Section B 4 marks will be awarded for correct answer and zero for marked for each review / unattempted/incorrect answer. Any textual, printed or written material, mobile phones, calculator etc. is not allowed for the students appearing for the test. All calculations / written work should be done in the rough sheet provided.

PHYSICS Section - A 1.

2.

A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. The frequency of vibration of the 8 kg block is : 1 1 Hz Hz (2) (1) 2 2 4 1 Hz (3) (4) 2Hz 2 A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net

ur field E at the centre O is j

(1) (3)

q 2

i

O 2

ˆj

4p e 0 r q - 2 2 ˆj 2p e0 r

(2) (4)

q 2

4p e 0 r 2 q 2

2p e 0 r 2

ˆj

ˆj

EBD_7308 JEE MAIN

MT-62

3.

If a bar magnet of magnetic moment 80 units be cut into two halves of equal lengths, the magnetic moment of each half will be (1) 80 units (2) 40 units (3) 160 units

4.

5.

where a, b, c, are constants and b2 < 4 ac. For what value of w does the resonance occur ? b a (3) w = c (4) w = 0 Three bars each of area of cross - section A and length L are connected in series. The thermal conductivities of their materials are K, 2K, 1.5K. If the temperatures of the external ends of the first and last bar are 200 & 18ºC, then the temperatures of both the junctions are

(1)

7.

(2) w =

(3) T1 = 132ºC, T2 = 98ºC (4) T1 = 164ºC, T2 = 62ºC 8.

In the following diagram the reading of the ammeter is (when the internal resistance of the battery is zero) 10V

A

4W

(2) Solid sphere

6.

b 2a

(2) T1 = 120ºC, T2 = 84ºC

(1) Ring (3) Hollow sphere

w=

(1) T1 = 116ºC, T2 = 74ºC

4l1 = 2l 2 = 2l3 = l 4 l1 = 2l 2 = 2l3 = l 4 l1 = l 2 = 4l3 = 9l 4 l1 = 2l 2 = 3l3 = 4l 4

A small object of uniform density rolls up a curved surface with an initial velocity v. It 3v2 reaches up to a maximum height of with 4g respect to the initial position. The object is

1 2

[aw - bw + c]

(4) 20 units

Hydrogen ( 1 H1 ) , Deuterium ( 1 H 2 ) , singly + ionised Helium ( 2 He4 ) and doubly ionised + + lithium ( 3 Li 6 ) all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are l1, l2, l3 and l4 respectively then approximately which one of the following is correct? (1) (2) (3) (4)

given by x =

5W

v

(4) Disc The amplitude of velocity of a particle acted on by a force F cos w t along the x direction, is

(1) (3)

40 A 29 5 A 3

(2)

10 A 9

(4) 2 A

Mock Test -6 9.

MT-63

A 2V battery is connected across AB as shown in the figure. The value of the current supplied by the battery when in one case battery’s positive terminal is connected to A and in other case when positive terminal of battery is connected to B will respectively be:

11.

5W

D1

In a uniform magnetic field of induction B a wire in the form of a semicircle of radius r rotates about the diameter of the circle with an angular frequency w. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R, the mean power generated per period of rotation is (1)

10 W

D2

( B pr w )2 2R

(2)

( B pr 2 w ) 2 8R

B pr 2 w ( B pr w 2 ) 2 (4) 2R 8R A steel wire is suspended vertically from a rigid support. When loaded with a weight in air, it extends by la and when the weight is immersed completely in water, the extension is reduced to lw. Then the relatively density of material of the weight is

(3)

12.

10.

A B (1) 0.4 A and 0.2 A (2) 0.2 A and 0.4 A (3) 0.1 A and 0.2 A (4) 0.2 A and 0.1 A A thin wire of length L and uniform linear mass density r is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX' is

(1)

(2)

(3)

(4)

rL3 8p

2

rL3 16p 2 5rL3 16p 2 3rL3

8p 2

X

(1)

X¢ O

13.

(2)

la la - l w

lw lw (4) la la - l w An electromagnetic wave of frequency 1 × 1014 hertz is propagating along z-axis. The amplitude of electric field is 4 V/m. If e0 = 8.8 × 10–12 C2/Nm2, then average energy density of electric field will be: (1) 35.2 × 10–10 J/m3 (2) 35.2 × 10–11 J/m3 (3) 35.2 × 10–12 J/m3 (4) 35.2 × 10–13 J/m3

(3)

90°

la lw

EBD_7308 JEE MAIN

MT-64

14.

The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t = 4 / 3 s is

V

D

x (cm)

1

A

0

4

8

12

t(s) D

(1) (3) 15.

3 2 p cm/s 2 32 p2 cm/s2 32

2

(2) (4)

–p cm/s 2 32

In a Young’s double slit experiment with light of wavelength l the separation of slits is d and distance of screen is D such that D >> d >> l. If the fringe width is b, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is: (1)

b 6

D

C

V A

B (3)

B

P

P

(4)

(1) Positive in all the cases (1) to (4) (2) Positive in cases (1), (2), (3) but zero in case(4) (3) Negative in cases (1), (2), (3) but zero in case (4) (4) Zero in all the cases Three masses m, 2m and 3m are moving in x-y plane with speed 3u, 2u and u respectively as shown in figure. The three masses collide at the same point at P and stick together. The velocity of resulting mass will be: y

2m, 2u 60°

(3)

16.

(2) P

b (2) 3

b b (4) 4 2 In diagrams (1 to 4), variation of volume with changing pressure is shown. A gas is taken along the path ABCD. The change in internal energy of the gas will be

B

P C

A

17.

C

A

V

3 2 – p cm/s 2 32

D

B

(1)

–1

V

C

m, 3u

(1) (3)

(

P

)

u ˆ i + 3jˆ 12 u ˆ -i + 3jˆ 12

(

x

60°

3m, u u ˆ i - 3jˆ (2) 12 u ˆ -i - 3jˆ (4) 12

(

)

(

)

)

Mock Test -6 18.

MT-65

A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is (1)

gR 2 R+ x

gR (2) R-x æ gR 2 ö 1/ 2 (4) ç ÷ è R + xø

(3) gx 19.

Section - B 21.

22.

The plots of intensity versus wavelength for three black bodies at temperatures T1, T2 and T3 respectively are as shown. Their temperature are such that I T1

(2) T1 > T3 > T2

(4) T3 > T2 > T1 l A Zener diode is connected to a battery and a load as shown below:

4 kW I 60 V

23.

24.

IL

A

C

T2

(3) T2 > T3 > T1

IZ 10 V = VZ

RL = 2kW

B The currents, I, IZ and IL are respectively. (1) 15 mA, 5 mA, 10 mA (2) 15 mA, 7.5 mA, 7.5 mA

The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 mT. What will be its value at the centre of loop (in mT)? One end of a massless rope, which passes over a massless and frictionless pulley P is tied to a hook C while the other end is free. Maximum tension that the rope can bear is 360 N. With what value of maximum safe acceleration (in ms-2) can a man of 60 kg climb on the rope? P

T3

(1) T1 > T2 > T3

20.

(3) 12.5 mA, 5 mA, 7.5 mA (4) 12.5 mA, 7.5 mA, 5 mA

25.

The velocity of projection of a body is increased by 2%, other factors remaining unchanged, what will be the percentage change in the maximum height attained? The insulation property of air breaks down at E = 3 × 106 volts/ metre. The maximum charge that can be given to a sphere of diameter 5 m is approximately ______ coulomb. A compound microscope has an objective and eye-piece as thin lenses of focal lengths 1 cm and 5 cm respectively. The distance between the objective and the eye-piece is 20 cm. The distance at which the object must be placed in front of the objective if the final image is located at 25 cm from the eye-piece, is numerically ______ cm.

EBD_7308 JEE MAIN

MT-66

26.

27.

28.

When light of 3000 Å is incident on sodium chloride, the stopping potential is 1.85 volt and when light of 4000 Å is incident, then the stopping potential becomes 0.82 volt. The threshold wavelength for sodium is ______ Å. (h = 6.6 × 10–34 J-sec.) In four complete revolution of the cap, the distance travelled on the pitch scale is 2 mm. If there are 50 divisions on the circular scale, then the least count of the screw gauge is _____ mm. An electric charge 10–3 λ C is placed at the origin

29.

30.

and B are situated at ( 2, 2) and (2, 0) respectively. The potential difference between the points A and B will be ______ volts. The Thallium-201 half-life is 74 hours. If the sample has an activity of 80 millicuries initially, what will be the activity (in mci) after 9.25 days ? An engine approaches a hill with a constant speed. When it is at a distance of 0.9 km, it blows a whistle whose echo is heard by the driver after 5 seconds. If the speed of sound in air is 330 m/s, then the speed of the engine (in m/s) is _____ .

(0, 0) of X – Y co-ordinate system. Two points A

CHEMISTRY Section - A 31.

32.

33.

Four species are listed below: H3 O+

i.

HCO3–

ii.

iii.

HSO4–

iv. HSO3F

Which one of the following is the correct sequence of their acid strength? (1) iv < ii < iii < i (2) ii < iii < i < iv (3) i < iii < ii < iv (4) iii < i < iv < ii Which one of the following pairs is mismatched (1) Fossil fuel burning - release of CO2 (2) Nuclear power - radioactive wastes (3) Solar energy - Greenhouse effect (4) Biomass burning - release of CO2 Which of the following undergoes hydrolysis by SN1 mechanism :

34.

(1) CH3CH2CH2CH2Cl (2) CH3CH2CH2Cl (3) CH3 – O – CH2Cl (4) CH3Cl Which of the following hydrocarbon can react with maleic anhydride ?

(i)

(ii)

(iii)

Mock Test -6

MT-67

39.

(iv)

35.

36.

37.

38.

(1) only (i) (2) (i) and (iv) (3) (i), (iii) and (iv) (4) all the four The process of the extraction of Au and Ag is based on their solubility in (1) NH3 or NH4OH (2) KCN or NaCN (3) HCl or HNO3 (4) H2SO4 Which of the following compounds having highest and lowest melting point respectively ? (1) CsF (2) LiF (3) HCl (4) HF Correct answer is (1) 2, 3 (2) 1, 4 (3) 4, 3 (4) 2, 1 Silver bromide when dissolve in hypo solution gives complex ..... in which oxidation state of silver is .... (1) Na3[Ag(S2O3)2], (I) (2) Na3[Ag(S2O3)3], (III) (3) Na3[Ag(S2O3)2], (II) (4) Na3[Ag(S2O3)4], (I) An organic compound (A) reacts with sodium metal and forms (B). On heating with conc. H2SO4, (A) gives diethyl ether, (A) and (B) are (1) C2H5OH and C2H5ONa (2) C3H7OH and CH3ONa (3) CH3OH and CH3ONa (4) C4H9OH and C4H9ONa

40.

41.

42.

Which one of the following esters cannot undergo Claisen self-condensation? (1) CH 3 - CH 2 - CH 2 - CH 2 - COOC 2 H 5 (2) C6H5COOC2H5 (3) C6H5CH2COOC2H5 (4) C6H11CH2COOC2H5 In recovery of silver from photographic film, you have decided to dissolve the silver ion with dilute nitric acid. Addition of dilute HCl to precipitate AgCl seems to result in unacceptable losses. You might improve recovery by addition of_______ in the latter step. (1) NaNO3 (2) NaCl (3) Ag2SO4 (4) sodium acetate One mole of fluorine is reacted with two moles of hot and concentrated KOH. The products formed are KF, H2O and O2. The molar ratio of KF, H2O and O2 respectively is (1) 1 : 1 : 2 (2) 2 : 1 : 0.5 (3) 1 : 2 : 1 (4) 2 : 1 : 2 In the reaction, HOCl

B

CH 2 < CH 2 ¾¾ ¾↑ A ¾¾ ↑ CH 2 OH |

CH2OH

The molecule ‘A’ and the reagent ‘B’ are (1) H C – CH and hot water 2

2

O (2) CH3CH2Cl and NaOH (3) CH3CH2OH and H2SO4 (4)

CH 2 – CH2 and NaHCO3 | | OH Cl

EBD_7308 JEE MAIN

MT-68

43.

44.

Alkali metals dissolve in liquid NH3 then which of the following observations is not true? (1) H2 gas is liberated. (2) Solution turns into blue due to solvated electrons. (3) It becomes diamagnetic. (4) Solution becomes conducting. At very high pressures, the compressibility factor of one mole of a gas is given by : (1) (3)

45.

Pb RT Pb 1RT

1+

(2)

(4) 1 -

I

(4) none of these. 47.

48.

b (VRT)

Among the following compounds (I-III) the correct order of reactivity with electrophile is

OCH3

46.

Pb RT

(3) mixing equal volumes of equimolar solutions of AgNO3 and KI.

49.

NO2

II

III

(1) II > III > I

(2) III < I < II

(3) I > II > III

(4) I = II > III

One desires to prepare a positively charged sol of silver iodide. This can be achieved by (1) adding a little AgNO3 solution to KI solution in slight excess. (2) adding a little KI solution to AgNO3 solution in slight excess.

50.

Which of the following is an artificial edible colour? (1) Melamine

(2) Saffron

(3) Carotene

(4) Tetrazine

Lassaigne’s test for nitrogen is positive for which compound? (1) NH2OH

(2) NH2NH2

(3) H2NCONH2

(4) All the three

It is always advisable not to keep egg yolk or mustard in silver cutlery because (1) silver reacts with water of egg yolk to form AgOH. (2) silver reacts with sulphur of egg yolk forming black Ag2S. (3) silver reacts with egg yolk forming Ag2SO4 which is a poisonous substance. (4) silver attracts UV light of the atmosphere, thereby spoiling the food. Which of the following complexes will give white precipitate with BaCl2 (aq)? (1) [Co(NH3)4SO4]NO2 (2) [Cr(NH3)4SO4]Cl (3) [Cr(NH3)5Cl]SO4 (4) Both (b) & (c)

Mock Test -6

MT-69

Section - B 51.

52. 53.

54.

55.

Calculate the molality of 1 litre solution of 93% H2SO4 (weight/volume). The density of the solution is 1.84 g/mL. Find out the number of stereoisomers obtained by bromination of trans-2-butene. The enthalpy of combustion of H2(g), to give H2O(g) is –249 kJ mol–1 and bond enthalpies of H – H and O = O are 433 kJ mol–1 and 492 kJ mol–1 respectively. The bond enthalpy of O – H is A certain metal when irradiated to light (v = 3.2× 1016 Hz) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light (v = 2.0 × 1016 Hz) The v0 (threshold frequency) of metal is x × 1015. Find the value of x. Malachite has the formula Cu2CO3(OH)2. What

56.

57.

58. 59.

60.

percentage by mass of malachite is copper? The freezing point of a solution, prepared from 1.25 g of a non-electrolyte and 20 g of water, is 271.9 K. If molar depression constant is 1.86 K mole–1, find the molar mass of the solute A solution of Ni (NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many mole of Ni will be deposited at the cathode? In O-2 , O2 and O 22 - molecular species. Find the total number of antibonding electrons. x × 1021 unit cells are present in a cube-shaped ideal crystal of NaCl of mass 1.00 g. Find the value of x. [Atomic masses: Na = 23, Cl = 35.5] The electrode potential of Mg2+/Mg electrode, in which conc. of Mg2+ is 0.01M, is (E o Mg 2 + / Mg = -2.36V)

MATHEMATICS 62.

Section - A y

61.

If x =

ò 0

dt 1+ t2

(1) y (3)

, then

d2y dx

2

is equal to :

(2) x 1+ y

2

1+ y

(4) y 2

2

The number of real roots of the equation 1 + a1 x + a2 x2 + …. + an xn = 0 1 and |an| < 2, is 3 n if n is even 0 for any natural number n 1 if n is odd None of these.

where |x| < (1) (2) (3) (4)

EBD_7308 JEE MAIN

MT-70

63.

If e (1) (2) (3)

-

p 2

p < q < , then 2 cos log q > log cos q cos log q < log cos q cos log q = log cos q

(4) cos log q = 64.

65.

66.

The value of

68.

w x +1 following equation w x + w2 w2

2 log cos q 3

s

n

r=0

s =1 r £s

å å

n

s

Cs Cr is

69.

(2) 3n + 1 (1) 3n – 1 n (3) 3 (4) 3(3n – 1) Number of ways to distribute 10 distinct balls to 3 persons. One getting 2, 2nd getting 3 and 3rd getting 5 balls is 10! 10! (1) (2) 2!.(3!) 2 .5! 2!.3!.5! 10! (3) (4) 10! 2!.5! The general solution of the differential equation,

æ dy ö sin 2x ç - tan x ÷ - y = 0 , is : è dx ø

67.

(1)

y tan x = x + c

(2)

y cot x = tan x + c

(3)

y tan x = cot x + c

(4)

y cot x = x + c

(1) a = b = c (2) c = a (3) a = b (4) b = c If w is a cube root of unity, then a root of the

In any triangle ABC, if cos A =

sin B , then 2sin C

70.

1

w2 1 =0 is x+w

(1) x = 1 (2) x = w 2 (3) x = w (4) x = 0. The lines lx + my + n = 0, mx + ny + l = 0 and nx + ly + m = 0 are concurrent if (1) l – m – n = 0 (2) l + m – n = 0 (3) l – m + n = 0 (4) l2 + m2 + n2 = lm + mn + nl If (x, y) are the co-ordinates of a point in the 3 4 2 plane , then 5 8 2 = 0 represent x y 2 (1) a. st. line || to y-axis (2) a st. line || to x-axis

71.

(3) a st. line (4) a circle The equation of the circle whose radius is 5 and which touches the circle x2 + y2– 2x – 4y–20 = 0 at the point ( 5,5) is (1) x2 + y2 + 18 x + 16 y + 120 = 0 (2) x2 + y2 – 18 x – 16 y + 120 = 0 (3) x2 + y2 – 18 x + 16 y + 120 = 0 (4) x2 + y2 + 18 x – 16 y + 120 = 0

Mock Test -6 72.

MT-71

æ 3ö ç 2, ÷ to the ellipse, è 2ø

The normal at

x2 y 2 + = 1 touch es a par abola, whose 16 3 equation is (1) y2 = – 104 x (2) y2 = 14 x

73.

(3) y2 = 26x If p : I study and q : I fail

76.

(4) y2 = – 14x

Then negation of 'I study or I fail' is

77.

(1) I do not study and I do not fail (2) I do not study or I do not fail (3) Either I study and I do not rail or I study and I do not fail (4) I study and I do not fail 74.

The general solution of the trigonometric equation sin x – cos x = 1 is given by

78.

(1) x = 2n p, n Î I p 4

p ,n ÎI 2 r r ˆ B = 4iˆ + 3jˆ + 4kˆ and If A = ˆi + ˆj + k, r C = ˆi + aˆj + bkˆ are linearly independent vectors r and C = 3 , then

(4) x = n

75.

(2) If A 3 + 2 A 2 + 3A + 5I = 0 ; then A is invertible. (3) If A2 = 0, then A = 0 (4) None of these If for a real number y, [y] is the greatest integer less than or equal to y, then the value of the 3p / 2

p 4

integral òp / 2 [2 sin x ]dx is

(2) x = n p + (–1)n + , n Î I p (3) x =2n p + , n Î I 2

(1) a = 1, b = –1 (2) a = 1, b = ± 1 (3) a = –1, b = ± 1 (4) a = ± 1, b = 1 p : Every quadratic equation has one real root and q : Every quadratic equation has two real roots, then truth value of p and q are (1) p is true and q is false (2) p is false and q is true (3) p and q both true (4) p and q both false Let A, B and C be n × n matrices. Which one of the following is a correct statement? (1) If AB = AC, then B = C

79.

(1) –p

(2) p

(3) p /2

(4) -p/2

If f(x) = ax2 + b, b ¹ 0, x £ 1; = bx2 + ax + c, x > 1, then f(x) is continuous and differentiable at x = 1 if (1) c = 0, a = 2b (3) a = b, c = 0

(2) a = b, c arbitrary (4) a = b, c ¹ 0

EBD_7308 JEE MAIN

MT-72

80.

The values of the parameter a such that the roots a, b of the equation 2x2 + 6x + a = 0 satisfy a b the inequality + < 2 are b a (1) a > 0 (2) a < 9/2 (3) a < 0 or a > 9/2 (4) None of these

a1 = a2 = 10, then value of m + n is ______ .

86.

é ù æ 3p ö 3 êsin 4 ç - a ÷ + sin 4 (3p + a ) ú è 2 ø ë û

Section - B 81.

82. 83.

It has been found that A and B play a game 12 times, A wins 6 times, B wins 4 times and they draw twice. A and B take part in a series of 3 games. The probability that they will win alternately is ______ .

x x + cos5 If period of the function sin = 2 5 kp, then value of x is ______ . If the least difference between the roots, in the pö æ first quadrant ç 0 £ x £ ÷ , of the equation 2ø è 3

4 cos x (2 - 3 sin 2 x ) + (cos 2x + 1) = 0 is

84.

then value of n is ______ . If [x] denotes the greatest integer £ x and lim

1

x ®3 n 3

([12 x] + [ 22 x] + [32 x] + ........+ [ n 2 x]) =

p , n

87.

88.

If f(x) =

x -3 2 , g (x) = x -3 x+4

2(2 x + 1) x 2 + x - 12

and

lim [f(x) +g(x) +h(x)] = k then value of | k | is x ®3 ______ .

89.

p , n

If P and Q be two given points on the curve uuur uuur 1 y=x+ such that OP.iˆ = 1 and OQ.iˆ = -1 x ˆ where i is a unit vector along the x–axis and the length of vector 2OP + 3OQ = n 5, then value of n is ______ .

For natural numbers m, n if (1 - y )m (1 + y ) n = 1 + a1 y + a2 y 2 + ¼ and

é ù æp ö -2 êsin 6 ç + a ÷ + sin 6 (5p - a ) ú is ______ . è ø 2 ë û r r r r r r The value of [A - B, B - C, C - A] where r r r | A |= 1, | B |= 2 and | C |= 3 is ______ .

h(x) = –

then value of n is ______ . 85.

The value of expression

90.

The area bounded by parabola y2 = x, st. line y = 4 and y-axis is ______ .

7 Time : 3 hrs.

Max. Marks : 300

INSTRUCTIONS 1. 2. 3. 4.

5.

6. 7.

This test will be a 3 hours Test. This test consists of Physics, Chemistry and Mathematics questions with equal weightage of 100 marks. Each question is of 4 marks. There are three parts in the question paper consisting of Physics (Q.no.1 to 30), Chemistry (Q.no.31 to 60) and Mathematics (Q. no.61 to 90). Each part is divided into two sections, Section A consists of 20 multiple choice questions & Section B consists of 10 Numerical value answer Questions. In Section B, candidates have to attempt only 5 questions out of 10. There will be only one correct choice in the given four choices in Section A. For each question 4 marks will be awarded for correct choice, 1 mark will be deducted for incorrect choice and zero mark will be awarded for unattempted question. For Section B 4 marks will be awarded for correct answer and zero for marked for each review / unattempted/incorrect answer. Any textual, printed or written material, mobile phones, calculator etc. is not allowed for the students appearing for the test. All calculations / written work should be done in the rough sheet provided.

PHYSICS Section - A 1.

2.

Using mass(M), length(L), time(T) and electric current (A) as fundamental quantities the dimensions of permittivity will be (1) MLT–1A–1 (2) MLT–2A–2 –1 –3 4 2 (3) M L T A (4) M2L–2T –2A2 é pt ù y = 2 (cm) sin ê + fú ë2 û What is the maximum acceleration of the particle executing the SHM?

(1)

p cm/s2 2

(2)

p2 cm/s2 2

p p2 cm/s2 (4) cm/s2 4 4 A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current I passes through the coil, the magnetic field at the center is

(3)

3.

(1) (3)

m o NI b

m 0 NI b ln 2 (b - a ) a

(2)

2 m o NI a

m 0 IN a (4) 2 b - a ln b ( )

EBD_7308 JEE MAIN

MT-74

4.

A cylindrical solid of length L and radius a 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 having emf V and negligible internal resistance. The electric field as a function of x is best described by r = r 0x

(3) 7.

x V

(1)

2

L V

(1) g 6.

x

2V

(2)

r0 L2

x

8.

x (4) None of these L2 A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is:

(3) 5.

2V

(2)

2 g 3

g 3

(3)

(4)

d o

10.

d o

(4) d T T + DT

(2) T T + DT Temperature (K)

X

d o

T T + DT Temperature (K)

o

X

T T + DT

X

Temperature (K)

Temperature (K)

A Hydrogen atom and a Li++ ion are both in the second excited state. If l H and l Li are their respective electronic angular momenta, and EH and ELi their respective energies, then (1) l H > l Li and |EH| > |ELi| (2) l H = l Li and |EH| < |ELi| (3) l H = l Li and |EH| > |ELi| (4) l H < l Li and |EH| < |ELi| The temperature of reservoir of Carnot’s engine operating with an efficency of 70% is 1000 kelvin. The temperature of its sink is (1) 300 K (2) 400 K (3) 500 K (4) 700 K A radar has a power of 1kW and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth = 6.4 × 106m) is : (1) 80 km (2) 16 km (3) 40 km (4) 64 km What is the electric potential at the centre of the square? –Q +Q a

Y

Y

(1)

9.

3 g 2

An ideal gas is initially at temperature T and volume V. Its volume is increased by DV due to an increase in temperature DT, pressure remaining constant. The quantity d = DV V DT varies with temperature as

Y

Y

C –Q

X

(1) zero (3) kq/ a2

a +Q (2) kq/a 2 (4) None of these

Mock Test -7 11.

MT-75

In the arrangement shown in the Fig, the ends P and Q of an unstretchable string move downwards with uniform speed U. Pulleys A and B are fixed. Mass M moves upwards with a speed (1) 2Ucos q A B (2) U/ cos q (3) 2U / cos q (4) U cos q

12.

14.

P

M

A transparent solid cylindrical rod has a 2 refractive index of . It is surrounded by air.. 3 A light ray is incident at the mid-point of one end of the rod as shown in the figure. q

Q

In Young’s double slit experiment, the slits are illuminated by white light. The distance between two slits is d and screen is D distance apart from the slits. Some wavelengths are missing on the screen in front of one of the slits. These wavelengths are

d2 2d 2 (2) λ = D D 2d 2 d2 (3) λ = (4) λ = 3D 2D A rectangular block of mass m and area of crosssection A floats in a liquid of density r. If it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T. Then 1 1 (1) T µ (2) T µ r A 1 (4) T µ r (3) T µ m There are two wires of the same length. The diameter of second wire is twice that of the first. On applying the same load to both the wires, the extension produced in them will be in ratio of (1) 1 : 4 (2) 1 : 2 (3) 2 : 1 (4) 4 : 1 (1) λ =

13.

qq

15.

The incident angle q for which the light ray grazes along the wall of the rod is -1 æ 2 ö ÷ (2) sin ç è 3ø

-1 æ 3 ö (1) sin ç 2 ÷ è ø

-1 æ 1 ö -1 1 ÷ (3) sin ç (4) sin 2 è 3ø 16. In the figure below, what is the potential difference between the point A and B and between B and C respectively in steady state

( )

3mF

1mF

B

3mF

1mF 1mF

20W

10W

100V

A

(1) VAB = VBC = 100 V (2) VAB = 75, VBC = 25 V (3) VAB = 25V, VBC = 75V (4) VAB = VBC = 50V

C

EBD_7308 JEE MAIN

MT-76

17.

A block is placed on a frictionless horizontal table. The mass of the block is m and springs are attached on either side with force constants K1 and K2. If the block is displaced a little and left to oscillate, then the angular frequency of oscillation will be (1)

(3)

æ K1 + K 2 çç m è

é K1K 2 ù 2 (2) ê ú ë m (K 1 + K 2 ) û

÷÷ ø

1 ù2

é K 1K 2 ê ú ë (K 1 - K 2 )m û

1 ù2

é ú (4) ê êë (K1 + K 2 )m úû K12

+ K 22

æ1ö A bullet looses ç ÷ of its velocity passing ènø through one plank. The number of such planks that are required to stop the bullet can be: 2

21.

1

1 ö2

th

18.

Section - B

22.

23.

24.

2

n 2n (2) 2n - 1 n -1 (4) n (3) infinite An ideal coil of 10H is connected in series with a resistance of 5W and a battery of 5V. 2second after the connection is made, the current flowing in ampere in the circuit is (1) (1 – e–1) (2) (1 – e) (3) e (4) e–1 A spacecraft of mass ‘M’, moving with velocity ‘v’ suddenly breaks into two pieces. After the explosion mass ‘m’ becomes stationary. What is the velocity of the other part of the spacecraft? Mv (1) (2) v M-m mv M-m v (3) (4) M m (1)

19.

20.

25.

26.

27.

In a series LCR circuit R = 200W and the voltage and the frequency of the main supply is 220V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30°. On taking out the inductor from the circuit the current leads the voltage by 30°. The power dissipated in the LCR circuit is ______ watt. In an isothermal expansion of 10g of gas from volume V to 2V the work done by the gas is 575J. What is the root mean square speed (in m/ s) of the molecules of the gas at that temperature? An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20m. If the car is going twice as fast i.e., 120 km/h, the stopping distance (in metre) will be At the centre of a circular current carrying coil of radius 5 cm magnetic field due to earth is 0.5 × 10–5 W/m2. What should be the current (in A) flowing through the coil so that it annuls the earth’s magnetic field? An air bubble of radius 0.1 cm is in a liquid having surface tension 0.06 N/m and density 103 kg/m3. The pressure inside the bubble is 1100 Nm–2 greater than the atmospheric pressure. At what depth (in metre) is the bubble below the surface of the liquid? (g = 9.8 ms–2) The escape velocity for a body of mass 1 kg from the earth surface is 11.2 kms–1. The escape velocity for a body of mass 100 kg would be ______ kms–1. The circuit shown here has two batteries of 8.0 V and 16.0 V and three resistors 3 W, 9 W and 9 W and a capacitor of 5.0 mF.

Mock Test -7

MT-77

I

5 mF

3W

9W

8.0 V I1

28.

9W

29.

16.0 V

I2

How much is the current (I) (in A) in the circuit in steady state? A beam of light of intensity 12 watt/cm2 is incident on a totally reflecting plane mirror of

30.

area 1.5 cm2, then the force (in newton) acting on the mirror will be A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of 25.5 kVm–1. The density of liquid is 1.26 × 103 kg m–3. The radius of the drop is ______ metre. (neglect buoyancy). Two waves of wavelengths 99 cm and 100 cm both travelling with velocity 396 m/s are made to interfere. The number of beats produced per second is

CHEMISTRY Section - A 31.

32.

Ethylene dichloride and ethylidine chloride are isomeric compounds. The false statement about these isomers is that they (1) react with alcoholic potash and give the same product. (2) are position isomers. (3) contain the same percentage of chlorine. (4) are both hydrolysed to the same product. By what ratio the average velocity of the molecule in gas changes when the temperature is raised from 50 to 200 ºC ? (1)

(2)

1.46 1

4 2 (4) 1 1 Which of the following would not give 2phenylbutane as the major product in a Friedel-

(3)

33.

1.21 1

34.

35.

Crafts alkylation reaction ? (1) 1-butene + HF (2) 2-butanol + H2SO4 (3) Butanoyl chloride + AlCl3 then Zn, HCl (4) Butyl chloride + AlCl3 Moissan boron is (1) amorphous boron of ultra purity (2) crystalline boron of ultra purity (3) amorphous boron of low purity (4) crystalline boron of low purity Which reaction characteristics are changed by the addition of a catalyst to a reaction at constant temperature ? (i) Activation energy (ii) Equilibrium constant (iii) Reaction enthalpy (1) (i) only (2) (iii) only (3) (i) and (ii) only (4) all of these

EBD_7308 JEE MAIN

MT-78

36.

37. 38.

39.

40.

41.

42.

A metal M reacts with N2 to give a compound ‘A’ (M3N). ‘A’ on heating at high temperature gives back ‘M’ and ‘A’ on reacting with H2O gives a gas ‘B’. ‘B’ turns CuSO4 solution blue on passing through it. M and B can be : (1) Al & NH3 (2) Li & NH3 (3) Na & NH3 (4) Mg & NH3 Asthma patient use a mixture of .... for respiration. (1) O2 and CO2 (2) O2 and He (3) O2 and NH3 (4) O2 and CO When ethanal reacts with CH3MgBr and C2H5OH/dry HCl, the product formed are : (1) ethyl alcohol and 2-propanol (2) ethane and hemi acetal (3) 2-propanol and acetal (4) propane and methyl acetate Amongst the following salts of iron, which is most unstable in aqueous solutions? (1) K3[Fe(CN)6] (2) Fe2(SO4).9H2O (3) FeSO4.7H2O (4) FeI3 Given the molecular formula of the hexacoordinated complexes (i) CoCl 3.6NH3, (ii) CoCl3.5NH3, (iii) CoCl3.4NH3 If the number of co-ordinated NH3 molecules in i, ii and iii respectively are 6, 5, 4, the primary valencies in (i), (ii) and (iii) are : (1) 6, 5, 4 (2) 3, 2, 1 (3) 0, 1, 2 (4) 3, 3, 3 Polyethylene is (1) Random copolymer (2) Homopolymer (3) Alternate copolymer (4) Crosslinked copolymer The reason for “drug induced poisoning” is : (1) Binding reversibly at the active site of the enzyme

43.

44.

45.

46.

47.

(2) Bringing conformational change in the binding site of enzyme (3) Binding irreversibly to the active site of the enzyme (4) Binding at the allosteric sites of the enzyme With a change in hybridization of the carbon bearing the charge, the stability of a carbanion increases in the order : (1) sp < sp2 < sp3 (2) sp < sp3 < sp2 3 2 (3) sp < sp < sp (4) sp2 < sp < sp3 Which of the following compounds are not arranged in order of decreasing reactivity towards electrophilic substitution? (1) Methoxy benzene > Toluene > Bromo benzene (2) Phenol > N-Propylbenzene > Benzoic acid (3) Chlorotoluene > Para Nitro toluene > 2-chloro-4-Nitrotoluene (4) Benzoic acid > Phenol > N-Propylbenzene Which of the given sets of temperature and pressure will cause a gas to exhibit the greatest deviation from ideal gas behaviour ? (1) 100 ºC & 4 atm (2) 100 ºC & 2 atm (3) –100 ºC & 4 atm (4) 0 ºC & 2 atm Which is not true for beryllium ? (1) Beryllium is amphoteric. (2) It forms unusual carbide, Be2C. (3) Be(OH)2 is basic. (4) Beryllium halides are electron deficient. Which of the following pairs show reverse properties on moving along a period from left to right and from up to down in a group : (1) Nuclear charge and electron affinity (2) Ionization energy and electron affinity (3) Atomic radius and electron affinity (4) None of these

Mock Test -7 48.

49.

50.

MT-79

An organic amino compound reacts with aqueous nitrous acid at low temperature to produce an oily nitrosoamine. The compound is: (1) CH3 NH2 (2) CH3CH2NH2 (3) CH3CH2NH.CH2CH3 (4) (CH3 CH2)3N A white sodium salt dissolves readily in water to give a solution which is neutral to litmus. When silver nitrate solution is added to the solution, a white precipitate is obtained which does not dissolve in dil. HNO3. The anion could be (1) CO 32 (2) Cl–

(3) SO 24 (4) S2– Energy of activation is lowest for which reaction? +

54.

55.

56.

57.

+

(1)

RCH 2 O H 2 ® R C H 2

(2)

R 2 CH O H 2 ® R 2 C H

+

+

+

(3) R 3C O H 2 ® R 3C + (4) All have same Eact

58.

59.

Section - B 51.

52.

53.

The pKa of HCOOH is 3.8 and pKb of NH3 is 4.8, fin d the pH of aqueous solution of 1M HCOONH4. The number of electrons passing per second through a cross-section of copper wire carrying 10–6 amperes of current is found to be x × 102. Find the value of x. Number H-atoms present in 0.046 g of ethanol is x × 1021. Find the value of x.

Calculate the radius (in pm) of the largest sphere which fits properly at the centre of the edge of a body centred cubic unit cell. (Given edge length is 100 pm) What is the molarity of H2SO4 solution if 25mL is exactly neutralized with 32.63 mL of 0.164 M, NaOH ? 1.500g of hydroxide of a metal gave 1.000g of its oxide on heating. What is the equivalent mass of the metal ? The enthalpy change on freezing of 1 mole of water at 5 °C to ice at –5 °C is : [Given DfusH = 6 kJ mol–1 at 0 °C, Cp(H2O, l) =75.3 J mol–1 K–1, Cp(H2O, s) = 36.8 J mol–1 K–1] The ground state energy of hydrogen atom is – 13.6 eV. Calculate the energy of second excited state of He+ ion in eV. The total number of basic groups in the following form of lysine is +

O CH

60.

C

O– H2N What quantity (in mL) of a 45% acid solution of a mono-protic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution?

EBD_7308 JEE MAIN

MT-80

MATHEMATICS 65.

Section - A 61.

62.

The function f (x) = tan–1(sin x increasing function in p (1) æç 0, ö÷ (2) è 2ø (3) æç p , p ö÷ (4) è 4 2ø

x2 y 2 = 1 meets 4 2 x-axis at P and y-axis at Q. Lines PR and QR are drawn such that OPRQ is a rectangle (where O is the origin). Then R lies on : 4 2 2 4 (1) (2) + =1 =1 2 2 2 x y x y2 2 4 4 2 + =1 (4) (3) =1 2 2 2 x y x y2

With the usual notation equal to (1)

64.

æ p pö çè - , ÷ø 2 2 æ p pö çè - , ÷ø 2 4

A tangent to the hyperbola

2

63.

+ cos x) is an

4+ 2 - 3

ò ( [x

2

)

] - [x]2 dx is

1

(2) 4 - 2 + 3

(3) 4 - 2 - 3 (4) None of these The standard deviation of a variate x is s. The ax + b standard deviation of the variable ; a, b, c c are constants, is a æaö s (1) ç ÷s (2) c c è ø æ a2 ö (3) ç 2 ÷ s (4) None èc ø

66.

Assume R and S are (non-empty) relations in a set A. Which of the following relation given below is false (1) If R and S are transitive, then R È S is transitive. (2) If R and S are transitive, then R Ç S is transitive. (3) If R and S are symmetric, then R È S is symmetric. (4) If R and S are reflexive, Then R Ç S is reflexive. Given that f and g are continuous functions on [0 a] satisfying f(a–x) = f(x) and g(x)+g(a–x) = 2. a

ò0 f (x) g (x) dx is equal to :

Then (1)

a

ò0 f (x) dx

(3) 0

67.

(2)

a

ò0 g(x) dx

(4) None of these x2 c = y = and If c1 = y = be two 2 1 + x2 2 curves lying in XY-plane, then 1 (1) area bounded by curve y = and 1+ x2 p y = 0 is 2 p (2) area bounded by c1 and c2 is - 1 2 p (3) area bounded by c1 and c2 is 1 2 1 and x(4) area bounded by curve y = 1+ x2 p axis is 2

1

Mock Test -7 68.

69.

Let f and g be functions from the interval [0, ¥) to the interval [0, ¥), f being an increasing and g being a decreasing function. If f {g(0)} = 0 then (1)

f {g( x )} ³ f {g(0)}

(2) (3) (4) dy dx

g{f ( x )} £ g{f (0)} f {g(2)} = 7 None of these

(1) (2) (3) (4)

70.

MT-81

73.

+ y = 2e 2 x then y is

ce - x +

2 2x e 3

(1 + x )e - x +

2 2x e +c 3

2 ce - x + e 2 x + c 3 2 2x -x e + e +c 3

74.

a+x a-x a-x If a - x a + x a - x = 0 then x is

(1) 0, 2a (2) a, 2a (3) 0, 3a (4) None of these The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is (1)

75.

76.

(2) x 2 + y 2 = 16a 2

(4) x 2 + y 2 = a 2 If a positive integer n is divisible by 9, then the sum of the digits of n is divisible by 9. So which statement is it contrapositive. (1) (sum of digits of n is divisible by 9) Þ (n is divisible by 9) (3)

72.

x 2 + y 2 = 9a 2

(1)

æ9 ö ç ÷ è 16 ø

(3)

æ3ö ç ÷ è5ø

77.

æ8 ö ÷ è 15 ø

7

(2) ç

7

(4) None of these

( -1 + 6)a

(2) ( 6 - 1)a

(3) a (4) None of these If n(A) = 1000, n(B) = 500 and if n(A Ç B) ³ 1 and n(A È B) = p, then (1) 500 £ p £ 1000 (2) 1001 £ p £ 1498 (3) 1000 £ p £ 1498 (4) 1000 £ p £ 1499 dy If y = log 2 {log 2 ( x)} , then is dx log 2 e 2.3026 (1) (2) x ln x x ln x ln 2

(3)

x 2 + y 2 = 4a 2

6

If a £ 0 then roots of x 2 - 2a x - a - 3a 2 = 0 is (1)

a-x a-x a+x

71.

(2) (sum of digits of n is not divisible by 9) Þ (n is not divisible by 9) (3) (sum of digits of n is divisible by 9) Þ (n is divisible by 9) (4) None of these Fifteen coupons are numbered 1, 2 ..... 15, respectively. Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on a selected coupon is 9, is

1 ln ( 2 x ) x

(4) None of these

f ( x ) = sin x . f (x ) is not differentiable at (1) x = 0 only (2) all x (3) multiples of p

(4) multiples of

p 2

EBD_7308 JEE MAIN

MT-82

78.

The angle between the pair of tangents drawn to the ellipse 3x2 + 2y2 = 5 from the point (1,2) is (1)

æ 12 ö tan -1 ç ÷ è5ø

79.

(

)

12 ö ÷÷ (4) tan -1 12 5 (3) tan ç è 5ø The least integral value a of x such that

x-5

80.

( )

(2) tan -1 6 5

-1 æç

84.

85.

> 1 , satisfies :

1 4 P(A) Ç P(B) Ç P(C) = 0, P(B Ç C) = 0 and 1 P (A Ç C) = , P (A Ç B) = 0 8

x 2 + 5 x - 14 (1) a2 + 3a – 4 = 0 (2) a2 – 5a + 4 = 0 (3) a2 – 7a + 6 = 0 (4) a2 + 5a – 6 = 0 The x satisfying

= Probability of C =

sin -1 x + sin -1 (1 - x ) = cos -1 x are (1) 1, 0 (2) 1, –1 1 (3) 0, (4) None of these 2 Section - B

81.

the probability that atleast one of the events A, B, C exists is ______ . 86.

1 + x 2 + 1 + y 2 = l( x 1 + y 2 - y 1 + x2 ) is

82.

______ . If the second term in the expansion

83.

n C3 æ a ö 13 5/2 ç a+ ÷ is 14a , then value of n ç ÷ C2 a -1 ø è is ______ . Given two independent events, if the probability

n

26 and the 49

If

n

r+2

å r + 1 n Cr =

r=0

87.

The degree of differential equation satisfying the relation

that exactly one of them occurs is

15 , then 49 the probability of more probable of the two events is ______ . a, b be the roots of x2 – 3x + a = 0 and g, d be the roots of x2 – 12x + b = 0 and numbers a, b, g, d (in order) form an increasing G.P. then value of a + b is ______ . The probability of A = Probability of B

probability that none of them occurs is

sin 3x , when x ¹ 0 sin x = k, when x = 0 for the function to be continuous, k should be ______ . f (x) =

p/3

88.

28 - 1 , then n is ______ . 6

If

ò

cos x + sin x

dx = kp then value of k is 1 + sin 2 x ______ . The S.D. of the following data is nearly 0

89.

x i 140 145 150 155 160 165 170 175 fi 4 6 15 30 36 24 8 2 90.

is ______ . A line passes through (2,2) and is perpendicular to the line 3x + y = 3, its y intercept is ______ .

8 Time : 3 hrs.

Max. Marks : 300 INSTRUCTIONS

1. 2. 3. 4.

5.

6. 7.

This test will be a 3 hours Test. This test consists of Physics, Chemistry and Mathematics questions with equal weightage of 100 marks. Each question is of 4 marks. There are three parts in the question paper consisting of Physics (Q.no.1 to 30), Chemistry (Q.no.31 to 60) and Mathematics (Q. no.61 to 90). Each part is divided into two sections, Section A consists of 20 multiple choice questions & Section B consists of 10 Numerical value answer Questions. In Section B, candidates have to attempt only 5 questions out of 10. There will be only one correct choice in the given four choices in Section A. For each question 4 marks will be awarded for correct choice, 1 mark will be deducted for incorrect choice and zero mark will be awarded for unattempted question. For Section B 4 marks will be awarded for correct answer and zero for marked for each review / unattempted/incorrect answer. Any textual, printed or written material, mobile phones, calculator etc. is not allowed for the students appearing for the test. All calculations / written work should be done in the rough sheet provided.

PHYSICS Section - A 1.

A given ideal gas with g =

Cp

Cv

2. = 1.5 at a

temperature T. If the gas is compressed adiabatically to one-fourth of its initial volume, the final temperature will be (1)

2 2T

(2) 4 T

(3) 2 T (4) 8 T A small bar magnet is being slowly inserted with constant velocity inside a solenoid as shown in figure. Which graph best represents the relationship between emf induced with time

EBD_7308 JEE MAIN

MT-84

(1) varies with time as x = A sin (wt + f) (1)

Time

(2) is constant and has value

(2) Time

5. (3) 3.

Time

(4) Time

There is a spherical cavity of radius R/2 in uniformly R charged spherical region R/2 having charge density r and radius R as shown in figure. r A small charged particle having q0 charge is released at point O. It will collide with the wall of cavity with kinetic energy (1)

rR 2 q 0 3 Î0

(2)

rR 2 q 0 6p Î0

6.

rR 2 q 0 rR 2 q 0 (4) 4p Î0 12 Î0 Two bodies of mass m and 2m connected by an unextended spring of spring constant ‘k’ are allowed to fall simultaneously in a uniform gravitational field g. The extension ‘x’ in the spring when the bodies A and B are falling

(3)

4.

8.

m g

2m

(3) is zero (4) increases linearly with time A radioactive nucleus of mass M emits a photon of frequency n and the nucleus recoils. The recoil energy will be (1) Mc2 – hn (2) h2n2 / 2Mc2 (3) zero (4) hn In Young’s double slit experiment, if intensity of maxima on screen is I (assume equal intensity of the source) and when the sources become incoherent, average intensity on the screen is I1 is I2 1 (2) (1) 2 (3) 1 (4) 0 2 Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the nth orbital will therefore be proportional to : (1) n 2 (2) n (3) n 1/2 (4) n 1/4 From a sphere of mass M and radius R, a smaller R sphere of radius is carved out such that the 2 cavity made in the original sphere is between its centre and the periphery (See figure). For the configuration in the figure where the distance between the centre of the original

I2, then the ratio

7.

2mg k

Mock Test -8

MT-85

sphere and the removed sphere is 3R, the gravitational force between the two sphere is:

holes and electrons respectively. The material is (1) an insulator

Ec

(2) a metal

Eg

(3) an n-type semiconductor (4) a p-type semiconductor

3R

(1) (3)

9.

41 GM 2 2

3600 R 59 GM 2

(2)

41 GM 2 450 R GM 2

(4) 450 R 2 225 R 2 Two rods of the same length and diameter having thermal conductivities K1 and K2 are joined in parallel. The equivalent thermal conductivity of the combination is (1)

K1 K 2 K1 + K 2

(2)

12.

2

13.

K1 + K 2

K1 + K 2 (4) K1 K 2 2 A photoelectric surface is illuminated successively by monochromatic light of l wavelengths l and . If the maximum kinetic 2 energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is :

(3)

10.

hc hc hc 3hc (2) (3) (4) 2l l 3l l In the energy band diagram of a material shown below, the open circles and filled circles denote

(1)

11.

14.

15.

Ev

The figure of merit of a galvanometer is defined as (1) the voltage required to produce unit deflection in the galvanometer (2) the current required to produce unit deflection in the galvanometer (3) the power required to produce unit deflection in the galvanometer (4) None of these The activity of a radioactive element decreases to one-third of the original activity I0 in a period of nine years. After a further lapse of nine years its activity will be 2 (1) I0 (2) I 0 3 I0 I0 (4) (3) 9 6 The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is (1) v0 + g /2 + f (2) v0 + 2g + 3f (3) v0 + g /2 + f/3 (4) v0 + g + f Which of the following graph represents the input characteristics of a common emmiter transistor correctly ?

EBD_7308 JEE MAIN

MT-86 Y Base current

(2) X

Base voltage

Y

Base current

(3)

(4) X

Base voltage

16.

X

Base voltage

Y

Base current

(1)

Base current

Y

18. Base Voltage

X

A charged ball B hangs from a silk thread S, which makes an angle q with a large charged conducting sheet P, as shown in the figure. The surface charge density s of the sheet is proportional to P

q

If Q1, Q2, Q3 indicate the heat absorbed by the gas along the three processes and DU1, DU2, DU3 indicate the change in internal energy along the three processes respectively, then (1) Q1 > Q2 > Q3 and DU1 = DU2 = DU3 (2) Q3 > Q2 > Q1 and DU1= DU2 = DU3 (3) Q1 = Q2 = Q3 and DU1 > DU2 > DU3 (4) Q3 > Q2 > Q1 and DU1> DU2 > DU3 A tennis ball (treated as hollow spherical shell) starting from O rolls down a hill. At point A the ball becomes air borne leaving at an angle of 30° with the horizontal. The ball strikes the ground at B. What is the value of the distance AB ? (Moment of inertia of a spherical shell of mass m 2 and radius R about its diameter = mR 2 ) 3 O

S

2.0 m

B

17.

(1) cot q (2) cos q (3) tan q (4) sin q An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram :

P

19.

A

2 3

V

30° 0.2 m

1 B

A

B

(1) 1.87 m (2) 2.08 m (3) 1.57 m (4) 1.77 m In an organ pipe, three successive resonance frequencies are observed at 425, 595 and 765Hz. The length of organ pipe is (Velocity of sound in air = 340 m/s). (1) 2.0 m (2) 1.5 m (3) 1.0 m (4) 2.5 m

Mock Test -8 20.

21.

22.

23.

24.

A ball is thrown from a point with a speed ' v0 ' at an elevation angle of q. From the same point and at the same instant, a person starts running ' v0 ' with a constant speed to catch the ball. 2 Will the person be able to catch the ball? If yes, what should be the angle of projection q ? (1) No (2) Yes, 30° (3) Yes, 60° (4) Yes, 45° Section - B An object is moving towards a concave mirror of focal length 24 cm. When it is at a distance of 60 cm from the mirror its speed is 9 cm/s. The speed of its image at the instant is ______ cm/s toward away from the mirror. An aluminium foil of relative emittance of 0.1 is placed in between two concentric spheres at temperature 300 K and 200 K respectively. What is the temperature in kelvin of this foil when steady state is reached? Assume the spheres to be black bodies. (s = 5.672 × 10–8) The space between the parallel plates of a capacitor is tightly filled with three dielectric slabs A,B, C of thickness 5mm, 3mm and 2mm with dielectric constants 2, 3 and 5 respectively. If a potential difference of 351 volts is applied to plates then electric field intensity in slab A is ______ v/m. Given that radius of earth is 6.4 × 106 m and a transmitting antenna at the top of a tower has a height 40 m and the height of the receiving antenna is 50 m. What is the maximum distance (in metre) between them for satisfactory communication in los mode?

MT-87

25.

26.

27.

A structural steel rod has a radius of 10 mm and length of 1.0 m. A 100 kN force stretches it along its length. Young’s modulus of structural steel is 2 × 1011 Nm–2. The percentage strain is about In the X-rays tube before striking the target we accelerate the electrons through a potential difference of V volt. For kV we will have X-rays of largest wavelength? A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is ______ N.

10N

28.

The effective resistance between points A and B of infinite network is .......... ohm. A

1W

30.

B

1W

1W 2W

29.

1W 2W

2W

While measuring the length of the rod by vernier callipers the reading on main scale is 6.4 cm and the eight division on vernier is in line with marking on main scale division. If the least count of callipers is 0.01 and zero error –0.04 cm, the length of the rod is ______ cm. (1) 5.25 cm (2) 6.52 cm (3) 5.52 cm (4) 6.44 cm A moving particle of mass m, makes a head on elastic collision with another particle of mass 2m, which is initially at rest. The percentage loss in energy of the colliding particle on collision, is close to ______ .

EBD_7308 JEE MAIN

MT-88

CHEMISTRY Section - A 31.

32. 33.

34.

35.

Match the items in Column I with its main use listed in Column II: Column I Column II (A) Silica gel (i) Transistor (B) Silicon (ii) Ion–exchanger (C) Silicone (iii) Drying agent (D) Silicate (iv) Sealant (1) (A)–(iii), (B)–(i), (C)–(iv), (D)–(ii) (2) (A)–(iv), (B)–(i), (C)–(ii), (D)–(iii) (3) (A)–(ii), (B)–(i), (C)–(iv), (D)–(iii) (4) (A)–(ii), (B)–(iv), (C)–(i), (D)–(iii) Which of the following has highest boiling point? (1) H2Se (2) H2Te (3) H2S (4) H2O In which of the following will the oxidation state change ? (1) Reaction of Cu2+with NaOH (2) Reaction of Cu2+ with iron metal (3) Reaction of Cu2+ with KI (4) both (2) and (3) Reduction of esters with sodium and alcohol is referred to as (1) Bouveault-Blanc reduction (2) Mendius reaction (3) Clemmensen’s reduction (4) MPV reduction p-Chloroaniline and anilinium hydrochloride can be distinguished by (1) Sandmeyer reaction (2) NaHCO3 (3) AgNO3 (4) Carbylamine test

36.

37.

Chemical reduction is not suitable for (1) conversion of bauxite to aluminium (2) conversion of cuprite into copper (3) conversion of haematite to iron (4) conversion of zinc oxide to zinc Which is the only group which is deactivating yet ortho-para directing when attached to benzene ring? (1) NO2 (2) CH3 (3) Cl (4) None of these C º C – H + CH3 - CH 2 MgBr gives

38.

(1)

C º CMgBr + CH – CH C º CMgBr + CH 3 – CH2

(2)

(3)

(4)

C º CMgBr + C2H 6

MgBr + CH3 – CH2 – C º CH

C 2H5 + HC º CMgBr

Mock Test -8 39.

MT-89

Which of the following will be most reactive towards nucleophillic substitution.

Br (1)

Cl (2)

OH (3) 40. 41.

42.

45.

OSO2CF3 (4)

46.

Ozone reacts with dry iodine to give (1) IO2 (2) I2O3 (3) I2O4 (4) I4O9 Dipole moment is shown by (1) 1, 4-dichlorobenzene (2) cis-1,2-dichloroethene (3) trans-1,2-dichloroethene (4) trans-2,3-dichloro-2 butene In what manner will increase of pressure effect the following equilibrium? C(s) + H 2 O(g)

43.

44.

CO(g) + H 2 (g)

(1) Shift in forward direction (2) Shift in reverse direction (3) Increase in yield of hydrogen (4) No effect Which of the following compounds is most reactive to an aqueous solution of sodium carbonate? (1)

(2)

(3)

(4)

47.

AgBr dissolves in Na2S2O3 solution and forms (A). On boiling an aquous solution of (A), a ppt. (B) is obtained. The colour of B is (1) Red (2) White (3) Black (4) Colourless Chloride of an element gives neutral solution in water. In the periodic table the element belongs to (1) third group (2) fifth group (3) first group (4) first transition series Arrange the following in order of decreasing order of reactivity towards Perkin reaction CHO CHO CH 3

I

II

CHO

CHO

N(CH 3 ) 2

NO 2

III IV (1) IV > III > I > II (2) II > III > IV > I (3) III > II > I > IV (4) IV > III > I > II In the preparation of p-nitro acetanilide from aniline titration is not done by nitrating mixture (a mixture of conc. H2SO4 and conc. HNO3) because (1) on nitration it gives o-nitro acetanilide (2) lt gives a mixture of o and p-nitro aniline (3) –NH2 group gets oxidised (4) it forms a mixture of o and p-nitro acetanilide.

EBD_7308 JEE MAIN

MT-90

48.

49.

Which of the following sequences represents the correct order of increasing stability of +3 oxidation state? (1)

Al < In < Ga

(3)

Bi < Sb < As

(2) As < Sb < Bi (4) none of these

An alloy of copper, silver and gold is found to have copper constituting the ccp lattice. If silver atoms occupy the edge centres and gold is present at body centre the alloy has a formula (1) Cu4Ag2Au (2) Cu4Ag4Au (3) Cu4Ag3Au (4) CuAgAu

OH 50.

TsCl

54.

OH (NH3)4 Co

55.

¾¾¾¾ ® A, Then A is LiAlH

56.

OLi

57.

4

(1)

(2)

(3)

(4)

Cl

58.

Section - B 51.

52. 53.

At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen (N2) at 4 bar. Find the molar mass of gaseous molecule How many stereoisomers are possible for 1, 3dichloro- cyclopentane? What is the work function (in eV) of the metal if the light of wavelength 4000Å generates photoelectrons of velocity 6×105 ms-1 from it ?

(Mass of electron=9×10–31 kg; Velocity of light = 3×108 ms–1; Planck’s constant=6.626×10–34 Js; Charge of electron =1.6×10–19 JeV–1) What is the value of x in the following complex?

59.

60.

OH

Co (NH3)2 (SO4)X

Consider the oxidation state of Cobalt in the complex same as that of iron in K3[Fe(CN)6]? The hydrogen electrode is dipped in a solution of pH = 3 at 25 ºC. Find the potential of the cell would be 50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia solution is 4.75. Find the pH of the mixture will be : When a solution containing w g of urea in 1kg of water is cooled to – 0.372ºC, 200g of ice is separated. If Kf for water is 1.86 K kg mol–1, calculate w. The rate of a reaction quadruples when the temperature changes from 300 to 310 K. Calculate the activation energy of the reaction (Assume activation energy and pre-exponential factor are independent of temperature; ln 2 = 0.693; R = 8.314 J mol–1 K–1) An ideal gas is allowed to expand from 1 L to 10 L against a constant external pressure of 1 bar. Calculate the work done in kJ. Calculate the amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess H2S in the presence of conc. HCl (assuming 100% conversion)

Mock Test -8

MT-91

MATHEMATICS (1) (–5, –7)

Section - A 61.

If f (x) = 1 - 1 - x 2 , then at x = 0,

65.

(1) f (x) is differentiable as well as continuous (2) f (x) is differentiable but not continuous

62.

In a D ABC, are such that

sin A sin(A - B) = , then a2, b2, c2 sin C sin(B - C)

(3) 66.

52C × 48 3

(4)

The domain of the function f (x) = exp( 5x - 3 - 2x 2 ) is (1) [3/2, ¥)

(2) [1, 3/2]

(2) they are in H.P.

(3) (–¥, 1]

(4) (1, 3/2)

2

2

(4) b = a + c

67.

2

å 50 Cr (2x - 3)r (2 - x) n -r

r =0

(1)

50C 25

(3) –50C25

is

(2) –50C24 (4)

50C 30

Solution of differential equation dy = sin x + 2 x , is dx

Coefficient of x25 in the expansion of expression 50

64.

13C × 4C × 48 3 3 52C × 48C 3 3

(2)

(1) they are in G.P. (3) they are in A.P. 63.

(3) (5, –7) (4) (–5, 7) The number of ways of selecting 4 cards out of 52 cards of an ordinary pack, so that exactly 3 of them are of same denomination is (1) 52 × 48

(3) f (x) is continuous but not differentiable (4) f (x) is neither continuous nor differentiable

(2) (5, 7)

68.

(1) y = x2 – cos x + c

(2) y = cos x + x2 + c

(3) y = cos x + 2

(4) y = cos x + 2 + c

If z = x + iy and w =

(1 - iz ) , then | w | = 1, show (z - i )

that in complex plane

The equation x3 + 5x2 + px + q = 0 and x3 + 7x2 +

(1) z is situated on imaginary axis

px + r = 0 have two roots in common. If their third roots are x1 and x2, then (x1, x2) =

(2) z is situated on real axis (3) z is situated on unit circle (4) None of these

EBD_7308 JEE MAIN

MT-92

69.

If z, wz and w z are the vertices of a triangle, then the area of the triangle will be (where w is cube root of unity) : (1)

70.

3 | z |2 2

(1) Continuous at x = –1 (2) Continuous at x = 0 (3) Discontinuous at x =

(2) –2

2 3

2 3

(4)

160 cm/sec 3 (4) 160 cm/sec.

(2)

(3) 10 cm/sec If

æx

1

ö

ò 1 + sin x dx = tan çè 2 + a ÷ø + b then

(1)

p a = - , bÎR 4

(3)

a=

5p , bÎR 4

p (2) a = , b Î R 4

(4) None of these

1 2

(4) All of these 74.

The altitude of a cone is 20cm and its semi-vertical angle is 30°. If the semi-vertical angle is increasing at the rate of 2° per second, then the radius of the base is increasing at the rate of– (1) 30 cm/sec

ì1 - [ x] ï , x ¹ -1 If f (x) = í 1 + x , then the value of ïî1 , x = -1 f ( 2 k ) will be (where [] shows the greatest

integer function]

3 | z |2 (4) None of these (3) 2 From a point (h, 0) common tangents are drawn to circles x2 + y2 = 1 and (x – 2)2 + y2 = 4, value of h is

(3)

72.

3 3 | z |2 2

(2)

(1) 2

71.

73.

75.

A variable circle passes through the fixed point A(p, q) and touches x-axis. The locus of the other end of the diameter through A is (1)

( y - q)2 = 4 px

(2) ( x - q)2 = 4 py

(3)

( y - p )2 = 4qx

(4) ( x - p)2 = 4qy

Let A = 2ˆi + kˆ , B = ˆi + ˆj + kˆ and C = 4ˆi - 3ˆj + 7kˆ . The vector R which satisfies the equations R ´ B = C ´ B and R . A = 0 is given by

(1)

- 2ˆi + kˆ

(2)

- ˆi - 8ˆj + 2kˆ

(3)

1 ˆ ˆ (i - j + 2kˆ ) 6

(4) None of these

Mock Test -8 65

76.

å

k =33

MT-93

2k p 2k p ö æ - i cos çè sin ÷= 8 8 ø

(1) 1 + i

(3)

(2) 1 – i

80.

1- i i (4) 2 2 –1 The complete solution set for sin (x) = 3 sin–1 (a) is (1)

78.

0£a £

1 2

(2) -

1 £a£0 2

1 1 (3) - £ a £ (4) None of these 2 2 The equation of the plane through the line of intersection of planes

(ab'-a ' b) x + (bc '-b' c) y + (ad'-a ' d) = 0

(2)

(ab'-a ' b) x + (bc'- b' c) y + (ad'-a ' d )z = 0

(3)

(ab'-a ' b) y + (ac'-a ' c)z + (ad'-a ' d) = 0

æ x 2 - y2 ö ÷ = e a where a is constant, If tan -1 ç 2 ç x + y2 ÷ ø è dy then dx

(1)

x y

(2)

y x

x2 The coordinates of the foot of perpendicular from the point (1, 0, 0) to the line

(1) (2, – 3, 8)

(2)

(1, – 1, – 10)

(3) (5, – 8, – 4)

(4)

(3, – 4, – 2)

81.

The smallest value of ‘p’ for which y2 + xy + px2 – x – 2y + p = 0 represent 2 straight lines is ____ .

82.

If the length of a focal chord of parabola y2 = 4x is 25 and has positive slope, then the slope will be 4 ______ .

83.

The roots of the equation x5 – 40x4 + px3 + qx2 + rx + s = 0 are in geometric progression and the sum of their reciprocal is 10. Then value of |S| is ______ .

(4) None of these 79.

y2

y2

Section - B

ax + by + cz + d = 0, a ' x + b' y + c' z + d' = 0 and parallel to the line y = 0, z = 0 is (1)

(4)

x - 1 y + 1 z + 10 = = are 2 -3 8

(3) 1 +

77.

x2

84.

Let | a |= 2 , | b |= 5 . The values of |k| for which the vectors

a + kb

an d

a - kb

are

perpendicular is ______ . 85.

From a pack of 52 cards two cards are drawn at random. The probability of both cards being spades is ______ .

EBD_7308 JEE MAIN

MT-94

86.

87.

If f(A) = 3, f ¢(A) = –2, g(A) = –1, g¢(A) = 4, then g ( x )f ( a ) - g ( a ) f ( x ) value of lim is ______ . x-a x ®a ex sin 2x tan x 2 If D ( x) = ln(1 + x ) cos x sin x =A+ Bx + Cx2 cos x 2 e x - 1 sin x 2

+ .... then value of B is ______ .

88.

The sides of a triangle ABC lie on the lines 3x + 4y = 0, 4x + 3y = 0 and x = 3. Let (h, k) be the centre of the circle inscribed in D ABC. The value of (h + k) is ______ .

89.

4 5 ù é2 ú ê If A = ê 4 8 10 ú . Then rank of A is êë- 6 - 12 - 15úû

90.

______ . If a coin is tossed n times, the probability that head appears odd no. of times is ______ .

9 Time : 3 hrs.

Max. Marks : 300

INSTRUCTIONS 1. 2. 3. 4.

5.

6. 7.

This test will be a 3 hours Test. This test consists of Physics, Chemistry and Mathematics questions with equal weightage of 100 marks. Each question is of 4 marks. There are three parts in the question paper consisting of Physics (Q.no.1 to 30), Chemistry (Q.no.31 to 60) and Mathematics (Q. no.61 to 90). Each part is divided into two sections, Section A consists of 20 multiple choice questions & Section B consists of 10 Numerical value answer Questions. In Section B, candidates have to attempt only 5 questions out of 10. There will be only one correct choice in the given four choices in Section A. For each question 4 marks will be awarded for correct choice, 1 mark will be deducted for incorrect choice and zero mark will be awarded for unattempted question. For Section B 4 marks will be awarded for correct answer and zero for marked for each review / unattempted/incorrect answer. Any textual, printed or written material, mobile phones, calculator etc. is not allowed for the students appearing for the test. All calculations / written work should be done in the rough sheet provided.

PHYSICS Section - A 1.

Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sources are R1 and R2 ( R2 > R1). If the potential difference across the source having internal resistance R2 is zero then (1) R = R1R2 /( R1 + R2 ) R = R1R2 /( R2 – R1 ) (3) R = R2 ´ ( R1 + R2 ) /( R2 - R1 ) (4) R = R2 - R1 Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in

the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B ? A B

(2)

2.

60°

(1) (2) (3) (4)

30°

4.9 ms–2 (in horizontal direction) 9.8 ms–2 (in vertical direction) Zero 4.9 ms–2 (in vertical direction)

EBD_7308 JEE MAIN

MT-96

3.

A capillary tube (A) is dipped in water. Another identical tube (B) is dipped in a soap -water solution. Which of the following shows the relative nature of the liquid columns in the two tubes. A

B

A

B

5.

(1)

(2)

(1) A

B

A

B

6.

(3)

(4) 4.

A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle q with the vertical is

(1)

3g 2g sin q sin q (4) 2l 2l The rectangular surface of area 8 cm × 4 cm of a black body at temperature 127°C emits energy E per second. If the length and breadth are reduced to half of the initial value and the temperature is raised to 327°C, the rate of emission of energy becomes

(3)

3g cos q 2l

(2)

2g cos q 3l

7.

3 81 E (2) E 8 16

(3)

9 81 E (4) E 16 64

T he figure shows the volume V versus temperature T graphs for a certain mass of a perfect gas at two constant pressures of P1 and P2. What inference can you draw from the graphs? (1) P1 > P2 V P2 (2) P1 < P2 P1 (3) P1 = P2 q2 (4) No interference q1 T can be drawn due to insufficient information. What should be the maximum acceptance angle at the air core interface of an optical fibre if n 1 and n2 are the refractive indices of the core and the cladding, respectively (1)

sin -1 (n 2 / n1 )

(2) sin -1 n12 - n 22

(3)

é -1 n 2 ù ê tan ú n1 û ë

é -1 n1 ù (4) ê tan n ú ë 2û

Mock Test -9 8.

MT-97

When Uranium is bombarded with neutrons, it undergoes fission. The fission reaction can be written as :

V R

235 + 0n1 ® 56Ba141 + 36Kr92 + 3x + Q(energy) 92 U

9.

10.

where three particles named x are produced and energy Q is released. What is the name of the particle x ? (1) electron (2) a-particle (3) neutron (4) neutrino Given that K = energy, V = velocity, T = time. If they are chosen as the fundamental units, then what is dimensional formula for surface tension? (1) [KV–2T–2] (2) [K2V2T–2] 2 –2 –2 (3) [K V T ] (4) [KV2T2] Three charges of (+ 2q), (– q) and (– q) are placed at the corners A, B and C of the equilateral triangle of side a as shown in the given figure. Then the dipole moment of this combination is

S1 C L

(1) Work done by the battery is half of the energy dissipated in the resistor (2) At t = t, q = CV/2 (3) At t = 2t, q = CV (1 – e–2) t (4) At t = , q = CV (1 – e–1) 2 12.

A + 2q

a

(3) qa

11.

–q B

(1) a

C –q

(2) zero 2

qa 3 In an LCR circuit as shown below both switches S1 and S2 are open initially. Now switch S1 is closed, S2 kept open. (q is charge on the capacitor and t = RC is capacitive time constant). Which of the following statements is correct ?

3

Torques t1 and t 2 are required for a magnetic needle to remain perpendicular to the magnetic fields at two different places. The magnetic fields at the those places are B1 and B2 respectively; then ratio

a D

(1) qa

(4)

S2

(3) 13.

B1 is B2

t2 t1 t1 + t 2 t1 - t 2

t1 t2 t -t (4) 1 2 t1 + t 2

(2)

The transverse displacement y (x, t) of a wave is given by y( x, t ) = e This represents a:

(

- ax 2 + bt 2 + 2 ab ) xt

)

(1) wave moving in – x direction with speed (2) standing wave of frequency

b

b a

EBD_7308 JEE MAIN

MT-98

(3) standing wave of frequency

b

(4) wave moving in + x direction speed 14.

16.

1 a b

A system of four gates is set up as shown. The ‘truth table’ corresponding to this system is :

Two identical short bar magnets, each having magnetic moment M, are placed a distance of 2d apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is (1)

A

(3) Y

B

(1)

(3)

15.

A B Y 0 0 1 0 1 0 1 0 0 1 1 1

(2)

A B Y 0 0 0 0 1 0 1 0 1 1 1 0

A B Y 0 0 1 0 1 0 1 0 1 1 1 0

(4)

A B Y 0 0 1 0 1 1 1 0 0 1 1 0

A particle of mass m is under a potential v(x) = (1/2)k1x2 for x > 0 and v(x) = k1x2 for x < 0. When disturbed a little from the position x = 0, it will (1) not execute SHM m 3k1 (3) execute SHM with T2 = 2p2m/(k12)

17.

18.

19.

(4) execute SHM with T =

æ 1 ö ÷÷ k1 çç1 + 2ø è

(2) (4)

m0 M ( 3) 3 4p d m0 M ( 5) 3 4p d

Inside a cylinder closed at both ends is a movable piston. On one side of the piston is a mass m of a gas, and on the other side a mass 2m of the same gas. What fraction of volume of the cylinder will be occupied by the larger mass of the gas when the piston is in equilibrium? The temperature is the same throughout. (1) 1/4 (2) 1/2 (3) 2/3 (4) 1/3 A hydrogen atom emits green light when it changes from n = 4 energy level to the n = 2 level. Which colour of light would the atom emit when it changes from n = 5 level to the n = 2 level? (1) red (2) yellow (3) green (4) violet As shown in the figure, a very thin sheet of aluminium is placed in between the plates of the condenser. Then the capacity Al strip

(2) execute SHM with T = 2p

p m

m0 M ( 2) 3 4p d 2 m M æ 0ö çè ÷ p ø d3

(1) (2) (3) (4)

Will increase Will decrease Remains unchanged May increase or decrease

Mock Test -9 20.

A coil having n turns and resistance RW is connected with a galvanometer of resistance 4RW. This combination is moved in time t seconds from a magnetic field W1 weber to W2 weber. The induced current in the circuit is n (W 2 - W1 ) (W2 - W1 ) (1) (2) 5 Rt Rnt (W2 - W1 ) n ( W 2 - W1 ) (4) (3) - 5 Rnt Rt Section - B

21.

22.

23.

24.

Radioactive material 'A' has decay constant '8 l' and material 'B' has decay constant 'l'. Initially 1 they have same number of nuclei. After time, xl the ratio of number of nuclei of material 'A' to 1 that 'B' is . Find the alue of x. e A ball projected from ground at an angle of 45° just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is ______metre. A particle of charge 16 × 10–16 C moving with velocity 10 ms–1 along x-axis enters ur a region where magnetic field of induction B is along the y-axis and an electric field of magnitude 104 Vm–1 is along the negative z-axis. If the charged particle continues moving along x-axis, the ur magnitude of B is ________ Wb m–2 A satellite is revolving round the earth with velocity v. The minimum percentage increase in its velocity necessary for the escape of satellite will be

MT-99

25.

26.

27.

28.

29.

30.

The work function of a metallic surface is 5.01 eV, photoelectrons are emitted when light of wavelength 2000Å falls on it. The minimum potential difference (in volts) required to stop the fastest photoelectrons is (h = 4.14 × 10–15 eV–s) A refrigerator takes heat from water at 0°C inside it and rejects it to the room at a temperature of 27°C. The latent heat of ice is 336 × 10 3 J kg–1. If 5 kg of water at 0°C is converted into ice at 0°C by the refrigerator, then the energy consumed by the refrigerator is close to ________ joule. The potential energy of a 1 kg particle free to move along the x-axis is given by æ x4 x 2 ö - ÷J. V ( x) = ç ç 4 2 ÷ø è The total mechanical energy of the particle is 2 J. Then, the maximum speed (in m/s) is The radiation corresponding to 3 ® 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10–4 T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to _____ eV. A stuntman plans to run across a roof top and horizontally jumps on to another roof 4.9 m below the first one and at a distance of 6.2 m away. What is the minimum velocity (in m/s) he must have before the jump ? A thin sheet of glass (m = 1.5) of thickness 6 microns introduced in the path of one of interfering beams in a double slit experiment shifts the central fringe to a position previously occupied by fifth bright fringe. Then the wave length of the light used is _______ Å.

EBD_7308 JEE MAIN

MT-100

CHEMISTRY CH3

Section - A 31.

32.

33.

34.

A hydrocarbon with five carbon atoms in the molecule, decolourizes alkaline KMnO4, but does not give a precipitate with ammonical Cu2Cl2 solution. The hydrocarbon is possibly (1) 1-pentyne (2) 1, 3-pentadiyne (3) 2-pentyne (4) 1, 4-pentadiyne The threshold frequency of a metal is 1 ´ 1015 s–1. The ratio of maximum kinetic energies of the photoelectrons when the metal is irradiated with radiations of frequencies 1.5 ´ 1015 s–1 and 2 ´ 1015 s–1 respectively would be (1) 3 : 4 (2) 1 : 2 (3) 2 : 1 (4) 4 : 3 A solution containing a group–IV cation gives a precipitate on passing H2S. A solution of this precipitate in dil.HCl produces a white precipitate with NaOH solution and bluish–white precipitate with basic potassium ferrocyanide. The cation is : (1) Co2+ (2) Ni2+ (3) Mn2+ (4) Zn2+ Identify the feasible reaction among the following: ® K 2 O + CO 2 K 2 CO3 ¾¾

(2)

® Na 2 O + CO 2 Na 2 CO3 ¾¾

35.

D

(2) O C–NH–CH3

NHCH3 (3) 36.

37.

(4)

Consider the reaction CaO(s) + CO2(g) in a closed CaCO3(s) container at equilibrium. What would be the effect of addition of CaCO3 on the equilibrium concentration of CO2 ? (1) Increases (2) Decreases (3) Data insufficient (4) Remains unaffected Consider the reaction sequence below : OCH3 Succinic anhydride AlCl3

Clemmenson 's reduction

¾¾¾¾¾¾¾® A ¾¾¾¾¾¾® X

is

D

® Li 2 O + CO 2 Li 2 CO 3 ¾¾

D (4) Rb 2 CO 3 ¾¾ ® Rb 2 O + CO 2 Which of the following gives N-nitrosoamine on reaction with nitrous acid?

N

(1)

D

(1)

(3)

NH2

CH3

OCH3 OH

(1)

OH

Mock Test -9

MT-101

OH

(2)

H3CO

40.

OCH3

OH

(3) (4) 38.

41. H3CO

,

Br CH–CH2Br

42.

Alc. KOH A ¾¾¾¾® NaNH2

NaNH2

¾¾¾¾® B CH 3CH 2Cl

A and B are CH=CH2

(1)

A=

CH2–CH3 ;B=

C º CH

(2)

A=

C º CCH2CH3 ;B=

C º CH

é d ( DG ) ù = -DS DG = DH - T DS and ê ë dT úû p The enthalpy of cell reaction, DH, is then given by é dE ù (1) T ê cell ú - Ecell ë dT û p

(2) CH2CH2C º CH

;B=

(3)

(4) None of these Which one of the following on treatment with 50% aqueous sodium hydroxide yields the

(4)

(3) 39.

43.

corresponding alcohol and acid? (1) C6H5CHO (2) CH3CH2CH2CHO O || (3) CH 3 - C - CH 3 (4) C6H5CH2CHO The complexes of Nickel (II) can be (1) Square planar, tetrahedral and octahedral (2) Square planar and octahedral (3) Tetrahedral and octahedral (4) Square planar only Which of the following represents the correct order of stability? (1) H – C º C – : > H 2 C = CH – : > H3 C – CH 2– : (2) H3C – CH 2– : > H 2C = CH – : > H – C º C – : (3) H 3C – CH 2– : > H 2C = CH –: = H – C º C – : (4) None of these Which of the metal is extracted by Hall-Heroult process? (1) Al (2) Cu (3) Ni (4) Zn

A=

é æ dE ö ù nF êT ç cell ÷ - Ecell ú êë è dT ø p úû é æ dE ö ù nF ê Ecell - T ç cell ÷ ú è dT ø p ú ëê û

æ dE ö nFT ç cell ÷ è dT ø p

EBD_7308 JEE MAIN

MT-102

44.

45.

Given E º

= 1.52 V and

Au 3+ / Au

E º 3+ = 1.36V . Au / Au + Point out the correct statement of the following (1) Au3+ disproportionates into Au4+ and Au2+ in aqueous solution (2) Au3+ disproportionates into Au4+ and Au+ in aqueous solution (3) Au+ disproportionates into Au3+ and Au in aqueous solution (4) Au+ disproportionates into Au2+ and Au in aqueous solution The major product of the following reaction is :

46.

47.

48.

OH

(i) K2CO3

¾¾¾¾¾® (ii) CH 3I (1. eq.)

The hybridization states of the central atoms in the complexes [Fe(CN)6]3– , [Fe(CN)6]4– and [Co(NO2)6]3– are (1) d2sp3, sp3 and d4s2 respectively (2) d2sp3, sp3d and sp3d2 respectively (3) d2sp3, sp3d2 and dsp2 respectively (4) all d2sp3 Among the acids given below CH3CH2COOH (X); CH2 = CHCOOH (Y) and CH º C-COOH (Z) The correct order of increasing acid strength is (1) X < Y < Z (2) X < Z < Y (3) Y < X < Z (4) Z < Y < X Which series of reactions correctly represents chemical reactions related to iron and its compound? dil . H 2SO 4 H 2SO 4 , O 2 (1) Fe ¾¾¾¾¾ ® FeSO 4 ¾¾¾¾¾ ® heat

® Fe Fe 2 ( SO4 )3 ¾¾¾

OH OCH3

(3)

2 Fe ¾¾¾¾® FeCl3 ¾¾¾¾®

O

Cl , heat

heat , air

Zn

FeCl 2 ¾¾® Fe

(2)

(4)

OH

O , heat

CO , 600°C

2 Fe ¾¾¾¾ ® Fe3O 4 ¾¾¾¾¾ ®

CO , 700°C

49. (3)

OCH3 OCH3

(4)

dil. H SO

2 2 4® Fe ¾¾¾¾ ® FeO ¾¾¾¾¾ heat ® Fe FeSO 4 ¾¾¾

(1) OH

O , heat

(2)

50.

® Fe FeO ¾¾¾¾¾

The form of BN which is as hard as diamond is (1) hexagonal form (2) Cubic form with ZnS structure (3) both of these (4) none of these A mixture of two aromatic compounds A and B is separated by dissolving in chloroform followed by extraction with aq. KOH solution. The alkaline aqueous layer gives a mixture of two isomeric

Mock Test -9

MT-103

compounds on treatment with carbon tetrachloride. The organic layer containing compound A gives an unpleasant odour on treatment with alcoholic solution of KOH. Compounds A and B respectively are CN

53.

OH

(1)

and

54.

NC

OH

(2)

and

55.

NH2

OH

(3)

56.

and OH

NH2

(4)

and

57.

Section - B 51.

52.

58.

Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below: 1

o

D H D eg Ho 1 2 diss ® Cl(g) ¾¾¾¾ ® Cl – (g) Cl 2 (g) ¾¾¾¾¾ 2

D

Ho

hyd ® Cl – (aq) ¾¾¾¾ (using the data, –1 D diss H°Cl = 240 kJ mol , 2

–1

D eg H °Cl = –349 kJ mol , –1

° – = –381 kJ mol ), DH will be _____. Dhyd H Cl

59. 60.

Bond distance in HF is 9.17 × 10–11 m. Dipole moment of HF is 6.104 × 10 –30 Cm. The percentage ionic character in HF will be ________. (electron charge = 1.60 × 10–19 C) The vapour pressure (at the standard boiling poin t of water ) of an aqueous solution containing 28% by mass of a non-volatile normal solute (molecular mass = 28) will be ________. Lithium chloride has a Cl– cubic structure as shown below. If the edge length is 400 pm., then the radii a/2 + of Cl– ions is Li At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen (N2) at 4 bar. Find the molar mass ofgaseous molecule. Normality of a mixed solution of sulphuric acid and hydrochloric acid is 0.6 N. 20 mL of this solution gives 0.4305 g of AgCl on reacting with AgNO3 solution. The strength of H2SO4 in g/ L in the mixed solution is ________ . What is the value of n in the molecular formula BenAl2Si6O18 ? The half life of a first order reaction is 10 min. If initial amount is 0.08 mol L-1 and concentration at some instant is 0.01 mol L-1. What is the time elapsed ? For soaps critical micelle concentration (CMC) is 10–x (min.) to 10–y (max.) mol/L. What is the value of x? A weak base BOH is titrated with a strong acid HA. When 10 ml of HA is added, the pH is found to be 9.00 and when 25 ml is added, pH is 8.00. Find the volume of the acid required to reach the equivalence point

EBD_7308 JEE MAIN

MT-104

MATHEMATICS Section - A 61.

(3)

If P = { x Î R :f (x) = 0 } and

(4)

Q = { x Î R :g (x) = 0 } then P È Q is

(1) (2)

62.

63.

{ x Î R : f (x) + g(x) = 0 } { x Î R : f (x) g(x) = 0 } 2 2 x Î R : ( f (x)) + ( g(x)) = 0

{

65.

}

(3) (4) None of these p+q If cosec q = ( p ¹ q ¹ 0 ) , then p-q æp qö cot ç + ÷ is equal to: è 4 2ø p q (1) (2) (3) pq (4) pq q p The expression satisfying the differential dy 2 + 2 xy = 1 is equation x - 1 dx (1) x 2 y - xy 2 = c

(

(2)

64.

òe (1) (2)

(log x + ax )

cos(bx 2 + c) dx is equal to

2 b e ax cos( bx 2 + c + tan -1 ) + A a a +b

2

1

2 b eax cos(bx 2 - c - tan -1 ) + A a 2 a +b

2

2

1 + sec x dx = 2(fog )( x ) + C, then

f ( x ) = sec x - 1 (2) f ( x) = 2 tan -1 x

(3) g( x ) = sec x - 1 (4) None of these If y2 = P (x) is a polynomial of degree 3, then

67.

d é 3 d2y ù êy ú dx êë dx 2 úû is equal to (1) P(x) + P¢ (x) (2) P(x) + P¢¢¢ (x) (3) P(x) P¢¢¢ (x) (4) None of these Which of the following is an even function? 2

68.

(1)

æ 1- x ö f ( x ) = log ç ÷ è1+ x ø

(2)

f ( x ) = log ( x + 1 + x 2 )

(3)

f ( x) =

x x

e -1

+

x +1 2

2x (4) f ( x ) = e + sin x A plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is (a, b, g ) . The equation of the plane is

1

2

2 b e ax cos(bx 2 + c - tan -1 ) + A 2 2 a a +b 2 1 b eax cos(bx 2 + c - tan -1 ) + A 2 2 a 2 a +b

66.

( y 2 - 1)x = y + c

2

ò

(1)

)

(3) ( x 2 - 1) y = x + c (4) None of these

If

1

(1) (3)

x y z + + =3 a b g x y z 1 + + = a b g 2

(2) ax+ by+ gz = 3abg (4) None of these

Mock Test -9 69.

Equation of the latus rectum of the hyperbola

MT-105

74.

(10x - 5) 2 + (10 y - 2) 2 = 9(3x + 4 y - 7) 2 is

(1)

1 3æ 1ö y - = - çx - ÷ 5 4è 2ø

(2)

1 3æ 1ö x - = - çy- ÷ 5 4è 2ø

(3)

y+

1 3æ 1ö = - çy + ÷ 5 4è 2ø The point (2a, a) lies inside the region bounded by the parabola x2 = 4y and its latus rectum. Then, (1) 0 £ a £ 1 (2) 0 < a < 1 (3) a > 1 (4) a < 0 If DABC is right angled at A then the inradius of the triangle is (1) 2(a + b – 2) (2) 2(b + c – a) b+c-a b+c-a (3) (4) 2 4 1 1 1 , , If are in A.P.. b+ c c+ a a+ b

(4)

70.

71.

72.

73.

1 3æ 1ö = - çx + ÷ 5 4è 2ø

-1 (2) 1/ 1 – x 2 esin x (3) sin –1 x (4) 1/(1 – x2) Which of the following functions is differentiable at x = 0? (1) cos(|x|) + |x| (2) cos (|x|) - |x| (3) sin (|x|) + |x| (4) sin (|x|) - |x|

(1)

75.

76.

x+

then 9 ax +1 ,9 bx +1 ,9cx +1 , x ¹ 0 are in : (1) G.P. (2) G.P. only if x < 0 (3) G.P. only if x > 0 (4) None of these p sin (2k - 1) x dx , then – If a k = ò sin x 0 (1) a1, a2, ....... are in A.P. (2) a1, a2, ....... are in G..P. (3) a1, a2, ....... are in H.P. (4) a1, a2, ....... form a constant sequence

Let f (x) = ex, g(x) = sin–1x and h (x) = f (g(x)), then h'(x)/h(x) =

77.

The largest interval lying in æç -p , p ö÷ for which è 2 2ø the function, 2 æx ö f ( x) = 4- x + cos -1 ç - 1÷ + log (cos x) is è2 ø defined, is p (1) é - p , p ö (2) éê 0, ö÷ ê 4 2÷ ë 2ø ë ø p p (3) [0, p] (4) æç - , ö÷ è 2 2ø If f (x) = sin x + cos x, g (x) = x2 – 1, then g (f (x)) is invertible in the domain

(1)

é pù ê0 , 2 ú ë û

é p pù (2) ê- , ú ë 4 4û

é p pù (4) [0, p] ê- 2 , 2 ú ë û Which of the following functions is NOT one-one ? (1) f : R ® R defined by f ( x ) = 6 x - 1.

(3)

78.

(2) (3) (4)

f : R ® R defined by f ( x ) = x 2 + 7. f : R ® R defin ed b y f ( x ) = x 3 . 2x + 1 . f : R - {7} ® R defined by f ( x) = x-7

EBD_7308 JEE MAIN

MT-106

79.

é1 ù 1 2 4 1 Lim ê sec 2 2 + 2 sec 2 2 ............ + sec2 1ú n n ® ¥ ë n2 n n n û

equals 1 (1) sec 1 2

80.

1 cosec 1 2 1 (4) (3) tan 1 tan 1 2 The differential equation of the family of curves

(2)

2 3 represented by c( y + c) = x is

84.

85.

é

86.

2

(1)

d2y æ dy ö y 2 - y 2 ç ÷ = 27 x è dx ø dx 2

87.

3

æ dy ö æ dy ö (2) 12 y ç ÷ = 8x ç ÷ - 27 x è dx ø è dx ø 3

(3)

2

æ dy ö æ dy ö 8 y ç ÷ = 12 x ç ÷ - 27 x è dx ø è dx ø 3

88.

(4)

81.

82.

83.

r r r r r r r r r r r ´ a = b ´ a, r ´ b = a ´ b and r × a = 30, then r magnitude of r is _____. The acute angle between two lines such that the direction cosines l, m, n, of each of them satisfy the equations l + m + n = 0 and l2 + m2 – n2 = 0 is (in degree) _____. In a two player game that always has a winner, A beats B with probability 2/3; B beats C with probability 2/3 & C beats A with the same probability. If B plays with C and then the winner plays with A, the chance that A will be the final winner is ______.

pù

= cot x and x-axis, x Î ê0, ú is log k, then value ë 2û of k is ________. From the point (15, 12) three normals are drawn to the parabola y2 = 4x. If centroid of triangle formed by three co-normal points is (m, n), then value of m + n is _____. If a + b + c = 3 and a > 0, b > 0, c > 0, and the greatest value of a2 b3 c2 is

2

æ dy ö æ dy ö æ dy ö çè ÷ø - çè ÷ø + çè ÷ø - y = 27 x . dx dx dx Section - B r ˆ ˆ ˆ r If a = i – 2 j + 3k , b = –3iˆ + ˆj - kˆ ,

r r r r r r r r If and a = i + j+k , b = 4i + 3 j + 4k r r r r c= i + a j + b k are linearly independent vectors and | rc | = 3 , then greatest value of a + b is __. If the area lying between the curves y = tan x, y

89.

90.

3p.2q 7r

, then value of

(p + q + r) is _____. If z1, z2, z3, z4 be the vertices of a quadrilateral taken in order such that z1 + z3 = z2 + z4 , æ z -z ö |z1 – z3 | = | z2 – z4 | and arg ç 1 2 ÷ = ± kp then è z3 - z2 ø value of k is ______. The function defined by ì 1 -1 ï 2 f (x) = í x + e 2 - x , x ¹ 2, is ï k x=2 î continuous from right at the point x =2, then k is _____. Consider a rectangle whose length is increasing at the uniform rate of 2 m/sec, breadth is decreasing at the uniform rate of 3 m/sec and the area is decreasing at the uniform rate of 5 m2/ sec. If after some time the breadth of the rectangle is 2 m then the length of the rectangle is ____m.

(

)

10 Time : 3 hrs.

Max. Marks : 300

INSTRUCTIONS 1. 2. 3. 4.

5.

6. 7.

This test will be a 3 hours Test. This test consists of Physics, Chemistry and Mathematics questions with equal weightage of 100 marks. Each question is of 4 marks. There are three parts in the question paper consisting of Physics (Q.no.1 to 30), Chemistry (Q.no.31 to 60) and Mathematics (Q. no.61 to 90). Each part is divided into two sections, Section A consists of 20 multiple choice questions & Section B consists of 10 Numerical value answer Questions. In Section B, candidates have to attempt only 5 questions out of 10. There will be only one correct choice in the given four choices in Section A. For each question 4 marks will be awarded for correct choice, 1 mark will be deducted for incorrect choice and zero mark will be awarded for unattempted question. For Section B 4 marks will be awarded for correct answer and zero for marked for each review / unattempted/incorrect answer. Any textual, printed or written material, mobile phones, calculator etc. is not allowed for the students appearing for the test. All calculations / written work should be done in the rough sheet provided.

PHYSICS Section - A 1.

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 :

2.

æ 2 Mg ö (1) L ç 1 + è 9 pYR 2 ÷ø

æ 1 Mg ö (2) L ç1 + è 9 pYR 2 ÷ø

æ 1 Mg ö (3) L ç 1 + è 3 pYR 2 ÷ø

æ 2 Mg ö (4) L ç1 + è 3 pYR 2 ÷ø

The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant k and compresses it by length

EBD_7308 JEE MAIN

MT-108

L. The maximum momentum of the block after collision is

same axial points. Then the potential at the common centre is

M

(1)

Q(R 2 + r 2 ) 4pe 0 (R + r )

(3) Zero

(1)

kL2 2M

(2)

6.

Mk L

ML2 (4) Zero k A mass M, attached to a horizontal spring, executes S.H.M. with amplitude A1. When the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with amplitude A2. The ratio

(3)

3.

1

(1) (3)

æ M + mö 2 çè ÷ M ø

(2)

æ M ö2 çè ÷ M + mø

(4)

M M +m

1

4.

5.

Three particles of equal mass m are situated at the vertices of an equilateral triangle of side L. What should be the velocity of each particle so that they move on a circular path without changing L? (1) Ö(GM/2L) (2) Ö(GM/L) (3) Ö(2GM/L) (4) Ö(GM/3L) A charge Q is distributed over two concentric hollow spheres of radii r and R (>r) such that the surface densities are equal and placed on the

q

æ1 ö (1) v cosç q ÷ è2 ø

æ A1 ö

M +m M

(4)

7.

æ1 ö (2) 2 v cosç q ÷ è2 ø

(3) v(1 + sin q) (4) v(1 + cos q) In the Bohr model an electron moves in a circular orbit around the proton. Considering the orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in nth excited state, is : (1)

æ e n2 h ö ç ÷ ç 2m 2p ÷ è ø

(2)

æ e ö nh ç ÷ è m ø 2p

2 æ e ö nh æ e ön h (4) ç ÷ ç ÷ è m ø 2p è 2m ø 2p From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path.

(3)

8.

Q R+r Q( R + r )

4pe 0 ( R 2 + r 2 ) A wheel is rolling straight on ground without slipping. If the axis of the wheel has speed v, the instantenous velocity of a point P on the rim, defined by angle q, relative to the ground will be P

of çè A ÷ø is: 2

(2)

Mock Test -10

MT-109

The relation between H, u and n is: (1) 2gH = n2u2 (2) gH = (n – 2)2 u2d (3) 2gH = nu2 (n – 2) (4) gH = (n – 2)u2 9.

The figure shows a system of two concentric spheres of radii r1 and r2 are kept at temperatures T1 and T2, respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to (1)

(3)

(4) 11.

ær ö In ç 2 ÷ è r1 ø

(2)

(r2 - r1 ) (r1 r2 )

(3)

( r2 - r1 )

r1 r2

T1 T2

r1 r2 (r2 - r1 ) An electron beam is accelerated by a potential difference V to hit a metallic target to produce Xrays. It produces continuous as well as characteristic X-rays.If lmin is the smallest possible wavelength of X-ray in the spectrum, the variation of log lmin with log V is correctly represented in :

12.

(4)

10.

(1)

(2)

A cell of e.m.f. E is connected across a reistance R. The potential difference between the terminals of the cell is found to be V. The internal resistance of the cell must be (1) [2(E–V)V]/R (2) [2(E–V)/V]R (3) [(E–V)/V]R (4) (E–V)R A cylindrical vessel of cross-section A contains water to a height h. There is a hole in the bottom of radius ‘a’. The time in which it will be emptied is: (1) (3)

13.

2A pa

2

h g

2 2A h pa 2 g

(2) (4)

2A h pa 2 g A 2pa

2

h g

Two straight long conductors AOB and COD are perpendicular to each other and carry currents I1 and I2 respectively. The magnitude of the magnetic induction at a point P at a distance ‘a’ from the point O where the two conductors intersect, in a direction perpendicular to the plane ABCD is (1)

m0 (I1 + I 2 ) 2 pa

(2)

(3)

m 0 é 2 2 ù 1/ 2 I1 + I 2 û 2pa ë

(4)

m0 (I1 – I 2 ) 2 pa m 0 æ I1I 2 ö ÷ ç 2pa çè I1 + I 2 ÷ø

EBD_7308 JEE MAIN

MT-110

14.

In a series resonant circuit, having L, C and R as its elements, the resonant current is i. The power dissipated in the circuit at resonance is i 2R

(2) zero

(1) æ 1 ö ç wL ÷ wC ø

è

15.

16.

(3) i2wL (4) i2R The co-ordinates of a moving particle at any time ‘t’are given by x = a t 3 and y = b t 3 . The speed of the particle at time ‘t’ is given by (1) 3t a 2 + b 2

(2) 3t 2 a 2 + b2

(3) t 2 a 2 + b2

(4)

a 2 + b2

The mutual inductance of a pair of coils, each of N turns, is M henry. If a current of I ampere in one of the coils is brought to zero in t second, the emf induced per turn in the other coil, in volt, will be MI (1) t

(2)

These materials are used to make magnets for elecric generators, transformer core and electromagnet core. Then it is proper to use : (1) A for transformers and B for electric generators. (2) B for electromagnets and transformers. (3) A for electric generators and trasformers. (4) A for electromagnets and B for electric generators. 18. 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 (1) v(cm)

O

(2)

MN MI (4) It Nt Hysteresis loops for two magnetic materials A and B are given below : B

v(cm)

NMI t O

(3)

17.

(3)

u(cm)

v(cm)

B

O

(4) H (A)

u(cm)

u(cm)

v(cm)

H (B)

O

u(cm)

Mock Test -10 19.

MT-111

U

U

(1)

22.

(2) O

O

V

U

V

U

(3)

(4) O

20.

times the time taken by the same body in slipping down a similar frictionless plane. The coefficient of dynamic friction between the body and the

A battery is connected across series combination of a capacitor and a resistor, at t = 0. If at an instant t potential difference across the capacitor be ‘V’ and energy stored in it be U then which of the following graph is correct?

O

V

V

A ray of light of intensity I is incident on a parallel glass-slab at a point A as shown in fig. it undergoes partial reflection and refraction. At each reflection 25% of incident energy is reflected. The rays AB and A¢B¢ undergo interference. The ratio Imax / Imin is

A

24.

B¢

B

I

23.

A¢

25.

C (1) 4 : 1 (3) 7 : 1

(2) 8 : 1 (4) 49 : 1 Section - B

21.

The time taken by a body in sliding down a rough inclined plane of angle of inclination 45° is n

26.

æ 1 ö plane is m = 1 – ç x ÷ . Find the value of x. èn ø A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes 100 W of power. The amplitude of the electric field (in V/m) associated with the electromagnetic radiation at a distance of 5 m from the lamp will be nearly ______ A Carnot’s engine works as a refrigerator between 250 K and 300 K. If it receives 750 calories of heat from the reservoir at the lower temperature, the amount of heat rejected at the higher temperature is ______ cal. The electric field in a region of space is given by, r E = Eoiˆ + 2Eoˆj where Eo = 100 N/C. The flux of

the field through a circular surface of radius 0.02 m parallel to the Y-Z plane is nearly ____ Nm2/C A gas molecule of mass M at the surface of the Earth has kinetic energy equivalent to 0°C. If it were to go up straight without colliding with 819k B any other molecules, it rises to h = . Find xMg the value of x. Assume that the height attained is much less than radius of the earth. (kB is Boltzmann constant). In a npn transistor 1010 electrons enter the emitter in 10–6 s. 4% of the electrons are lost in the base. The current transfer ratio will be

EBD_7308 JEE MAIN

MT-112

27.

28. 29.

56 tuning forks are so arranged in series that each fork gives 4 beats per sec with the previous one. The frequency of the last fork is 3 times that of the first. The frequency of the first fork is ______ Hz. For sky wave propagation of a 10 MHz signal, what should be the minimum electron density (in m–3) in ionosphere? In an experiment the angles are required to be measured using an instrument, 29 divisions of

30.

the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half- a degree (= 0.5°), then the least count of the instrument is ______ minute. The counting rate observed from a radioactive source at t = 0 second was 1600 counts per second and at t = 8 seconds it was 100 counts per second. The counting rate observed, as counts per second at t = 6 seconds will be

CHEMISTRY Section - A 31.

32.

If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of He+ is : (1)

5A 9

(2)

9A 5

(3)

36A 5

(4)

36A 7

The value of (1) 2(2n – 2)

(2) 2(2n –2)–1

(4) None of these 2 n -1 - 1 Urea and hydrazine react to form ammonia gas along with compound X which reacts with aldehydes and ketones to form specific crystalline derivatives. X is

(3) 33.

8n -1 - 1

t 0.875 for nth order reaction is t 0.50

34.

(1) phenylhydrazine

(2) semicarbazide

(3) biuret

(4) acetylurea

The IUPAC name of CH 2 - CH - C H 2 is |

CN

|

CN

|

CN

(1) 1, 2, 3-propanetrinitrile (2) 1, 2, 3-tricyanopropane (3) 3-cyano-1, 5-dinitrilepentane 35.

(4) 1, 2, 3-propanetricarbonitrile The catenation tendency of C, Si and Ge is in the order Ge < Si < C. The bond energies (in kJmol– l) of C – C, Si – Si and Ge –Ge bonds are respectively; (1) 348, 297, 260 (2) 297, 348, 260 (3) 348, 260, 297 (4) 260, 297, 348

Mock Test -10 36.

The repeating units of acrilan are

MT-113

41.

H | (1) H 2 C = C - COOCH 3 H | (2) H 2 C = C - COOC 2 H 5 H | (3) H 2 C = C - CN

37.

38.

39.

40.

CH3 | (4) H 2 C = C - COOCH3 A metal X on heating in nitrogen gas gives Y. Y on treatment with H2O gives a colourless gas which when passed through CuSO4 solution gives a blue colour. Y is (1) Mg(NO3)2 (2) Mg3N2 (3) NH3 (4) MgO The turbidity of a polymer solution measures (1) the light scattered by solution (2) the light absorbed by a solution (3) the light transmitted by a solution (4) None of these Which of the following has a shape different from others? (1) [Zn(NH3)4]2+ (2) Ni(CO)4

(3) [Cd(CN)4]2+ (4) [Cu(NH3)4]2+ Di-n-propyl ether and diallyl ether can be distinguished by (1) Acetic acid (2) Sodium Metal (3) Cold dilute KMnO4 solution (4) PCl5

42.

43.

In which of the following cases, the stability of two oxidation states is correctly represented (1) Ti3+ > Ti4+ (2) Mn2+ > Mn3+ 2+ 3+ (3) Fe > Fe (4) Cu+ > Cu2+ Which one of the following oxides of chlorine is obtained by passing dry chlorine over silver chlorate at 90°C ? (1) Cl2O (2) ClO3 (3) ClO2 (4) ClO4 The main product of the following reaction is conc.H SO

2 4¾ C6 H5CH 2CH(OH)CH(CH3 )2 ¾¾¾¾¾ ®?

(1)

H 5C 6

(2)

C6H5CH2 H

(3)

44.

CH(CH3)2

C=C

H5C6CH2CH 2 H3C

(4)

H

C=C

H

CH3 CH3

C = CH2

C6H5 CH(CH3)2 C=C H H Zirconium phosphate [Zr 3(PO4)4] dissociates into three zirconium cations of charge + 4 and four phosphate anions of charge – 3. If molar solubility of zirconium phosphate is denoted by S and its solubility product by Ksp then which of the following relationship between S and Ksp is correct? (1) S = [Ksp/ (6912)1/7] (2) S = [Ksp/ 144]1/7 (3) S = [Ksp/ 6912]1/7 (4) S = [Ksp/ 6912[7

EBD_7308 JEE MAIN

MT-114

45.

The final product of the following sequence of reactions

O

offensive smelling compound ‘C’. Th e compound ‘C’ is

NaOH , H O

2¾® ¾¾ ¾ ¾ ¾

H

D

LiCu (CH 3 ) 2

LiAlH 4

H 2O

H 2O

¾® C ¾¾ ¾¾® D B ¾¾ ¾ ¾ ¾

(1) 3-ethyl-2-methyl-1-hexanol

49.

CH3CH 2 NH2

(2)

CH 3 CH 2 N ® =C

(3)

CH3 C º N

(4) CH 2 CH 2 OH The reaction of

(2) 2, 3-dimethyl-1-pentanol

CH3 CH = CH

(3) 2-ethyl-3-methyl-1-hexanol

with HBr gives

(4) 3, 3-dimethyl-1-pentanol 46.

(1)

A solution containing As3+, Cd2+, Ni2+ and Zn2+ is made alkaline with dilute NH4OH and treated with H2S. The precipitate obtained will consist of

OH

(1) CH3CH = CH

Br

(2) CH3CH2CHBr

OH

(3) CH3CHBrCH2

Br

(4) CH3CH2CHBr

Br

(1) As2S3 and CdS (2) CdS, NiS and ZnS (3) NiS and ZnS (4) Sulphide of all ions 47.

The intermediate formed during the addition of HCl to propene in the presence of peroxide is ·

(1) CH3CHCH2Cl ·

(3) CH3CH 2CH 2 48.

(2) CH 3 CHCH3

+ (4) CH3CH 2 CH 2

An organic compound ‘A’ having molecular formula C2H3N on reduction gave another compound ‘B’. Upon treatment with nitrous acid, ‘B’ gave ethyl alcohol. On warming with chloroform and alcoholic KOH, B formed an

50.

Final product (D) formed in the following reaction sequence is D

CH2 = CHCH = CH2 +

2NBS

¾¾® A ¾¾¾® B alc.KOH

Pd ,( -H )

2 ®D ¾¾ ¾¾® C ¾¾ ¾¾ ¾ (1) Naphthalene (2) Tetralin (3) Benzene (4) Anthracene

Mock Test -10

MT-115

Section - B 51.

55.

M2SO4 (M+ is a monovalent metal ion) has a Ksp of 3.2 ´ 10–5 at 298 K. The maximum concentration of SO4–2 ion that could be attained in a saturated solution of this solid at 298 K is x × 10–2 M. Find the value of x.

56.

The colour of KMnO4 solution is decolourised by Fe2+ solution, one mole of Fe2+ reacts with x moles of KMnO4. Find x. For the coagulation of 500 mL of arsenious sulphide sol, 2 mL of 1 M NaCl is required. What is the flocculation value of NaCl? The standard reduction potential for Cu2+/Cu is + 0.34. Calculate the reduction potential at pH = 14 for the above couple. (Ksp Cu(OH)2 = 1 × 10–19) A current of 2.0A when passed for 5 hours through a molten metal salt deposits 22.2g of metal of atomic weight 177. Find the oxidation state of the metal in the metal salt A gaseous mixture of three gases A, B and C has a pressure of 10 atm. The total number of moles of all the gases is 10. If the partial pressures of A and B are 3.0 and 1.0 atm respectively and if ‘C’ has mol. wt of 2.0 what is the weight of ‘C’ in g present in the mixture

Determine enthalpy of formation for H2O2(l), using listed enthalpies of reaction : N 2 H 4 ( l ) + 2H 2 O2 ( l ) ¾¾ ® N 2 ( g ) + 4H 2 O ( l ) ; D r H1° = -818 kJ / mol

N 2 H 4 ( l ) + O 2 ( g ) ¾¾ ® N 2 ( g ) + 2H 2 O ( l ) ; D r H°2 = -622 kJ / mol

57.

H 2 ( g ) + 1/ 2O2 ( g ) ¾¾ ® H 2O ( l ) ; 52.

53.

54.

D r H3° = -285 kJ / mol A metallic element exists as cubic lattice. Each edge of the unit cell is 2.88 Å. The density of the metal is 7.20 g cm–3. How many unit cell will be present in 100 g of the metal in the multiple of 1023?

The solution containing 4.0 % PVC in one litre of dioxane was found to have osmotic pressure of 6.0 × 10–4 atm at 300 K. Find the molecular mass of polymer in terms of x × 105. For a chemical reaction X ® Y, the rate of reaction increases by a factor of 1.837 when the concentration of X is increased by 1.5 times. Find the order of the reaction with respect to X

58.

59.

60.

MATHEMATICS Section - A 61.

n+4C – nC – 3.nC n n r r r–1 – 3. Cr–2 – Cr–3 is equal to n+1 n+2 (1) Cr–1 (2) Cr–1 (3) n+3Cr–1 (4) n+4Cr–1

62.

Let a/(a–1) and b/(b–1) be the roots of x2 + ax + b = 0. Then 1/a and 1/b are the roots of (1) bx2 + ax + 1 = 0 (2) bx2 – ax + 1 = 0 (3) bx2 + (a + 2b)x + a + b + 1 = 0 (4) bx2 – (a + 2b) x + a + b + 1 = 0

EBD_7308 JEE MAIN

MT-116

63.

64.

65.

66.

For any real q, the maximum value of cos2(cos q) + sin2(sin q) is (1) 1 (2) 1 + sin21 2 (3) 1 + cos 1 (4) 1/2 If tan(A/2), tan(B/2), tan(C/2) are in A.P. then sec A, sec B, sec C are in (1) A.P. (2) G..P. (3) H.P. (4) None The equation of the common tangent to the parabolas y2 = 4ax and x2 = 4by is given by (1) xa1/3 + yb1/3 + a2/3 b2/3 = 0 (2) xb1/3 + ya1/3 + a2/3 b2/3 = 0 (3) xa1/3 + yb1/3 – a2/3 b2/3 = 0 (4) None of these If tangents be drawn to the cricle x2 + y2 = 12 at its points of intersection with the circle x2 + y2 – 5x + 3y – 2 = 0, then the tangents intersect at the point 18 ö æ (1) ç – 6, ÷ 5ø è

67.

68.

1

lim (4 - 3 sin x - 2 cos 2 x ) 2 sin x -1 is

x ®p / 6

69.

70.

(1) 1 (2) e (3) Öe (4) e–1/2 The function f (x) = 3 cos4x + 10 cos3x + 6 cos2x – 3, (0 £ x £ p) is – æ p 2p ö (1) Increasing in çè , ÷ø 2 3 æ p ö æ 2p ö (2) Increasing in çè 0, ÷ø È çè , p÷ø 2 3 æ p 2p ö (3) Decreasing in çè , ÷ø 2 3 (4) All of the above Value of

p

ò 0 sin

n

x cos 2m+1 x dx

( where m, n Î N) is

æ 18 ö (2) ç 6, ÷ è 5ø

18 ö 18 ö æ æ (3) ç – 6,– ÷ (4) ç 6,– ÷ 5ø 5ø è è 2 Let R = {(x, y) : x, y Î N and x – 4xy + 3y2 = 0}, where N is the set of all natural numbers. Then the relation R is : (1) reflexive but neither symmetric nor transitive. (2) symmetric and transitive. (3) reflexive and symmetric, (4) reflexive and transitive.

The value of

71.

(1)

(2m + 1)! n!

(3)

ò 0 cos

p

2m +1

(2) x dx

(2m + 1)!

(4) None of these

If f ¢( x) = f ( x ) and f (0) = 2, then f ( x)

ò 3 + 4f (x ) dx = (1) log(3 + 8e x ) + C 1 log(3 + 8e x ) + C 4 1 log( 3 + 8e x ) + C (3) 2 (4) None of these

(2)

n2

Mock Test -10 72.

MT-117

The area under the curve y = | cos x – sin x |, p 0 £ x £ , and above x-axis is : 2 (1) 2 2

(2)

73.

74.

75.

76.

2 2 -2

(3) 2 2 + 2 (4) None of these Let coordinates of a point ‘p’ with respect to the r r r system of non-coplanar vectors a , b and c is (3, 2, 1). Then coordinates of ‘p’ with respect to r r r r r r the system of vectors a + b + c , a – b + c r r r and a + b – c is (1)

æ3 1 ö ç , ,1÷ è2 2 ø

(2)

æ3 1ö ç ,1, ÷ è2 2ø

(3)

æ1 3 ö ç , ,1÷ è2 2 ø

(4) None

Equation of the line through the point (2,3,l) and parallel to the line of intersection of the planes x – 2y – z + 5 = 0 and x + y + 3z = 6 is x - 2 y - 3 z -1 (1) = = -5 -4 3 (2) x - 2 = y - 3 = z - 1 5 -4 3 x - 2 y - 3 z -1 (3) = = 5 4 3 x - 2 y - 3 z -1 (4) = = 4 3 2 If x + 2 = 6, then x = 4. So, which statement is its converse? (1) If x ¹ 4, then x + 2 ¹ 6 (2) If x = 4, then x + 2 ¹ 6

77.

(3) If x = 4, then x + 2 = 6 (4) If, x ¹ 4 then x + 2 = 6 If y = f(x) be a monotonically increasing or decreasing function of x and M is the median of variable x, then the median of y is (1) f (M) (2) M/2 (3) f –1 (M) (4) None of these A fair die is tossed eight times. The probability that a third six is observed on the eighth throw is (1) (3)

78.

7

7

C2 C6

55 68

(2)

7

C3

53 68

56

(4) None of these 68 uur uur uur If a , b , c are non coplanar vectors and l is a real number then uur uur uur uur uur uur uur uur [l (a + b ) l 2 b l c ] = [a b + c b ] for (1) exactly one value of l (2) no value of l (3) exactly three values of l (4) exactly two values of l

æ ö 1 1 ö -1 æ 79. S = tan -1 ç ÷ + tan ç 2 ÷ + ... 2 è n + 3n + 3 ø è n + n +1 ø æ ö 1 + tan -1 ç ÷ , then tan S is 1 + ( n + 19)( n + 20) è ø equal to : 20 n (2) (1) 2 401 + 20 n n + 20n + 1 n 20 (3) (4) 2 401 + 20 n n + 20n + 1

EBD_7308 JEE MAIN

MT-118

80.

The middle term in the expansion of

85.

n

æ 1ö n çè1 - ÷ø (1 - x ) in powers of x is x (1) – 2nCn–1 (2) – 2nCn 2n (3) Cn – 1 (4) 2n Cn

86. 87.

Section - B 81.

The value of S =

5 12 × 4 2

+

that 11 2

4 ×7

2

+

17 2

7 ×10 2

+

83.

The line 2x + y = 3 cuts the ellipse 4x2 + y2 = 5 at P and Q. If q be the angle between the normals at these points, then value of tan q is ______ . r – 1ö æ r ÷÷ , Matrix M r is defined as M r = çç è r –1 r ø

r Î N. If value of det(M1) + det(M2) + det(M3) + .... + det(M2007) = (k)2, then value of k is ____ . 84.

æ 1 -1 1 ö ÷ ç Let A = ç 2 1 - 3 ÷. and ç1 1 1 ÷ø è 2 2ö æ 4 ÷ ç (10) B = ç - 5 0 a ÷. ç 1 -2 3÷ ø è

If B is the inverse of matrix A, then value of a is ______ .

Z2 is a purely imaginary number, then Z1

2Z1 + 3Z 2 2Z1 - 3Z 2 is ______ .

¼ ¥ is ______ . 82.

The derivative of tan –1[(3x2 – 1)/(3x – x3)] with respect to sin–1[(x2 –1)/( x2 + 1)] is ______ . If 2a + 3b + 6c = 0 and at least one root of the equation ax2 + bx + c = 0 lies in the interval (m, n), then value of m + n is ______ . If Z1 ¹ 0 and Z2 be two complex numbers such

88.

89.

90.

If the straight lines x = 1 + s , y = -3 - l s , z = 1 + l s t and x = , y = 1 + t , z = 2 - t , with parameters 2 s and t respectively, are co-planar, then value of l is ______ . A and B are two independent events. The probability that both A and B occur is 1/6 and the probability that neither of them occurs is 1/3. The most probability of occurrence of A is ______ . If X and Y are independent binomial variates æ 1ö æ 1ö B ç 5, ÷ and B ç 7, ÷ , then P (X + Y = 3) is è 2ø è 2ø ______ .

Mock Test-1 (4) (4) (3) (2) (4) (1) (3) (4) (1) (1) (1) (2) (2) (2) (2)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

(1) (3) (1) (1) (2) (170) (6.67) (6.275) (20) (0.25) (4) (40.05) (5.3) (23) (2)

16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

ANSWER KEY (1) 46 (1) 47 (2) 48 (2) 49 (4) 50 (3) 51 (4) 52 (3) 53 (3) 54 (3) 55 (4) 56 (4) 57 (4) 58 (2) 59 (3) 60

(1) (1) (1) (2) (3) (1.24) (2.25) (0.5) (1.75) (5.97) (400) (210) (0.04) (6) (0.241)

61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

(3) (2) (2) (2) (3) (1) (3) (1) (1) (4) (1) (2) (1) (4) (1)

(3) (1) (4) (4) (1) (0) (0.80) (0.25) (0.33) (7.00) (1.00) (3.00) (1.50) (14.00) (3.20)

76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

Solutions PHYSICS 1.

also tan q » sin =

x q2 = Hence, 2l 4p Î0 x 2 ´ mg

(4) q

2q 2 l 4p Î0 mg 1/3

Tsin q q

x mg

In equilibrium, Fe = T sin q

mg = T cos q tan q =

Þ x3 =

Tcos q

q q

x/2 l

Fe q2 = mg 4p Î0 x 2 ´ mg

Fe

2.

æ q2l ö x = çç ÷÷ \ è 2p Î0 mg ø Therefore x µ l1/3 (4) Case I : T – N = 40 a and 20g – T = 20 a Also N = 20 a After simplifying, we get

T T 20g

T N

A

EBD_7308 JEE MAIN

MT-120

a=

Magnetic field at the centre of the ring is

g 4

Acceleration of block B, = 2a = Case II :

g 2 2

m 0 qf m0I = 2R 2R (3) P-V indicator diagram for isobaric

B=

.

7.

T

P T

slope A

dP =0 dV

20g

V

T =40 a and 20 g – T =20 a After simplifying above equation, we get a =g/3

P-V indicator diagram for isochoric process

P

g/2 2 3 = Ratio = . g /3 2 2

3.

(3)

Vin =

2 -GM é æ r ö ù ê3 - ç ÷ ú , 2R ê è R ø ú ë û

Vsurface

4.

(2)

slope

V P-V indicator diagram for isothermal process

-GM -GM = , Vout = R r

P

T ¯ (300K to 70K )

slope =

T ¯ R metal ¯ R ­ semi - conductor

(Al)

5.

A(T1 - T2 )t A(T1 - T2 ) t = d1 d 2 (d 1 + d 2 ) / K + K1 K1

d + d 2 d1 d 2 ; \ 1 = + K K1 K 2

\K =

6.

(d1 + d 2 ) d1 d 2 + K1 K 2

(1) When the ring rotates about its axis with a uniform frequency f Hz, the current flowing in the ring is I=

q = qf T

dP - P = dV V

(Si)

(4) When two rods are connected in series Q=

dP =¥ dV

V 8.

(4) Let 'S' be the distance between two ends 'a' be the constant acceleration As we know v2 – u2 = 2aS v2 - u2 2 Let vc be velocity at mid point.

or, aS =

2 2 Therefore, vc - u = 2a

vc2 = u 2 + aS

vc2 = u 2 +

vc =

v2 - u 2 2

u 2 + v2 2

S 2

Solutions-Mock Test-1 9.

MT-121

(1) Angular retardation,

10.

(1) Given : k A = 300 N / m, k B = 400 N / m Let when the combination of springs is compressed by force F. Spring A is compressed by xA. Therefore compression in spring B

15.

73 ´ 10 100 (2) As mv > mr therefore, vv < vr.

16.

(1) a =

x B = 8.75 - 5 = 3.75cm 17.

14.

(1) From F < v(t) Þ m dv < R v(t) dt t 2 t2 dv Rdt 1 It is not possible.

75.

f ( x ) = xe x(1- x) , x Î R

f ' ( x ) = e x(1- x) . é1 + x – 2 x 2 ù ë û

2

As a ¹ 1 \ç ÷ >1 è 2a ø

74.

(3)

p 2

M

2

y A

x

B

10 cm

Solutions-Mock Test-1

MT-129

We know that the volume of cone 1 = p (radius)2 × height 3

log p 2 = =e

When x = log p , t = e 2 log

1 \ V = px 2 y. Let ÐBAD = a 3 BD 5 1 Þ tan a = = = . AD 10 2

\I =

Again, from right angled DAMR, we have MR x 1 x y tan a = = ; Þ = ; \x = . AR y 2 y 2

79.

10 x Putting y = in 4 x 2 - 9 y 2 = 36 9 gives imaginary roots resulting in no tangents.

80.

(1)

I =ò

log p log p 2

ö æ1 e 2 x sec 2 ç e 2 x ÷ dx 3 ø è

Put e 2 x = t Þ 2 e 2 x dx = dt When x = log p 2 , t = e 2 log

p2

=p

dt

3é 1 ù ê 3ú= 3 2ë 3û (0) If the G.P be a, ar, ar2, .... then an = arn – 1 =

81.

log a + ( n - 1) log r

D=

log a + n log r

log a + (n + 1) log r

log a + n log r log a + (n + 1) log r log a + ( n + 2) log r log a + ( n + 1) log r log a + ( n + 2) log r log a + ( n + 3) log r

R3 ® R3 – R2 and R2 ® R2 – R1 gives,

dV p 2 dy p dy dy 16 = .3y . ; \ 4 = y2. ; \ = . dt 12 dt 4 dt dt py2

When y = 6 cm, dy 16 4 = = cub.cm. / min . 2 dt p6 9p (4) Slope of the tangent to 4x2 – 9y2 = 36 is given by 4x dy dy 4x or m1 = 8x - 18y = 0Þ = 9y dx dx 9 y Slope of the straight line, 5x + 2y – 10 = 0 is 5 m2 = 2 Therefore, for the perpendicularity, m1m2 = –1 4 x -5 10x Now, . ´ = -1 Þ y = 9y 2 9

ò

p1 æ1 ö sec 2 ç t ÷ p2 è3 ø 2 p

p

t ù 1 1é = . ê tan .ú 3 ûp2 2 1ë 3 3 é p pù = ê tan - tan ú 2 ë 3 6û

2

p 1 1 æ yö \V = p.x2y = p.ç ÷ .y = y3 ......(1). 3 3 è 2ø 12 By question, the rate of change of volume dV = = 4 cub.cm./min. dt We have to find out the rate of increase of dy . water-level i.e. dt Differentiating (1) with respect to t, we get

p 2

log a + (n - 1)log r log a + n log r log a + (n + 1)log r

=

log r

log r

log r

log r

log r

log r

= 0, since R2 and R3 are identical. 82. (0.80) Total no. of arrangements of the letters of 10! . 2! No. of arrangements when both I's are together = 9! So. the no. of ways in which 2 I’s do not

the word UNIVERSITY is

10! - 9! 2! \ Required probability

together =

10! - 9! 10!- 9! 2! 2! = = 10! 10! 2!

=

10 ´ 9!- 9!.2! 9![10 - 2] = 10! 10 ´ 9!

=

8 4 = = 0.80 10 5

EBD_7308 JEE MAIN

MT-130

83.

84.

(0.25) n(S) = the area of the circle of radius r r n(E) = the area of the circle of radius 2 2 ærö pç ÷ n( E ) 1 2 = è 2ø = . \ The probability = n( S ) 4 pr

88.

4

199

æ2ö è3ø

-199

89.

3 ù 30 5 é æ 5ö ê1 + ç ÷ + ......¥ ú = 36 ê è 6 ø úû 91 ë (7.00) Given equation of ellipse is

=

85.

x2 y 2 + =1 16 b 2

eccentricity = e = 1 foci: ± ae = ± 4 1 -

b2 16

b2 16

2 2 Equation of hyperbola is x - y = 1 144 81 25

x2 y2 =1 144 81 25 25 81 25 81 = 1+ Eccentricity = e = 1 + ´ 144 25 144

Þ

225 15 = 144 12 12 15 foci: ± ae = ± ´ = ± 3 5 12 Since, foci of ellipse and hyperbola coincide

=

b2 = ± 3 Þ b2 = 7 16 (1.00) For x > 10, f (x) = x – 2. Therefore, g(x) = x – 2 – 2 = x – 4 \ g¢(x) = 1. a+b+c (3.00) Q A.M. ³ G.M. Þ ³ (abc)1/3 3 Þ a + b + c ³ 3(abc)1/3 (1) But given : ab2c3, a2b3c4, a3b4c5 are also in AP. (Q abc ¹ 0) Þ 2abc = 1 + a2b2c2 Þ (abc – 1)2 = 0 \ abc = 1

\ ± 4 1-

86.

87.

æ3ö ´ç ÷ è2ø

200

200

3200-199 3 æ 3ö æ3ö ´ = = = 1.50 ç ÷ ÷ 2200-199 2 è2ø è 2ø (14.00) Since 0 < y < x < 2y x x \ y> Þ x- y < 2 2 \ x – y < y < x < 2x + y y+x = 10 Hence median = 2 ...(i) Þ x + y = 20 And range = (2x + y) – (x – y) = x + 2y But range = 28 \ x + 2y = 28 ...(ii) From equations (i) and (ii), x = 12, y = 8 \ Mean ( x - y ) + y + x + (2 x + y ) 4 x + y = = 4 4 y 8 = x + = 12 + = 14 4 4 dx (3.20) ò cos3 x 4sin x cos x dx =ò 4 2 cos x tan x Let tan x = t2 Þ sec2 x = 1 + t4 sec2 x dx = 2t dt = ç

8

5 1 æ 5ö 1 æ 5ö 1 + ...¥ ´ +ç ÷ ´ +ç ÷ 6 è 6ø 6 6 6 è 6ø

putting x = 1 in the expression

So, sum of coefficients = ç ÷

(0.33) P(E E) + P(E E E E E ) + ...¥ =

Now from equation (1), a + b + c ³ 3 (1)1/3 Þ (a + b + c) ³ 3 Hence, minimum value of is a + b + c is 3. (1.50) Sum of coefficients is obtained by simply

90.

= =

sec 4 x dx

ò2

ò

tan x

=

ò

sec 2 x (sec2 x dx )

2 tan x

4

(1 + t )2t dt = ò (1 + t 4 ) dt 2t

= t+

t5 +k 5

1 tan x + tan 5 / 2 x + k ét = tan x ù ë û 5 1 1 5 A= ,B= ,C= 5 2 2 16 A+ B+C= = 3.20 5

=

Mock Test-2 1 2 3 4 5 6 7 8 9 10 11 12 13

(1) (3) (4) (1) (1) (1) (3) (4) (3) (1) (3) (3) (4)

16 17 18 19 20 21 22 23 24 25 26 27 28

(1) (4) (1) (1) (1) (45°) (7.6) (2) (45.5) (3) (5) (0.5) (6)

14

(3)

29

(2

15

(3)

30

-2 3

(5)

)

31 32 33 34 35 36 37 38 39 40 41 42 43

ANSWER KEY (3) (3) 46 (3) (2) 47 (2) (2) 48 (4) (3) 49 (1) (2) 50 (4) (546) 51 (4) (8) 52 (4) (1.47) 53 (3) (2) 54 (2) (0.2303) 55 (4) (4) 56 (2) (2) 57 (4) (964) 58

61 62 63 64 65 66 67 68 69 70 71 72 73

(2) (1) (3) (3) (3) (1) (1) (1) (1) (1) (1) (1) (3)

76 77 78 79 80 81 82 83 84 85 86 87 88

(2) (3) (4) (3) (2) (16) (3) (0.5) (2008) (18) (12) (0.75) (0.38)

44

(2)

59

(60)

74

(1)

89

(336.00)

45

(3)

60

(10.85)

75

(1)

90

(9)

Solutions Therefore gravitational field due to the left over part of the sphere

PHYSICS 1.

2.

(1) Least count of vernier callipers = value of one division of main scale – value of one division of vernier scale Now, N × a = (N + 1) a' (a' = value of one division of vernier scale) N a¢ = a N +1 a \ Least count = a - a ¢ = N +1 (3) Let mass of smaller sphere (which has to be removed) is m R Radius = (from figure) 2 M m = 4 3 4 æ R ö3 pR pç ÷ 3 3 è2ø M Þm= 8 Mass of the left over part of the sphere M 7 M' = M - = M 8 8

=

GM ' x2

=

7 GM 8 x2

3. 4.

(4) (1) Charge conservation is violated in [2, 3, 4], nucleon conservation is violated in (4), (1) works.

5.

(1)

6.

7.

1 æ1ö C = sin –1 ç ÷ and m µ l èmø Yellow, orange and red have higher wavelengths than green, so m will be less for these rays, consequently critical angle for these rays will be high, hence if green is just totally internally reflected then yellow, orange and red rays will emerge out. (1) The conductivity of semiconductor s = e (heµe + hhµh) = 1.6 × 10 –19(5 × 1018 ×2 + 5 × 1019 × 0.01) = 1.6 × 1.05 = 1.68

(3)

1 1 E1 = kx 2 , E 2 = ky 2 , 2 2

EBD_7308 JEE MAIN

MT-132

E=

8.

1 k(x + y) 2 = E1 + E 2 + 2 E1E 2 2

= 2 + 8 + 2 16 = 18J (4) From figure, sin a = d/R a

10.

æ A Lö æA 3 ö V ´ D ´ g = ç ´ L÷ ´ d ´ g + ç ´ ÷ ´ 2d ´ g è5 4 ø è 5 4ø A´ L ´ d ´ g D d æA ö Þ = Þ çè ´ L÷ø ´ D ´ g = 4 5 4 5

R

\D=

a d

Þ R= \

mv qB

sin a =

11.

mv 2 = qvB R

And we know,

dqB mv

9.

A

b ( m - 1) l

b ( m - 1) l

t1 -

b ( m - 1) l

t2

( t1 - t2 )

ml 2 ; 12 Restoring torque

(3)

I=

= qElsin q » qEl q =

d 2q

I dt 2 qEl qEl ´12 2p \w = = = I T ml 2

1 2ù é êQ qV = 2 mv ú ë û

l

(3)

5 d 4

(3) Shift = =

12.

q sin a = Bd 2mV

(1) Weight of cylinder = upthrust due to both liquids

ml 3qE (4) When object is kept at a distance ‘a’ from thin convex lens \T = p

Y

13.

Wire (1)

O 3A

Y

I1 v

l/3 Wire (2)

a 1 1 1 By lens formula : v – u = f

As shown in the figure, the wires will have the same Young’s modulus (same material) and the length of the wire of area of cross-section 3A will be l/3 (same volume as wire 1). F/A For wire 1, Y = D x/l

For wire 2 , Y =

F '/ 3 A Dx /( l / 3)

From (i) and (ii) , Þ F ' = 9F

...(i) ...(ii) F l F' l ´ = ´ A Dx 3 A 3Dx

1 1 1 – = V (– a) f

1 1 1 = – ...(i) v f a Mirror forms image at equal distance from mirror

or,

I1

v

v

I2

Solutions-Mock Test-2

MT-133

18.

Now, again from lens formula

I3

I2

a/3 v 3 1 1 – = a V f

14.

15.

16.

17.

3 1 1 1 – + = [From eqn. (i)] a f a f Hence, a = 2f (3) The equation of wave moving in negative xdirection, assuming origin of position at x = 2 and origin of time (i.e. initial time) at t = 1 sec. y = 0.1 sin (2pt + 4x) Shifting the origin of position to left by 2m, that is, to x = 0. Also shifting the origin of time backwards by 1 sec, that is to t = 0 sec. y = 0.1 sin (2p (t – 1) + 4 (x – 2)) v–u (3) Acceleration (a) = = t (0 , 50) < ,5 m / s 2 (10 , 0) u = 50 m/s \ v = u + at = 50 – 5t Veocity in first two seconds t = 2 v(at t 2a and x < 0 2a - x So, curve does not lie in the portion x > 2a and x < 0, therefore curve lies in 0 £ x £ 2a .

x xö æ 2 ç sin + cos ÷ dx = -4cos(ax + b) + C 2 2ø è

æ

+

But

\ Equation of line QM is (y – 0) = – 3 (x – 0)

3x Þ

x

-4

From (i) we have, y2 =

Let QM bisects the ÐPQR ,

Þ y= –

a -16 / 3

= a -16 / 3 (a 4 ) = a -4 / 3

76.

2p / 3 P (–1, 0)

(1) 4x3 + 4y3 (dy/dx) = 0 Þ dy/dx = –x3/y3 Equation of tangent, Y – y = – x3/y3 (X–x) Þ y3Y + x3X = x4 + y4 = a4 X Y Þ + =1 4 3 4 a /x a / y3

3pa 2 sq. unit 2 (3) Rewrite the given differential equation as follows : =

77.

dy 2x 1 + y= 2 , which is a linear form dx x 2 - 1 x -1 The integrating factor I.F.

=e

ò

2x x 2 -1

dx

= e ln ( x

2 -1)

= x 2 -1

EBD_7308 JEE MAIN

MT-140

Thus multiplying the given equation by dy + 2 xy = 1 ( x - 1), we get ( x - 1) dx

Þ

(2) For non-trivial solution,

-1

+ cos B [sin B + cos B] = 0

Þ -a 2 + sin 2B + cos 2B = 0 Þ a Î[-1,1]

Þ x3 – 25x2 + 175x – 375 = 0 81.

(x – 5) (x2 – 20x + 75) = 0

(16.00)

(x – 5)2 (x – 15 ) = 0 Þ x = 5, 15 (3) Let I1 =

ò0

C(a, 80)

A (0, 20)

sin -1 t dt

M

2 Put t = s in u Þ d t = 2 sin u c o s u d u Þ dt = sin 2udu

\I1 =

O(0, 0)

x

ò0 u sin 2u du

Let I2 =

cos 2 x

ò0

N

cos -1 t dt

Line OC and AB intersect at M.

Put t = cos v Þ dt = - 2 cos v sin v dv

To find: Length of MN.

Þ dt = - sin 2 v dv

æ 80 - 0 ö x Eqn of OC: y = ç è a - 0 ÷ø

x

x

2

2

\I 2 = ò p v ( - sin 2v ) dv = - ò p v sin 2v dv = -ò

Þ y=

u sin 2u du [change of variable]

\ I = I 1 + I2 =

x

ò0

x

u sin 2u du - ò p u sin 2u du 2

p 2

ò

x

x

æ 20 - 0 ö Eqn of AB: y = çè ÷ ( x - a) 0-a ø

-20 ( x - a) a At M: (1) = (2) Þ y=

-20 80 x= ( x - a) a a

...(2)

òp

2

2

Þ

80 -20 a x + 20 Þ x = x= a a 5

\

y=

= u sin 2u du = 0

...(1)

òp

p 2

ò

80 x a

Þ

= u sin 2u du + u sin 2u du - u sin 2u du 0

B (a, 0)

We put one pole at origin. BC = 80 m, OA = 20 m

2

x p 2

=0

sin B - cos B

(4) f2 (x) = f (f (x)) = f (x) = x

sin 2 x

cos B sin B

Þ a 2 [- cos 2 B - sin 2 B] - sin B [ - cos B + sin B]

f3 (x) = f (f2 (x)) = f (x) = x

79.

2a sin B 1 cos B

D=

d [ y( x 2 - 1)] = 1 dx

On integrating we get y( x 2 - 1) = x + c 78.

80.

2

2

p [Integrate by parts] 4

80 a ´ = 16 a 5

Solutions-Mock Test-2 82.

MT-141

3 (3.00) x + iy = cos q + i sin q + 2

n

85.

Þ

1 cos q + i sin q + 2 = x + iy 3

Þ

Þ

x - iy 1 = [(cos q + 2) + i sin q] ( x + iy) ( x - iy) 3

or

Þ

x 2

x +y 2

=

2

x

Þ

2

2 1 = cos q 3 3

-

2

1 2

x + y2

2

(3 - 4 x ) + 1 = 0

86.

Þ 4x - x 2 - y 2 = 3

(0.50) Let A be the event such that sum is Rs. 20 or more \ P (1) = 1 – P (Total value is < 20) 6

C2 - 2C2 8

C2

= 1-

=

1 2

k +1 1 = n-k 2 2k + 2 = n – k n – 3k = 2 n

Ck +1

n

Ck + 2

...... (1)

=

2 3

n! (k + 2)!(n - k - 2)! 2 . = n! 3 (k + 1)! (n - k - 1)!

Þ 3 - 4x = - x 2 - y 2

= 1-

Ck

C k +1

Similarly,

æ x 2 ö æ - y ö÷ 1 ç = - ÷ +ç ç x 2 + y2 3 ÷ ç x 2 + y2 ÷ 9 ø è ø è Þ

n

(k + 1)!(n - k - 1)! 1 n! = k! (n - k)! n! 2

1 (cos q + 2) 3

x +y y 1 = sin q and - 2 2 3 x +y Squaring and adding, we get

83.

(18.00)

1 1 14 = 1- = 28 2 2

2 k+2 = n - k -1 3 3k + 6 = 2n – 2k – 2 2n – 5k = 8 ...... (2) From (1) and (2) n = 14 and k = 4 \ n + k = 18 (12.00) Let the observations be x1, x2, x3, x4, x5 and x6, so 6

their mean x =

å xi i =1

6

= 8 Þ å xi = 48

6 i =1 On multiplying each observation by 3, we get the new observations as 3x1, 3x2, 3x3, 3x4, 3x5 and 3x6. Now, their mean 6

84.

(2008.00) q1 + q2 =

p 2

dq \ I= ò = æp ö q1 1 + tan ç - q÷ è2 ø

and also I = ò

q2

q1

\ 2I =

ò dq = q2 - q1 =

q1

q2

tan q dq

ò 1 + tan q

Hence, K = 2008.

6

1002p 501p ÞI= 2008 2008

i =1

å (3x i - 24) 2

q1

dq 1 + tan q

q2

3 ´ 48 = 24 6 6 Variance of new observations

=x =

q2

å 3x i

6

32 å (x i - 8) 2

= i =1 6 6 = 9 × Variance of old observations = 9 × 42 = 144 Thus, standard deviation of new observations

=

=

i =1

=

Variance = 144 = 12

EBD_7308 JEE MAIN

MT-142

87.

(0.75) lim

x ®0

= 1×

88.

(tan x)3/ 2 [1 - (cos x)3/ 2 ]

lim

x

1 - cos3 x

x ®0

x

2

3/ 2

.

.x

2

90.

(9.00) 2 sec 4C + sin22A + sin B = 0 A = 45°, B = 90° and C = 45° C

1 1 + (cos x)

3/2

P

3 1 1 2 = . (1 + cos x + cos x) = = 0.25 2 2 4 (0.38) n = 3, P (success) = P (HT or TH) = 1/2

Þ

p=q=

1 and r = 2 2 2

89.

.

1 3 æ 1ö P (r = 2) = 3C2 ç ÷ . = = 0.38 è 2ø 2 8 (336.00) Let S = {1, 2, 3} Þ n(S) = 3 Now, P (S) = set of all subsets of S total no. of subsets = 23 = 8 \ n[P(S)] = 8 Now, number of one-to-one functions from S ® 8! P(S) is 8P3 = = 8 × 7 × 6 = 336. 5!

Q

A

R,B

Let AQ = a, then BP = PQ =

a , 2

a and QR = a 2

\ PR = a 2 + \ 1: a : b =

a2 5a = 4 2

5a a :a : = 1: 2 : 5 2 2

\ a = 2 and b = 5 \ a 2 + b2 = 9

Mock Test-3 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

(3) (1) (1) (2) (1) (1) (2) (3) (2) (1) (3)

16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.

(4) (2) (1) (1) (2) (1600) (0.025) (19039) (0.33) (1.25) (72)

31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41.

12.

(3)

27.

13. 14. 15.

(2) (1) (3)

28. 29. 30.

(4.8 × 10 ) (0.2) (8) (10)

8

ANSWER KEY (3) 46. (4) (3) 47. (4) (3) 48. (2) (1) 49. (4) (4) 50. (4) (2) 51. (2) (2) 52. (14.28) (4) 53. (2) (1) 54. (80) (1) 55. (5.0) (3) 56. (1)

61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71.

(4) (1) (2) (2) (1) (4) (4) (3) (2) (1) (3)

76. 77. 78. 79. 80. 81. 82. 83. 84. 85. 86.

(4) (1) (3) (1) (4) (2.33) (9.00) (32.00) (343.00) (3.00) (0.53)

42.

(1)

57.

(395)

72.

(2)

87.

(2.00)

43. 44. 45.

(2) (1) (1)

58. 59. 60.

(3) (3) (1.14)

73. 74. 75.

(2) (1) (1)

88. 89. 90.

(3.00) (0.50) (0.50)

Solutions PHYSICS 1.

(3) For total internal reflection on face AC q > critical angle (C) and sinq ³ sinC sin q ³

2.

Ei =

and orbital velocity, v0 =

=

5GMm 6R (1) The forces acting on the block are shown. Since the block is not moving forward for the maximum force F applied, therefore F cos 60° = f = µN... (i) (Horizontal Direction) Note: For maximum force F, the frictional force is the limiting friction = µN] and F sin 60° + mg = N... (ii) From (i) and (ii) K=

4 mw sin q ³ Þ sin q ³ 3 3 mg 2 8 \ sin q ³ . 9 (1) As we know,

Ef =

Ei = E f Therefore minimum required energy,

1 wm g

Gravitational potential energy =

-GMm +K R

3.

-GMm r

N

GM / R + h

1 GMm 1 GM GMm mv02 = m 2 3R 2 3R 3R

GMm æ 1 ö - GMm ç - 1÷ = 3R è 2 ø 6R

F cos 60o f

mg

60

F sin 60o

o

F

EBD_7308 JEE MAIN

MT-144

F cos 60° = µ [F sin 60° + mg] Þ F=

6.

m mg cos 60° - m sin 60° 1

mv mv 3 ...... (i) + 2 2 Horizontal component, 2 mv' cos q = mv sin 60° – mv cos 45° mv mv ...... (ii) 2mv'cos q = 2 2

´ 3 ´ 10 5 2 3 = = 20 N = 1 1 3 1 ´ 4 2 2 3 2

2mv 'sin q =

(2) If student will use angular momentum = mvr. He/she may conclude answer (a) as r is decreasing angular momentum must decrease hence (a) is incorrect. v

Y

m q mg

5.

1 3 + 2+ 3 2 2 tan q = = 1 1 - 2 1 2 2

7.

n0 =

(l + 2Dl) - (l - 2Dl) ù = v éê ú ë l 2 - 4 Dl 2 û

U = KT 4 V 1 4 So, P = KT 3

4 pR 3T 3 = constant 3 1 Therefore, Tµ R

v , 2l v v , n2 = n1 = 2 ( l / 2 - Dl) 2 ( l / 2 + Dl) Beat frequency = n 1 – n 2

(2)

1 ù é 1 Þ vê ë l - 2Dl l + 2 Dl úû

But

uRT 1 = KT 4 [As PV = u RT] V 3

45° X' – v cos 45° For particle B

Dividing eqn (i) by eqn (ii),

1æUö (1) As, P = ç ÷ 3èVø

or

v sin 45°

X v cos 60° For particle A

///////////////////////////////////////// O d

Y'

30° 60°

r

But the magnitude of angular momentum of particle about O = mvd Since speed v of particle increases, its angular momentum about O increases. Magnitude of torque of gravitational force about O = mgd Þ constant Moment of inertia of particle about O = mr2 Hence MI of particle about O decreases. v sin q Angular velocity of particle about O = r Q v and sin q increases and r decreases. \ angular velocity of particle about O increases.

A B

v sin 60°

4.

(1) For particle C, According to law of conservation of linear momentum, verticle component, 2 mv' sin q = mv sin 60° + mv sin 45°

8 Dlv 8Dln 0 = l 2l l l - 4Dl (3) Wavelength for which maximum obtained at the hole has the maximum intensity on passing.

=v

8.

4D l

2

So, x =

l=

2

=

n lD d

xd 1 ´ 10-3 ´ 0.5 ´ 10-3 = nD n ´ 50 ´ 10-2

Solutions-Mock Test-3

MT-145

1 ´ 10-6 1000nm = n n n = 1, l = 1000 nm ® Not in the given range n = 2, l = 500nm I C

9.

(2)

12.

q I

I

L

I I

13.

=

(3) From question, Horizontal velocity (initial), ux =

or,

1 uy × 2 + (–10) ×4 2 50 = 2uy – 20

or,

uy =

Þ

70 = 35m / s 2 u y 35 7 tan q = = = u x 20 4

R1= 5 W A R 2= 15 W

4A

4A

R 3= 1.25 W

4 – I1

We know that I =

E = 20 = 4 A R eq 5

Potential difference across R1 and R2 are same (Parallel combination) I1R1 = (4 – I1)R2

14.

Þ 5I1 = (4 – I1) ´ 15 Þ I1 = 12–3I1 Þ I1 = 3A Thus reading of ammeter = 3A Voltage across 1.25W = I ´ R = 4 ´ 1.25 = 5V (Reading of voltmeter) (1) Let initial temperature and volume be T0 and V0. Since the process is adiabatic, the first temperature and volume is TV g -1 = T V g -1 (g 0 0 = 5/3 for monoatomic gas)

40 = 20m/s 2

Vertical velocity (initial), 50 = uy t +

\

I1

3 mF 8

And 4mF (12 & 6 mF) and 4mF in parallel = 4 +4 = 8mF 1 8 8mF in series with 1mF = + 1 Þ mF 8 9 8 8 32 Now Ceq = ∗ < 9 3 9 1 1 9 32 With C, C = C + 32 = 1 Þ C = 23 μF eq

R 1R 2 ( Parallel combination); R1 + R 2

5 ´ 15 125 75 5 + + =5 W = 5 + 15 100 20 4

increasing or æç 1 LI 2 ö÷ gives that I is increasing. è2 ø (1) Capacitors 2mF and 2mF are parallel, their equivalent = 4 mF 6mF and 12 mF are in series, their equivalent = 4 mF Now 4mF (2 and 2 mF) and 8mF in series =

11.

(2) Req =

RNet = Reg+R3(Series combination) RNet of the circuit

I

As it can be easily seen by the direction of I that Q is decreasing thus, energy of capacitor is decreasing and hence, energy of inductance is

10.

7 4 (3) As insect moves along a diameter, the effective mass and hence the M.I. first decreases then increases so from principle of conservation of angular momentum, angular speed, first increases then decreases.

Þ Angle q = tan–1

=

1 2 gt 2

æ V0 ö \ T = T0 ç è V0 / 8 ÷ø

15.

2/3

= 4T0

Q Flux going in pyramid = 2e 0 Which is divided equally among all 4 faces

(3)

Q \ Flux through one face = 8e 0

EBD_7308 JEE MAIN

MT-146

16.

17.

(4) From the figure it is clear that liquid 1 floats on liquid 2. The lighter liquid floats over heavier liquid. Therefore we can conclude that r1 < r2 Also r3 < r2 otherwise the ball would have sink to the bottom of the jar. Also r3 > r1 otherwise the ball would have floated in liquid 1. From the above discussion we conclude that r1 < r3 < r2.

20.

w=

(0.1) 2 k = m

2

M.I. (I¢) =

æ lö M ' = m ´ ç ÷ ´ 3 = ml = M è 3ø

3.6 = 6 rad / s 0.1

21.

18.

(1) First, the length of wire goes on increasing ie., area decreases and finally at breaking stress the wire breaks.

19.

(1)

r

O

mg cos a mg

2 s. 9 3 (1600) Given, R1 = 100 W, r' = r/2, R2 = ? rl Resistance of wire, R = A Q Area × length = volume rV Hence, R = 2 A Since, r ® constant, V ® constant 1 Rµ 2 A 1 or R µ Q A = pr2 r4 R2 = 16 Þ R2 = 16 ´ 100 = 1600 W R1 (0.025) At equilibrium, weight of the given block is balanced by force due to surface tension, i.e., 2L. S = W \T'=

p \ Eqn. y = 0.1 sin æç 6t + ö÷ è 4ø

A

1 æ m öæ l ö I ç ÷ç ÷ ´ 3 = 9 12 è 3 ø è 3 ø

We have, Magnetic moment (M) = Pole strength (m) × l \ New magnetic moment,

= 3.6 ;

F1

I I = 2p MB M ´B

1 ml 2 12 When the magnet is cut into three pieces the pole strength will remain the same and

A = 0.1m, m = 0.1 kg, KEmax = 18 × 10–3 J, p f= 4 36 ´ 10 -3

T = 2p

where I =

(2)

k=

(2)

a R y

Bowl

h mg sin a

The insect crawls up the bowl upto a certain height h only till the component of its weight along the bowl is balanced by limiting frictional force. For limiting condition at point A R = mg cosa ...(i) F1 = mg sina ...(ii) Dividing eq. (ii) by (i) F 1 tan a = = 1 = m [ As F1 = mR ] cot a R

1é 1 ù Þ tan a = m = êQ m = ( Given ) ú 3ë 3 û \ cot a = 3

22.

or S = 23.

T

=

1.5 ´ 10 -2 N W = = 0.025 Nm -1 2L 2 ´ 0.3 m

(19039) Given:

dN 0 = 20 decays/min dt

dN = 2 decays/min dt T1/2 = 5730 years As we know,

N = N 0 e -lt

Solutions-Mock Test-3

log \

MT-147

N0 = lt N t= =

N 1 log 0 l N 2.303 ´ T1/2 N ´ log10 0 0.693 N 27.

28.

dQ DT (T2 - T1 ) KA = = 3x dt 3x KA

2

1 ætö f t2 fç ÷ = 2 è6ø 72

(4.8 × 108) Optical source frequency

= (3.8 ´1012 ) /(8 ´103 ) ~ 4.8 ´108 (0.2) The current voltage relation of diode is I = (e1000 V /T - 1) mA (given)

When, I = 5mA, e1000 V /T = 6mA

1 ì A(T2 - T1 ) K ü = í ý 3î x þ

1000 T (By exponential function)

Also, dI = (e1000

1 3

V /T

)´

1000 ´ (0.01) = 0.2 mA 300 (8) Force acting on conductor B due to conductor A is given by relation m 0 I1 I 2 l F= 2 pr l-length of conductor B r-distance between two conductors

= (6 mA) ´

=

ne eAve

Þ

v 5 7 7 ve Þ e = = ´ vh 4 4 5 vh

25.

(1.25)

26.

(72) Distance from A to B = S =

Ih

t 6

c = 3 ´ 108 /(800 ´10 -9 ) = 3.8 ´1014 Hz l Bandwidth of channel (1% of above) = 3.8 × 1012Hz Number of channels = (Total bandwidth of channel) / (Bandwidth needed per channel) Number of channels for audio signal

x 4x x 2 x 3x + = + = KA 2KA KA KA KA

Ie

............ (ii)

f=

2.303 ´ 5730 ´1 0.693 = 19039 years (0.33) The thermal resistance is given by

\f =

1 2 f t1 = S 2

Þ S=

\ t=

\

............. (i)

Dividing (i) by (ii), we get t1 =

dN 0 N 20 = 10 But dt = 0 = dN N 2 dt

24.

Þ f t1t = 12 S

nh eAvh

29.

1 2 ft 1 2

Distance from B to C = (ft 1 ) t 2

(ft ) u = 1 2a 2(f / 2)

Distance from C to D = = ft 12 = 2S

A f B t1

t 15 S

Þ S + f t1t + 2S = 15 S

C f /2 D 2t 1

4p´ 10-7 ´ 10 ´ 2 ´ 2 = 8 × 10–5 N 2 ´ p´ 0.1 (10) By Newton’s law of cooling

\F=

2

30.

q1 - q2 é q + q2 ù = -K ê 1 - q0 ú t 2 ë û

where q0 is the temperature of surrounding. Now, hot water cools from 60°C to 50°C in 10 minutes,

60 - 50 é 60 + 50 ù = -K ê - q0 ú 10 ë 2 û

...(i)

EBD_7308 JEE MAIN

MT-148

The coordination number exhibited by beryllium is 4 and not 6 so statement (b) is incorrect. Both BeCl2 and AlCl3 exhibit bridged structures in solid state so (c) is correct statement.

Again, it cools from 50°C to 42°C in next 10 minutes.

50 - 42 é 50 + 42 ù - q0 ú = -K ê 10 ë 2 û Dividing equations (i) by (ii) we get

...(ii) 34.

55 - q0 1 = 0.8 46 - q0

(1)

Anhydrous HI

O

¾¾¾¾¾® SN2

10 55 - q 0 = 8 46 - q 0

SN2

2q0 = 20

CHEMISTRY

36.

(2) nCH 2 = CH - CH = CH 2 1, 3-Butadiene

æ 1 1 ö n = RcZ ç 2 - 2 ÷ çn ÷ è 1 n2 ø

CH = CH2

1 æ 1 n 3 = RcZ2 ç 2 - 2 è2 ¥

\ n1 - n2 = n3

32.

(– CH2 – CH = CH – CH2 – CH – CH2 –)n Butadiene Styrene copolymer (SBR or BUNA - S)

37.

(3) ICl, 2 Þ 2 bp + 3lp 3 (thus, sp d hybridisation) = linear BrF2∗ Þ 2 bp + 2lp

38.

(thus, sp3 hybridisation) = pyramidal ClF4, Þ 4 bp + 2lp (thus, sp3d2 hybridisation) = square planar

33.

Styrene

2

2 ö RcZ = ÷ 4 ø

AlCl, 4 Þ 4 bp + 0lp (thus sp3 hybridisation) = tetrahedral (3) The correct formula of inorganic benzene is B3N3H6 so (d) is incorrect statement. OH |

Na, Heat Polymerisation

¾¾¾¾¾®

+n

1 ö æ1 n1 = RcZ2 ç 2 - 2 ÷ = RcZ2 è1 ¥ ø ö 3RcZ ÷= 4 ø

(II)

(4) One mole of a substance contains the number of molecules which is independent of pressure.

2

æ1 1 n 2 = RcZ2 ç 2 - 2 è1 2

HO

I

35.

q0 = 10°C

(3)

(I)

Conc. HI

460 - 10q0 = 440 - 8q0

31.

HO

I

(2) In electrolysis of NaCl when Pt electrode is taken, then H2 liberated at cathode, while with Hg cathode it forms sodium amalgam because more voltage is required to reduce H+ at Hg than at Pt. (4) (1) For tetrahedral d6 ion,

4 unpaired electrons (2) For [Co(NH3)6]3+,

0 unpaired electrons (3) For square planar d7 ion,

Boric acid (H3BO3 or B - OH ) is a Lewis |

OH acid so (a) is incorrect statement.

1 unpaired electrons

Solutions-Mock Test-3 (4) B.M. =

MT-149

n(n + 2) , n = unpaired electrons

39.

(1) (2) (3) (4)

5.92 B.M. = n(n + 2) n = 5 unpaired electrons

40.

(1)

Conc. HI CH3CH2O CH(CH3)2 ¾¾¾® CH3CH2I

+

(1) SiC Þ Covalent carbide Be2C + 4H2O ¾¾ ® 2Be(OH)2 + CH4 ­ CaC2 + 2H2O ¾¾ ® HC º CH ­ + Ca(OH)2 Mg2C3 + 2H2O ¾¾ ® 2Mg(OH)2 + H3C – C º CH ­ (CH3)2CHI

(B)

(A)

NaOH

CH3CH2OH

(C) NaOH

(CH3)2CHOH [O]

[O]

CH3COOH hence the IUPAC name of compound is 2

1

(CH3)2CO ® CdS ¯ (2) SnS2 ¯ + (NH 4 ) 2 S ¾¾ Yellow

CH3 — CH 2 — O — CH — CH3 ½ 3CH3

(NH ) S

4 2 2 ® (NH ) SnS ¾¾¾¾¾ 4 2 3

Soluble

2 -ethoxy propane

41.

(3) Cd

(3) (1)

2+

+ (NH 4 ) 2 S ¾¾ ® CdS ¯ Yellow

(NH 4 ) 2 S2

CdS ¯ + (NH 4 ) 2 S ¾¾ ® CdS ¯ Yellow

¾¾¾¾¾ ® CdS ¯

Yellow

Insoluble

(NH 4 ) 2 S2

¾¾¾¾¾ ® CdS ¯

Insoluble

2+ ® SnS ¯ (4) Sn + (NH 4 ) 2 S ¾¾ Brown

(NH ) S

4 2 2 ® (NH ) SnS ¾¾¾¾¾ 4 2 3

(Amm. thiostannate) Soluble

42. 43.

(1) The compounds of the type M(AA)2B2 exhibit both geometrical and optical isomerism. (2) COOK aq. KOH

Me – CH = CH2 + CHCl3 ¾¾¾® CH3 CH – CH3

COOH

COOK | CH3 – CH – CH2 ¾¾® CH3 – CH – CH3

CCl 2

Cl

HO

Cl C

OH

O

OH C

aq. KOH

CH3 – CH – CH2 ¾¾¾® CH3 – CH

OH – CH2 ¾¾¾® – H2O

CH3 – CH – CH2

EBD_7308 JEE MAIN

MT-150

44.

(1)

47.

(4) 3CH4 + 2O3 ¾® 3CH2= O + 3H2O Formaldehyde

® 2N 2O 4 (CCl4 ) + O 2 (g) 2N 2 O5 (CCl4 ) ¾¾ t=0

a mol

t = 500 min.

a–x

t=¥

0

CH 2 = CHCH = O CH 3 COONO 2

0

x mol 2 a mol 2

a = 100 x = 90

Acrolein

48.

CH = O

2.303 1 a k = ln = 500 t a-x

45.

46.

CH = O

(1) The order of basic character of the transition metal monoxide is TiO > VO > CrO > FeO because basic character of oxides decreases with increase in atomic number. (4) NH4ClO4 + HNO3 (dilute) ¾¾® NH4NO3 + HClO4

49.

(X)

NH4NO3 ¾¾¾ ® N2O + 2H2O 50.

(Y)

CH2OH (B)

(4) (1) The concentration of sodium thiosulphate solution should always be less than the concentration of the potassium iodide solution. (2) Freshly prepared starch solution should be used (3) Experiments should be performed with the fresh solutions of H2O2 and KI.

(4) +

+

Ring expansion

H+

1, 2 – CH3 shift

+

(3° Carbocation stable)

51.

(2) b 3° b 2° a

a

53.

b 3°

+ N

N

No. of O 2-

(2aH) 52.

0.25–x 0.25–x

Kp = 0.16 =

x

x

x 54.

2

(0.25 - x) 2

x Þ 0.4 = Þ 0.1 - 0.4x = x 0.25 - x x = 0.0714

Mole% of CO (g) =

1 ´4 = 2 2 1 1 = 8´ ´6´ = 4 8 2

(AB2O4) \ value of n = 2

(14.28) H2 (g) + CO2 (g) ƒ CO (g) + H2O (g) At eqm

1 (2) No. of A 2+ = ´ 8 = 1 8

No. of B3+ =

¾¾®

0.0714 ´ 100 = 14.28 0.50

– + CO2 K

Conc. KOH ¾¾¾¾® Cannizzaro

(A)

heat

(X)

Peroxyacetyl nitrate (PAN)

(2)

(80)

P° - P n2 = P° n1 + n2

640 - 600 2.5 m = 39 78 640

m=

640 ´ 78 ´ 2.5 = 80 39 ´ 40

Solutions-Mock Test-3

55.

(5.0)

pH = pKa + log

MT-151

[CH3COO - ] [CH 3COOH]

-

pKa = – log (1.8 × 10–5) = 4.7447 [CH3COO–] = 2 × [(CH3COO)2Ba] = 0.2 M [CH3COOH] = 0.1 M

5 dMnO4– dI 2 = 2 dt dt On substituting the given value

\ -

0.2 = 5.046 » 5.0 0.1 (1) H3PO2 is named as hypophosphorous acid. It is monobasic as it contains only one P – OH bond, its basicity is one.

pH = 4.7447 + log

O || P H | OH H (395)Since, work is done against constant pressure and thus, irreversible. Given, DV = (6 – 2) = 4 L; P = 1 atm

\

57.

61.

(since 0.0821 L-atm = 1.987 cal) = – 96.81 cal = – 96.81 × 4.184 J

(Q1 cal = 4.184 J )

58.

59.

m=

\ E. wt =

a1 + a2 =

c -b , a1a2 = . a a

..(1)

-q r , b1b2 = ...(2) p p Since the given system of equations has a non-trivial solution

and b1 + b2 =

a1 a 2 = 0 i.e. a1b2 – a2b1 = 0 b1 b 2

m ´ 96500 22.2 ´ 96500 = = 60.3 Q 2 ´ 5 ´ 60 ´ 60

or

a1 a 2 a1 + a 2 = = = b1 b 2 b1 + b 2

At wt. 177 = =3 Eq. wt . 60.3

dMnO4– = 4.56 × 10–3 Ms–1 dt From the reaction given, -

\ The lines are x – y = 0, x + 7y = 0 (1) Since a1, a2 and b1, b2 are the roots of ax2 + bx + c = 0 and px2 + qx + r = 0 respectively, therefore

\

(1.14) Given -

1 dMnO 4 – 4.56 ´ 10 -3 Ms -1 = 2 dt 2

2

m 3 + 2 2 = 2 Þ m = 1, –1/7 m2 + 1

E.wt ´ Q ; 96500

Oxidation state =

60.

62.

= – 405.05 J Now from Ist law of thermodynamics q = DU – W 800 = DU + 405.05 \ DU = 395 J (3) Three, these are CH3CH2OCH2CH3 (I), CH3OCH2CH2CH3 (II) and CH3OCH(CH3)2 (III). Here I and II, I and III are pairs of metamers. (3)

æ 1 3ö Centre of the circle is çè , - ÷ø . 2 2 Its distance from the line x + y – 1 = 0 is Let the required line be mx – y = 0

(4)

\

1 ´ 4 ´ 1.987 W = – 1 × 4 L-atm = – cal 0.0821

\

dI 2 4.56 ´ 10-3 ´ 5 = = 1.14 × 10–2 M/s dt 2

MATHEMATICS

|

56.

1 dMnO 4– 1 dI 2 = 2 dt 5 dt

pc b 2 ac Þ = ra q 2 pr (2) Suppose, there are two points x1 and x2 in (a, b) such that f '(x1) = f '(x2) = 0. By Rolle's theorem applied to f ' on [x1, x2], there must be a c Î ( x1, x2 ) such that f ''(c) = 0. This contradicts

Þ

63.

a1a 2 b1b 2

pb = qa

the given condition f ''( x ) < 0, " x Î (a, b).

EBD_7308 JEE MAIN

MT-152

Hence, our assumption is wrong. Therefore, there can be atmost one point in (a, b) at which f '(x) is zero.

x = 2 2 - 3 Þ |10a | = [| 20 2 - 30 |] = 30 - 20 2

x Î [0, 1]

1 + x2 = 2x 2 1 + x2 = 2x Þ x = 1 Þ | 10a | = 10

16

64.

(2)

x+

ò f (t) dt

I=

0

x4

ò f (t) dt Þ g(0) = 0

Consider g(x) =

0

LMVT for g in [0, 1] gives, some a Î (0, 1) such g(1) - g(0) = g ¢ (a ) that ..... (1) 1- 0 Similarly, LMVT in [1, 2] gives, some g(2) - g(1) = g ¢ (b) b Î (1, 2) such that .... (2) 2 -1 Eq. (1) + Eq. (2) ; g ¢(a ) + g ¢ (b) = g(2) - g(0) {

67.

68.

2 cos q ± 4 cos 2 q - 4 = cos q ± i sin q 2 Let x = cos q + i sin q \ x2n – 2xn cos nq + 1 = cos 2nq + i sin 2nq – 2 (cos nq + i sin nq) cos nq + 1 = cos 2nq + 1 – 2 cos2 nq + i (sin 2nq – 2 sin nq cos nq) =0+i0=0 x=

zero

but g' (x) = f (x4) . 4x3 \ 4 éëa 3f (a 4 ) + b3f (b4 ) ùû =

16

ò f (t) dt 0

65.

(1)

3u 2

2u 3

2

3

3v

3w

2

2v

2w

1 1 =0

3

| 10a | = 10, | 20 2 - 30 | Þ [ | 10a | ] = 1, 10 (4) Each point (x, y) has an image in line y = 0 as (x, –y). So, replacing y by –y in the given equation, we get the image as ax2 – 2hxy + by2 = 0. (3) x2 – 2x cos q + 1 = 0,

1

69.

1

(2)

ò

u+v

Þ v+w

2

u + v + vu w3

2

u+w+v

Þ v+w

66.

w3

2

1+ x = -2x 2 x2 + 6x + 1 = 0 x+

1

2

" 0 < x < 1, x > x

1

1

Þ

ò

1

2

ò

3

3

2 x dx > 2 x dx Þ I1 > I 2 .

0

0

0

Also " 1 < x < 2 2

1

Þ (v2 + w2 + vw) – (v + w) [v + w + u] = 0 Þ uv + vw + wu = 0 (4) x Î [–1, 0]

3

ò

1

v2 + w 2 + vw 0 = 0

w2

2

2

ò

0

R1 ® R1 – R2 1

0

I3 = 2 x dx, I 4 = 2 x dx

v2 + w 2 + vw 0 = 0

w2

3

ò

0

R1 ® R1 – R2 and R2 ® R2 – R3 2

1

2

I1 = 2 x dx , I 2 = 2 x dx ,

2

3

ò

x 4

Þ m Î (-¥, - 3) È ( 3, ¥)

represents a parabola with vertex (0,2) axis as y–axis and concavity downwards. Both the curves are plotted in the figure and the required area is shown by the shaded region. The points A and C are the points of intersection of y 2 = x 2 with x 2 + y - 2 = 0 . Solving the two equations, we get

D P

Þ x 2 = -( y - 2)

y 2 + y - 2 = 0 [putting value of x2 = y2]

Þ ( y + 2)( y - 1) = 0 giving y = –2 and 1, but y = –2 is discarded as the required area is above the x-axis.

\y = 1 Þ x = ± 1 The points A and C are respectively (–1, 1) and (1, 1) now due to symmetry Area of the bounded region OABCO

Solutions-Mock Test-3

MT-155

1

[

r Now, | A |2 = 4 + 9 + 36 = 49

]

= 2 × Area OBCO = 2 × ò ( 2 - x ) - x dx 2

2 3 6 r r r [A B C] = 1 1 -2

0

[Since y = 2 –x2 is the upper curve and y = x is the lower curve]

1 2

1

é x3 x 2 ù é 1 1ù 7 = 2ê2x ú = 2 ê2 - - ú = = 2.33 3 2 ë 3 2û 3 úû 0 ëê 82. (9.00) Number of digits are 9 Select 2 places for the digit 1 and 2 in 9C2 ways from the remaining 7 places select any two places for 3 and 4 in 7C2 ways and from the remaining 5 places select any two for 5 and 6 in 5C2 ways Now, the remaining 3 digits can be filled in 3! ways \ Total ways = 9C2 . 7C2 . 5C2 . 3! 9! 7! 5! . . .3! = 9.7! = k. 7! = 2!.7! 2!.5! 2!.3! Þ k=9 r 83. (32.00) n = 7iˆ + 2jˆ - kˆ is normal to plane

85.

86.

C1(–1,0) B

3 C2(3,0)

Area = 3 × 3 tan 30° = 3 3 = l 3 Þ l=3 6R 5B (0.53) Box 4W

n (E) = 6C 2 · 5C 1 · 4C 1 + 5C 2 · 6C 1 · 4C 1 + 4C 2 · 6C 1 · 5C 1 n (S) = 15C4 \ P (E) =

720 × 4! 48 = = 0.53 15 ×14 ×13 × 12 91

87.

(2.00)

88.

æ 3q ö æ q ö 2sin ç ÷ cos ç ÷ sin q+ sin 2q è 2 ø è 2 ø = tan æ 3q ö = ç ÷ cos q+ cos2q æ 3q ö æ q ö è2ø 2cos ç ÷ cos ç ÷ è 2 ø è2ø 2p p =k Hence period = 3 3 Þ k = 2. (3.00) The equation of the tangent at

uuur AP = 6iˆ + 3jˆ - 16kˆ \ Distance

æ r r r r r r r rö r r r r r ´(A.B)A- (A.A)A´B÷.C = – | A |2 [A B C] = çA 1 4 2 4 3 ç ÷ è zero ø

1

P(E) = P(R R B W or B B R W or W W R B)

A(2,–1,4)

84.

?

\ coordinates of A are (–3, 0) We have sin q = 1/2 \ q = 30°

n

k 6 = Þ k = 32 . 9 (343.00) r r r r r r r r r V = A ´ éë (A.B)A - (A.A)B ùû .C

(3.00) A divides C1C2 externally in the ratio 1 : 3. A(–3,0)

P(8,2,–12)

uuur r AP.n 42 + 6 + 16 64 64 64 6 32 6 = = = = r = |n| 18 9 49 + 4 + 1 54 3 6

=7

r 2 r r r \ - | A | [A B C] = 49 ´ 7 = 343

(Assuming n = aiˆ + bjˆ + ckˆ and using r uuur r uuur r uuur n.AB = 0, n.BC = 0, n.AC = 0 ) P = (8, 2, – 12)

d=

1

ö æ 16 çç 4 cos q, sin q ÷÷ to the ellipse 11 è ø

16x2 + 11y2 = 256 is æ 16 ö 16x (4 cos q) + 11y çç sin q ÷÷ = 256 è 11 ø

EBD_7308 JEE MAIN

MT-156

or 4x cos q + 11y sin q = 16 This touches the circle (x – 1)2 + y2 = 42 if 4 cos q - 16 16 cos 2 q + 11sin 2 q

89.

cos q =

1 p p Þq= ± =± 2 3 k

1 100

B2 : person does not suffer \ P(B2) = P(A/B1) =

99 1 , P(A/B2) = 100 100

p /2

(0.50) a n =

90.

ò

(1 - sin t) n sin 2t dt

0

Let 1 – sin t = u Þ – cos t dt = du 1 1 æ1 ö = 2ò u n (1 - u) du = 2 ç ò u n du - ò u n +1du ÷ = 2 æç 1 - 1 ö÷ è n +1 n + 2ø

Þ k= 3 (0.50) A: blood result says positive about the disease B1: Person suffers from the disease \ P(B1) =

P(B1 ).P(A / B1 ) P(B1).P(A / B1 ) + P(B2 ).P(A / B2 )

1 99 × 99 1 100 100 = = = 0.50 = 1 99 99 1 2 × 99 2 × + × 100 100 100 100

=4

Þ (cos q – 4)2 = 16cos2q + 11sin2q Þ 15cos2q + 11sin2q + 8cos q – 16 = 0 Þ 4 cos2q + 8 cos q – 5 = 0 Þ (2 cosq – 1) (2 cos q + 5) = 0 Þ

P (B1 / A) =

è0

0

ø

an æ 1 1 ö Hence, n = 2 çè n(n + 1) - n(n + 2) ø÷ n

å n ®¥ lim

1

99 100

0

1 ö 1 æ1 1 öö an æ æ1 = 2 ç åç ÷ - åç ÷ n n 1 2 n n + + 2 ø ÷ø n ø è è è

æ n 1 1 öö n æ 1 1 ö = 2 ç å æç ç è n n + 1 ÷ø ÷÷ å çè n n + 2 ÷ø è 1 ø 1 éæ

1ö

æ1

1ö

æ 1 1ö

ù

3

1

= 2(1) - êç 1 - ÷ + ç - ÷ + ç - ÷ + .....ú = 2 - = 2 2 ëè 3 ø è 2 4 ø è 3 5 ø û =0.50

Mock Test-4 (3) (2) (4) (2) (3)

1 2 3 4 5

16 17 18 19 20

(4) (4) (4) (1) (2)

31 32 33 34 35

ANSWER KEY (2) 46 (2) 47 (4) 48 (4) 49 (4) 50

(1) (2) (2) (2) (2)

(3) (2) (3) (2) (2)

61 62 63 64 65

(4) (2) (2) (2) (4)

76 77 78 79 80

6

(3)

21

(3)

51

(600)

66

(2)

81

(4.00)

(2) (2) (1) (2) (4) (3) (1) (3)

22 23 24 25 26 27 28 29

(10 2 ) (83.3) (3) (70.7) (10.6) (1) (2) (2) (2)

36

7 8 9 10 11 12 13 14

37 38 39 40 41 42 43 44

(4) (1) (4) (3) (2) (3) (4) (3)

52 53 54 55 56 57 58 59

(10.93) (8) (68.4) (2) (4) (16) (69.6) (3.8)

67 68 69 70 71 72 73 74

(2) (4) (4) (4) (3) (3) (3) (1)

82 83 84 85 86 87 88 89

(0.13) (0.20) (75) (23.00) (1.00) (1.00) (3.50) (5.00)

15

(2)

30

(5 2 )

45

(1)

60

(2)

75

(3)

90

(2.25)

Solutions PHYSICS 1.

(3)

2.

(2) After 2 sec the pulses will overlap completely. The string becomes straight and therefore does not have any potential energy and it entire energy must be kinetic.

3.

(4) mg = 2TL Þ pr2 Ldg = 2TL Þ pr2 dg = 2T..

R +q A

–q C 2L

B

L

D

Potential at C = VC = 0 Potential at D = VD

2 Kq æ -q ö Kq == K çè ÷ø + L 3L 3 L Potential difference -2 Kq 1 æ 2 qö = VD – VC = ç- . ÷ 3 L 4p Î0 è 3 L ø

Þ Work done = Q(VD – VC) 2 1 qQ -qQ =- ´ = 3 4pe 0 L 6pe 0L

4.

This relation is independent of L. (2) When either of A or B is 1 i.e. closed then lamp will glow. In this case, Truth table Inputs

Output

A

B

Y

0

0

0

0

1

1

1

0

1

1

1

1

This represents OR gate.

EBD_7308 JEE MAIN

MT-158

5.

(3) Heat required to change the temperature of vessel by a small amount dT – dQ = mCpdT Total heat required

8.

3

æ T ö ÷ dT è 400 ø

4

– Q = m ò20 32 ç

4

=

100 ´ 10 –3 ´ 32 é T 4 ù ê ú ë 4 û 20 (400)3

Þ Q = 0.001996 kJ Work done required to maintain the temperature of sink to T2 W = Q1 – Q2 =

T –T Þ W = æç 1 2 è T2 For T2 = 20 K

Q1 – Q2 æT ö Q2 = ç 1 –1÷ Q2 T Q2 è 2 ø

ö ÷ Q2 ø

9.

10.

300 – 20 ´ 0.001996 = 0.028 kJ 20 For T2 = 4 K

W1 =

6.

W2 = 300 – 4 ´ 0.001996 = 0.148 kJ 4 As temperature is changing from 20k to 4 k, work done required will be more than W1 but less than W2. l (3) As we know, time period, T = 2p g When additional mass M is added then

TM = 2p

l + Dl g

11. 2

l + Dl l + Dl TM æ TM ö = or ç ÷ø = è T l l T 2

or,

7.

Mg æ TM ö çè ÷ø = 1 + T AY

(4) Energy in joule (E) = charge × potential diff. in volt Eelectron = qeV and Eproton = qp 4V de-Broglie wavelength l =

Mgl ù é êëQ Dl = AY úû

le =

2 ù 1 éæ T ö A \ = êç M ÷ -1ú Y êëè T ø úû Mg

(2) More the initial temperature more is the rate of cooling. Hence, T3 > T2 > T1 or The rate of cooling decreases with decrease in temperature difference between body and surrounding.

(2) We have, F = kx where, F, x and k are force, length and constant respectively. ......(1) \ 5 = kx and 7 = ky ......(2) Multiplying eq (2) by 2 14 = 2ky ......(3) Subtracting eq (1) from (3), 14 – 5 = 2ky – kx or 9 = k(2y – x) Hence, required length = 2y – x (1) From the graph, it is clear that for the same value of load, elongation is maximum for wire OA. Hence OA is the thinnest wire among the four wires (2) Resistance between P and Q 5 r´ r æ r rö 6 = 5r rPQ = r P ç + ÷ = è 3 2ø 5 11 r+ r 6 Resistance between Q and R r 4 ´ r r r 4 rQR = P (r + ) = 2 3 = r r 4 3 11 2 + r 2 3 Resistance between P and R r 3 ´ r 3 r ær ö 3 2 = r rPR = P ç + r÷ = 3 è 2 ø r + 3 r 11 3 2 Hence, it is clear that rPQ is maximum

12.

h

lP = 2me eV and

h = P

h 2mE

h 2mP e4V

(Q qe = qP) h 2m e eV 2m P e4V mP l \ e = = =2 h 2me eV me lP 2m P e4V m I m Isin q (3) B = 0 (sin q + sin q) = 0 4p a 2pa

Solutions-Mock Test-4 B at any point on Y axis is inversely proportional to a. a Y

q q

P 13.

2

19.

(1)

W = MB (cos q1 - cos q2 ) = MB (cos 0° - cos 60°) 1 MB = MB(1 - ) = 2 2

O X a/2 = mg.

–a/2 I (1) f max = mmg, a max If A is the amplitude a max 2

MT-159

\ t = MB sin q = MB sin 60° = 3 20.

(2)

O

2

= Aw = 4p AV = mg . Therefore, A =

14.

15.

16.

17.

mg

q

.

æ n +n2 ö d d d ÷= + = dçç 1 ÷ n ; 2n 1 2n 2 2 n n 1 2 ø eff è 3 n 2 = 3n 1 Þ n eff = n 1 2 (2) On the screen, we have four amplitudes pair wise coherent.

mg In equilibrium, Fe = T sin q mg = T cos q tan q =

(A1 + A2) + (A3 + A4) = A12 + A34 However, if A12 and A34 have equal magnitude because of random phase of A12 and A34, no fringes will be seen. (4) For good demodulation, 1 1 > f f

\ x=

21.

2

Since, g monoatomic > g diatomic Hence, curve 2 corresponds to helium and curve 1 corresponds to oxygen.

kq kq + = 4 kmg tan q x/2 x/2

( 10 2 ) 4 m/sec2 Car

Bus

2 m/sec2

200 m Given, uC = uB = 0, aC = 4 m/s2, aB = 2 m/s2 hence relative acceleration, aCB = 2 m/sec2 1 Now, we know, s = ut + at 2 2 1 200 = ´ 2t 2 Q u = 0 2 Hence, the car will catch up the bus after time

R1 - R2 ( R1 - R2 )T = l 0.693 i.e., µ (R1 – R2)T (4) In the graph given, slope of curve 2 is greater than the slope of curve 1.

æ gP ö æ gP ö çè ÷ø > çè ÷ø Þ g 2 > g1 V 2 V 1 g He > gO

q2 4p Î0 tan q mg

V=

=

18.

Fe q2 = mg 4p Î0 x 2 ´ mg

Electric potential at the centre of the line

0.693

2. R = lN t t1/ 2 Radioactivity at T1 is R1 = l N1, Radioactivity at T2 is R2 = l N2 \ Number of atoms decayed in time (T1–T2) = (N1–N2)

Tcosq q C Tsinq q x Fe

q

4p 2 V 2 (3) Total time taken to travel distance d is :

(4) 1. l =

MB = 3W 2

t = 10 2 second 22.

(83.3) Power of source = EI = 240 × 0.7 = 166 Þ Efficiency =

140 Þ h = 83.3% 166

EBD_7308 JEE MAIN

MT-160

23. 24.

(3) In the given equation [r] = [b][x]; \[b] =[r]/[x]. But r is mass per unit length and x is distance, therefore [b]=ML–1/L=ML–2T0 (70.7) Electric field intensity at the centre of the disc. s E= (given) 2 Î0 Electric field along the axis at any distance x from the centre of the disc æ ö x ç1 ÷ ç 2 2 ÷ + x R è ø From question, x = R (radius of disc)

Divide (1) by (2), 27.

æ R \ E ' = s ç1 2 2 Î0 çè R + R2

=

(2) The angle for which the ranges are same is complementary. Let one angle be q, then other is 90° – q

T1 =

28.

ö ÷ ÷ ø

s æ 2R - R ö ç ÷ 2 Î0 çè 2R ÷ø

4 ö æ çè E - E÷ø ´ 100 1000 14 % ; 70.7% = = E 14 (10.6) Work done in going from a distance r 1 to a distance r 2 away from centre of the earth, by a body of mass m, is, W = GMm (1/r1 – 1/r2), For our case we should have 1/2 mv2 = GMm [ (1/Re) – (1/10Re)] = (GMm/Re) ´ (9/10) v = Ö[(2GM/Re) ´ ( 9/10)] = Ö(9/10) ´ escape velocity = Ö(9/10) ´ 11.1 km/s = 10.6 km/s

(1)

mvr =

29.

30.

2p w = l c

k=

w 6 ´ 108 = = 2 m -1 c 3 ´ 108

æv ö w rod = w point = ç rel ÷ , è r ø vrel represents the velocity of one point w.r.t. other. 3v - v = and ‘r’ being the distance between r them. 2v = r (5 2 ) V0 100 V 100 = = 10 iR = 0 = = 5 , iL = X L 10 R 20 (2)

V0 100 = =5 XC 20

Current, i = i 2R + (iC - i L )2 nh (where 2p

= 52 + 52 = 5 2 amp.

n = 1)

CHEMISTRY

1´ h 2p r = mv

....(1);

h mv

....(2)

l=

k=

and i C =

nh h ; , l= 2p mv

Using the two concept we get, mvr =

=

w = 6 × 108,

4 E 14 \ % reduction in the value of electric field

26.

4u 2 sin q cos q

2R u 2 sin 2q R = (Q ) g g2 g Hence it is proportional to R. (2) On comparing the given equation to r E = a0iˆ cos (wt – kz)

=

25.

2u cos q 2u sin q , T2 = g g

T1T2 =

s E' = 2 Î0

2pr h ´ mv 1 = = = 1:1 l mv ´ h 1

31.

(2) The molecule 2,3 - pentadiene does not have any chiral C but at the same time it does not have any mirror plane which makes the molecule chiral.

Solutions-Mock Test-4 CH3 32.

MT-161

CH3

CH3 Br

Br 2

¾¾¾®

FeBr3

NO2

NO2

HCl

NO2

N = NCl CH3 Br

33.

(4) The compound A is cyclic acetal, so it should have an aldehyde and a diol as the two starting compounds.

38.

5Cl2 + 6H 2 O + I2 ® HIO3 + 10HCl Colourless

O [A]

39.

CH2COOH ¬¾¾ CH3

(4) Tertiary halides on treatment with base, such as sodium methoxide, readily undergo elimination resulting in the formation of alkenes. (Williamson’s Synthesis)

CH3 | +– E2 CH3 — C — Cl + N a O CH2CH3 ¾¾® | CH3 CH3 — C = CH2 t-Butyl chloride | CH3

CH3 CH O

+ HO

OH CH2COOH CH3

34.

(4) In N2+ , there is one unpaired electron hence it is paramagnetic.

35.

(4) In Ag2O (O.N. of Ag +1) in Ag the O.N. is O. There is gain of electrons, hence H2O2 is acting as reducing agent.

36.

(3)

37.

HBr CuBr

Br (1) 3Cl2 + 2Nal ® 2NaCl + I2 I2 gives violet colouration in CHCl3.

CH3 O

Br

NaNO

2 ¾¾¾®

Sn/HCl

¾¾®

(2)

CH3

Br

H 2C – CH 2+RMgX ¾® CH2 – CH 2 | | O OMgX R CH 2 – CH 2 H 2O ¾¾¾¾® | | –Mg(OH)X R OH (4) Total energy of a revolving electron is the sum of its kinetic and potential energy.

Isobutylene

40.

(3) Performic acid causes hydroxylation of the double bond ; the two –OH groups add in antimanner. Hence cis-isomer gives racemic mixture while the trans-isomer gives meso.

41.

(2)

C6 H5 NH 2 + NaNO2 + HCl ¾¾¾ ® 0°C +

OH N2Cl +

¾® OH

Total energy = K.E. + P. E.

N= N

2 e 2 æ e2 ö = +ç- ÷ = - e 2r è r ø 2r Red dye

-

C6 H5 N º N Cl

EBD_7308 JEE MAIN

MT-162

42.

43. 44.

45. 46. 47.

(3) Multiple bonds formation tendency with carbon and nitrogen decreases from sulphur to tellurium. CS2 (S = C = S) is moderately stable, CSe2 (Se = C = Se) decomposes readily whereas, CTe2 (Te = C = Te) does not exist. (4) Liquation is the principle based on difference in melting points. (3) In NO 2 + odd (unpair ed) electron is removed. In peroxides (O22– ) no unpaired electrons are persent as the antibonding p M.O.’s acquired one more electron each for pairing. AlO-2 containing Al3+ (2s2p6) configuration and 2 oxides (O2– ) ions each of which does not contain unpaired electron. Superoxide O 2– has one unpaired electron in p antibonding M.O. (1) The two solutions are isotonic hence there will be no movement of H2O. (1) Melamine plastic crockery is a copolymer of HCHO and melamine. (2) Hydrated CoCl2. 6H2O is pink coloured and contains octahedral [Co(H2O)6]2+ ions. If this is partially dehydrated by heating, then blue coloured tetrahedral ions [Co(H2O)4]2+ are formed. [Co(H 2 O)6 ]2+ ƒ [Co(H 2 O) 4 ]2 + + 2H 2O pink

48. 49.

50.

51.

i.e. r + + r - =

52.

53.

54.

blue

(2) (2) • Li does not form peroxide or superoxide due to its small size. • Solubility of carbonates and biocarbonates increases on moving down the group. • The increasing order of size of hydrated ions of alkali metals is Li+ > Na+ > K+ > Rb+ > Cs+ • Cesium used in photoelectric cells due to its low I.E. Hence statements (2) is correct. (2) (i) A3+ + e– ¾® A2+, DG1 = – 1 F y2 (ii) A2+ + 2e– ¾® A, DG2 = –2F(–y1) = 2Fy1 Add, (i) and (ii), we get A3+ + 3e– ¾® A ; DG3 = DG1 + DG2 –3FE° = –Fy2 + 2Fy1 –3FE° = –F (y2 – 2y1) y - 2 y1 E° = 2 3

(600) In a fcc lattice, the distance between the cation and anion is equal to the sum of their radii, which is equal to half of the edge length of unit cell, a 2

(where a = edge length)

r+ = 100 pm, r– = 200 pm Edge length = 2r+ + 2r– = (2 × 100 + 2 × 200) pm = (200 + 400) pm = 600 pm. (10.93) DH = (-40.8) ´ 1.8 = -4.08 kJ ; 18 3 DH -4.08 ´ 10 DS = = = -10.93 JK -1 T 373.15 (8) n = 5 means l = 0, 1, 2, 3, 4 since ml = + 1 hence total no. of electrons will be = 0(from s) + 2(from p) + 2(from d) + 2(from f) + 2(from g) =0+2+2+2+2=8 (68.4) As we know, No.of moles of sugar Molarity = Volume of solution (in L) No.of moles of sugar 0.1 = 2L So, no. of moles of sugar = 0.2 mol \ Mass of sugar = No. of moles of sugar ×

55.

Molar mass of sugar = 0.2 × 342 = 68.4 g (2) The reaction involves Cope eliminiation (heating of a 3º amine oxide to form an alkene with the elimination of a 2º hydroxylamine).

There are two unsaturated carbon in the product.

Solutions-Mock Test-4 56.

57.

MT-163

(4) According to question C – Terminal must be alanine and N – Terminal do have chiral carbon means it should not be glycine. So possible sequence is : Val Phe Gly Ala Val Gly Phe Ala Phe Val Gly Ala Phe Gly Val Ala (16)

rO2 rH 2

=

nO2 nH 2

=

MATHEMATICS 61.

(3) Given equation is (p – q) x2 + (q – r) x + (r – p) = 0 By using formula for finding the roots

M H2

w / 32 w = = M O2 4 / 2 64

2 1 = 32 4 w = 64/4 = 16 g

59.

P° - PS w/m = P° W/M (640–600)/640 = wM / mW 40/640 = 2.175 ´ 78/m ´ 39.08 m = 2.175 ´ 78 ´ 640 / 39.08 ´ 40 = 69.458 @ 69.60

As pH = 6.0 [H3O]+ =10–6 é H 3 O+ ù é HCO3– ù ûë û (H O is in excess, K=ë 2 2 [CO 2 ][ H 2 O ]

60.

62.

(69.6)

é H 3 O+ ù é HCO3– ù ûë û (3.8) K = ë 2 [ CO 2 ][ H 2 O ]

therefore its conc. remains constant) éHCO3– ù K 3.8 ´ 10 -6 ë û = = = 3.8 10 -6 [ CO 2 ] é H 3 O+ ù ë û 0.69 , t 3/14 = t 75% (2) t1/2 = k 2.303 a t 3/4 = log 3a ö k æ çè a - ÷ø 4 2.303 2.303 0.69 ´ 2 log 4 = ´ 2 ´ 0.3010 = = k k k t 3/4 0.69 ´ 2 k = ´ Þ t 3/4 = 2t1/2 t1/2 k 0.69

-b ± b2 - 4ac , we get 2a

x=

( r - q) ± ( q - r )2 - 4( r - p)( p - q) 2( p - q)

Þ x=

=

58.

x=

(2)

(r - q) ± (q + r - 2 p) r - p = ,1 2( p - q) p-q

(a 2 + b 2 + c 2 ) p 2 - 2(ab + bc + cd ) p + b2 + c2 + d 2 £ 0

Þ (a 2 p 2 - 2abp + b 2 ) + (b 2 p 2 - 2 bcp + c 2 ) + (c 2 p 2 - 2cdp + d 2 ) £ 0

Þ (ap - b)2 + (bp - c) 2 + (cp - d) 2 £ 0

Þ ap - b = 0, bp - c = 0 & cp - d = 0 b c d = = Þ a, b, c and d are in G.P a b c Also ad = bc (3) 1 2 (1) log( a + 2 b) = log( a + 2 b) 2 1 2 2 = log(a + 4b + 4ab) 2 1 = log(12ab + 4ab) 2 1 = log( 2 4 . ab ) 2 1 = (4 log 2 + log a + log b) 2 log x log y log z = = =k (2) Let b-c c-a a -b Þ

63.

Þ log x = k (b - c), log y = k(c - a ), log z = k(a - b) \ x a . y b . z c = p k[ a (b -c))+ b (c - a )+ c(a - b)] k ( 0) =1 =p where p is any arbitrary base of the log.

EBD_7308 JEE MAIN

MT-164

67.

(3) Given expression

Putting z = w we get (w - 1)(w - a1 )(w - a 2 )(w - a3 )(w - a 4 )

= log xyz xy + log xyz yz + log xyz zx = log xyz ( xy . yz . zx) 2

2

= w5 - 1

Putting z = w we get (w2 – 1)(w2 – a1)(w2 – a2)(w2 – a3)(w2 – a4) = w10 – 1

2

2

= log xyz ( x . y . z )

= 2 log xyz ( xyz) = 2 ´ 1 = 2 64.

p p Þ a + g = – b. 2 2 æp ö so that cot ( a + g ) = cot ç - b÷ è2 ø

(2)

Þ

(2) We have z5 – 1 = (z – 1)(z – a1) … (z – a4)

\

a+b+ g =

(w –1)(w – a1 )(w – a 2 )(w – a3 )(w – a 4 )

2

(w –1)(w2 – a1)(w2 – a 2 )(w2 - a3 )(w2 – a 4 )

=

1 cot a cot g - 1 = cot a + cot g cot b

w 5 –1 w10 –1

Þ

Þ cot a cot g - 1 = 2 Þ cot a cot g = 3.

(since cot a + cot g = 2cot b)

65.

given by T = S1 where, T =

S1 =

x2 a2

-

y2 b2

68.

= 1 is

a

x12

2

-

-

y12

yy1 b2

- 1 and

and AA-1 = I Þ det( AA-1 ) = det( I ) Þ det(A ) det(A -1 ) = 1 Þ det(A -1 ) =

\

æ px ö 2 2 (2) sin ç ÷ = x - 2 3x + 4 = (x - 3) + 1 è2 3ø Q RHS ³ 1 so, the solution exists

If and only if x - 3 = 0 Þ x = 3 and then equation is obviously satisfied

(4) (1) We have | AB | = | A | | B |

(2) Since A is invertible, therefore A–1 exists

5x 3y 25 9 [Using (i)] – – = 25 16 25 16 Þ 125x – 48y = 625 – 144 Þ 125x – 48y = 481

66.

= (w + 1)2 = w4 = w

\ | 3AB |= 33 | A || B |= 27( -1)(3) = -81

-1

x2 y 2 =1 16 25

(w –1) 2

because each element of the matrix A is multiplied by k and hence in this case we will have k3 common

a 2 b2 According to the question, (x1, y1) = (5, 3) and a2 = 16, b2 = 25 as 25x2 – 16y2 = 400 Þ

(w 2 –1) 2

Also for a square matrix of order 3, | kA |= k 3 | A |

..... (i) xx1

(w 2 – a1 )(w 2 – a 2 )(w 2 – a 3 )(w2 – a 4 ) =

(2) The equation of the chord, having mid-point as (x1 , y1), of the hyperbola

(w – a1 )(w – a 2 )(w – a3 )(w – a 4 )

(3)

1 det(A)

(A + B) 2 = (A + B)(A + B) 2 2 = A + AB + BA + B

2 2 = A + 2AB + B if AB = BA .

69.

ì – x + 1, x < 1 (4) f (x) = | x – 1| = í î x – 1, x ³ 1 Consider f (x2) = (f (x))2

Solutions-Mock Test-4 If it is true it should be " x \ Put x = 2 LHS = f (22) = |4 – 1| = 3 RHS = ( f (2))2 = 1 \ (2) is not correct Consider f (x + y) = f (x) + f (y) Put x = 2, y = 5 we get f (7) = 6; f (2) + f (5) = 1 + 4 = 5 \ (b) is not correct Consider f (| x |) = | f (x) | Put x = – 5 then f (| –5 |) = f (5) = 4 | f (– 5) | = | – 5 – 1| = 6 \ (3) is not correct. 70.

MT-165

Now x + 3 ³ 0 and -1 < cos x + 3 £ 1

x + 3 ¹ (2n - 1) p, n Î N So, range of fogoh is [1, ¥) 72.

(3) Since | x | is not diff. at x = 0

|x – 1| 1

(4) Let a = tan q and b = tan f

Þ | x - 1 | is not diff at x = 1.

2a ù -1 é 2 tan q ù \ sin -1 é êë1 + a 2 úû = sin êë 1 + tan 2 q úû

x n | x | is n times diff. at x = 0

-1 = sin -1 [ sin 2q] = 2q = 2 tan a

é 2b ù é 2 tan f ù and sin -1 ê = sin -1 ê 2ú 2 ú ë1 + b û ë 1 + tan f û

Þ ( x - 1) 2 | x - 1 | is twice diff. at x = 1

but not thrice diff. at x = 1 73.

(3) f(x) =

-1 -1 = sin [ sin 2f] = 2f = 2 tan b

tan–1 \ 71.

x=

x=

tan–1

a+b 1 - ab

a+b 1 - ab

(3) Let f image of goh image, fogoh = F (x) = f [goh (x)] = f [g ( x + 3)] = f (cos x + 3)

F(x) =

2 cos x + 3 + 1

( x - 1) 2 , which is not ( 2 x - 1)( x - 2)

defined at x = 1/2, 2. Hence the set of points where (gof) (x) is discontinuous is {1/2, 1, 2}

é 2b ù sin -1 ê = 2 tan–1 b ë 1 + b2 úû é 2a ù é 2b ù sin -1 ê + sin -1 ê \ ë 1 + a 2 úû ë1 + b2 úû –1 –1 = 2 tan a + 2 tan b Þ tan–1 x = tan–1a + tan–1b

1 is discontinuous at x =1. x -1

(gof) (x) = g(f(x)) = -

é 2a ù Thus, sin -1 ê = 2 tan–1 a and 2ú ë1 + a û

2 tan–1x =

x

1

74.

(1)

m

å

r =0

n+r

m

Cn = å

n+ r

r =0

Cr

(Q n + r Cn =n + rCn + r - n )

= nC0 +

n + 1C 1

+

n + 2C 2

+

n + 3C 3

Using, nC0 = 1 = n + 1C0

+......+

n + mC m

= (n+1C0 + n + 1C1) + n + 2C2 + n + 3C3 +....+ n + mC m Using, nCr + nCr+1 = n+1Cr+1 = (n + 2C1 + n + 2C2) + n + 3C3 +.....+ n + mCm Using this again and again, we are left with = n + mCm – 1 + n + mCm = n + m + 1Cm = n + m + 1Cn + 1

EBD_7308 JEE MAIN

MT-166

seeing the options

(an -1) / n

75.

ò

I=

(3)

1n

x a-x + x

(an -1) / n

ò

Then, I =

1n

dx

a-x x + a-x

....(i)

ò

2I =

1.dx =

1n

76.

(4)

dx ....(ii)

79.

an - 2 æ an - 2 ö Þ I =ç ÷ n è 2n ø

I=

p /4 3 2 ò (x | x |+ sin x + x tan x + 1) dx ¯ ¯ -p /4 ¯ odd f odd f odd f

I=

p/4 p ò dx = 2 -p / 4

2

1 æ7 ö y1 = ç – 3÷ = . è2 ø 4 æ 7 1ö Hence the point is ç , ÷ . è 2 4ø

ò

[Q f (x)dx = 0, as f (x) is an odd function] -a

(2) (1 – x – 2x2)6 = (1 + x)6 (1 – 2x)6 = 1+ a1x + a2x2 + …….. + a12x12 Putting x = 1/2, we have 0 = 1 + a 1/2 + a2/22 + a3/23 + a4/24 + ……..+ a12/212 .....(1) Putting x = –1/2, we have 1 = 1 – a1/2 + a2/22 – a3/23 + a4/24 – ….. + a12/212 ....(2)

Þ

Þ m 2 (8m 2 + 2) = 1

2 tan A 2

2x 1- x

2

=-

1 y = mx + and y = mx + 8m 2 + 2 m 1 = 8m 2 + 2 m

+

2 tan B

2 tan C

+

1 - tan A 1 - tan B 1 - tan 2 C 2

2

.

2 tan B

81.

+

2y 1- y

2

2

.

2 tan C

1 - tan A 1 - tan B 1 - tan 2 C Put the value of tanA , tanB, tanC, we get

y 2 = 4x &

Comparing

2 tan A

=

x 2 y2 + =1 8 2 Equation of tangent to above curves are respectively.

(2)

(4) x = tanA, y = tanB, – z = tanC. Then (x + y – z) = –xyz. Þ tanA + tanB + tanC = tanA tanB tanC Þ A + B + C = p Þ 2A + 2B = 2p – 2C Þ tan(2A + 2B) = tan(2p – 2C) = – tan2C Þ tan2A + tan2B + tan2C = tan2A.tan2B.tan2C

Þ

Adding (1) and (2), we have 1 = 2 (1 + a2/22 + a4/24 + …..+ a12/212) Þ a2/22 + a4/24 + a6/26 …..+ a12/212 = –1/2 78.

1 x ± 2 Þ 2y = ± x ± 4 2 i.e. 2y = x + 4 & x + 2y + 4 = 0 (2) Let the point be ( x1, y1). Therefore y1 = (x1 – 3)2 ...(i) \ Now slope of the tangent at ( x1, y1) is 2(x1 – 3), but it is equal to 1.

Therefore, 2(x1 – 3) = 1 Þ x1 = 7 2

80.

a

77.

1 satisfy the equation 2

Þy=±

Adding (i) and (ii), we get (an -1)/n

m=±

-

2z 1- z 2

8xyz 2

(1 - x )(1 - y 2 )(1 - z 2 )

(4.00) Set A = {a1, a2, ....., a20} has 20 distinct elements. We have to select 5-element subset. \ Number of 5-element subsets = 20C5

Solutions-Mock Test-4

MT-167

Solving (1) and (2), we get C is (16, 7)

According to question

(

)

Area ABCD = 2 (Area of D ABC)

20 C = 19 C .k 5 4

82.

Þ

æ 19! ö 20! = k. ç 5! 15! è 4! 15!÷ø

Þ

20 = k Þk=4 5

5 +1 4

(cos

æ 5 + 1ö =ç ÷ è 4 ø

2

60° - sin 2 18°

Since, system is consistent and has infinitely many solutions \ (adj. A) B = 0

)

æ 3a - 25 15 - 2a 1 ö æ 6 ö æ 0 ö a - 6 -2 ÷ ç 9 ÷ = ç 0 ÷ Þ ç 10 - a ç -1 -1 1 ÷ø çè b ÷ø çè 0 ÷ø è

2 é æ ö ù ê1 - 5 -1 ú ê 4 çè 4 ÷ø ú ë û

(

)

=

é 5 +1 ê 116 ê êë

=

5 +1 é 4 - 6 + 2 5 ù 1 ê ú = = 0.13 16 ë 4 û 8

2ù 5 -1 ú ú 4 úû

86.

B(I – A) = (I + A + A2 + . . . + Ak – 1) (I – A) = I – A + A – A2 + A2 – A3 +. . . –Ak – 1 + Ak – 1– Ak = I – Ak = I, since Ak = 0 Þ B = (I – A)–1

cos(cos -1 x + sin -1 x + sin -1 x)

Hence (I – A)–1 = I + A + A2 + . . . + Ak–1 Thus

1 æp ö = cos ç + sin -1 x ÷ = - sin(sin -1 x) = - x = 5 è2 ø

84.

(75.00)

[Using cos

-1

x + sin

-1

p x= ] 2

|p| = |–1| = 1 87.

(1.00) As x ®

1 ; {x + 1} ® {1 + 1 / 3} ® 1 / 3 3

Similarly {x + 2} ®

B (1, 7) (1, 2) A

Þ – 6 – 9 + b = 0 Þ b = 15 and 6(10 – a) + 9(a – 6) – 2(b) = 0 Þ 60 – 6a + 9a – 54 – 30 = 0 Þ 3a = 24 Þ a = 8 Hence, a + b = 8 + 15 = 23. (1.00) Let B = I + A + A2 + . . . + Ak – 1 Now multiply both sides by ( I – A ), we get

(0.20) The given expression is equal to

1 = = 0.20 5

AB × BC = 5 × 15 = 75 units (23.00) Given system of equations can be written in matrix form as AX = B where æ1 2 3ö æ6ö A = ç 1 3 5 ÷ and B = ç 9 ÷ ç2 5 a÷ çb÷ è ø è ø

(0.13) cos36°cos42°cos78° = cos36°cos(60° – 18°)cos(60° + 18°) =

83.

85.

1 = 2× AB × BC 2

C

D (4, – 2) Here, centre is A (1,2), and Tangent at B (1,7) is x.1 + y.7 – 1 (x + 1) – 2 (y + 7) – 20 = 0 or y = 7 ...(1) Tangent at D (4,–2) is 3x – 4y – 20 = 0 ...(2)

Þ

88.

lim f (x) x ®1/ 3

1 1 as x ® 3 3

= lim

x ®1/ 3

x - 1/ 3 =1 x - 1/ 3

(3.50) Given differential equation is ( 2 + sin x ) . dy = cos x (1 + y ) dx which can be rewritten as

dy cos x dx = 1 + y 2 + sin x

EBD_7308 JEE MAIN

MT-168

where a is a constant. Q f(0) = 5 Þ 0 + a = 5 Þ a = 5

Integrate both the sides, we get

ò

cos x dx dy =ò 1+ y 2 + sin x

Þ log (1 + y) = log (2 + sin x) + log C Þ 1 + y = C (2 + sin x) Given y(0) = 2 3 Þ 1 + 2 = C[2 + sin 0] Þ C = 2 æ pö Now, y çè ÷ø can be found as 2 1+y= Þ y=

3æ pö 9 ç 2 + sin ÷ Þ 1 + y = 2è 2ø 2

7 2

æ pö 7 Hence, y ç ÷ = è 2ø 2 89.

(5.00) Differentiating w.r.t.x, we get 1 1 = 2f (x )f ¢( x) f ( x ) (f ( x )) 2

1 1 Þ f '(x) = Þ f (x) = x + a 2 2

1 a

90.

b

(2.25) Given, in DABC 1 c a = 0 1 b c 1(c2 – ab) – a(c – a) + b(b – c) = 0 a2 + b2 + c2 – ab – bc – ca = 0 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 0 (a2 + b2 – 2ab) + (b2 + c2 – 2bc) + (c2 + a2 – 2ca) = 0 Þ (a – b)2 + (b – c)2 + (c – a)2 = 0 Here, sum of squares of three members can be zero if and only if a = b = c Þ DABC is equilateral. Þ Þ Þ Þ

Þ ÐA = ÐB = ÐC = 60°

\ sin 2 A + sin 2 B + sin 2 C = (sin 2 60° + sin 2 60° + sin 2 60°) 2

æ 3ö 9 = = 2.25 = 3´ç ç 2 ÷÷ 4 è ø

Mock Test-5 ANSWER KEY 1

(3)

16

(4)

31

(2)

46

(2)

61

(3)

76

(3)

2

(2)

17

(1)

32

(1)

47

(3)

62

(3)

77

(4)

3

(1)

18

(1)

33

(4)

48

(3)

63

(1)

78

(3)

4

(3)

19

(4)

34

(2)

49

(2)

64

(1)

79

(4)

5

(3)

20

(4)

35

(2)

50

(1)

65

(1)

80

(1)

6

(3)

21

(4)

36

(4)

51

(19.55)

66

(4)

81

(3.00)

7

(1)

22

(3)

37

(3)

52

(8)

67

(3)

82

(2.00)

8

(2)

23

(20)

38

(1)

53

(128)

68

(3)

83

(0.60)

9

(1)

24

(7)

39

(2)

54

(0.32)

69

(4)

84

(1.00)

10

(3)

25

(9)

40

(4)

55

(0.4)

70

(3)

85

(17.67)

11

(2)

26

(6.5)

41

(2)

56

(68.13)

71

(4)

86

(325.00)

12

(3)

27

(5)

42

(1)

57

(5.19)

72

(2)

87

(0.50)

13

(4)

28

(8.66)

43

(3)

58

(4.0)

73

(3)

88

(1.50)

14

(2)

29

(2)

44

(2)

59

(4.0)

74

(1)

89

(2.00)

15

(2)

30

(1.25)

45

(2)

60

(100)

75

(3)

90

(6.00)

Solutions PHYSICS 1.

(3) For the man standing in the lift, the acceleration of the ball r r r abm = ab - am Þ abm = g – a

\ Total charge in the spherical region from centre to r (r < R ) is r

æ5 xö q = ò dq = 4 pr0 ò ç - ÷ x 2 dx è4 Rø 0

Where 'a' is the acceleration of the mass (because the acceleration of the lift is 'a' ) For the man standing on the ground, the acceleration of the ball r r r abm = ab - am Þ abm = g – 0 = g 2.

3.

(2) For a diamagnetic material, the value of µr is less than one. For any material, the value of Îr is always greater than 1. (1) Let us consider a spherical shell of radius x and thickness dx. Charge on this shell

æ5 x ö 2 dq = r.4px 2 dx = r0 ç 4 - R ÷ .4px dx è ø

x

dx

é 5 r3 1 r 4 ù 3æ5 r ö = 4pr0 ê 4 . 3 - R . 4 ú = pr0r ç - ÷ è3 Rø ëê ûú \ Electric field at r,

E=

1 q . 2 4p Î0 r

=

1 pr0 r 3 æ 5 r ö . ç - ÷ 4pÎ0 r 2 è 3 R ø

8.

r0 r æ 5 r ö = 4Î ç 3 - R ÷ ø 0è

4.

(3)

Mg = P0 A P0V0 g = PV g Mg = P0A

P0 Ax0 g = PA( x0 - x ) g

P=

9.

g P0 x0

( x0 - x )g

(2) Q The E.M. wave are transverse in nature i.e., r r k ´E r =H = …(i) mw r r where H = B m r r r k ´H … (ii) and = -E r rwe r r k is ^ H and k is also ^ to E r r r r r or In other words X || E and k || E ´ B (1) For conservation of momentum, we have, m1v1 = (m1 + m2) v or v = m1/(m1 + m2) v1 Now the loss of energy =

Piston

\ x

x0

Let piston is displaced by distance x

= 1 – [(m1+m2) / m1] ´ ( v2/v12) = 1 – [(m1+m2)/m1]×[ m12 v12/[(m1+m2)2] × (1/v12) = 1 – [ m1/[(m1+m2)] = m2 /(m1+m2)

P0 x0g

æ ö Mg - ç ÷ A = Frestoring g çè ( x - x) ÷ø 0

10

p pö æ E = 100 sin ç wt + + ÷ 2 3ø è

gP0 Ax x0

5.

6. 7.

1 2p

gP0 A 1 x0 M = 2p

I = I0 sin ( wt + f ) ; I = 4 sin wt ;

\ Frequency with which piston executes SHM. f =

(3)

E = E0 sin ( wt + f) ;

[ x0 >>> x ] F=-

Fraction of energy lost 1 1 m1v12 - (m1 + m 2 )v 2 2 2 = 1 m1v12 2

Cylinder containing ideal gas

æ ö x0g P0 A ç1 ÷ = Frestoring çè ( x - x )g ÷ø 0

1 1 m1v12 – (m1+ m2)v2 2 2

gP0 A2 MV0

(3) Inside the sphere, V is constant and is equal to that on the surface of the sphere. Outside the sphere it comes out to be, V = –Gm/r i.e. |V| µ 1/ r. Hence the graph (c) is correct. (3) (1) Rate of cooling µ surface area and for a given mass, surface area of sphere is minimum.

Phase diff. between I and E = 11.

(2) Mutual inductance = [Henry] =

12.

[ MT -1Q -1 L2 ] -1

p p 5p + = 3 2 6

f BA = I I

= ML2 Q -2

[QT ] (3) Magnetic field due to current in wire 1 at point P distant r from the wire is m i B = 0 1 [ cos q + cos q ] 4p r

EBD_7308

JEE MAIN

MT-170

Solutions-Mock Test -5

MT-171

(v = 5gl to complete motion along vertical circle)

q i1

2 m1v1 2m1v1 = mv Þ v = 3 m 3

i2 P dl

r

or

v1 =

q

15. m 0 i1 cos q (directed perpendicular to the B= 2p r plane of paper, inwards) The force exerted due to this magnetic field on current element i2 dl is dF = i2 dl B sin 90° \ dF = i2 dl

13.

16.

é m 0 i1 cos q ù m0 ê 2 p r ú = 2 pr i1 i2 dl cos q ë û (4) Let a1 = a, I1 = a12 = a2 a2 = 2a, I2 = a22 = 4a2 I2 = 4I1 Ir = a12 + a22 + 2a1a2 cos f

= I1 + I 2 + 2 I1 I2 cos f

Ir = 5I1 + 4 I1 cos f 2

… (1) 2

I Imax = 9I1 Þ I1 = max 9 Substituting in equation (1) 5 Imax 4 I max + cos f 9 9 I = max [ 5 + 4 cos f] 9 Imax é f ù = 5 + 8 cos2 - 4 ú ê 9 ë 2 û

Imax é fù 1 + 8 cos 2 ú ê 9 ë 2û (2) Using conservation of momentum, v m1v1 = m1 1 + mv 3 m v Þ m1v1 - 1 1 = mv 3 Ir =

14.

×

×

×

×

×

×

×

×

× l

×

×

×

×

×

×

×

×

×

V

×

dA d (l ´ x ) = -B dt dt

dx = - Blv dt (1) Here, fc = 1.5 MHz = 1500 kHz, fm = 10 kHz

\ e = - Bl

17.

\ Low side band frequency = fc – fm = 1500 kHz – 10 kHz = 1490 kHz

Ir =

Ir

×

\ e = –B

Now, I max = ( a1 + a2 ) = ( a + 2a ) = 9 a 2

Ir

(2) The p-n junction diode is a half wave rectifier which produces output in forward biased mode only. Thus, there will be no output corresponding to –5V input. Hence, output will be obtained corresponding to +5V only. (4) The induced emf is r r -d f d ( B. A) - d ( BA cos 0º ) e= == dt dt dt

X

Ir = I1 + 4 I1 + 2 4 I12 cos f Þ

3 m 5gl 2 m1

Upper side band frequency = fc + fm = 1500 kHz + 10 kHz = 1510 kHz 18. 19.

20.

(1) A + d = i + i ' Þ i ' = A + d – i = 30 + 30 – 60 = 0° (4) As l is increased, there will be a value of l above which photoelectrons will be cease to come out so photocurrent will become zero. (4) K.E. gained by charged particle of charge q when accelerated under a pot. diff. V will be Ek = qV; For a given V, E µ q. For proton, deutron and a-particle, the ratio of charges is 1 : 1 : 2.

(4) 6W and 2W are in parallel combination

C'1 =

B

2W

6W

C1

1.5W 6V

3W

6V

3W

1.5W

3W

Hence, R AB = V = IR Þ

6´2 = 1.5W 6+2

B 12V

Charge on C'1 = 9mC;

3´3 9 = = 1.5W] 3+ 3 6

(3)

23.

and ycm = 0 (20) Number of undecayed atom after time t2 ; N0 = N 0 e -lt2 3

4 ×12 = 16mC 3 Hence, charge on C1 and C3 is 9mC, as both are in series combination. (6.5) Heat is extracted from the source in path DA and AB is

Charge on C'2 =

1 ´ 0 + 1´ PQ + 1´ PR PQ + PR = 1+ 1+1 3

22.

26.

3 æ p v ö 5 æ 2p v ö DQ = R ç 0 0 ÷ + R ç 0 0 ÷ 2 è R ø 2 è R ø

...(i)

Number of undecayed atom after time t1; 2N0 = N 0 e -lt1 3

From (i), e

-lt2

æ 1ö Þ–lt2 = loge çè ÷ø 3

25.

...(iv)

1 = 12 – (1000 ´ 10- 6 ) ´103 ´10 = 12 - 5 = 7 N 2 (9) As S2 is open, hence C1 & C3 are in series, also C2 & C4 are in series combination. 3 3 ×V = × 12 = 9mC; 4 4

E=

13.6Z 2

n2

; According to question,

æ 5ö Þ 13.6Z2 ç ÷ = 47.2 è 36 ø

Solving (iii) and (iv), we get t2 – t1 = 20 min (7) Apparent wt. = Real wt.–Upthrust

Q=

(5)

13.6Z2 13.6Z2 = 47.2 4 9

2 3

æ 2ö

24.

27.

...(iii)

Þ–lt1 = loge çè 3 ÷ø

æ 13 ö 3 5 p0v0 + 2 p0v0 = ç ÷ p0v0 è 2ø 2 2

Þ

...(ii)

1 = 3

From (ii) – e -lt1 =

2´ 4 4 = F 2+ 4 3

C2

6 = I Þ I = 4A 1.5 [Q R eq =

xcm =

C'2 =

3W

1.5W

28.

Þ Z 2 = 25 Þ Z = 5 (8.66) From the figure it is clear that range is required

R=

u 2 sin 2q (10)2 sin(2 ´ 30°) = =5 3 g 10 u 30° Range R

10m

A 6V

1´ 3 = 3 / 4F 1+ 3

Tower

10m

21.

EBD_7308

JEE MAIN

MT-172

Solutions-Mock Test -5 29.

(2) Here, Imax = I1 + I 2 + 2 I1 I2 cos 0º

\ 30.

MT-173

38.

(1)

Imax + Imin = 2(I1 + I2 )

[NiCl 4 ]2- ; Ni = +2 ; [Co Cl 4 ]2- ; Co = + 2

-1

(1.25) Yc ´ ( DLc / Lc ) = Ys ´ (DLs / Ls )

-1

39.

(2)

(Black)

HgCl 2 + 2NH 4 OH ¾¾ ® Hg(NH 2 )Cl + (B)

36.

DS increases.

(4) Reaction between diborane and alkene are carried out in dry ether under an atmosphere of N2 because B2H6 and the products are very reactive. The products further treated with alkaline H2O2 to convert into alcohols. alkaline H 2O 2

B2 H 6 + 6C2 H 4 ¾¾ ® B(C2 H 4 )3 ¾¾¾¾ ® reactive

37.

(White)

NH 4Cl + H 2 O

(2) Correct order of increasing basic strength is NH3 > PH3 > AsH3 > SbH3 > BiH3 (1) Aluminium has greater affinity for oxygen and the reaction is highly exothermic. (4) Smaller the size and higher the charge more will be polarising power of cation. Since the order of the size of cation is K + > Ca 2 + > Mg 2+ > Be 2 + . So the correct order of polarising power is : K+ < Ca2+ < Mg2+ < Be2+ (2) The more the reduction potential, the more is the deposition of metals at cathode. Cation having E° value less than – 0.83 V (reduction potential of H2O) will not deposit from aqueous solution. (2) DH = nCpDT = 0 means DH constant. æv ö DS = nR ln ç 2 ÷ ³ 0 è v1 ø

Hg 2Cl 2 + 2NH 4 OH ¾¾ ®

[H 2 N - Hg - Cl + Hg] ¯ + NH 4 Cl + 2 H 2 O

CHEMISTRY

35.

-1

(A)

= DLc + DLs = 1 mm + 0.25 mm = 1.25 mm

34.

-1

[Fe(CN)6 ]3- Fe = +3 ; [Cu(CN) 4 ]2 - Cu = +2

0.5 ´10-3 = 0.25 mm \ DLs = 2 Therefore, total extension of the composite wire

33.

2-

(Highest oxidation state)

æ 1´10-3 ö æ DL ö ÷ = 2 ´ 1011 ´ ç s ÷ ç 1 ÷ è 0.5 ø è ø

32.

-1

Imin = I1 + I2 + 2 I1 I 2 cos180º

Þ 1´1011 ´ ç

31.

2-

Cr O 2 Cl2 ; Cr = +6; MnO 2 ; Mn = +4

3CH3CH 2OH + H3BO3 (3) Ions I and IV are the same (trans), with mirror plane through en groups.

HgCl2 + 2KI ¾¾ ® HgI 2 ¯ +2KCl (Red)

HgI 2 + 2KI(excess) ¾¾ ® K 2 (Hg I 4 )

40. 41.

(4) In option (4) oxidation number changes from + 4 in NO2 to + 3 in HNO2 and + 5 in HNO3. (2) In neutral and alkaline medium MnO 4- + 2H 2 O + 3e - ¾¾ ® MnO 2 + 4OH -

In acidic medium: MnO 4- + 8H + + 5e - ¾¾ ® Mn 2 + + 4H 2 O

42.

(1)

O

3 CH 2 = CH – CH2 – CH = CH 2¾¾®

HOOC – CH 2 – COOH + 2HCOOH Zn / H2O

2HCHO + OHC – CH2 – CHO ¬¾¾¾¾ Reduction

43.

(3) The compound is diethyl ether (CH3CH2)2O which is resistant to nucleophilic attack by hydroxyl ion due to absence of double or triple bond, whereas all other compounds given are unsaturated. O C2 H5 OC2 H5 Ether

||

CH3 - C - OCH3 Methyl acetate

O CH3 - C º N Acetonitrile

||

CH3 - C - NH 2 Acetamide

44.

(2) 2NaI + 2 NaNO 2 + 4 CH 3COOH ¾ ¾®

O

C º CH

I 2 + 2NO + 4 CH 3COONa + 2 H 2 O

45.

The colour of CCl4 layer turns purple due to liberated I2. (2) Enantiomers of C4H10O are |

|

O

H3 CH 2 C - C - CH3 H3C - C - CH 2CH3 |

|

H

H OH

H C = CHOH

H SO , Hg

O

H C = CHOH

H SO , Hg

|

2+

Due to electronegative nature of > C = O, p electrons are transferred toward benzene ring

HO

OH

H2SO4, Hg

CH2–CHO

H SO

2 4 CH3 CH 2- CH - CH3 ¾¾¾¾ ®

[B]

(A)

CH 3 CH = CH - CH 3 + CH 2 = CHCH 2 CH 3

48.

Minor ( C )

Major ( B )

(3) Sandmeyer’s reaction NH2

HBr / Peroxide

CH 3CH = CH – CH3 ¾ ¾ ¾ ¾ ¾ ®

NaNO

2 ¾¾¾®

(B)

HCl

Br alkaline | hydrolysis CH3 – CH – CH2 – CH3 ¾ ¾ ¾ ¾ ® OH | CH3 – CH – CH2 – CH3 (A)

NH2+Cl–

Cl CuCl ¾¾®

49.

(2) C6H5ONa + CH2 = CH – CH2 – Br ¾¾ ® C6H5 – O – CH2 – CH = CH2 + NaBr

50.

(1) Alkyl or Aryl cyanide react with grignard reagent to form ketones CºN

HBr / Peroxide

CH2 = CHCH 2CH 3 ¾¾¾¾¾®

Allyl phenyl ether

(C)

alkaline hydrolysis

BrCH 2 – CH 2 – CH 2 – CH 3 ¾ ¾ ¾ ¾ ®

OH – CH2 – CH2CH 2CH3 Isomer of (A)

46.

(2) In addition homopolymers such as teflon, empirical formula resembles with monomer. C º CH

47.

CH CH

(3)

2+

H2SO4, Hg

O

OH H SO , Hg H SO , Hg

C – CH3

C = CH2 [A]

Ether + C6H 5MgBr ¾¾®

O C H3O+ C6H5 – C = NMgBr ¾¾® | C6H5 or C6H5 – C = O + MgBrNH2 | C6H5 51. (19.55) Energy of photon corresponding to second line of Balmer series for Li2+ ion

EBD_7308

JEE MAIN

MT-174

Solutions-Mock Test -5

MT-175

1ù 2é 1 = (13.6 ) ´ ( 3 ) ê 2 - 2 ú ë2 4 û

56.

27 16 Energy needed to eject electron from n = 2 level in H-atom;

= 13.6 ´

1 ù 13.6 2 é 1 = 13.6 ´1 ´ ê 2 - 2 ú Þ 4 ë2 ¥ û K.E. of ejected electron

9 ´ 3 13.6 æ 27 - 4 ö = 13.6 ´ ç ÷ 16 4 è 16 ø Þ 19.55 eV

= 13.6 ´

52.

53.

(8)

O ||

H - O - S- O - H ; || O

57.

6s & 2p

d=

(128) For fcc, 2a a = = 0.3535a 4 2 2 given a = 361 pm r = 0.3535 × 361 = 128 pm (0.32) Cell reaction :

58.

r=

54.

0.0591 [Reduced state] log n [Oxidised state]

0.421 = E 0 -

0.0591 0.001 log 3 (0.01)3

E0 = 0.48 E

0

= E 0Ag + / Ag

0 - EM 3+ /M

E 0M 3+ / M = 0.8V - 0.48V = 0.32 volt

55.

(0.4) 6.4 g of CaC2 º 0.1 mol of CaC2 0.1 mol contains 0.1 NA molecules of CaC2 Q 1 molecules have 4p electron \ 0.1 NA molecule will have 0.1 NA × 4 = 0.4 NA

n´M 3

a ´ NA

=

2 ´ 100 -8

(4 ´ 10 cm)3 ´ 6.02 ´ 1023

= 200/38.528 = 5.19 g/cc (4.0) Flask I: 1M

t1/2 = 4 hr

\ t1/2 = 4 hr.

0.5 M

t1/2 = 4 hr

0.25 M

t = 8 hr

Flask II: ® 0.3 M 0.6 M ¾¾¾¾¾

® M 3+ + 3Ag (s) M + 3Ag + (aq) ¾¾

E = E0 -

(68.13) Let atomic masses of A and B be a and b amu respectively \ Molar mass of AB2 = (a + 2b) g mol–1 and Molar mass of AB4 = (a + 4b) g mol–1 For compound AB2 DTb = Kb´ WB ´ 1000/ WA ´ MB 2.3 = 5.1 ´ 1 ´ 1000/ 20.0 ´ (a + 2b)....I For compound AB4 1.3 = 5.1 ´ 1 ´ 1000/ 20.0 ´ (a + 4b)....II Solving (I) and (II), a = 25.49 b = 42.64 Sum of atomic masses = 25.49 + 42.64 = 68.13 (5.19) For bcc lattice, number of atoms per unit cell = 2 Now

t1/2 = 4 hr

A first order reaction is independent of the initial concentration of the reactant. 59.

(4.0) pOH = pK b + log [salt ] [base]

= - log10-10 + log1 = 10 ; pH = 14 - 10 = 4

60.

(100)

C (graphite) ® C (diamond) DH = DU + P . DV 12 Vm ( diamond ) = mL 3 12 Vm ( graphite ) = mL 2 DH – DU = (500 × 103 × 105 N/m2) æ 12 12 ö -6 ç - ÷ ´ 10 = – 100 kJ/mol è 3 2ø DU – DH = + 100 kJ/mol

MATHEMATICS 61.

h=

y (3) (1) x2y2 = 16a4 ; LST = x Þ xy = 4a2 T

y + xy' = 0 Þ y ¢ =

-y ; x

y = x Þ LST = 2a y/x \ (1) is true. LST =

(3)

æ x 2 + 1ö (x 4 + 3x 2 + 1) tan -1 ç ÷ è x ø (Dividing Num. and Den. by x2) 1ö æ çè1 - 2 ÷ø dx x I=ò 1ö 1ö æ 2 -1 æ çè x + 3 + 2 ÷ø tan çè x + x ÷ø x dt =ò t 2 + 1 tan -1 t

(

)

æ 1 1ö ö æ çè where t = x + x Þ dt = èç1 - 2 ø÷ dx ÷ø x æ x 2 + 1ö = log | tan -1 t | + C = log tan -1 ç ÷ +C è x ø

x +1 x (1) Clearly P is (a cos q, b sin q) and Q is

Þ f (x) =

63.

2

(-a sin q, b sin q) so the mid point (h, k) of PQ will be given by

4k 2 2

=2Þ

h2

k2

2

+

2

=

Þ (x 2 - 4a 2 )(x 2 + 8a 2 ) = 0 Þ x = ± 2a A(0, 2a)

(–2a, a)

\

(2a, a)

O

Required

é 2a 8a 3 ù 2a x 2 area = 2 ê ò dx dx ò 0 4a úú 2 2 êë 0 x + 4a û

(x 2 - 1) dx

I =ò

2

+

x2 8a 3 = Þ x 4 + 4a 2 x 2 - 32a 4 = 0 4a x 2 + 4a 2

dy x = = 1 i.e. x = 2 dx 2 \ (2, 1) is the point on the curve x2 = 4y at which the normal is : y – 1 = – 1 (x – 2) i.e. x + y = 3 \ (b) is true (3) y = – 4x2, y = e–x/2 The curves are non-intersecting \ curves are not orthogonal i.e. (3) is false. 2 (4) y = x3 – 2ax2 +2x+ 5 3 dy = 2x2 – 4ax + 2 = 2 (x – a)2 + 2 – 2a2 > 0 dx [Q a Î (–1, 0) Þ 0 < a2 < 1] \ (4) is true

(2)

62.

4h 2

1 2 a b a b (1) The curve of y (x2 + 4a2) = 8a3 is symmetrical about y-axis and cuts it at A (0, 2a). Tangent at A is parallel to x-axis. x-axis is asymptote. This curve meets x2 = 4ay Where,

\

64.

a cos q - a sin q b sin q + b cos q and k = 2 2

a2 (6p - 4). 3 r rr r r (1) p r + ( r . b ) a = c ...(1) rr rr rr rr p( r .b) + ( r.b)(a.b) = c.b rr r r c.b rr Þ r.b = , since a.b = 0 , putting in (1), p rr r r rr r r c (c.b) r c b.c a a = – Þr= p p2 p p2 (4) For f (x) to be defined, we must have =

65.

( )

66.

x - 1 - x 2 ³ 0 or \

x ³ 1 - x2 > 0

2 x2 ³ 1 – x2 or x ³

Also, 1 – x2 ³ 0 or x2 £ 1.

1 . 2

1 ö æ 1 ö æ 1 Þ çè x ÷ø çè x + ÷ ³0 2 2ø 2 1 1 Þ x£or x ³ 2 2 Also, x2 £ 1 Þ (x – 1) ( x + 1) £ 0 Þ –1 £ x £ 1 2 Now, x ³

EBD_7308

JEE MAIN

MT-176

Solutions-Mock Test -5 2 Thus, x > 0, x ³

67.

68.

MT-177

2 éæ x ö 3x 2 ù Þ ( x 1 - x 2 ) êç x 1 + 2 ÷ + 2 ú = 0 . 2 ø 4 ú êëè û Þ x1 – x2 = 0, because the other factor is nonzero. Þ x1 = x2 \ f is one-one. f is onto. Let k ÎR any real number.

1 and x2 £ 1 2

é 1 ù ,1ú ÞxÎê ë 2 û (3) A = {1, 2, 3}, B = {4, 5} Then A Ç B = f A × B = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)} B × A = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)} (A × B) Ç (B × A) = f

(3)

1/ 3

æk+7ö f ( x ) = k Þ 4x 3 - 7 = k Þ x = ç ÷ è 4 ø

æ k + 7ö Now ç è 4 ÷ø

2

f ( x) = x [ x]

–1

Y

O

1

1/ 3

Î R, because k Î R and

é æ k + 7 ö 1/ 3 ù f êç ÷ ú=4 êë è 4 ø úû æ k + 7ö =4ç -7 = k è 4 ÷ø

X

y = –x2

3

é æ k + 7 ö 1/ 3 ù êç ÷ ú –7 êë è 4 ø úû

1/ 3

æk+7ö ÷ \ k is the image of ç è 4 ø \ f is onto. \ f is a bijective function.

ìï- x 2 , -1 < x < 0 Þ f ( x) = í 0 £ x 1 Þ x2 > 1, then 1 –x2 < 0 2

b b2 1 + b3 c

If | x | < 1 Þ x < 1, then 1–x > 0 2

(3)

a2 1+ a3

= (1 + abc)(a - b)(b - c)(c - a ) = 0 ....(1)

1

2

78.

2 2

(1 + x )

2(1 - x 2 )

dy 2 x +1 log 2 2 x log 2 = dx 1 + 2 2 ( x +1) 1 + 2 2 x

æ2 1ö æ 1ö æ dy ö = (log 2) ç - ÷ = (log 2) ç - ÷ \ç ÷ è5 2ø è 10 ø è dx ø x = 0

æ 2x ö y = sin -1 çç ÷÷ è 1+ x2 ø

(3)

é 2 x +1 - 2 x ù é 2 x (2 - 1) ù tan -1 ê ú ú = x x +1 x x +1 êë1 + 2 .2 úû ëê1 + 2 .2 ûú

(4) y = tan -1 ê

79.

1 c c2 coplanar = (a - b)(b - c)(c - a ) ¹ 0 ...(2) From (1) & (2), abc = –1 (4) A and B will agree in a certain statement if both speak truth or both tell a lie. We define following events

E1 = A and B both speak truth Þ P(E1) = xy E2 = A and B both tell a lie Þ P (E2) = (1 – x ) (1 – y) E = A and B agree in a certain statement

Clearly, P(E / E1 ) = 1 and P(E / E 2 ) = 1 The required probability is P(E1 / E). Using Baye’s theorem P ( E1 / E )

=

P(E1 )P(E / E1 ) P(E1 )P(E / E1 ) + P(E 2 )P(E / E 2 )

=

xy xy.1 = xy.1 + (1 – x )(1 – y).1 1 – x – y + 2 xy

EBD_7308

JEE MAIN

MT-178

Solutions-Mock Test -5 80.

81.

MT-179

(1) | x1z1 – y1z2 |2 + | y1z1 – x1z2 |2 = | x1z1 |2 + | y1z2|2 – 2Re(x1y1z1z2) + | y1z1 |2 + | x1z2 |2 + 2Re (x1y1z1z2) = x12 | z1 |2 + y12 | z2 |2 + y12 | z1 |2 + x12 | z2 |2 = x12 | z1 |2 + y12 | z2 |2 + y12 | z1 |2 + x12 | z2 |2 = 2(x12 + y12) (42) = 32(x12 + y12) (3.00)

m( x) =

2

84.

1 , which is discontinous at x = 1 x -1 1

f (u) =

=

2

1 , (u + 2) (u - 1)

u +u -2 which is discontinous at u = – 2, 1

which Þ

1 1 when u = – 2, then = -2 Þ x = x -1 2 1 =1 Þ x = 2 when u = 1, then x -1 Hence given composite function is discontinous 1 at three points, x = 1, and 2. 2

82.

Þ

Þ f ¢( x ) = 2Pe 2 x + Qe x + R

Þ 8P + 2Q + R = 31

......... (i)

Also, 0 = P + Q

......... (ii)

& Þ

log 4

ò0

log 4

ò0

(Pe

(f (x) - Rx)dx =

2x

39 2

=

P 2 log 4 P 39 e + Q elog 4 - - Q = 2 2 2 ......... (iii) Þ 15P + 6Q = 39 Solving (i), (ii) and (iii), we get P = 5, Q = – 6, R = 3 Þ P + Q + R = 5 – 6 + 3 = 2. (0.60) Given, x1 + x2 + .... + x10 = 12 2 and x12 + x22 + .... + x10 = 18

(p + 2)3

p 1

(p + 1) 1

(p + 2) = 0 1

p3

(p + 1)3 - p3

(p + 2)3 - p 3

p

1

2

1

0

0

=0

Þ 2 (p3 +1 + 3p2 + 3p) – 2p3 – (p3 + 8 + 12p + 6p2) + p3 = 0 Þ 2p3 + 2 + 6p2 + 6p – 2p3– p3 – 8 – 12p – 6p2 + p3 = 0 Þ – 6 – 6p = 0 Þ | p | = |– 1| = 1. (17.67) Let f (x) = 3x10 – 7x8 + 5x6 – 21x3 + 3x2 – 7 f ¢ (x) = 30x9 – 56x7 + 30x5 – 63x2 + 6x f ¢ (1) = 30 – 56 + 30 – 63 + 6 = 66 – 63 – 56 = – 53 a®0

f (1 – a ) – f (1) a3 +3a

f ¢ (1– a )(–1)–0 a®0 3a 2 +3 (By using L’hospital rule) f ¢(1 – 0)(–1) – f ¢ (1) 53 = = = 3 3 = 17.67 3(0)2 + 3

=

39 2

Þ

(p + 1)3

Consider lim

39 + Qe )dx = 2

éP ù Þ ê e 2x + Qe x ú ë2 û0

83.

85.

x

log 4

p3

Þ 2 (p + 1)3 – 2p3 – (p + 2)3 + p3 = 0

(2.00) f ( x ) = Pe 2 x + Qe x + Rx

Þ 31 = 2Pe 2 log 2 + Qelog 2 + R

1 æ Sx ö s 2 = Sx 2 - ç ÷ n è n ø 2 18 æ 12 ö 9 36 9 = -ç ÷ = - = 10 è 10 ø 5 25 25 3 \ SD = = 0.60 5 (1.00) The given system of equations are : p3x + (p +1)3 y = (p +2)3 ...(1) px + (p +1)y = (p +2) ....(2) x +y = 1 ....(3) This system is consistent, if values of x and y from first two equation satisfy the third equation.

86.

lim

(325.00) F a e

f g c B

b H d

87.

a + e + f + g = 285, b + d + f + g = 195 c + d + e + g = 115, e + g = 45, f + g = 70, d + g = 50 a + b + c + d + e + f + g = 500 – 50 = 450 we obtain a + f = 240, b + d = 125, c + e = 65 a + e = 215, b + f = 145; b + c + d = 165 a + c + e = 255; a + b + f = 335 Solving we get b = 95, c = 40, a = 190 , d = 30, e = 25, f = 50 and g = 20 Desired quantity = a + b + c = 325 (0.50) sin (cot–1(x + 1)) = cos (tan –1x) Þ cos( p / 2 - cot -1 ( x + 1)) = cos(tan -1 x)

p - cot -1 ( x + 1) = 2np ± tan -1 x Put n = 0 2 Þ p / 2 - cot -1 (x + 1) = ± tan -1 x = tan -1 (± x)

Þ

-1

Þ p / 2 = tan (± x ) + cot

88.

-1

( x + 1)

Þ x + 1 = ± x Þ 2x + 1 = 0 ; 1 1 | x | = - = = 0.50 2 2 (1.50) 1 - cos 3 x x ® 0 x sin x cos x

lim

(1 - cos x) (1 + cos x + cos 2 x) = lim x sin x cos x x ®0 æ xö 2sin 2 ç ÷ è 2ø (1 + cos x + cos2 x) = lim ´ cos x æ xö æ xö x ®0 x.2sin ç ÷ cos ç ÷ è 2ø è 2ø

æ xö sin ç ÷ è 2 ø 1 + cos x + cos 2 x = lim ´ æ xö x®0 æ x ö cos ç ÷ cos x 2ç ÷ è 2ø è 2ø

1 3 = ´ 3 = = 1.50 2 2

89.

(2.00) é æ 1 öù f {f [f ( x)]} = f êf ç ÷ú ë è 1 - x øû

æ ç 1 =fç ç1- 1 ç è 1- x

ö ÷ x -1 ö ÷ = f æç ÷ ÷ è x ø ÷ ø

æ \ f (x) is not defined for x = 1; f ç

1 ö ÷ is not è1- x ø

90.

defined for x = 0. \ f {f [f(x)]} is discontinuous at x = 0 and 1 i.e., there are two points of discontinuity. (6.00) We have cos x + cos 2 x + cos 3x = 0 or (cos3x + cos x) + cos 2x = 0 or cos 2x(2cos x + 1) = 0

or cos 2x (2 cos x + 1) = 0 We have, either cos 2x = 0 or 2 cos x + 1 = 0 If cos 2x = 0, then 2x = (2m + 1)

p 2

p or x = (2m + 1) , m Î I 4

If 2 cos x + 1 = 0, then cos x = -

...(i) 1 2p = cos 2 3

2p ...(ii) , nÎI 3 So, the required general solution are \ Values of x in the interval (0 < x < 2p) in equation (i) Put m = 0, 1, 2, 3

Þ

x = 2 np ±

p 3p 5p 7 p So the value is x = , , , 4 4 4 4 2 p 4p , 3 3 Hence, 6 roots of the given equation.

In equation (ii) Put, n = 0, 1 x =

EBD_7308

JEE MAIN

MT-180

Mock Test-6 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

(3) (3) (2) (3) (4) (1) (1) (4) (1) (4) (2) (2) (3) (4) (3)

16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

(4) (4) (4) (2) (4) (250) (4) (4) –3 (2 × 10 ) (1.06) (5451) (0.01) (0) (80) (30)

31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

ANSWER KEY (3) 46 (2) (3) 47 (2) (3) 48 (3) (2) 49 (2) (2) 50 (3) (1) 51 (10.43) (1) 52 (1) (1) 53 (464) (2) 54 (8) (2) 55 (57.5) (2) 56 (105.7) (4) 57 (0.05) (2) 58 (21) (1) 59 (2.57) (3) 60 (2.42)

61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

(1) (2) (1) (1) (3) (4) (2) (4) (4) (3) (2) (1) (1) (2) (4)

76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

(4) (2) (4) (1) (3) (0.14) (10) (6) (3) (80) (1) (0) (0.29) (1) (21.33)

Solutions j +

+ + + dE

dq cos q

q

dl + +

+

1.

1 k = 1 Hz (3) Frequency of spring (f ) = 2p m

+ +

PHYSICS

O 2 Þ 4p =

k m

m = 1 kg

If block of mass m = 1 kg is attached then, k = 4p2 Now, identical springs are attached in parallel with mass m = 8 kg. Hence, keq = 2k F=

2.

1 2p

dE dE sin q q æ ö q dq = ç ÷ dl = (rd q) pr è pr ø

k 8 kg

(3) Let us consider a differential element dl. charge on this element.

(Qdl = rd q)

æqö = ç ÷dq è pø Electric field at O due to dq is dE =

k´2 1 = Hz m 2

i

1 dq 1 q . . dq = 2 4 p Î0 pr 2 4p Î0 r

The component dE cos q will be counter balanced by another element on left portion. Hence resultant field at O is the resultant of the component dE sinq only.

EBD_7308 JEE MAIN

MT-182

\ E = ò dE sin q = =

= =

q

4p2 r 2 Î0 q 4p2 r 2 Î0 q

p

q

ò 4 p2 r 2 Î 0

p

sin qd q

0

[ - cos q]0 (+1 + 1)

2p2 r 2 Î0 The direction of E is towards negative y-axis. r q ˆj \ E=- 2 2 2p r Î0

3.

4.

5.

(2) Let pole strength = m l \ M = m ´ l when l' = 2 ml M M' = m ´ l' = = = 40 units 2 2 é 1 1 ù 1 (3) Wave number = RZ 2 ê - ú 2 l êë n1 n 2 úû 1 Þlµ Z2 By question n = 1 and n 1 = 2 Then, l1 = l2 = 4l3 = 9l4 (4) From conservation of mechanical energy æ 3v2 ö 1 1 æ v2 ö mv 2 + I ç 2 ÷ = mg ç ÷ 2 èR ø 2 è 4g ø after solving I = X=

mR 2 Which is for disc. 2 1

6.

(1)

7.

aw 2 - bw + c becomes minimum. \ Differential above term and equating it to zero. b \ 2aw - b = 0 Þ w = . 2a (T - T ) (200 - T1 ) = (2K)A 1 2 (1) Q = KA L L 1 =(1.5K) A (T2-18) × L 200 – T1= 2T1 – 2T2 = 1.5 T2- 27, Solving T1 = 116ºC and T2= 74ºC

8.

a w 2 - bw + c At resonance X becomes maximum and

(4)

4i1 = 5i 2

5 i2 4 i1 = current through 4 W i2 = current through Ammeter 5 ´ 4 20 = Total R = 9 9 10 ´ 9 9 5 9 I= = Þ i 2 + i 2 = Þ i 2 = 2A 20 2 4 2 (1) When positive terminal connected to A then diode D1 is forward biased, current, 2 I= = 0.4 A 5 When positive terminal connected to B then diode D 2 is forward biased, current, \ i1 =

9.

2 = 0.2 A 10 (4) Mass per unit length of the wire = r Mass of L length, M = rL and since the wire of length L is bent in a form I=

10.

of circular loop therefore 2pR = L Þ R = Moment of inertia of loop about given axis 3 2 = MR 2 2

3 æ Lö 3rL3 rL ç ÷ = 2 è 2p ø 8p2 r r (2) f = B. A ; f = BA cos wt

=

11.

df wBA = wBA sin wt ; i = sin wt dt R 2 æ wBA ö Pinst = i 2 R = ç ´ R sin 2 wt ÷ è R ø

e=-

T

ò Pinst ´ dt

Pavg =

0

T

ò dt

0 T

(wBA) = R

2

ò sin

2

wtdt

0

T

ò dt

=

1 ( wBA) 2 2 R

0

(w B pr 2 ) 2 \ Pavg = 8R

é pr 2 ù êA = ú 2 ûú ëê

L 2p

Solutions-Mock Test-6 12.

MT-183

(2) Let V be volume of the load, and let its relative density be r then FL VrgL = Y= . Al a Al a

At t =

d2x

......(1)

dt 2

when the load is immersed in liquid, the net weight = weight - upthrust \Y =

F' L (Vrg - V ´ l ´ g )L = Al w Al w

......(2)

r r -1 Equating (1) and (2), we get = la lw which gives r = 13.

la la - l w

(3) Given: Amplitude of electric field, E0 = 4 V/m Absolute permitivity, e0 = 8.8 × 10–12 C2/N-m2 Average energy density uE = ? Applying formula, Average energy density uE =

1 e0 E 2 4

1 ´ 8.8 ´ 10 -12 ´ (4) 2 4 = 35.2 × 10–12 J/m3 (4) From the graph it is clear that the amplitude is 1 cm and the time period is 8 second. Therefore the equation for the S.H.M. is

Þ uE =

14.

æ 2p ö æ 2p ö x = a sin ç ÷ ´ t = 1sin ç ÷ t èT ø è 8ø p x = sin t 4 The velocity (v) of the particle at any instant of time ‘t’ is v=

dx d é æ p ö ù p æ pö = êsin ç ÷ t ú = cos ç ÷ t è 4ø dt dt ë è 4 ø û 4

The acceleration of the particle is 2

d x dt

2

2

æ pö æ pö = - ç ÷ sin ç ÷ t è 4ø è 4ø

4 s we get 3 2

p 4 -p2 p æ pö = - ç ÷ sin ´ = sin è 4ø 4 3 16 3

- 3p 2 cm / s 2 32 2 æ Df ö (3) 2I0 = 4I0 cos ç ÷ è 2 ø =

15.

But, Df =

2p Dx l

here, Df = so, Dx =

p 2

l 4

dy l = ....(i) D 4 lD =b ....(ii) d Multiplying equation (i) and (ii) we get, b y= 4 16. (4) Internal energy is a state function hence in a cyclic process change in internal energy is zero. 17. (4) From the law of conservation of momentum we know that, m1u1 + m2u2 + .... = m1v1 + m2v2 + .... Given m1 = m, m2 = 2m and m3 = 3m and u1 = 3u, u2 = 2u and u3 = u ®

Let the velocity when they stick = v Then, according to question, Y 2m, 2u

j sin 60° j m, 3u

P

60° 60° i

X i cos 60° (–j) sin 60° –3m, u

()

m × 3u iˆ + 2m × 2u

( -ˆi cos 60° - ˆjsin 60°) + 3 m × u ® ( -ˆi cos 60° + ˆjsin 60°) = (m + 2m + 3m) v

EBD_7308 JEE MAIN

MT-184

ˆi ˆi æ 3 ˆö Þ 3muiˆ - 4mu - 4mu ç j÷ - 3mu 2 2 è 2 ø ® æ 3 ˆö + 3mu çç j ÷÷ = 6m v è 2 ø ® 3 3 mujˆ = 6m v Þ muiˆ - muiˆ 2 2 ® 1 ˆ 3 ˆ muj = 6m v Þ - mui 2 2 ® u ˆ - i - 3jˆ v= Þ 12 18. (4) Gravitational force provides the necessary centripetal force.

(

\

21.

(250) The magnetic field at a point on the axis of a circular loop at a distance x from centre is, B=

2( x 2 + a 2 )3 / 2 m i B' = 0 2a

\ B' =

)

B.( x 2 + a2 )3/ 2

22.

T

1 and T from the figure (lm)1 < (lm)3 < (lm)2 therefore T1 > T3 > T2.

23.

24.

(4) Here, R = 4 kW = 4 × 103 W Vi = 60 V Zener voltage Vz = 10 V RL = 2 kW = 2 × 103 W VZ 10 = = 5 mA RL 2 ´ 103

V - VZ Current through R, I = i R

=

60 - 10

=

25.

u 2 sin 2 q or H µ u 2 2g DH Du = 2. = 2(+2%) = 4% increased H u 1 q (2 × 10–3) Electric Intensity = ´ 4 p Î0 r 2 1 r2E or qmax = 9 9 ´ 10 1 = × 2.5 × 2.5 × 3 × 106 9 ´10 9 or qmax = 2 × 10–3 coulomb

(4)

H=

20 cm

(1.06) u0

v0

ue

50

= 12.5 mA 4 ´ 10 4 ´ 103 Fom circuit diagram, I = IZ + IL Þ 12.5 = IZ + 5 Þ IZ = 12.5 – 5 = 7.5 mA 3

mg

mg

1/ 2

(2) According to Wien’s law l m µ

Load current, IL =

®

20.

a

T

R2 mv 2 \ = mg ( R + x) ( R + x) 2

19.

54(53 ) = 250 µT 3´ 3´ 3

(4) mg – T = ma T 360 \ a = g - = 10 = 4m / s 2 m 60

n! mv 2 æ GM ö R 2 \ = mç è R 2 ÷ø ( R + x )2 r !( n - r ) ! ( R + x)

æ gR 2 ö gR v = \ v2 = Þ ç ÷ R+ x è R + xø

a3

Put x = 4 & a = 3 Þ B ' =

GmM mv 2 GM = also g = ( R + x ) ( R + x )2 R2

2

m 0i a 2

objective

Ve=25 cm

Eye piece

Solutions-Mock Test-6

MT-185

Final image distance L = v0 + ue = 20

The distance of point A

1 1 1 + for objective = f0 v0 u 0 for eye piece

Now, potential at A, VA =

25 95 v0 = 20 - = 6 6 1 1 95 89 \ = 1=1- = u0 v0 6 95

Potential at B, VB =

VA – VB =

1 ì hc ü í - fý eîl þ

=

Put l1 = 3 ´10 -7 m, V1 = 1.85 , l 2 = 4 ´ 10 -7 m, V2 = 0.82 in the above equation respectively hc f = 1.216 ´ 10 -19 J = Þ l 0 = 5451Å l0 Distance travelled on pitch scale (0.01) Pitch = Number of rotation 2mm = = 0.5 mm 4 Least count Pitch = Number of division on circular scale

= 28.

(0)

0.5mm = 0.01 mm 50

O

Q æ 1 1 ö Q æ 1 1ö ç ÷ = ç - ÷ 4p Î0 è OA OB ø 4p Î0 è 2 2 ø =

29.

30.

1 Q 1 Q . . 4p Î0 OA 4p Î0 OB

Q ´ 0 = 0. 4p Î0

(80) 9.25 days is equal to three half-lives for Tl201. The fraction remaining is then : 1/2 × 1/2 × 1/2 = 1/8. Thus 1/8 of 80 mCi remains.

ENGINE

(30)

A

0.9 km C

Let after 5 sec engine at point C t=

AB BC + 330 330

0.9 ´ 1000 BC + 330 330 \ BC = 750 m Distance travelled by engine in 5 sec = 900 m – 750 m = 150 m Therefore velocity of engine

A(Ö2,Ö2)

(0,0) ® r2 B (2,0)

1 Q . 4p Î0 (OB)

5=

Y

® r1

1 Q . 4p Î0 (OA)

\ Potential difference between the points A and B is given by

95 \u0 = - ; 1.06 89

27.

from the

The distance of point B(2, 0) from the origin, uur 2 2 OB = | r2 | = (2) + (0) = 2 units.

25 cm 6 \ v o = 20 - u e

V=

)

ur 2 2 OA = | r1 | = ( 2) + ( 2) = 4 = 2units.

1 1 1 =- + fe 25 u e

\ue =

(5451)

2, 2

origin,

1 1 1 or + = 5 25 u e

26.

(

X

=

150 m = 30 m/s . 5sec

H B I L L

EBD_7308 JEE MAIN

MT-186

37.

CHEMISTRY 31.

(3) The correct order of acidic strength of the given species is : HSO 3 F > H 3O + > HSO 4– > HCO 3– (iv)

32. 33.

(iii)

(ii)

(i)

(i) < (iii) < (ii) < (iv) It corresponds to choice (c) which is correct answer. (3) Solar energy is not responsible for green house effect instead it is a source of energy for the plants and animals. (3) CH3OCH2Cl is a 1° halide but still hydrolysis takes place through SN1 mechanism

thiosulphate

38.

39.

40.

41. O

CH — CO

3 2

36.

(2) Ag2S + 2NaCN

8

9

42.

CO

O Na2S + 2AgCN

¾® Na[ Ag ( CN ) 2 ] AgCN + NaCN ¾ 1 (1) M.P. µ (HCl having more covalent f character) M.P. µ Lattice energy (Ionic compd.) LiF is ionic having highest Lattice energy.

(4)

2

1 2

2

1

HOCl

CH 2 < CH 2 ¾¾ ¾↑ ClCH 2 CH 2 OH aq. NaHCO

1

3 ↑ HOCH CH OH ¾¾¾ ¾¾ 2 2

CH — CH

CO 35.

1

7

4

2C2 H5ONa ∗ H 2 ←

Thus, (A) is ethanol and (B) is sodium ethoxide. (2) Claisen self condensation is given by esters which contain at least one hydrogen on an alpha carbon atom (e.g., CH3COOC2H5). Compound, C6H5COOC2H5 will not give this reaction. All remaining compounds will give this reaction. (2) The step involved is a precipitation step. Increasing the Cl– concentration will reduce the concentration of Ag+ in solution. Remember Ksp = [Ag+ ] [Cl–] and the source of the Cl– is irrelevant. Thus increased concentration of Cl – must result in decreased concentration of Ag+ in order to maintain the solubility product constant. Addition of Ag2SO4 would probably be counterproductive, since additional Ag+ is being added, and probably all will not be recovered. (2) 2F2 + 4KOH ® 4 KF + O 2 + 2H 2 O for 1 mole of F2 the molar ratio. F2 KOH KF O2 H2O

6

5

2C2 H5OH ∗ 2Na ¾¾ ↑

Sod. ethoxide (B) conc. H2SO4 2C2 H5OH ¾¾¾¾¾¾ ® C2 H 5 OC2 H5 - H 2O Diethyl ether

+

+

(1)

Ethanol (A)

+

CH — CO

Sodium argento (soluble complex)

Q x + 2 (–2) = –3 x = +1

because CH 3O - C H 2 is stabilized by resonance C H3O CH 2Cl « CH 3– O – CH2 « + CH3– O = CH2 34. (2) It is an example of Diel’s-Alder reaction in which a dienophile (maleic anhydride, here) reacts with a conjugated diene to form cyclic adduct. Here although structures (i), (iii) and (iv) have conjugated system of double bonds, hence theoretically all the three can undergo Diel’sAlder reaction, but structure (iii) does not undergo this reaction because this will lead to larger ring (9-membered) which is unstable.

(1) AgX + 2 Na2S2O3 ® Na3[Ag(S2O3)2] + NaX

43. 44.

(2) All alkali metals dissolve in liquid ammonia giving deep blue solution. (1) According to van der Waal's equation for one mole of gas æ a ö çççP ∗ 2 ÷÷÷ (V , b) < RT è V ø

at very high pressure P ==

a V2

Solutions-Mock Test-6 So,

a

is negligible. V2 P (V – b) = RT PV – Pb = RT on dividing RT on both sides. Pb compressibility factor.. RT (3) For electrophilic substitution reaction, the order of reactivity among the given compounds is as follows : C6H5OCH3 > benzene > C6H5NO2 (2) KI ∗ AgNO 3 (slight excess) ¾ ¾ ↑ AgI ∗ KNO3 ∗ ,; AgNO3 ¾¾ ↑ Ag ∗ NO3

MT-187

52.

(1) H H

CH3

CH3 Br

H

[ Z < 1∗

45.

46.

47. 48.

49.

50.

51.

AgI(s) ∗ Ag ∗ ¾¾ ↑ [AgI]Ag ∗ (2) Saffron is an artificial edible colour. (3) Hydroxylamine and hydrazine, both do not have carbon, hence NaCN will not be formed in Lassaigne’s extract leading to negative test for nitrogen. (2) Both egg yolk and mustard contain large quantity of sulphur (as its compounds) which react with Ag. 2Ag + S ® Ag2S (Black colour) (3) In (c), sulphate ion is present outside the coordination sphere so it can form white ppt of BaSO4 with BaCl2 (aq).

(10.43) Mass of H2SO4 in 100ml of 93% H2SO4 solution = 93g \ Mass of H2SO4 in 1000 ml of the H2SO4 solution = 930g Mass of 1000 ml H2SO4 solution = 1000 × 1.84 = 1840g Mass of water in 1000 ml of solution = 1840 – 930 = 910 g Wt. of H 2SO4 930 = Moles of H2SO4 = Mol Wt. of H 2SO4 98 \ Moles of H2SO4 in 1 kg of water

930 1000 ´ = 10.43 mol kg–1 = 98 910 \ Molality of 1 litre solution = 10.43

Br

¾ ¾®

CH3

H

CH3

+

CH3

Br H

CH3

H

Br –

¾¾ ®

Br

CH3

Br Br

H H

Br

CH3

(meso)

53.

54.

55.

1 (464) H 2 (g) + O2 (g) ® H 2 O(g) ΔH = – 249 2 Let the bond enthalpy of O - H is x. Then DH = SB.E. of reactant - S B.E.of product 1 –249 = 433 + ´ 492 - 2 x 2 Þ x = 464 kJ mol-1. (8) KE = hv – hv0 hv1 – hv0 = 2(hv2 – hv0) v0 = 2 v2 – v1 = 2 (2.0 × 1016) – (3.2 × 1016) = 8 × 1015 s–1 = 8 × 1015 Hz (57.5) Percentage by mass of copper in malachite

= 56.

2 ´ 63.5 = 57.5% 221

K f ´ w ´ 1000 DTf ´ W Given : Kf = 1.86, w = 1.25 g, W = 20 g, DTf = 273 – 271.9 = 1.1K Therefore, molar mass of solute

(105.7) Molar mass of solute =

= 57.

1.86 ´ 1.25 ´ 1000 = 105.7 1.1´ 20

(0.05) According to the Faraday’s law of electrolysis, nF of current is required for the deposition of 1 mole. According to the reaction, Ni ( NO3 ) 2 ¾¾ ® Ni 2 + + 2NO 3-

2 F of current deposits = 1 mol \ 0.1 F of current deposits = = 0.05 mol

EBD_7308 JEE MAIN

MT-188

58.

(21) Molecular orbital electronic configuration of these species are : O 2- (17e - ) : s1s 2 , s *1s 2 , s 2s 2 , s *2s2 , s 2p 2z ,

62.

p2p2x = p 2p2y , p * 2p2x = p *2p1y O2 (16e - ) : s1s 2 , s *1s 2 , s 2s 2 , s *2s 2 , s2p 2z , O22 - (18e

59.

p2p2x = 2

6.023 ´ 10 6.023 ´ 10 atoms constitute = 58.5 58.5 ´ 4 = 2.57 × 1021 unit cells. o = -2.36V (2.42) Given : E 2+

E

Mg 2 + / Mg

\

61.

(1) x =

ò 0

dt

1+ t2

Þ 1=

0.059 [Mg 2+ ] + log 2 [Mg]

}

< log q < log

p 2

p p p < log q < log 0 2 2 But 0 < cos q < 1, \ log ( cos q) < log 1 = 0

i.e., log (cos q) < 0. Hence, cos (log q) > log ( cos q) s

d d ù f( x ) ú y ( x ) - f {f( x )} . dx dx û

dy = 1 + y2 dx

p 2

\–

dy . 1+ y 2 dx

{

-

p pù é p êQ 2 < e \ log 2 < log e = 1 and 1 < 2 ú û ë

64.

}

p (1) Since, e 2 < q < 2

i.e., –

1

y ( x) é dI(x) êQ If I( x) = ò f (t ) dt , then = f {y ( x)}. dx ê f ( x ) ë

{

| a 1x + ….. + anxn | < 1 for all n . – 1 < a1x + ….. + anxn < 1 for all n

\ log e

Given [Mg2+] = 0.01 M; [Mg] = 1 0.059 log(0.01) Hence, E Mg2+/Mg = -2.36 + 2 = -2.42V MATHEMATICS y

1 3 . =1 3 2

p

63.

/ Mg

Mg 2 + / Mg

2| x| 2| x| 2 1 < [1- | x | n ] < 1- | x | 1- | x | 3 1- | x |

Þ 1 + a1x + .. + a nxn ¹ 0 for all n

23

(aq) + 2e ® Mg(s)

=E

dy y . 1+ y2 = y = 2 dx 1+ y

Therefore, | a1x + ….. + anxn | < 2.

-

o

2 1+ y

.2y .

Þ 1 + a1x + a2x2 + … + anxn > 0 for all n

23

2+

2

(2) | a1 x + a2x2 + … + anxn | £ | a1 | | x |+| a2 | | x | 2 + | a3 | | x | 3 + ……. + | an | | x | n

=

6.023 ´ 10 23 atoms 58.5 4 atoms constitute 1 unit cell

Mg

1

£ 2 [| x | + | x |2 + ….. + | x |n ] [Q a n < 2]

Hence number of antibonding electrons are 7, 6 and 8 respectively. (2.57) Since in NaCl type of structure 4 formula units form a cell. 58.5 g of NaCl = 6.023 × 1023 atoms

Mg

dx

= p * 2p1y ) : s1s , s *1s , s 2s , s *2s 2 , s2p 2z , p 2p2x : p2p2y , p * 2p x2 = p * 2p 2y

\

=

2

p 2p 2y , p * 2p1x 2 2

1 g of NaCl =

60.

d2 y

Þ

n

å å

(1)

r=0

n

Cs s Cr

s =1 r£s

n

(

= å nCs s C 0 + s C1 + s C2 + ... + s Cs s =1

=

n

å

s =1

n

Cs 2 s =

= 3 –1 n

n

)

å n Cs 2s – n C0 2 0 =(1 + 2) – 1

s=0

n

Solutions-Mock Test-6 65.

MT-189

(3) Number of ways of making 3 sets of 10 balls

we get,

10! and further they 2!.3!.5! can be given to 3 persons one each in 3! ways.

having 2, 3 and 5 balls

66.

1 - log|tan x|

y (I. F.) = \ y cot x = \ y cot x =

log

tan x

w

w2

x + 1+ w + w2

x + w2

1

1

x+w

x + 1+ w + w

10! ´ 3! Total number of ways = 2!3!5! æ dy ö (4) Given, sin 2 x çè - tan x ÷ø - y = 0 dx y dy + tan x = or, dx sin 2 x dy or, ...(1) - y cosec2 x = tan x dx - cosec2x Now, integrating factor (I.F) = e ò

( ) = e or, I.F = e 2 1 = cot x = tan x Now, general solution of eq. (1) is written as

x + 1+ w + w2

Þ x x + w2 x 1 1

69.

ò Q(I.F.) dx + c ò ò 1.dx + c

tan x . cot x dx + c

68.

2

b b +c -a = 2c 2bc 2 2 Þc =a Þ c=a (4) Operate C1+C2 +C3 ,

x+w

l

m

n

(4) Lines are concurrent if m n

n

l =0

l

m

(C1 ® C1 + C2 + C3 )

1 m

n

Þ (l + m + n) 1 n 1

l =0

l

m

Þ (l + m + n )( mn + nl + lm - l 2 - m 2 - n 2 ) = 0 Þ l + m + n = 0; l 2 + m 2 + n 2 - lm - mn - nl = 0

cos A =

2

=0Þx =0

1

l+ m+n m n Þ l+ m+n n l = 0 l+ m+n l m

(2) From cosine and sine formula, we have

2

w2

w

2 Þ x 1 x+w 1 1

-1

b2 + c2 - a 2 and 2bc sin A sin B sin C = = = k. a b c Given, in any DABC, sinB cos A = 2sin C From the above given formula, we have sin B = bk, sin C = ck bk b \ cos A = = 2ck 2 c Put the value of cos A in the formula, which gives

=0

1 x+w

[Q 1 + w + w2 = 0]

\ y cot x = x + c

67.

=0

w2

w

x

2

\ l2 + m2 + n2 = lm +mn + nl 70.

(3)

3 4 2

3

4

2

5 8 2 =

2

4

0

x

71.

y 2

x -5 y -8 0

(R 2 ® R 2 - R1 , R 3 ® R 3 - R 2 ) = 2 (2y – 16 – 4x + 20 ) = 2 (2y – 4x + 4 ) \ given determinant = 0 Þ 2y – 4x + 4 = 0 Þ 2x – y – 2 = 0 which represents st. line. (2) Centre of the given circle is (1,2) and its radius = 1 + 4 + 20 = 5 . Since the radii of the two circles are equal, therefore these will touch externally and the point of contact will lie mid way between the two centres. If (h,k) is the centre of the circle, then

EBD_7308 JEE MAIN

MT-190

h +1 k+2 = 5, = 5 \ h = 9, k = 8 2 2 \ its equation is ( x –9)2 + ( y – 8)2 = 25 i.e., x2 + y2 – 18x – 16y + 120 = 0

72.

(1) Ellipse is

x2 y 2 + =1 16 3

Now, equation of normal at (2, 3/2) is 16 x 3 y = 16 - 3 2 3/ 2 13 Þ 8x – 2y = 13 Þ y = 4 x 2 13 Let y = 4 x - touches a parabola 2 y2 = 4ax. We know, a straight line y = mx + c touches a parabola y2 = 4ax if a – mc = 0 \

73.

æ 13ö a - ( 4) ç - ÷ = 0 Þ a = – 26 è 2ø

76.

77.

a0 + a1 x + a2 x 2 + ....... + a n x n = 0,

where a0 ¹ 0 then A is invertible. Since A, B and C are n × n matrices and A satisfies the equation x3 + 2x2 + 3x + 5 = 0 as A 3 + 2 A 2 + 3A + 5I = 0 , therefore, A is invertible.

78.

75.

(4)

3p / 2 p/2

1 1 1 3 4 =0 4 if 1 a b

we get, 4

p/ 2

5p / 6

3p / 2

+ ò [2 sin x ]dx + ò [ 2 sin x ]dx p

7p / 6

5p / 6

p

7p /6

p/2

5p / 6

p

= ò 1 dx + ò 0 dx + ò (-1) dx 3p / 2

+

ò

(-2) dx

7p / 6

7 p ö æ 7 p 3p ö p æ æ 5p p ö - ÷==ç - + 0 + çp ÷ + 2ç è 6 2 ÷ø 6 ø è 6 2 ø 2 è

79.

(1) L f ¢(1)

f ( x ) - f (1) ax 2 + b - a - b = Lt x -1 x -1 x ®1 x ®1

= Lt

a ( x 2 - 1) = Lt a ( x + 1) = 2a x ®1 x - 1 x ®1

= Lt

Rf ¢(1)

f ( x) - f (1) bx 2 + ax + c - a - b = Lt x -1 x -1 x ®1 x ®1 c = Lt [b ( x + 1) + a ] + x -1 x ®1 = Lt

Operate C3®C3 – C1 ; C2®C2 – C1 1

p

7p / 6

pö p æ (2) sin x – cos x = 1 Þ sin ç x - ÷ = sin Þ x 4 4 è ø

p p p p – = np + (–1)n Þ x = np + (–1)n + 4 4 4 4 r r ˆ B = 4iˆ + 3jˆ + 4kˆ (4) Vector A = ˆi + ˆj + k, r And C = ˆi + aˆj + bkˆ are linearly independent,

5p / 6

ò [ 2 sin x]dx = ò [2 sin x ]dx + ò [2 sin x ]dx

Hence, required equation of parabola is y2 = 4 (– 26)x = – 104 x (1) I study or I fail = p Ú q Now, ~ (p Ú q) º~ p Ù (~ q) Hence, negation of 'I study or I fail' is I do not study and I do not fail.

74.

r Again, | C | = 1 + a 2 + b 2 = 3 Þ 1 + a2 + b2 = 3 Þ 1 + a2 + 1 = 3 \ a2 = 1 Þ a = ± 1 Hence, a = ± 1 , b = 1 (4) We know that every quadratic equation has exactly two roots which are either, both real or both are imaginary. So, any quadratic equation has neither exactly one real root nor booth roots are always real. (2) We have a theorem that if a square matrix A satisfies the equation;

0

0

-1

0

1 a -1 b -1

Þ–(b–1)=0Þb=1

=0

c = 2b + a if c = 0 x -1 Since, f is diff. at x = 1\ 2 a = 2 b + a Þ a = 2 b. Thus, result holds if a = 2 b, c = 0. = 2b + a +

Solutions-Mock Test-6

MT-191

80. (3) We have a + b = – 3 and ab = a/2

a a2 + b 2 b Now + 1 Þ x2 + 5x – 14 < x – 5 (1) 2 x + 5 x - 14 Þ x2 + 4x – 9 < 0 Þ a = – 5, – 4, – 3, – 2, – 1, 0, 1 a = – 5 does not satisfy any of the options a = – 4 satisfy the option (a) a2 + 3a – 4 = 0

=

79.

80.

(3)

Þ

77.

2 (12)2 - (9)(-4) 2 h 2 - ab = 9 + ( -4) a+b

æp ö sin -1 (1 - x ) = ç - sin -1 x ÷ - sin -1 x 2 è ø p (Q cos–1x = - sin -1 x ) 2

p - 2 sin -1 x 2 Taking sum of both sides sin -1 (1 - x) =

æp ö 1 - x = sin ç - 2 sin -1 x ÷ = cos( 2 sin -1 x ) è2 ø

= cos 2q, where sin -1 x = q

81.

1 - x = 1 - 2 sin 2 q = 1 - 2 x 2 or x(1 – 2x) = 0 or 1 x = 0, 2

(1.00)

1 + x 2 + 1 + y 2 = l( x 1 + y 2 – y 1 + x 2 )

Þ

At x = 0, for f (x) to be continous lim f (0 - ) = f (x = 0) = lim f (0 + )

x® 0

Þ

x® 0

1 + x 2 (1 + ly ) = 1 + y 2 ( lx - 1)

1 + x2

1+ y2

=

lx - 1 ly + 1

x2 + 1

l 2 x 2 - 2l x + 1 = y 2 + 1 l 2 y 2 + 2l y + 1

f (x) = 0 at x = 0 RHL = lim sin( x + h ) = sin h > 0

Þ

L. H. L. = lim sin(x - h ) = sin (-h ) < 0

Þ ( y 2 + 1)(l 2 x 2 - 2lx + 1)

x ®0

x ®0

= ( x 2 + 1)(l 2 y 2 + 2ly + 1)

Hence, not differentiable at x = 0 Þ

l 2 x 2 y 2 - 2lxy 2 + y 2 + l 2 x 2 - 2lx + 1

Solutions-Mock Test-7

MT-203

= l 2 x 2 y 2 + 2lx 2 y + x 2 + l 2 y 2 + 2ly + 1

15 49 15 Þ 1 – b – a + ab = 49 34 Þ a + b – ab = 49 From (i) and (ii), 6 a+b= 7 8 and ab = 49

Þ (1 – a) × (1 – b) =

l 2 ( x 2 - y 2 ) - 2l( xy 2 + x 2 y + x + y ) = 0

Þ

Þ l 2 ( x + y )( x - y ) - 2l [ xy( x + y) + ( x + y )] = 0 Þ l ( x + y) [ l ( x - y ) - 2 xy - 2] = 0

Þ ( x + y) [ l ( x - y) - 2 xy - 2] = 0 Þ l ( x - y ) - 2 xy - 2 = 0 2 xy + 2 xy + 1 l =l Þ = Þ x- y x- y 2 æ dy ö æ dy ö ç x + y ÷ ( x - y ) - ( xy + 1) ç1 - ÷ dx è ø è dx ø = 0 Þ 2 ( x - y) This is the first order differential equation and dy clearly degree of is 1. Hence degree of the dx differential equation is 1.

82.

(4.00) We have T2 = 14

Þ

1 n -1 3 + na 13 2

5 2

84.

5

= 14a 2 n

83.

3 2

C3

14

42 42 4 ´ 8 ´ = 49 49 49

14 196 \ a–b= ...(iv) 49 2401 From (iii) and (iv), 4 2 a= ,b= 7 7 Hence probability of more probable of the two 4 events = = 0.57 7 (34.00) a + b = 3; ab = a ; g + d = +12 ; gd = b a, b, g, d are in increasing G.P..

b = ax , g = ax 2 , d = ax 3 a + b = a + ax = 3 = a(1 + x)

C3

12 = = =4 Þ n = 14 Þ n 14 3 C2 C2 (0.57) Let the probability of occurrence of first event A, be ‘a’ i.e., P(A) = a \ P(not A) = 1 – a And also suppose that probability of occurrence of second event B, P(B) = b, \ P(not B) = 1 – b Now, P(A and not B) + P(not A and B) 26 = 49 26 Þ P(A) × P(not B) + P(not A) × P(B) = 49 26 Þ a × (1 – b) + (1 – a) b = 49 26 Þ a + b – 2ab = ...(i) 49 15 And P(not A and not B) = 49 15 Þ P (not A) × P(not B) = 49

...(iii)

(a – b)2 = (a + b)2 – 4ab =

5 a2

) ( a ) = 14a Þ C ( n

1 n -1 a13

...(ii)

2

3

.....(1) 2

g + d = ax + ax = 12 = ax (1 + x) .....(2) a(1 + x ) 3 = 12 a x 2 (1 + x ) 1 1 = or or x = 2 4 x2 Þ b = 2a and a + 2a = 3 Þ a = 1 and b = 2 \a = ab = 2

Divding

g = ax 2 = 1´ 2 2 = 4; d = ax 3 = 1´ 23 = 8

\ b = g d = 4 ´ 8 = 32 Þ a + b = 2 + 32 = 34. 85.

P (A Ç B)

(0.63)

P(B) P(A)

P (A Ç B Ç C)

P (B Ç C)

P (A Ç C) P(C)

EBD_7308 JEE MAIN

MT-204

\ Probability that atleast one of the events A, B, C exists is given by the shaded region. Req. prob. = P(A) + P(B) + P(C) - P(A Ç B)

86.

p/3

=

0

- P(B Ç C) - P(C Ç A) + P(A Ç B Ç C) 1 1 1 1 5 = + + - 0 - 0 - + 0 = = 0.63 4 4 4 8 8 n n r+2 n r + 1+1 n (5.00) å r + 1 C r = å r + 1 C r r=0 r =0 =

n n

n

r =0 n

=2 +

å

1 n n +1 r=0

= 2n +

1 (2 n +1 - 1) n +1

1 [(n + 1)2 n + 2 n +1 - 1] n +1 1 = [2 n (n + 3) - 1] n +1 =

lim f ( x ) x®0

= f (0 )

f(0) = k sin 3x 3× 3x sin lim lim 3x = 3 = Lim x ® 0 f (x) = x ® 0 sin x x ®0 sin x x é lim sinq ù êQx®0 q = 1ú ë û Þ k= 3 p/3

88.

(0.33)

ò 0

cos x + sin x 1 + sin 2x

dx

2

dx =

p

ò dx = 3 = kp 0

fi

ui =

X i - 155 5

u i2

fiui

f i u i2

–3 –2 –1 0 1 2 3 4

9 4 1 0 1 4 9 16

–12 –12 –15 0 36 48 24 8 77

36 24 15 0 36 96 72 32 311

æ f u2 æ f u 2 ç å i i –ç å i i s = \ Variance, ç n ç n è è

Given,

87.

(cos x + sin x)

4 140 145 6 150 15 155 30 160 36 165 24 170 8 175 2 Total 125

å n +1C r+1

(n + 3)2 n - 1 28 - 1 (5 + 3) . 25 - 1 = = n +1 6 5 +1 Þ n=5 (3.00) For f (x) to be continuous,

p/3

cos x + sin x

dx

1 = 0.33 3 (7.26) Let us assume an arbitary mean a = 155. Following table is constructed :

Xi

C r +1 n +1

= 2n +

sin x + cos 2 x + 2 sin x cos x

Þ k=

r =0

r =0

ò 0

n +1

n

cos x + sin x 2

p/3

=

89.

C

å n C r + å r + 1r

ò

90.

\ S.D. = variance = 7. 26 (1.33) Line is ^ to 3x + y = 3 1 \ Slope of line, m = 3 Equation is, y = mx + c = x + c 3 2 It passes through (2, 2) Þ 2 = + c 3 4 Þ c= 3 x Þ y - = 4 / 3 Þ 3y - x = 4 3 \ y-intercept = 4 / 3 = 1.33

ö ÷÷ ø

2

ö ÷ ÷ ø

Mock Test-8 ANSWER KEY 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

(3) (3) (4) (3) (2) (1) (3) (1) (3) (1) (4) (2) (3) (3) (1)

16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

(3) (1) (2) (3) (3) (4) (264) (4.5 × 104) (48) (0.16) (10) (2) (2) (6.52) (90)

31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

(1) (4) (4) (1) (3) (1) (3) (2) (4) (4) (2) (2) (4) (3) (3)

(3) (3) (3) (3) (3) (112) (3) (2.1) (2) (0.177) (9.25) (9.6) (107.2) (–0.9) (0.125)

46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

(3) (3) (3) (1) (1) (2) (1) (2) (2) (2) (2) (1) (4) (4) (2)

76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

(4) (3) (3) (2) (4) (0.25) (1.33) (32.00) (0.40) (0.60) (10.00) (0) (0) (1.00) (0.50)

Solutions PHYSICS 1.

(3)

r æRö ç ÷q 0 rq 0 R F 3 Î0 è 2 ø = a= = m m 6m Î0

TV g -1 = constant g -1

T1V1g -1 = T2V2

rq R R \ v2 = 2a(R / 2) = 2. 0 . 6mÎ0 2

1

1 æV ö 2 Þ T (V ) 2 =T2 ç ÷ è 4ø

Vù é êëQ g = 1.5, T1 = T ,V1 = V and V2 = 4 úû

K.E. =

4.

1

2.

3.

æ 4V ö 2 \ T2 = ç ÷ T = 2T èV ø (3) Initially, fB increases as magnet approaches the solenoid \ e = – ve and increasing in magnitude. When magnet is moving inside the solenoid, increase in fB slow down and finally fB starts decreasing \ emf is positive and increasing. Only graph (c) shows these characteristic. (4) Acceleration

5.

2 1 1 q 0 rR mv 2 = 2 12 Î0

(3) Both fall with equal acceleration g, have equal displacements in time t; therefore extension = 0. (2) Momentum E hn = c c Recoil energy

Mu =

1 1 M2u 2 1 æ hn ö Mu 2 = = 2 2 M 2M çè c ÷ø

=

h 2 n2 2Mc2

2

EBD_7308 JEE MAIN

MT-206

6.

(1) I1 = 4I0

G

I2 = Aaverage ;

=

= [ I0 + I0 + 2 I0 I0 cos f]ev = 2I0 4I I \ 1 = 0 =2 I 2 2I 0

7.

9.

qB =

or,

nh r = 2pqB

or,

r=

K=

K=

10.

2

nh 2pqB

hc ,f ...(ii) l/2 (for monochromatic light of wavelength l/2) From question, æ hc ö hc , f < 3çç , f÷÷÷ K.E.l / 2 < 3(K.E.l ) Þ ç è ø l/2 l K.E.l / 2
0 x p p or 0 £ £ 2 and - < x < 2 2 2 p p or 0 £ x £ 4 and - < x < 2 2 é pö \ x Î ê 0, ÷ ë 2ø (2) f (x) = sin x + cos x, g (x) = x2 – 1 Þ g (f (x)) = (sin x + cos x)2 – 1 = sin 2x

\ ‘f’ cannot be one-one. (3) f (x) = x3, x Î R. Let f (x1) = f(x2), x1, x2 Î R Þ x13 = x 2 3 Þ x13 – x 2 3 = 0 Þ ( x1 - x 2 ) ( x1 2 + x1 x 2 + x 2 2 ) = 0

Þ x1 - x 2 = 0 Þ x1 = x 2 , because the other factor cannot be zero \ ‘f’ is one-one (4) We have f ( x) =

2x + 1 , x ÎR - {7}. x-7

Let f ( x1 ) = f ( x2 ), x1 , x2 Î R - {7}. 2x + 1 2x 2 + 1 Þ 1 = x1 - 7 x2 - 7 Þ - 15x1 = -15x 2 Þ x1 = x 2 . \ f is one-one. é1 2 1 2 2 4 3 2 9 ù ê 2 sec 2 + 2 sec 2 + 2 sec 2 ú n n n n n ú 79. (4) lim ê n n®¥ ê 1 2 ú +.... + sec 1ú êë n û

r

r2

n®¥ n

n2

is equal to lim

sec2 2

EBD_7308 JEE MAIN

MT-228

81.

2 1 r 2 r = lim . sec 2 n®¥ n n n

(1.67) We have

r ´a + r ´b = a ´b + b ´a = 0 Þ r ´ (a + b ) = 0

Þ Given limit is equal to value of 1

Þ r = l (a + b ) = l (–2ˆi – ˆj + 2 kˆ )

integral ò x sec 2 x 2 dx 0

1

1

0

0

Þ rˆ =

1 1 or ò 2 x sec2 x 2dx = ò sec2 t dt 2 2

ur ur r × a = 6l = 30 Þ l = 5 ur ˆ \ r = 5(–2iˆ – ˆj + 2k)

[put x 2 = t Þ 2xdx = dt] = 80.

1 1 1 (tan t ) = tan1 . 0 2 2

\

(2) The equation c ( y + c) 2 = x3 ...(i) has one arbitrary constant, so the equation must be of the first order. Differentiating (i) we get

2c ( y + c )

dy = 3x 2 dx

82.

...(ii) 2

æ dy ö Þ 4c 2 ( y + c )2 ç ÷ = 9 x 4 è dx ø Divide (iii) by (i), we get

...(iii)

2

9 x æ dy ö 4c ç ÷ = 9 x Þ c = è dx ø 4 æ dy ö 2 çè ÷ø dx Put the value of c in (ii), we get é ù ê ú 9 x ê 9 x ú æ dy ö 2 + 2× y ç ÷ = 3x 4 æ dy ö 2 ê 4 æ dy ö 2 ú è dx ø çè ÷ø ê çè ÷ø ú dx ë dx û ù é ú ê 9 x ú æ dy ö = 6x 2 ç ÷ Þ 9x ê y + ê 4 æ dy ö 2 ú è dx ø ê ç ÷ ú êë è dx ø úû 2

27 æ dy ö æ dy ö Þ 3y ç ÷ + x = 2x ç ÷ è dx ø è dx ø 4 2

æ dy ö æ dy ö Þ 12 y ç ÷ = 8 x ç ÷ - 27 x è dx ø è dx ø

ur r = 5 / 3 = 1.67

(60) Let l1, m1, n1 and l2, m2, n2 be the d.c of line 1 and 2 respectively, then as given l1 + m1 + n1 = 0 and l2 + m2 + n2 = 0 and l12 + m12 – n12 = 0 and l22 + m22 – n22 = 0 (Q l + m + n = 0 and l2 + m2 – n2= 0) Angle between lines, q is cos q = l1l2 + m1m2 + n1n2 ...(1) 2 2 2 As given l + m = n and l + m = – n Þ (– n)2 – 2lm = n2 Þ 2lm = 0 or lm = 0 So l1m1 = 0, l2m2 = 0 If l1 = 0, m1 ¹ 0 then l1m2 = 0 If m1 = 0, l1 ¹ 0 then l2m1 = 0 If l2 = 0, m2 ¹ 0 then l2m1 = 0 If m2 = 0, l2 ¹ 0 then l1m2 = 0 Also l1l2 = 0 and m1m2 = 0 l2 + m2 – n2 = l2 + m2 + n2 – 2n2 = 0 Þ 1 – 2n2 = 0 Þ n = ± \ n1 = ±

3

3

1 (–2ˆi – ˆj + 2kˆ ) 3

83.

1

1 2

1

, n2 = ± 2 2 1 \ cos q = q = 60° (acute angle) 2 (0.56) If A will be final winner

Þ (A beats B ) (B beats C) + (A beats C) (C beats B)

Solutions-Mock Test-9

MT-229

2 2 æ 2 öæ 2 ö 4 1 5 = 0.56 = ´ + ç1 - ÷ç1 - ÷ = + = 3 3 è 3 øè 3 ø 9 9 9 84.

r r r (2.00) If a, b, c are linearly independent vectors, then rc should be a linear combination r of ar and b . r r r Let c = pa + qb

) (

(

i.e. ˆi + aˆj + bkˆ = p ˆi + ˆj + kˆ + q 4iˆ + 3jˆ + 4kˆ

87.

a a b b b c c , , , , , , , we get 2 2 3 3 3 2 2

)

r r r Equating the coefficient i , j, k , we get 1 = p + 4q, a = p + 3q and b = p + 4q.

From first and third, b = 1. Now | rc | = Þ

2 2 3 Þ1+ a + b =3

a 2 = 1, \ a = ± 1.

88.

Hence , a = ± 1, b = 1. Þ a + b = 2 or 0. (2) Required area =

p/4

ò0

p/2

tan x dx + ò

p/4

z1 + z3 z 2 + z 4 = 2 2

D (z4)

C (z3)

A (z1)

B (z2)

cot x dx

y=

t an

x

Y tx co

O

p/4

p/4

= [ log sec x ] 0

x = p/2

X

p/2

+ [ log sin x ] p / 4

= log 2 + log 2 = 2 log 2 = log 2 = logk Þ k = 2 86.

1 a b c 2. + 3. + 2. ìïæ a ö 2 æ b ö3 æ c ö 2 üï 7 2 3 2 ³ ç ÷ ç ÷ ç ÷ ý í 7 ïîè 2 ø è 3 ø è 2 ø ïþ 10 4 37 a 2 b 3c 2 2 b3 c2 £ 3 .2 ³ Þ a Þ 7 7 22.33.2 2 77 310.2 4 3P.q q = r \ greatest value of a2 b3 c2 = 7 77 Þ P + q + r = 10 + 4 + 7 = 21 (0.50) Since z1 + z3 = z2 + z4

Þ

y=

85.

\ 12 = – 15t + 2t + t3 i.e., t3 – 13t – 12 = 0 \ t = – 1, –3, 4 \ The points are (1, –2), (9, – 6), (16, 8) 26 \ Centroid is æç , 0ö÷ = (m, n). è 3 ø 26 Þ m+n = = 8.67 3 (21.00) Taking A.M. and G.M. of 7 numbers

(8.67) Let the equation of any normal be y = – tx + 2t + t3 Since it passes through the point (15, 12)

æ Midpoint of ö æ Midpoint of ö Þ ç ÷=ç ÷ è z1 and z3 ø è z2 and z 4 ø Þ Quadrilateral ABCD is a parallelogram. Also, | z1 – z3 | = |z2 – z4 | æ Distance ö æ Distance ö i.e. çç Between ÷÷ = çç Between ÷÷ ç z and z ÷ ç z and z ÷ è 1 3ø è 2 4ø

EBD_7308 JEE MAIN

MT-230

Þ Length of diagonals are equal. \ The parallelogram ABCD is a rectangle

1 é ù 2 2 -( 2 + h ) ú ê Þ k = lim ê( 2 + h ) + e ú h ®0 êë úû

æ z1 - z 2 ö Now, arg ç z - z ÷ i.e. angle between z1, z2 and è 3 2ø p which is angle between 2 sides AB and BC (see figure) 1 p \± kp = ± Þ k = = 0.50 2 2

Þ k = lim é4 + h 2 + 4h + e -1/ h ù û h ®0 ë

z3 (taken in order) is ±

-1

89.

1 ù é 2- x ú 2 ê (0.25) f ( x ) = x + e and f (2) =k ê ú êë úû If f (x) is continuous from right at x =2 then lim f ( x ) = f ( 2 ) = k

x ®2 +

1 ù é Þ lim ê x 2 + e 2 - x ú ú x ® 2+ ê ë û Þ k = lim f ( 2 + h ) h® 0

-1

=k

90.

-1

-1

-1 1 Þ k = é 4 + 0 + 0 + e-¥ ù Þ k = = 0.25 ë û 4 (3) Let A be the area, b be the breadth and l be the length of the rectangle.

dA dl db = -5 , = 2, = -3 dt dt dt We know, A = l × b

Given :

dA db dl = -3l + 2b = l. + b. dt dt dt Þ – 5 = – 3l + 2b. When b = 2, we have

Þ

– 5 = – 3l + 4 Þ l =

9 = 3m 3

Mock Test-10 ANSWER KEY 1

(3)

16

(2)

31

(4)

46

(4)

61

(3)

76

(1)

2

(2)

17

(2)

32

(3)

47

(2)

62

(4)

77

(1)

3

(3)

18

(3)

33

(2)

48

(2)

63

(2)

78

(2)

4

(2)

19

(1)

34

(4)

49

(2)

64

(3)

79

(3)

5

(4)

20

(4)

35

(1)

50

(1)

65

(1)

80

(4)

6

(2)

21

(2)

36

(3)

51

(187)

66

(4)

81

(0.33)

7

(3)

22

(2.68)

37

(2)

52

(5.82)

67

(4)

82

(0.60)

8

(3)

23

(900)

38

(1)

53

(1.6)

68

(4)

83

(2007)

9

(4)

24

(0.125)

39

(4)

54

(1.5)

69

(1)

84

(5.00)

10

(3)

25

(2)

40

(3)

55

(2)

70

(4)

85

(1.50)

11

(3)

26

(0.96)

41

(2)

56

(0.2)

71

(2)

86

(1.00)

42

(3)

57

(4)

72

(2)

87

(1.00)

43

(1)

58

(0.22)

73

(1)

88

(2.00)

12

(2)

27

(110)

13

(3)

28

(1.2 × 10 )

12

14

(4)

29

(1)

44

(3)

59

(3)

74

(1)

89

(0.50)

15

(2)

30

(200)

45

(3)

60

(12)

75

(3)

90

(0.05)

Solutions PHYSICS 1.

(3) Consider a small element dx of radius r, r
cos 2 1

64.

Þ

Þ cos(C/2)sin(A/2 – B/2) = cosA/2 sin (B/2 – C/2)

\ 6 mole of C will weigh = 2 ´ 6 = 12

Þ 2cos B = cos A + cos C Þ secA, secB, secC are in H.P. (1) Any tangent to the parabola y2 = 4ax is

(3)

65.

y = mx +

C r - (n C r +3.n C r -1 +3.n C r - 2 + n C r -3 )

= n + 4 C r - (n C r +n C r -1 ) + (n C r -1 + n C r -2 ) =

n+4

C r - n+1C r +n+1C r-1+n+1C r-1+n+1C r-2

= n + 4 C r - n + 2 C r + n + 2 C r -1 = n + 4 C r - n +3Cr = n + 3 C r -1

f (q) = cos 2 (cos q) + sin 2 (sin q) Now, put q = p /2 & 0

...(1)

16b2m4 + 16abm = 0 or m3 = -

(4) Let a' = a/(a – 1), b' = b/(b – 1) Þ a = a'/(a'– 1), b = b'/(b' – 1) Þ 1/a = (a' – 1)/a', 1/b = (b' – 1)/b' Then equation whose roots are 1/a, 1/b, is x2 – (1/a + 1/b)x + 1/ab = 0 Þ x2 – [(a' – 1)/a' + (b' – 1)/b'] x + [(a' – 1) (b' –1)] / a'b' = 0 2 ' Þ a'b'x – [2a'b' – (a + b')]x + a'b' – (a' + b') + 1 =0 Þ bx2 – (a + 2b)x + a + b + 1 = 0 (2)

a m

aö æ meets x2 = 4by where x2 = 4b ç mx + ÷ mø è 2 2 or mx – 4bm x – 4ab = 0 ...(2) If the line (1) be a tangent to the second parabola, then roots of (2) must be equal. The condition for this is

+ (n C r -1 + n C r - 2 ) + ( n C r - 2 +n C r -3 )

63.

sin A / 2 sin B / 2 sin B / 2 sin C / 2 = cos A / 2 cos B / 2 cos B / 2 cos C / 2

Þ cos B - cos A = cosC - cosB

n +4

62.

p p p < q< &1> ] 4 2 4 Now, seeing the options 1 + sin21 is greater than all other options. (3) tanA/2, tanB/2, tanC/2 are in A.P. Þ tanA/2 – tanB/2 = tanB/2 – tanC/2 [Q sin q > cos q "

1 mole of C weighs = 2 g. MATHEMATICS 61.

MT-239

1/ 3

\

æaö m= -ç ÷ èbø

=-

a b

a1/ 3 b1 / 3

Substituting this value of m in (1), we get æ b1 / 3 ö a1/ 3 y = - 1 / 3 x + a çç - 1 / 3 ÷÷ b è a ø

66.

i.e. xa1/3 + yb1/3 + a2/3 b2/3 = 0 which is the required equation of the common tangent. (4) Let the point of contact be (h,k); equation of chord of contact is T = 0

f (0) = cos 2 1 + 0 = cos 2 1

Þ xh + yk – 12 = 0

f (p / 2) = 1 + sin 2 1

Equation of common chord C1 – C2 = 0

Þ 5x – 3y –10 = 0

………..(1) ………. (2)

EBD_7308 JEE MAIN

MT-240

71.

(1) and (2) represent the same line

67.

68.

Þ h/5 = k/(–3)= –12/(–10) Þ h = 6, k = –18/5 (4) R = {(x, y) : x, y Î N and x2 – 4xy + 3y2 = 0} Now, x2 – 4xy + 3y2 = 0 Þ (x – y) (x – 3y) = 0 \ x = y or x = 3y \ R = {(1, 1), (3, 1), (2, 2), (6, 2), (3, 3), (9, 3),......} Since (1, 1), (2, 2), (3, 3),...... are present in the relation, therefore R is reflexive. Since (3, 1) is an element of R but (1, 3) is not the element of R, therefore R is not symmetric Here (3, 1) Î R and (1, 1) Î R Þ (3, 1) Î R (6, 2) Î R and (2, 2) Î R Þ (6, 2) Î R For all such (a, b) Î R and (b, c) Î R Þ (a, c) Î R Hence R is transitive.

72.

Y f(x) = cos x

Clearly, f ' (x) > 0 for

X

O

p 2p or