NTA CUET (UG) 10 Mock Test Sample Question Papers Physics (2024) 9789357288187, 935728818X

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For 2024 Exam

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PHYSICS Section II (Domain Specific Subject) Strictly ar per the Latest Examination Pattern issued by NTA

The ONLY book you need to Ace CUET (UG)

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Crisp Revision

Valuable Exam Insights

100% Exam Readiness

Extensive Practice

Concept Clarity

With On-Tips Notes & Updated Mind Maps

With Latest Solved Paper 2023

With 10 Solved Sample Question Papers

With 650+ NCERT - based MCQs

With 450+ Explanations & Smart Answer Keys

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3rd EDITION

ISBN SYLLABUS COVERED

YEAR 2023-24 “9789357288187”

CUET (UG) CERTIFICATE OF COMMON UNIVERSITY ENTRANCE TEST

PUBLISHED BY

C OPYRIG HT

RESERVED BY THE PUBLISHERS

All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without written permission from the publishers. The author and publisher will gladly receive information enabling them to rectify any error or omission in subsequent editions.

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This book is published by Oswaal Books and Learning Pvt Ltd (“Publisher”) and is intended solely for educational use, to enable students to practice for examinations/tests and reference. The contents of this book primarily comprise a collection of questions that have been sourced from previous examination papers. Any practice questions and/or notes included by the Publisher are formulated by placing reliance on previous question papers and are in keeping with the format/pattern/ guidelines applicable to such papers. The Publisher expressly disclaims any liability for the use of, or references to, any terms or terminology in the book, which may not be considered appropriate or may be considered offensive, in light of societal changes. Further, the contents of this book, including references to any persons, corporations, brands, political parties, incidents, historical events and/or terminology within the book, if any, are not intended to be offensive, and/or to hurt, insult or defame any person (whether living or dead), entity, gender, caste, religion, race, etc. and any interpretation to this effect is unintended and purely incidental. While we try to keep our publications as updated and accurate as possible, human error may creep in. We expressly disclaim liability for errors and/or omissions in the content, if any, and further disclaim any liability for any loss or damages in connection with the use of the book and reference to its contents”.

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Preface National Testing Agency (NTA) has been established in November 2017 under the Societies Registration Act (1860) by the Ministry of Education as a premier, specialist, autonomous, and self-sustained testing organization to conduct entrance examinations for admission/fellowship in higher educational institutions. The Common University Entrance Test (CUET (UG) - 2022) is being introduced for admission into all UG Programmes in all Central Universities for the academic session 2023-24 under the Ministry of Education, (MoE). The Common University Entrance Test (CUET) will provide a common platform and equal opportunities to candidates across the country, especially those from rural and other remote areas, and help establish a better connection with the Universities. A single examination will enable the Candidates to cover a wide outreach and be part of the admissions process to various Central Universities. CUET – UG Computer Based Test (CBT) for the Central Universities is to be conducted by the National Testing Agency (NTA). The curriculum for CUET is based on the National Council of Educational Research and Training (NCERT) syllabus for class 12 only. CUET scores are mandatorily required while admitting students to undergraduate courses in 44 central universities. A merit list will be prepared by participating Universities/organizations. Universities may conduct their individual counselling on the basis of the scorecard of CUET (UG)-2023 provided by NTA.

A few benefits of studying from Oswaal Sample Question Papers • • • • •

Crisp Revision with On-Tips Notes & Updated Mind Maps Valuable Exam Insights with Latest Solved Paper 2023 100% Exam Readiness with 10 Solved Sample Question Papers Extensive Practice with 650+ NCERT-based MCQs Concept Clarity with 450+ Explanations & Smart Answer Keys

Our Heartfelt Gratitude! Finally, we would like to thank our authors, editors, and reviewers. Special thanks to our students who send us suggestions and constantly help improve our books. We promise to always strive towards ‘Making Learning Simple’ for all of you. Wish you all Happy Learning!

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-Team Oswaal Books

Oswaal Books Expert Tips to Crack CUET (UG) in the First Attempt Excited about your UG but unsure if you will get admission to your preferred university? In a major announcement by the chairman of the University Grants Commission, the Naonal Tesng Agency will be conducng the Common Universies Entrance Test (CUET (UG) 2022) for undergraduate programs in Central Universies for the upcoming academic session. However, the UGC Chairperson also stated that CUET (UG) will not just be limited to admissions to Central Universies. Many prominent private universies have indicated that they would also like to adopt a common entrance exam for undergraduate admissions and take admissions on the basis of CUET (UG) scores. This makes CUET (UG) a very important examinaon in itself and hence it becomes mandatory to be aware of the ps & tricks that could help you ace the exam on the first a† empt.

The first step is to understand The pa† ern of the examinaon. CUET includes three secons, secon 1 includes queson based on languages, secon 2 includes 27 domain-based subjects and secon 3 includes General Test. The syllabus of the upcoming Common University Entrance Test, CUET 2022, will be completely based on the syllabus of class 12 th . No queson will be asked from class 11th syllabus.

While preparing for the exam, it is i m p o r t a n t t o i d e n f y t h e important topics and pracce important quesons from those t o p i c s . P r a c c e i m p o r t a n t q u e s o n s t h r o u g h O s w a a l Q u e so n B a n k a n d S a m p l e Q u e s o n P a p e r s , L i s n g topics also helps in idenfying the weak areas that need special effort and me. The aspirants can start preparing to focus on the areas that they consider to be tough, followed by the ones that are their strengths.

Devote a sufficient amount of me to all the secons of the examinaons. This requires a wellmade plan and an honest adherence to the said plan. Priorize the most important topics or the topics that the aspirants are not familiar with to be able to master them in me.

Make a habit of preparing notes f ro m t h e b e g i n n i n g o f t h e preparaon. It will not only help in making the study systemac but also make the revision of the syllabus easy even when you might have limited me to revise.

Collecng and preparing from the appropriate study material cannot be ignored as irrelevant. The books chosen by the aspirants to study from should be on the lines of the current syllabus and the ones that could help you with swi• revision before the examinaon.

Make sure to revise as much as possible. The revision will help the aspirants in keeping the concepts fresh in their minds unl the day of the final examinaons. They may refer to a few good pracce quesons and concise revision notes to achieve their desired results.

With this said, an important queson that is gaining ground amongst students who will be appearing for this exam is if they should take coaching to get themselves ready for the exams. The answer is a simple no, the exam will simply not require any coaching as it is completely based on the Class 12th syllabus which will be quite fresh in students' minds as they will be just out of school. All they need is a good revision and pracce of quesons from Oswaal Queson Bank and Sample Queson Papers for CUET (UG) preparaons.

Contents l Oswaal Books Expert Tips to Crack CUET (UG) in the first Attempt l Latest Syllabus l CUET Solved Paper 2023 (26th May 2023)

iv - iv

vi - vii

2 - 10 l CUET Solved Paper 2022 (5th August-Slot-2, 6th August-Slot-2, 8th August-Slot-1, 23rd August Slot-1, 26th August Slot-2, 30th August Slot-2) 11 - 75 l CUET Solved Paper 2021 (23rd SEP. 2021–Slot-2 UIQP02)

76 - 80

Sample Question Papers

1 - 5 6 - 9 10 - 14 15 - 20 21 - 25 26 - 29 30 - 33 34 - 37 38 - 41 42 - 45

l Sample Question Paper - 1 l Sample Question Paper - 2 l Sample Question Paper - 3 l Sample Question Paper - 4 l Sample Question Paper - 5 l Sample Question Paper - 6 l Sample Question Paper - 7 l Sample Question Paper - 8 l Sample Question Paper - 9 l Sample Question Paper - 10

Solutions 46 - 49 50 - 53 54 - 58 59 - 63 64 - 68 69 - 72 73 - 76 77 - 80 81 - 84 85 - 88

l Sample Question Paper - 1 l Sample Question Paper - 2 l Sample Question Paper - 3 l Sample Question Paper - 4 l Sample Question Paper - 5 l Sample Question Paper - 6 l Sample Question Paper - 7 l Sample Question Paper - 8 l Sample Question Paper - 9 l Sample Question Paper - 10 

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Latest Syllabus PHYSICS -322 Note: There will be one Question Paper which will have 50 questions out of which 40 questions need to be attempted. Unit I : Electrostatics Electric charges and their conservation. Coulomb’s law – force between two point charges, forces between multiple charges; superposition principle, and continuous charge distribution. Electric field, electric field due to a point charge, electric field lines; electric dipole, electric field due to a dipole; torque on a dipole in a uniform electric field. Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet, and uniformly charged thin spherical shell (field inside and outside). Electric potential, potential difference, electric potential due to a point charge, a dipole and system of charges; equipotential surfaces, the electrical potential energy of a system of two point charges, and electric dipoles in an electrostatic field. Conductors and insulators, free charges, and bound charges inside a conductor. Dielectrics and electric polarization, capacitors and capacitance, the combination of capacitors in series and in parallel, the capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor, Van de Graaff generator. Unit II : Current Electricity Electric current, the flow of electric charges in a metallic conductor, drift velocity and mobility, and their relation with electric current; Ohm’s law, electrical resistance, V-I characteristics (linear and non-linear), electrical energy and power, electrical resistivity and conductivity. Carbon resistors, colour code for carbon resistors; series and parallel combinations of resistors; temperature dependence of resistance. The internal resistance of a cell, potential difference, and emf of a cell, combination of cells in series and in parallel. Kirchhoff ’s laws and simple applications. Wheatstone bridge, metre bridge. Potentiometer – principle, and applications to measure potential difference, and for comparing emf of two cells; measurement of internal resistance of a cell. Unit III : Magnetic Effects of Current and Magnetism Concept of the magnetic field, Oersted’s experiment. Biot - Savart law and its application to current carrying circular loop.

Ampere’s law and its applications to infinitely long straight wire, straight and toroidal solenoids. Force on a moving charge in uniform magnetic and electric fields. Cyclotron. Force on a current-carrying conductor in a uniform magnetic field. The force between two parallel currentcarrying conductors – definition of ampere. Torque experienced by a current loop in a magnetic field; moving coil galvanometer – its current sensitivity and conversion to ammeter and voltmeter. Current loop as a magnetic dipole and its magnetic dipole moment. The magnetic dipole moment of a revolving electron. Magnetic field intensity due to a magnetic dipole (bar magnet) along its axis and perpendicular to its axis. Torque on a magnetic dipole (bar magnet) in a uniform magnetic field; bar magnet as an equivalent solenoid, magnetic field lines; Earth’s magnetic field and magnetic elements. Para-, dia- and ferromagnetic substances, with examples. Electromagnets and factors affecting their strengths. Permanent magnets. Unit IV : Electromagnetic Induction and Alternating Currents Electromagnetic induction; Faraday’s law, induced emf and current; Lenz’s Law, Eddy currents. Self and mutual inductance. Alternating currents, peak and rms value of alternating current/voltage; reactance and impedance; LC oscillations (qualitative treatment only), LCR series circuit, resonance; power in AC circuits, wattless current. AC generator and transformer. Unit V : Electromagnetic Waves Need for displacement current. Electromagnetic waves and their characteristics (qualitative ideas only). Transverse nature of electromagnetic waves. Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays) including elementary facts about their uses. Unit VI : Optics Reflection of light, spherical mirrors, mirror formula. Refraction of light, total internal reflection, and its applications, optical fibres, refraction at spherical surfaces, lenses, thin lens formula, lens maker’s formula. Magnification, power of a lens, combination of thin lenses

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Contd... in contact combination of a lens and a mirror. Refraction and dispersion of light through a prism.

details should be omitted; only the conclusion should be explained.)

Scattering of light–blue colour of the sky and reddish appearance of the sun at sunrise and sunset.

Unit VIII : Atoms and Nuclei

Optical instruments: Human eye, image formation, and accommodation, correction of eye defects (myopia and hypermetropia) using lenses. Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers. Wave optics: Wavefront and Huygens’ principle, reflection, and refraction of plane wave at a plane surface using wavefronts. Proof of laws of reflection and refraction using Huygens’ principle. Interference, Young’s double hole experiment and expression for fringe width, coherent sources, and sustained interference of light. Diffraction due to a single slit, width of central maximum. Resolving the power of microscopes and astronomical telescopes. Polarisation, plane polarised light; Brewster’s law, uses of plane polarised light and Polaroids. Unit VII : Dual Nature of Matter and Radiation  Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation – particle nature of light. Matter waves – wave nature of particles, de Broglie relation. Davisson-Germer experiment (experimental

Alpha - particle scattering experiment; Rutherford’s model of atom; Bohr model, energy levels, hydrogen spectrum. Composition and size of nucleus, atomic masses, isotopes, isobars; isotones. Radioactivity – alpha, beta, and gamma particles/rays, and their properties; radioactive decay law. Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number; nuclear fission and fusion. Unit IX : Electronic Devices Energy bands in solids (qualitative ideas only), conductors, insulators, and semiconductors; semiconductor diode – I-V characteristics in forward and reverse bias, diode as a rectifier; I-V characteristics of LED, photodiode, solar cell, and Zener diode; Zener diode as a voltage regulator. Junction transistor, transistor action, characteristics of a transistor; transistor as an amplifier (common emitter configuration) and oscillator. Logic gates (OR, AND, NOT, NAND and NOR). Transistor as a switch. Unit X: Communication Systems Elements of a communication system (block diagram only); bandwidth of signals (speech, TV, and digital data); bandwidth of transmission medium. Propagation of electromagnetic waves in the atmosphere, sky, and space wave propagation. Need for modulation. Production and detection of an amplitude-modulated wave. 

Don't Stop Reading !

You never know what might be asked in the exam. To download Chapter-wise Mind Maps scan the code below

To download On Tips Notes scan the code below

SCAN

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0808

BARIPADA

UTTAR PRADESH DEHRADUN

(1)

CUET (UG) Exam Paper 2023 National Testing Agency Held on 26th May 2023

PHYSICS Solved

[This includes Questions pertaining to Domain Specific Subject only] Max. Marks: 200 

Time allowed: 45 Minutes

General Instructions: 1. The test is of 45 Minutes duration. 2. The test contains 50 questions out of which 40 questions need to be attempted. 3. Marking Scheme of the test: a. Correct answer or the most appropriate answer: Five marks (+5). b. Any incorrectly marked option will be given minus one mark (−1). c. Unanswered/Marked for Review will be given zero mark (0).

1. A beam of electron is used in Young's double slits

Ans. Option (4) is correct.

experiment. The slit width is d. When velocity of electron is increased, then: (1) No interference is observed (2) Fringe width increases (3) Fringe width decreases (4) Fringe width remains same

Explanation: A simple microscope is a magnifying glass that has a bi convex lens with a short focal length. A compound microscope uses multiple lenses to magnify an image for an observer. It is made of two convex lenses. The first, the ocular lens, is close to the eye and the other lens is objective lens. Reflecting telescopes use mirrors while refracting telescopes use lenses.

Ans. Option (3) is correct. h D , mv d So, the higher the velocity, the lesser the fringe width. Explanation: For electron,  

3. Match List - I with List - II List - I (Field lines)

2. Match List - I with List - II. List - I

List - II

(A) Simple Microscope

(I)

(B) Compound Microscope

(II) Two convex lenses both of small focal length

(C) Refracting Telescope

(III) Two convex lenses, one of small focal length other of large focal length

(D) Reflecting Telescope

(IV) Single convex lens of small focal length

List - II (Charge configuration)

(A)

(I)

q0

(D)

(IV) Unequal charges

A concave mirror is used as an objective

Choose the correct answer from the options given below: (1) (A)-(IV), (B)-(II), (C)-(I), (D)-(III) (2) (A)-(IV), (B)-(II), (C)-(III), (D)-(I) (3) (A)-(II), (B)-(IV), (C)-(III), (D)-(I) (4) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Solved Paper - 2023 Choose the correct answer from the options given below: (1) (A)-(I), (B)-(III), (C)-(II), (D)-(IV) (2) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) (3) (A)-(III), (B)-(I), (C)-(IV), (D)-(II) (4) (A)-(I), (B)-(III), (C)-(IV), (D)-(II)

Ans. Option (3) is correct. Explanation: Electric field lines always point away from a positive charge and towards a negative point. In fact, electric fields originate at a positive charge and terminate at a negative charge. For unequal charges the field lines will not be proportionate at the two sides of the charge, that is the number of field lines will differ.

4. The emfs and resistances in the given circuit have the following values

3

will be a decrease in the barrier potential and also the resistance.

6. A wire carries a current of 10 A in south to north direction. The magnetic field due to 1 cm piece of wire at a point 200 cm north-east from the piece: (1) 0.9 × 10−9 downwards (2) 0.9% × 10−9 T upwards (3) 1.8 × 10−9 T south to north (4) 1.8 × 10−9 T downwards

Ans. Option (4) is correct. Explanation: As the distance of point of observation being much larger than the length of the wire we may write  idl sin  dB  0  4  r2 or, dB  (10 )7 

10  10 2  sin 45° ( 200  10 2 )2

∴ dB = 1.8 × 10−9 T. By right hand thumb rule, the direction is vertically downward.

E1 = 4.2 V, E2 = 1.9 V r1 = 2.0 W, r2,=1.6 W, R = 6.0 W

7.

Ans. Option (3) is correct.

Figure

What is the current in the circuit ? (1) 383 mA (2) 635 mA (3) 240 mA (4) 958 mA Explanation: Applying Kirchhoff ’s law, Let the current in the circuit be i, then 4.2 + 2i + 6i + 1.6i – 1.9 = 0 2.3 ∴|i| = = 0.23958 A = 239.58 mA ≈ 240 mA 9.6

5. During forward bias in a p–n junction: (1) Width of the depletion resistance increases (2) Width of the depletion resistance decreases (3) Width of the depletion resistance decreases (4) Width of the depletion resistance increases

layer increases and layer decreases and layer increases and layer decreases and

Ans. Option (2) is correct. Explanation: During forward bias of p–n junction diode, we are aware of the fact that the p type semiconductor is connected to the positive terminal and the n type semiconductor is connected to the negative terminal. This causes the holes as well as the electrons to migrate towards the junction from the p and n regions, respectively. As they come closer and closer, the width of the depletion layer decreases. The distance between the diffused electrons and holes starts to get reduced and this causes a decrease in the electric field in the depletion area. Thus we can conclude that there

shows a circuit contains three identical resistors with resistance R = 9.0 Ω each, two identical inductors with inductance, L = 2.0 mH each, and an ideal battery with emf e = 18 V. In the given circuit if the switch is closed. In steady state the current will be: (1) 2 A (2) 27 A (3) 6 A (4) 0.67 A

Ans. Option (3) is correct. Explanation: In the steady state the inductor resistance will be zero (closed circuit), so, all the resistances will be parallel. Equivalent resistance will be R/3 =9/3 = 3 ohm V=IR or I =V/R=18/3=6 A

8. Identify the wrong nuclear equation: B  e  

(1)

11 11 6C  5

(2)

210 211 83 B  84

(3)

242 238 94 Pu  92

(4)

 120 54 Xe  e

P0  e   U  42 He

120 53 I  

Ans. Option (2) is correct. Explanation: Electron release will not change the mass number it will remain unchanged so option (2) is wrong.

OSWAAL CUET (UG) Sample Question Papers, PHYSICS

4

9. A potentiometer is supplied with a constant voltage of 5 V. A cell of emf 1.4 V is balanced by the voltage drop across 280 cm of the potentiometer wire. The total length of the potentiometer wire will be: (1) 5 m (2) 4 m (3) 10 m (4) 8 m

Ans. Option (3) is correct. Explanation: Let the total length be l so E/E' =l/l' or l = El'/E' or l = 5×

E =

below: (1) (A) B2 (2) B2 > B1 (3) B1 = B2 (4) B1 = B2 = 0 16. To convert a galvanometer to ammeter a shunt S is to be connected with the galvanometer. The effective resistance of the ammeter then is (1) GS/(G+S) (2) (G+S)/GS (3) G+S (4) None of the above 17. A galvanometer can be converted into a voltmeter by connecting a (1) high resistance in series. (2) high resistance in parallel. (3) low resistance in parallel. (4) low resistance in series. 18. When a charged particle moves through a magnetic field perpendicular to its direction. Then (1) linear momentum changes (2) kinetic energy remains constant (3) both (1) and (2) (4) both linear momentum and kinetic energy varies 19. Current sensitivity of a galvanometer is given by (1) Cq/nBA (2) nBA/C (3) nBA/CG (4) CG/nBA 20. A coil of N turns and radius R carries a current I. It is unwound and rewound to make a square coil of side a having same number of turns (N). Keeping the current I same, the ratio of the magnetic moments of the circular coil and the square coil is

R2 (1) p 2 a (3)

R2 2 a

(2) p

a2 R2

(4) None of the above

21. A magnetic dipole moment is a vector quantity directed from:

Sample Question Papers

(1) South to North (3) East to West

(2) North to South (4) West to East

22. A ferromagnetic substance is heated above its curie

temperature. Which of the following statements is correct?           (1) Ferromagnetic domains get perfectly arranged. (2) Ferromagnetic domains get randomly arranged. (3) Ferromagnetic domains are not at all influenced. (4) Ferromagnetic material transforms into diamagnetic substance.



23. Which of the following relation is correct? (1) B = BV × BH

(2) B = BV / BH

(3) B = BV + BH

(4) B =

B2V + B2H

24. Three waves A, B and C of frequencies 1,600

kHz, 5 MHz and 60 MHz, respectively are to be transmitted from one place to another. Which of the following is the most appropriate mode of communication? (1) A is transmitted via space wave while B and C are transmitted via sky wave. (2) A is transmitted via ground wave, B via sky wave and C via space wave. (3) B and C are transmitted via ground wave while A is transmitted via sky wave. (4) B is transmitted via ground wave while A and C are transmitted via space wave.

25. A square of side L meters lies in the x-y plane

in a region where the magnetic field is given by B = B0 (2ˆi + 3ˆj + 4 kˆ) Tesla, where B0 is constant. The magnitude of flux passing through the square is (1) 2B0L2 Wb (2) 3B0L2 Wb 2



(4) 29B0 L Wb 26. The polarity of induced emf is defined by (1) Ampere’s circuital law. (2) Biot-Savart law. (3) Lenz’s law. (4) Fleming’s right hand rule.



27. Which of these sets of logic gates are designated as





(3) 4B0

L2

Wb

universal gates? (1) NAND (3) EX-OR

(2) NOX (4) EX-NOR

28. Magnetic field energy stored in a coil is (1) Li2 (2) ½ Li (3) Li (4) ½ Li2



L1 L2

(2) M = k



(3) M = k L1 + L2

(4) None of the above

30. In which of the following application, eddy current

has no role to play? (1) Electric power meters (2) Induction furnace

31. If the rms current in a 50 Hz AC circuit is 5 A, the value of the current



(1) 5 2 A



(3)

5 A 6

1 s is: 300

3 (2) 5 A 2 5 A (4) 2



32. When a voltage measuring device is connected



33. Which of the following combinations should be

to AC mains, the meter shows the steady input voltage of 220 V. this means (1) input voltage cannot be AC voltage, but a DC voltage. (2) maximum input voltage is 220 V. (3) The meter reads not v but (v2) and is calibrated 2 to read ( v ) . (4) The pointer of the meter is stuck by some mechanical defect.



selected for better tuning of an L-C-R circuit used for communication? (1) R = 20 W, L = 1.5 H, C = 35 mF (2) R = 25 W, L = 2.5 H, C = 45 mF (3) R = 15 W, L = 3.5 H, C = 30 mF (4) R = 25 W, L = 1.5 H, C = 45 mF



34. The sharpness of tuning of a series LCR circuit at resonance is measured by Q factor of the circuit which is given by



(1) Q =

1 L R C

(2) Q =

1 C R L



(3) Q =

1 R L C

(4) Q =

1 R C L

35. The underlying principle of transformer is (1) resonance. (2) mutual induction. (3) self induction. (4) none of the above.

36. The power factor of series LCR circuit at resonance

is (1) 0.707 (3) 0.5

together, their mutual inductance becomes (1) M = k L1L2



(3) LED lights (4) Magnetic brakes in trains



29. If two coils of self inductance L1 and L2 are coupled



33



(2) 1 (4) 0

37. A linearly polarized electromagnetic wave given

as E = E0 i cos (kz–wt) is incident normally on a perfectly reflecting infinite wall at z = a. Assuming that the material of the wall is optically inactive, the reflected wave will be given as (1) Er = – E0 i cos (kz-wt) (2) Er = E0 i cos (kz +wt) (3) Er = – E0 i cos (kz+wt) (4) Er = E0 i sin (kz-wt)

38. In vacuum, the wavelength of the electromagnetic wave of frequency 5 × 1019 Hz is (1) 6 × 10–12 m (2) 3 × 10–8 m 11 (3) 1.6 × 10 m (4) 15 × 1027 m

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

4

39. Semiconductors behave like insulators at ______ (1) 0°C (2) 0 K (3) 273 K (4) None of the above

40. In Figure, Vo is the potential barrier across a p-n



(1) 1 and 3 both correspond to forward bias of junction (2) 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junction (3) 1 corresponds to forward bias and 3 corresponds to reverse bias of junction. (4) 3 and 1 both correspond to reverse bias of junction.



junction, when no battery is connected across the junction

41. The relationship between angle of incidence i,



prism of angle A and angle of minimum deviation for a triangular prism is (1) A + dm = i (2) A + dm = 2i (3) A + dm/2 = i (4) 2A + dm = i



42. Which of the following graphs shows the variation



of de-Broglie wavelength with potential through which a particle of charge q and mass m is accelerated? (1) (2)



(3)

(4)







43. Taking the Bohr radius as a0 = 53 pm, the radius of



44. Convex mirrors are preferred over plane mirrors as

Li++ ion in its ground state, on the basis of Bohr’s model, will be about (1) 53 pm. (2) 27 pm. (3) 18 pm. (4) 13 pm. rear view mirror in automobile since (1) the image formed is magnified. (2) the image formed is real. (3) the field of view is large. (4) it is light weight.

45. Assertion (A): Density of all the nuclei is same. Reason (R): Radius of nucleus is directly proportional to the cube root of mass number. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is true II. Read the following text and answer the following questions on the basis of the same: Diffraction in a hall: A and B went to purchase a ticket of a music programme. But unfortunately only one ticket was left. They purchased the single ticket and decided that A would be in the hall during the 1st half and B during the 2nd half. Both of them reached the hall together. A entered the hall and found that the seat was behind a pillar which creates an obstacle. He was disappointed. He thought that he would not be able to hear the programme properly. B was waiting outside the closed door. The door was not fully closed. There was a little opening. But surprisingly, A could hear the music programme. This happened due to diffraction of sound. The fact we hear sounds around corners and around barriers involves both diffraction and reflection of sound. Diffraction in such cases helps the sound to "bend around" the obstacles. In fact, diffraction is more pronounced with longer wavelengths implies that we can hear low frequencies around obstacles better than high frequencies. B was outside the door. He could also hear the programme. But he noticed that when the door opening is comparatively less he could hear the programme even being little away from the door. This is because when the width of the opening is larger than the wavelength of the wave passing through the gap then it does not spread out much on the other side. But when the opening is smaller than the wavelength more diffraction occurs and the waves spread out greatly – with semicircular wavefront. The opening in this case functions as a localized source of sound.

55

Sample Question Papers



46. A and B could hear the music programme due to phenomenon named (1) interference. (3) diffraction.

(2) scattering. (4) dispersion.

47. Diffraction is more pronounced with ______ wavelengths. (1) Longer (3) fluctuating

(2) Shorter (4) all

48. The minimum and maximum frequencies in the

musical programme were 550 Hz and 10 kHz. Which frequency was better audible around the pillar obstacle? (1) 10 kHz (2) 550 kHz (3) Mid frequency



(4) The complete frequency range

49. Diffraction of sound takes place more when :

(1) sound is diffracted through an opening having width equal to the wavelength of the sound. (2) sound is diffracted through an opening having width more than the wavelength of the sound. (3) sound is diffracted through an opening having width less than the wavelength of the sound. (4) diffraction of sound does not depend on the width of the opening.

50. How the waveform will look like outside the door of the hall? (1) Sound repeater (2) Sound reflector (3) Localized sound source (4) None of the above



SAMPLE

Question Paper Maximum Marks : 200

2

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50. (ii) Correct answer or the most appropriate answer: Five marks (+5). (iii) Any incorrect option marked will be given minus one mark (– 1). (iv) Unanswered/Marked for Review will be given no mark (0). (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted.

Section - II



PHYSICS

(1) (a)

1. Hole is

(1) an anti-particle of electron. (2) a vacancy created when an electron leaves a covalent bond. (3) absence of free electrons. (4) an artificially created particle. 2. When a forward bias is applied to a p-n junction, it (1) raises the potential barrier. (2) reduces the majority carrier current to zero. (3) lowers the potential barrier. (4) None of the above 3. Plastic rod rubbed with fur and glass rod rubbed with silk (1) repel each other (2) mix up with each other (3) attract each other (4) None of the above 4. A hemisphere is uniformly positively charged. The electric field at a point on a diameter away from the centre is directed (1) perpendicular to the diameter. (2) parallel to the diameter. (3) at an angle tilted towards the diameter. (4) at an angle tilted away from the diameter.

5. Truth table for the given circuit figure is A

B

(2) (b)





(c) (3)



(4) (d)

A

B

E

0

0

1

0

1

0

1

0

1

1

1

0

A

B

E

0

0

1

0

1

0

1

0

0

1

1

1

A

B

E

0

0

0

0

1

1

1

0

0

1

1

1

A

B

E

0

0

0

0

1

1

1

0

1

1

1

0

C

D





Sample Question Papers

6. The work done to move a charge along an



7. The electric potential at a point on the equatorial

equipotential surface from A to B (1) cannot be defined. (2) is a negative quantity. (3) is zero. (4) is a positive quantity.

line of a electric dipole is (1) directly proportional to the square of the distance. (2) indirectly proportional to the square of the distance. (3) directly proportional to the charge. (4) None of the above

8. A parallel plate capacitor is charged by connecting



9. A capacitor of 4 mF is connected as shown in the

it to a battery. Which of the following will remain constant if the distance between the plates of the capacitor is increased in this situation? (1) Energy stored (2) Electric field (3) Potential difference (4) Capacitance circuit Figure. The internal resistance of the battery is 0.5 W. The amount of charge on the capacitor plates will be :

(1) 0 mC (3) 16 mC

(2) 4 mC (4) 8 mC

10. The magnitude of electric force, F is (1) directly proportional to the multiplication of both charges. (2) directly proportional to the distance between both charges. (3) directly proportional to the square of the distance between both charges. (4) constant. 11. The Electric field at a point is (1) always discontinuous. (2) discontinuous if there is a positive charge at that point. (3) discontinuous only if there is a negative charge at that point. (4) discontinuous if there is a charge at that point.

12. Which of the following I-V characteristic represents

the characteristic of an Ohmic conductor? (1)



77

(2)

(3)



(4)



13. Wheatstone Bridge is not suitable for measurement



14. In a metre bridge, what is the effect on null



15. A 100 m long antenna is mounted on a 500 m tall

of (1) very high value resistances. (2) very low value resistances. (3) both (1) and (2). (4) medium value resistances.

deflection of galvanometer, when the radius of the meter bridge wire is doubled? (1) There will be no change (2) Null point will shift to L1 / 2 point (3) Null point will shift to 2L1 point (4) Null point will not be available



building. The complex can become a transmission tower for waves with l (1) ~400 m. (2) ~25 m. (3) ~150 m. (4) ~2,400 m.



16. A current carrying circular loop of radius R is



17. Magnetic field due to a solenoid at any point



18. A magnetic dipole moment is a vector quantity

placed in the x – y plane with centre at the origin. Half of the loop with x > 0 is now bent so that it now lies in the y – z plane. (1) The magnitude of magnetic moment now diminishes. (2) The magnetic moment does not change. (3) The magnitude of B at (0,0,z), z >>R increases. (4) The magnitude of B at (0,0,z), z >>R is unchanged. inside it is B = m0ni. Magnetic field at the end of the solenoid is (1) B (2) B/2 (3) 2B (4) B/4



directed from: (1) South to North (3) East to West

(2) North to South (4) West to East

8

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

19. The self-inductance L of a solenoid of length l and

area of cross-section A, with a fixed number of turns N increases as (1) l and A increase. (2) l decreases and A increases. (3) l increases and A decreases. (4) both l and A decrease.

20. An inductor and a bulb are connected in series with



21. There are two coils A and B as shown in figure. A

a dc source. A soft iron core is then inserted in the inductor. What will happen to intensity of the bulb? (1) Intensity of the bulb remains the same. (2) Intensity of the bulb decreases. (3) Intensity of the bulb increases. (4) The bulb ceases to glow. current starts flowing in B as shown, when A is moved towards B and stops when A stops moving. The current in A is counter clockwise. B is kept stationary when A moves. We can infer that

(1) there is a constant current in the clockwise direction in A. (2) there is a varying current in A. (3) there is no current in A. (4) there is a constant current in the counterclockwise direction in A.

22. When frequency of applied alternating voltage is

very high then (1) A capacitor will tend to become SHORT (2) An inductor will tend to become SHORT (3) Both (1) and (2) (4) No one will become short

23. Relation between r.m.s. voltage and instantaneous voltage of an AC (1) V0 = Vrms / √2 (2) Vrms = V0 / √2 (3) Vrms = 0.707V0 (4) Both (2) and (3)

24. The core of a transformer is laminated as (1) it improves the ratio of voltage in the primary and secondary. (2) it checks rusting of the core may be stopped. (3) it reduces energy losses due to eddy currents. (4) it increases flux linkage. 25. Quantity that remains unchanged in a transformer is (1) voltage. (2) current. (3) frequency. (4) none of these.

1 3 r 1 (3) r (1)



27. An EM wave radiates outwards from a dipole

antenna, with E0 as the amplitude of its electric field vector. The electric field E0 which transports significant energy from the source falls off as

1 r2

(4) remains constant

28. Find the true statement. (1) A Van de Graff generator produces large voltage and less current (2) A Van de Graff generator produces large resistance and less voltage (3) A Van de Graff generators produces large current and large resistance (4) A Van de Graff generators produces large current and less voltage

29. The phenomena involved in the reflection of radio



30. The focal length of a concave mirror is f. An



31. If m1 and m2 be the linear magnifications of the



32. Phase difference between any two points of a

waves by ionosphere are similar to (1) reflection of light by a plane mirror. (2) total internal reflection of light in air during a mirage. (3) dispersion of light by water molecules during the formation of a rainbow. (4) scattering of light by the particles of air. object is placed at a distance x from the focus. The magnification is (1) (f+x)/f (2) f/x (3) x/f (4) f/(f+x)

objective and eyepiece of a compound microscope, then the magnifying power of the compound microscope is (1) m1 + m2 (2) m1 – m2 (3) m1 × m2 (4) (m1+m2)/2 wavefront is (1) p (3) 0

(2) p/2 (4) p/4

33. Huygens' theory could not explain (1) photoelectric effect. (2) reflection of light. (3) diffraction of light. (4) interference of light.

34. A Young’s Double slit experiment is performed in air and in water. Which of the following relationship is true regarding fringe width b? (1) bAIR > bWATER (2) bWATER > bAIR (3) bAIR = bWATER (4) bWATER = 0

35. The wavelength of a photon needed to remove



a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly (1) 1.2 nm (2) 1.2 × 10–3 nm (3) 1.2 × 10–6 nm (4) 1.2 × 10 nm



36. Kinetic energy of electrons emitted in photoelectric

26. Light with an energy flux of 20 W/cm2 falls on a

non-reflecting surface at normal incidence. If the surface has an area of 30 cm2, the total momentum delivered (for complete absorption) during 30 minutes is (1) 36 × 10–5 kg m/s. (2) 36 × 10–4 kg m/s. (3) 108 × 104 kg m/s. (4) 1.08 × 107 kg m/s.

(2)

effect is (1) directly proportional to the intensity of incident light. (2) inversely proportional to the intensity of incident line. (3) independent of the intensity of incident light. (4) independent of the frequency of light.

99

Sample Question Papers 37. A set of atoms in an excited state decays. (1) in general, to any of the states with lower energy. (2) into a lower state only when excited by an external electric field. (3) all together simultaneously into a lower state. (4) to emit photons only when they collide.

38. The simple Bohr model cannot be directly applied



39. Heavy stable nucleus have more neutrons than



40. Nuclear force is a ___________ and ___________



41. Consider the following reaction:

to calculate the energy levels of an atom with many electrons. This is because (1) of the electrons not being subject to a central force. (2) of the electrons colliding with each other. (3) of screening effects. (4) the force between the nucleus and an electron will no longer be given by Coulomb’s law.

(3) A is true but R is false (4) A is false and R is true I. Read the following text and answer the following questions on the basis of the same: Photocell: A photocell is a technological application of the photoelectric effect. It is a device whose electrical properties are affected by light. It is also sometimes called an electric eye. A photocell consists of a semi-cylindrical photo-sensitive metal plate C (emitter) and a wire loop A (collector) supported in an evacuated glass or quartz bulb. It is connected to the external circuit having a high-tension battery B and micro ammeter (mA) as shown in the figure.

protons. This is because of the fact that (1) neutrons are heavier than protons. (2) electrostatic force between protons are repulsive. (3) neutrons decay into protons through beta decay. (4) nuclear forces between neutrons are weaker than that between protons. force. (1) Strong, long-range (2) Strong, short range (3) Weak, long-range (4) Weak, short-range

A A A −4 A −4 Z X → Z +1Y → Z −1 Z → Z −1 Z Radioactive radiation emitted in the following sequence: (1) β, α, γ (2) a, b, γ (3) g, b, a (4) β, g, a



42. When the conductivity of a semiconductor is

due to rupture of its covalent bond only then the semiconductor is called (1) Intrinsic (2) Extrinsic (3) Donor (4) Acceptor

43. The magnetic flux linked with a coil is given by an

equation f = 5t2 + 2t + 3. The induced e.m.f. in the coil at the third second will be                                                       (1) 32 units (2) 54 units (3) 40 units (4) 65 units

44. A galvanometer of 100 W resistance gives full scale



45. Assertion (A): Self inductance may be called the

deflection for 10 mA current. To use it as an ammeter of 10 A range, the resistance of the shunt required is (1) 10 W (2) 0.10 W (3) 0.01 W (4) 0.001 W inertia of electricity. Reason (R): Due to self inductance, opposing induced e.m.f. is generated in a coil as a result of change in current or magnetic flux linked with the coil. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A

Sometimes, instead of the plate C, a thin layer of photosensitive material is pasted on the inside of the bulb. A part of the bulb is left clean for the light to enter it. When light of suitable wavelength falls on the emitter C, photoelectrons are emitted. These photoelectrons are drawn to the collector A. Photocurrent of the order of a few microampere can be normally obtained from a photo cell. A photocell converts a change in intensity of illumination into a change in photocurrent. This current can be used to operate control systems and in light measuring devices.

46. Photocell is an application of



47. Photosensitive material should be connected to

(1) thermoelectric effect. (2) photoelectric effect. (3) photoresistive effect. (4) None of the above

(1) –ve terminal of the battery. (2) +ve terminal of the battery. (3) any one of (1) or (2). (4) connected to ground.

48. Which of the following statement is true? (1) The photocell is totally painted black. (2) A part of the photocell is left clean. (3) The photocell is completely transparent. (4) A part of the photocell is made black.

49. The photocurrent generated is in the order of



50. A photocell converts a change in ___ of incident

(1) ampere (2) milliampere (3) microampere (4) None of the above

light into a change in ________ (1) intensity, photovoltage (2) wavelength, photovoltage (3) frequency, photocurrent (4) intensity, photocurrent



SAMPLE

Question Paper Maximum Marks : 200

3

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50. (ii) Correct answer or the most appropriate answer: Five marks (+5). (iii) Any incorrect option marked will be given minus one mark (– 1). (iv) Unanswered/Marked for Review will be given no mark (0). (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted.



Section - II PHYSICS

1. The conductivity of a semiconductor increases

with increase in temperature because (1) number density of free current carriers increases. (2) relaxation time increases. (3) both number density of carriers and relaxation time increase. (4) number density of current carriers increases; relaxation time decreases but effect of decrease in relaxation time is much less than the increase in number density. 2. In an unbiased p-n junction, holes diffuse from the p-region to n-region because (1) free electrons in the n-region attract them. (2) they move across the junction by the potential difference. (3) hole concentration in p-region is more as compared to n-region. (4) All of the above. 3. In the circuit shown in figure, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is

(1) 1.3 V (3) 0 V

(2) 2.3 V (4) 0.5 V



4. ___________ has the mass closest to the mass of



5. Suppose we consider a large number of containers



6. Two nuclei have mass number in the ratio 1 : 2. The



7. The binding energy of a H-atom, considering an

positron. (1) Proton (2) Neutron (3) Electron (4) Neutrino

each containing initially 10,000 atoms of a radioactive material with a half-life of 1 year. After 1 year, (1) all the containers will have 5,000 atoms of the material. (2) all the containers will contain the same number of atoms of the material but that number will only be approximately 5,000. (3) the containers will in general have different numbers of the atoms of the material but their average will be close to 5,000. (4) none of the containers can have more than 5000 atoms. ratio of their nuclear densities is (1) 1 : 2 (2) 2 : 1 (3) 1 : 1 (4) Cannot be defined from mass number ratio

electron moving around a fixed nuclei (proton), is

Sample Question Papers me 4 (m = electron mass). If one decides 8n 2 ε02 h 2 to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be B=−

B=−

Me 4 (M = proton mass) . 8n 2 ε02 h 2

This last expression is not correct because (1) n would not be integral. (2) Bohr-quantisation applies only to electron (3) the frame in which the electron is at rest is not inertial. (4) the motion of the proton would not be in circular orbits, even approximately. 8. Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced, is (1) 10.20 eV. (2) 20.40 eV. (3) 13.6 eV. (4) 27.2 eV. 9. An electron is moving with an initial velocity v = vo i and is in a magnetic field B = B j . Then, its o

de-Broglie wavelength (1) remains constant. (2) increases with time. (3) decreases with time. (4) increases and decreases periodically. 10. A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to (1) H  (2) H1/2 (3) H0 (4) H–1/2 11. A Young’s double slit experiment is performed with blue (wavelength 460 nm) and green light (wavelength 550 nm) respectively. If y is the distance of 4th maximum from the central fringe then (1) yB = yG (2) yB > yG (3) yG > y B (4) yB/yG = 550/460 12. In Young’s double slit experiment, the distance between the slits is reduced to half and the distance between the slits and the screen is doubled. The fringe width (1) will be double. (2) will be half. (3) will remain same. (4) will be four times. 13. The main condition for diffraction to be observed is (1) size of obstacle should be comparable to the wavelength of the wave (2) size of obstacle should be much larger than the wavelength of the wave (3) size of obstacle should be much smaller than the wavelength of the wave (4) for any size of obstacle

14. What is the order of potential difference built up

by the Van de Graff generator? (1) Potential difference of the order of hundreds (2) Potential difference of the order of several million volts

11 11

(3) Potential difference of the order of thousands (4) Potential difference of the order of tens 15. _________ mirror has real focus. (1) Concave (2) Convex (3) Plane (4) all of the above

16. In a concave mirror, an object is placed at a distance



17. The optical density of turpentine is higher than

x1 from the focus. Image if formed at a distance x2 from the focus. The focal length of the mirror is (1) x1x2. (2) x1 + x2. (3) x1/x2. (4) None of these



that of water while its mass density is lower. Figure shows a layer of turpentine floating over water in a container. For which one of the four rays incident on turpentine in Figure, the path shown is correct?

(1) 1 (3) 3

(2) 2 (4) 4

18. Which of the following graphs is the correct angle of incidence vs. angle of deviation graph?



(1)



(2)



(3)

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

12

(4)



26. At resonance, the impedance in series LCR circuit is



27. If rotational velocity of an armature is doubled,



28. A loop, made of straight edges has six corners at

(1) maximum. (3) infinity.

(2) zero. (4) minimum.

emf generated in a generator will be (1) half. (2) two times. (3) four times. (4) unchanged.



A(0, 0, 0), B(L, 0, 0) C(L, L, 0), D(0, L, 0), E(0, L, L) and F(0, 0, L). A magnetic field B=B0 (iˆ + kˆ) Tesla is present in the region. The flux passing through the loop ABCDEFA (in that order) is

19. In a compound microscope, image produced by



objective is ______ and the image produced by eyepiece is ______. (1) Real, real (2) Virtual, virtual (3) Real, virtual (4) Virtual, real





20. How many entries will be in the truth table of 2





input of NAND gate? (1) 8 (3) 4

21. The electric field intensity produced by the

(3)

E 2



(4) 2E

22. In electromagnetic waves, the phase difference



between magnetic and electric field vectors is (1) zero (2) p (3) p/2 (4) p/4



23. An alternating current generator has an internal resistance Rg and an internal reactance Xg. It is used to supply power to a passive load consisting of a resistance Rg and a reactance XL. For maximum power to be delivered from the generator to the load, the value of XL is equal to (1) zero

(2) Xg

(3) −Xg

(4) Rg



24. The heat produced in a given resistance in a given time by the sinusoidal current I0 sin wt will be the same as heat produced by a steady current of magnitude

(1) 0.707 I0

(3)

(2) 6 (4) 10

radiations coming from 100 W bulb at a 3 m distance is E. The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is E (1) (2) 2E 2

(3) I0

(2) 1.412 I0 (4) √I0

(1) B0L2 Wb.

(2) 2B0L2 Wb. (4) 4B0L2 Wb.

2

2B0 L Wb .

29. The coil A is made to rotate about a vertical axis as

shown in figure. No current flows in B if A is at rest. The current in coil A, when the current in B (at t = 0) is counter-clockwise and the coil A is as shown at this instant, t = 0, is A

w

B

(1) constant current clockwise. (2) varying current clockwise. (3) varying current counter-clockwise. (4) constant current counter-clockwise.

30. A 1 kW signal is transmitted using a communication channel which provides attenuation at the rate of –2 dB per km. If the communication channel has a total length of 5 km, the power of the signal P received is [Gain (in dB) = 10 log  0  ] P 1



(1) 900 W. (3) 990 W.

(2) 100 W (4) 1010 W.

31. A cylindrical bar magnet is rotated about its axis in

the figure. A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then axis A N bar magnet

w

25. An A.C. source is connected to a resistive circuit. Which of the following statements is true?

(1) Current leads the voltage in phase

(2) Current lags the voltage in phase



(3) Current and voltage are in same phase



(4) Either (1) or (2) depending on the value of resistance.

S

(1) a direct current flows in the ammeter A. (2) no current flows through the ammeter A. (3) an alternating sinusoidal current flows through the ammeter A with a time period 2p/w.

Sample Question Papers (4) a time varying non-sinusoidal current flows through the ammeter A.

32. Which of the following is most suitable for the core

of an electromagnet? (1) Soft iron (3) Alnico

(2) Steel (4) Copper



33. Relative permeability of a magnetic material is 0.5.



34. A toroid of n turns, mean radius R and cross-

The material is (1) diamagnetic. (2) ferromagnetic. (3) paramagnetic. (4) not a magnetic material.

sectional radius a carries current I. It is placed on a horizontal table taken as x-y plane. Its magnetic moment m (1) is non-zero and points in the z-direction by symmetry. (2) points along the axis of the toroid (m = mf). (3) is zero, otherwise there would be a field falling 1 as 3 at large distances outside the toroid. r

13 13

shifts by 50 cm. What is the value of the unknown resistance? (1) 250 W (2) 10 W (3) 16.7 W (4) None of the above

40. The shape of equipotential surfaces due to an

isolated charge is (1) Concentric spherical shells and the distance between the shells increases with the decrease in electric field. (2) Concentric spherical shells and the distance between the shells decreases with the decrease in electric field. (3) Equi-spaced concentric spherical shells. (4) Changes with the polarity of the charge.



41. Figure shows some equipotential lines distributed in space. A charged object is moved from point A to point B.

(4) is pointing radially outwards.

35. An ammeter gives full scale deflection when



36. The strength of the magnetic field at distance r

current of 1.0 A is passed in it. It is converted into a 100 A range ammeter, what will be the ratio of the shunt resistance and its resistance ? (1) 1 : 9 (2) 9 : 1 (3) 1 : 11 (4) 11 : 1

from a long straight current carrying wire is B. The field at a distance r/2 will be (1) B (2) 2B (3) B/2 (4) B/4

37. Biot-Savart law indicates that the moving electrons (velocity v) produce a magnetic field B such that

(1) B ^ v. (2) B ||v. (3) it obeys inverse cube law. (4) it is along the line joining the electron and point of observation.

(1) The work done in Figure (i) is the greatest. (2) The work done in Figure (ii) is least. (3) The work done is the same in Figure (i), Figure (ii) and Figure (iii). (4) The work done in Figure (iii) is greater than Figure (ii), but equal to that in Figure (i). 42. In Huygens theory, light waves (1) are transverse waves and require a medium to travel. (2) are longitudinal waves and require a medium to travel. (3) are transverse waves and require no medium to travel. (4) are longitudinal waves and require no medium to travel.

38. A length L of wire carries a steady current I. It is



bent first to form a circular plane coil of one turn. A current I flowing through it produces a magnetic field B at the centre of the coil. The same length is now bent more sharply to form a double loop of smaller radius. The magnetic field at the centre caused by the same current is (1) B (2) 2B (3) 4B (4) B/2



39. Consider a metre bridge whose length of wire is 2m. A resistance of 10 W is connected across one gap of the meter bridge and an unknown resistance is connected across the other gap. When these resistances are interchanged, the balance point



43. The radius of curvature of the curved surface of a

plano-convex lens is 20 cm. If the refractive index of the material of the lens be 1.5, it will (1) act as a convex lens only for the objects that lie on its curved side. (2) act as a concave lens for the objects that lie on its curved side. (3) act as a convex lens irrespective of the side on which the object lies. (4) act as a concave lens irrespective of side on which the object lies.

44. The current density (number of free electrons

per m3) in metallic conductor is of the order of (1) 1028 (2) 1023 20 (3) 10 (4) 1015

14

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

45. Assertion (A): Principle of operation of AC

generator is electromagnetic induction. Reason (R): Resistance offered by inductor for AC is zero. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A

(3) A is true but R is false (4) A is false and R is true I. Read the following text and answer the following questions on the basis of the same:

Losses of transformer There are 4 types of losses in a transformer: Core loss, Ohmic loss,  Stray load loss and dielectric loss.

This loss is also called variable or ohmic losses because this loss changes based on the load. (3) Stray Loss These types of losses in a transformer occur because of the occurrence of the leakage flux. As compared with copper and iron losses, the percentage of stray losses are less, so these losses can be neglected. (4) Dielectric Loss This loss mainly occurs within the oil of the transformer. Oil is an insulating material. Once the oil quality in the transformer deteriorates then the transformer’s efficiency is affected. Efficiency of Transformer It is the ratio of output power and input power. Efficiency = Output Power / Input Power. The transformer is a highly efficient device which ranges between 95% – 98.5%.

(1) Core loss

46. What is the relationship among core loss, hysteresis loss and eddy current loss? Core loss  has two components - hysteresis loss  and eddy current loss. These together are called (1) Eddy current loss = Core loss + Hysteresis loss no-load losses of a transformer and are calculated (2) Core loss = Hysteresis loss + eddy current loss by open circuit test. (3) Hysteresis loss = Core loss + eddy current loss (1) Hysteresis loss: This loss mainly depends (4) Core loss = Hysteresis loss X eddy current loss on the core material used in the transformer. To 47. Which of the following losses in transformer is also reduce this loss, the high-grade core material can known as no-load loss? be used. CRGO- Cold rolled grain oriented Si steel (1) Copper loss (2) Stray loss is commonly used for this purpose. (3) Dielectric loss (4) Core loss (2) Eddy current loss: This loss can be reduced by 48. Which of the following losses in transformer is also designing the core using slight laminations.  known as variable loss? These losses are present even when no load is (1) Copper loss (2) Stray loss connected. So, these are also known as no-load (3) Dielectric loss (4) Core loss loss. (2) Copper Loss Copper losses occur because of the Ohmic resistance in the windings of the transformer. If the currents in primary and secondary windings of the transformer are I1 and I2, and if the resistances of these windings are R1 & R2 then the copper losses that occurred in the windings are I12R1 & I22R2 respectively. So, the entire copper loss will be I1 2 R 1 + I 2 2 R 2 .



49. How hysteresis loss can be reduced? (1) Using core of Si Steel (2) Using laminated core (3) Using core of non-ferro magnetic material (4) Using oil of higher dielectric constant

50. Specify the range of transformer efficiency. (1) 10-15% (2) 95-98% (3) 50-60% (4) 40-50% 

SAMPLE

Question Paper Maximum Marks : 200

4

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50. (ii) Correct answer or the most appropriate answer: Five marks (+5). (iii) Any incorrect option marked will be given minus one mark (– 1). (iv) Unanswered/Marked for Review will be given no mark (0). (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted.



Section - II PHYSICS

1. Five charges q1, q2, q3, q4, and q5 are fixed at their



2. Figure shows electric field lines in which an electric dipole p is placed as shown. Which of the following statement is correct?

positions as shown in figure. S is a Gaussian surface. The Gauss’s law is given by: ∫ E.ds = q / e0

-q

Which of the following statement is correct? (1) E on the LHS of the above equation will have a contribution from q1, q5 and q3, while q on the RHS will have a contribution from q2 and q4 only. (2) E on the LHS of the above equation will have a contribution from all charges, while q on the RHS will have a contribution from q2 and q4 only. (3) E on the LHS of the above equation will have a contribution from all charges, while q on the RHS will have a contribution from q1, q3 and q5 only. (4) Both E on the LHS and q on the RHS will have contributions from q2 and q4 only.

P

+q

(1) The dipole will not experience any force. (2) The dipole will experience a force towards right. (3) The dipole will experience a force towards left. (4) The dipole will experience a force upwards.

3. The electrostatic potential on the surface of a

charged conducting sphere is 100 V. Two statements are made in this regard : S1 : At any point inside the sphere, electric intensity is zero. S2 : At any point inside the sphere, the electrostatic potential is 100 V. Which of the following is a correct statement? (1) S1 is true, but S2 is false. (2) Both S1 and S2 are false. (3) S1 is true, S2 is also true and S1 is the cause of S2. (4) S1 is true, S2 is also true but the statements are independent.

16

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS



4. A positively charged particle is released from rest



5. Two spheres are separately charged and then



6. 4 capacitors, each of 2 mF, are connected as shown.

in an uniform electric field. The electric potential energy of the charge (1) remains constant because the electric field is uniform. (2) increases because the charge moves along the electric field. (3) decreases because the charge moves along the electric field. (4) decreases because the charge moves opposite to the electric field. brought in contact, so (1) total charge on the two spheres is conserved. (2) total energy of the two spheres is conserved. (3) Both (1) and (2) (4) None of the above



What will be the equivalent capacitor across the points A, B?

(1) 0.5 mF (3) 8 mF

(2) 2 mF (4) 4 mF

7. In the circuit shown in figure, initially key K1 is closed and key K2 is open. Then K1 is opened and K2 is closed. Then

(1) Voltage across C1 = Voltage across C2 (2) Voltage across C1 > Voltage across C2 , if C1 > C 2 (3) Charge on C1 = charge on C2 (4) None of the above

8. Electric charges under action of electric forces is

called (1) electrostatic (3) electric field

e e (4) The charge to mass ratio satisfy:   +   = 0  m 1  m 2

13. When a charge of 1C moving with velocity 1 m/s



15. The ratio of contributions made by the electric field

normal to a magnetic field experiences a force 1 N , then the magnitude of the magnetic field is (1) 1 Gauss (2) 1 Tesla (3) 1 Oersted (4) None of the above   14. If E and B represents electric and magnetic field vectors of the electromagnetic wave, the direction of propagation of electromagnetic wave is along   (1) E (2) B     (3) (4) E × B B×E

and magnetic field components to the intensity of an EM wave is

(1) c : 1

(2) c2 : 1

(3) 1 : 1

(4) c : 1



16. The direction of ray of light incident on a concave

mirror is shown by PQ while directions in which the ray would travel after reflection is shown by four rays marked 1, 2, 3 and 4 (figure). Which of the four rays correctly shows the direction of reflected ray? 1

(2) electric flux (4) electric field lines

Q

2

9. Which of the following characteristics of electrons

determines the current in a conductor? (1) Drift velocity alone (2) Thermal velocity alone (3) Both drift velocity and thermal velocity (4) Neither drift nor thermal velocity.

(1) source of emf. (2) electric field produced by charges accumulated on the surface of wire. (3) the charges just behind a given segment of wire which push them just the right way by repulsion. (4) the charges ahead. 11. Kirchhoff ’s laws are valid for _______. (1) only passive circuits (2) only linear circuits (3) only non-linear circuits (4) both (2) and (3). 12. Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field B = B0 k . (1) They have equal z-components of momenta. (2) They must have equal charges. (3) They necessarily represent a particleantiparticle pair.

10. Consider a current carrying wire (current I) in the

shape of a circle. Note that as the current progresses along the wire, the direction of j (current density) changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for is

C 3 (1) 1 (3) 3

F

P

(2) 2 (4) 4

4

Sample Question Papers

17. The phenomenon involved in the reflection of

radio waves by ionosphere are similar to (1) reflection of light by a plane mirror. (2) total internal reflection of light in air during a mirage. (3) dispersion of light by water molecules during the formation of a rainbow. (4) scattering of light by the particles of air.



i

22. Figure shows a standard two-slit arrangement

with slits S1, S2. P1, P2 are the two minima points on either side of P shows in figure. At P2 on the screen, there is a hole and behind P2 is a second screen, 2-slit arrangement with slits S3 and S4 and a second screen behind them.

S2

i r

1

2



(4)

S3

P2

S4

Second Screen

(1) There would be no interference pattern on the second screen but it would be lighted. (2) The second screen would be totally dark. (3) There would be a single bright point on the second screen. (4) There would be a regular two slit pattern on the second screen.

2

(3)

P



1 r



P1 S



Screen

S1

2

(2)

1

2

23. A body is negatively charged means (1) It has only negative charges. (2) Positive charges have been neutralized by negative charges. (3) The quantity of negative charges present is more than the quantity positive charges present. (4) The positive are displaced from their original positions.



19. The magnifying power of a telescope is M. If

the focal length of the eyepiece is halved, the magnifying power will become (1) M/2 (2) 4M (3) M/4 (4) None of the above



1

i



(4) According to Maxwell’s electromagnetic theory light is treated as a wave in nature and require no medium to travel. According to Huygens theory light is treated as a wave in nature and require medium to travel.

laboratories which have a negative refractive index (figure). A ray incident from air (medium 1) into such a medium (medium 2) shall follow a path given by :

r





18. There are certain materials developed in

(1)

17 17

20. Which of the following is a digital device?

24. A point positive charge is brought near an isolated

conducting sphere in Figure. The electric field is best given by :

(i)

+q



(ii)

(1) Mechanical switch (2) Fan regulator (3) Traffic light (4) Microphone

21. Which of the following statement is true? (1) According to both Maxwell’s electromagnetic theory and Huygens wave theory light is treated as a wave in nature and require medium to travel. (2) According to both Maxwell’s electromagnetic theory and Huygens wave theory light is treated as a particle in nature and require medium to travel. (3) According to both Maxwell’s electromagnetic theory and Huygens wave theory light is treated as a wave in nature and does not require medium to travel.

+q

(iii)

+q



(iv)

+q

18

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

(1) Fig (i) (3) Fig (ii)



(2) Fig (iii) (4) Fig (iv)

25. An infinitely long straight conductor is bent into the shape as shown in the figure. Current in it is i and the radius of the circular loop is r. The magnetic field at its centre is

(1) Zero i (3) 0    1 2 r

(2) Infinite i (4) 0    1 2 r

26. When was Van de Graff generator invented and by

whom? (1) 1944, Robin Van de Graff (2) 1932, Robert Van de Graff (3) 1933, Robin Van de Graff (4) 1933, Robert Van de Graff

(2) mass (4) momentum

28. If the back e.m.f. induced in a coil, when current

changes from 1A to zero in one millisecond, is 5 volts, the self-inductance of the coil is  (1) 5 H (2) 1 H (3) 5 × 10–3 H (4) 5 × 103 H 29. When a body is charged by conduction, its mass (1) remains same. (2) increases. (3) decreases. (4) increase or decrease.

30. Law stating that “force is directly proportional to

product of charges and inversely proportional to square of separation between them” is called (1) Newton’s law. (2) Coulomb’s law (3) Gauss’s law. (4) Ohm’s law

(3) (E g )C > (E g )Si > (E g )Ge



= = (4) (E g )C (E g )Si (E g )Ge



33. In Figure, assuming the diodes to be ideal,

(1) D1 is forward biased and D2 is reverse biased and hence current flows from A to B (2) D2 is forward biased and D1 is reverse biased and hence no current flows from B to A and vice versa. (3) D1 and D2 are both forward biased and hence current flows from A to B. (4) D1 and D2 are both reverse biased and hence no current flows from A to B and vice versa.

34. The output of the given circuit in Figure

27. Lenz’s law is consequence of the law of

conservation of (1) Charge (3) energy



(1) would be zero at all times. (2) would be like a half wave rectifier with positive cycles in output. (3) would be like a half wave rectifier with negative cycles in output. (4) would be like that of a full wave rectifier.





another particle of mass m will be given by Newton’s law: M.m F = G 2 , where r is in km and r



= M mproton + melectron (1)



B (B = 13.6 eV). (2) M =mproton + melectron − 2 c (3) M is not related to the mass of the hydrogen atom.



(4) M =mproton + melectron −

31. A speech signal of 3 kHz is used to modulate a



carrier signal of frequency 1 MHz, using amplitude modulation. The frequencies of the side bands will be (1) 1.003 MHz and 0.997 MHz. (2) 3,001 kHz and 2,997 kHz. (3) 1,003 kHz and 1,000 kHz. (4) 1 MHz and 0.997 MHz.



32. Carbon, silicon and germanium have four valence

electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (E g )C ,(E g )Si and(E g )Ge Which of the following statements is true?



(1) (E g )Si < (E g )Ge < (E g )C



(2) (E g )C < (E g )Ge > (E g )Si

35. The gravitational force between a H-atom and

V ( V = magnitude of c2 the potential energy of electron in the H-atom).



36. X amount of energy is required to remove an



37. Which of the following material will be the best for

electron from its orbit and Y amount of energy is required to remove a nucleon from the nucleus. (1) X = Y (2) X > Y (3) Y > X (4) X ≥ Y nuclear fusion? (1) Lighter element (2) Heavier element (3) Both of the above (4) None of the above

Sample Question Papers



38. O2 molecule consists of two oxygen atoms. In the

molecule, nuclear force between the nuclei of the two atoms (1) is not important because nuclear forces are short-ranged. (2) is as important as electrostatic force for binding the two atoms. (3) cancels the repulsive electrostatic force between the nuclei. (4) is not important because oxygen nucleus have equal number of neutrons and protons.

39. For the ground state, the electron in the H-atom

has an angular momentum = h, according to the quantity simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true, (1) because Bohr model gives incorrect values of angular momentum. (2) because only one of these would have a minimum energy. (3) angular momentum must be in the direction of spin of electron. (4) because electrons go around only in horizontal orbits.

40. Consider a beam of electrons (each electron with



41. The phenomenon which shows quantum nature



42. When an object is placed between f and 2f of a

energy E0) incident on a metal surface kept in an evacuated chamber. Then, (1) no electrons will be emitted as only photons can emit electrons. (2) electrons can be emitted but all with an energy, E0 . (3) electrons can be emitted with any energy, with a maximum of E0– f (f is the work function). (4) electron can be emitted with energy, with a maximum of E0. of electromagnetic radiation is: (1) photoelectric effect. (2) tyndall effect. (3) interference. (4) reflection and refraction.



concave mirror, the image formed is (1) Real, diminished (2) Real, magnified (3) Virtual, diminished (4) Virtual, magnified



43. In a Young's double-slit experiment the source is



white light. One of the holes is covered by a red filter and another by a blue filter. In this case, (1) there shall be alternate interference patterns of red and blue. (2) there shall be an interference pattern for red distinct from that for blue. (3) there shall be no interference fringes. (4) there shall be an interference pattern for red mixing with one for blue.



19 19

44. Consider the two idealized systems : (i) a parallel

plate capacitor with large plates and small separation and (ii) a long solenoid of length L, R, radius of cross-section. In (i), E is ideally treated as a constant between plates and zero outside. In (ii), magnetic field is constant inside the solenoid and zero outside. These idealized assumptions, however, contradict fundamental laws as below : (1) Case (i) contradicts Gauss’s law for electrostatic fields. (2) Case (ii) contradicts Gauss’s law for magnetic fields.

∫ E.dl = 0 (4) Case (ii) contradicts ∫ H.dl = I

(3) Case (i) agrees with

s

en

45. Assertion (A): A double convex air bubble is

formed within a glass slab. The air bubble behaves like a converging lens. Reason (R): Refractive index of glass is more than the refractive index of air. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is true I. Read the following text and answer the following questions on the basis of the same: TOROID A toroid is a coil of insulated or enamelled wire wound on a donut-shaped form made of powdered iron. A toroid is used as an inductor in electronic circuits, especially at low frequencies where comparatively large inductances are necessary. A toroid has more inductance , for a given number of turns, than a solenoid with a core of the same material and similar size. This makes it possible to construct high-inductance coils of reasonable physical size and mass. Toroidal coils of a given inductance can carry more current than solenoidal coils of similar size, because larger-diameter wires can be used, and the total amount of wire is less, reducing the resistance . In a toroid, all the magnetic flux is contained in the core material. This is because the core has no ends from which flux might leak off. The confinement of the flux prevents external magnetic fields from affecting the behaviour of the toroid, and also prevents the magnetic field in the toroid from affecting other components in a circuit. Standard toroidal transformers typically offer a 95% efficiency, while standard laminated transformers typically offer less than a 90% rating. One of the most important differences between a toroidal transformer and a traditional laminated transformer is the absence of gaps. The leakage flux through the gaps contributes to the stray losses in the form of eddy currents (which is also expelled in the form of heat).

20

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

A toroidal core doesn’t have an air gap. The core is tightly wound . The result is a stable, predictable toroidal core, free from discontinuities and holes. Audible vibration or hum in transformers is caused by vibration of the windings and core layers from the forces between the coil turns and core laminations. The toroidal transformer’s construction helps quiet this noise. In audio, or signal transmitting applications, unwanted noise will affect sound quality, so a transformer with low audible vibration is ideal. For this reason, many sound system engineers prefer to use a toroidal transformer instead of a traditional laminated transformer. 46. Toroid is a (1) fixed value resistor. (2) capacitor. (3) inductor. (4) variable resistor.

47. A toroid has _____ inductance, for a given number of turns, than a solenoid with a core of same material and similar size.

(1) same (3) less

(2) more (4) variable

48. Why inductance of solenoid is more than the



inductance of a solenoid having same number of turns, core of same material and similar size? (1) Core is endless hence there no leakage of flux. (2) Resistance of wire is less hence magnitude of current flow is more (3) Number of turns per unit length is more. (4) Both (1) and (2)



49. Why sound system engineers prefer to use toroidal



50. Efficiency of toroidal transformer is around ______



transformer? (1) It is cheaper. (2) It is lighter. (3) It is compact. (4) It does not create vibration or hum.



% which is ______ than laminated core transformer. (1) 95, lower (2) 95, higher (3) 50, lower (4) 80, higher 

SAMPLE

Question Paper Maximum Marks : 200

5

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50. (ii) Correct answer or the most appropriate answer: Five marks (+5). (iii) Any incorrect option marked will be given minus one mark (– 1). (iv) Unanswered/Marked for Review will be given no mark (0). (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted.



Section - II PHYSICS

1. In given figure, two positive charges q2 and q3 fixed



along the y axis, exert a net electric force in the + x direction on a charge q1 fixed along the x axis. If a positive charge Q is added at (x, 0), the force on q1

3. Two batteries of emf e1 and e2 (e2 > e1) and internal resistances r1 and r2 respectively are connected in parallel as shown in figure :

(1) The equivalent emf eeq of the two cells is between e1 and e2, i.e. e1< eeq < e2. (2) The equivalent emf eeq is smaller than e1. (3) The eeq is given by eeq = e1 + e2 always. (4) eeq is independent of internal resistances r1 and r2.

(1) shall increase along the positive x-axis. (2) shall decrease along the positive x-axis. (3) shall point along the negative x-axis. (4) shall increase but the direction changes because of the intersection of Q with q2 and q3.

2. A point charge +q, is placed at a distance d from an

isolated conducting plane. The field at a point P on the other side of the plane is (1) directed perpendicular to the plane and away from the plane. (2) directed perpendicular to the plane but towards the plane. (3) directed radially away from the point charge. (4) directed radially towards the point charge.



4. A resistance R is to be measured using a meter

bridge. Student chooses the standard resistance S to be 100 W. He finds the null point at l1 = 2.9 cm. He is told to attempt to improve the accuracy. Which of the following is a useful way? (1) He should measure l1 more accurately. (2) He should change S to 1000 W and repeat the experiment. (3) He should change S to 3 W and repeat the experiment. (4) He should give up hope of a more accurate measurement with a meter bridge.

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

22

5. An electron is projected with uniform velocity

along the axis of a current carrying long solenoid. Which of the following is true? (1) The electron will be accelerated along the axis. (2) The electron path will be circular about the axis. (3) The electron will experience a force at 45° to the axis and hence execute a helical path. (4) The electron will continue to move with uniform velocity along the axis of the solenoid.

6. A solenoid of 1.5 metre length and 4.0 cm diameter

has 10 turn per cm. A current of 5 A ampere is flowing through it. The magnetic field at axis inside the solenoid is (1) 2p × 10–3 T (2) 2p × 10–3 G (3) 2p × 10–7 T (4) 2p × 10–7 G 7. The deflecting torque acting on the coil of a galvanometer is (1) inversely proportional to number of turns. (2) inversely proportional to current flowing. (3) inversely proportional to area of the coil. (4) directly proportional to the magnetic field strength. 8. A message signal of frequency wm is superposed on a carrier wave of frequency wc to get an amplitude modulated wave (AM). The frequency of the AM wave will be (1) wm (2) wc c  m   m (3) (4) c 2 2

9. The magnetic field of Earth can be modelled by

that of a point dipole placed at the centre of the Earth. The dipole axis makes an angle of 11.3° with the axis of Earth. At Mumbai, declination is nearly zero. Then, (1) the declination varies between 11.3° W to 11.3° E. (2) the least declination is 0°. (3) the plane defined by dipole axis and Earth axis passes through Greenwich. (4) declination averaged over Earth must be always negative. 10. Let the magnetic field on Earth be modelled by that of a point magnetic dipole at the centre of Earth. The angle of dip at a point on the geographical equator (1) is always zero. (2) is always positive (3) is always negative (4) can be positive or negative or zero. 11. In which of the following circuit power dissipation is maximum? (1) Pure capacitive circuit (2) Pure inductive circuit (3) Pure resistive circuit (4) LR or CR circuit





purification and eye surgery is (1) Infrared (2) Microwave (3) X-rays (4) None of the above 14. Which one of the following statement is correct? (1) X-rays are suitable for radar system and aircraft navigation. (2) Water molecules readily absorb infrared radiation and their thermal motion increases. (3) Microwaves are produced in Coolidge tube (4) Gamma radiations generate due to electron transitions between upper and lower energy levels of heavy element when excited by electron bombardment

15. A car is moving with at a constant speed of 60 km



16. Radius of curvature of human eye is 0.78 cm. For



an object at infinity, image is formed at 3 cm behind the refracting surface. The refractive index of eye is (1) 1.35 (2) 3 (3) 6.2 (4) 1



17. In the circuit shown in the figure, the input voltage

h–1 on a straight road. Looking at the rear-view mirror, the driver finds that the car following him is at a distance of 100 m and is approaching with a speed of 5 km h–1. In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car after every 2 s till the other car overtakes. If the the following statement(s) is/are correct? (1) The speed of the car in the rear is 65 km/h. (2) In the side mirror the car in the rear would appear to approach with a speed of 5 km h–1 to the driver of the leading car. (3) In the rear view mirror the speed of the approaching car would appear to decrease as the distance between the cars decreases. (4) In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases.

Vi is 20V, VBE = 0 and VCE = 0. The values of IB, IC and β are given by : 20 V

RC

(3)

2p LC

(4)

1 L 2p C

4 k C

RB Vi

12. When a capacitor C is charged to a certain potential

and connected to an inductor L, then frequency of energy oscillation is given by 1 1 (1) (2) LC 2p LC

13. The electromagnetic radiations used for water

500 k

B E



(1) IB = 20 μA, IC = 5 mA, β = 250 (2) IB = 25 μA, IC = 5 mA, β = 200 (3) IB = 40 μA, IC = 10 mA, β = 250 (4) IB = 40 μA, IC = 5 mA, β = 125

Sample Question Papers

18. Angular width of central maxima of a single slit diffraction pattern is independent of (1) slit width (2) frequency of the light used (3) wavelength of the light used (4) distance between slit and screen



20. At stopping potential, the kinetic energy of emitted

photoelectron is (1) minimum. (2) maximum. (3) zero. (4) cannot de predicted

21. Choose the correct alternative from the clues given

at the end of each statement : (1) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no different from/much less than) (2) In the ground state of .........., electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson’s model/Rutherford’s model) (3) A classical atom based on ............. is doomed to collapse. (Thomson’s model/Rutherford’s model) (4) An atom has a nearly continuous mass distribution in ............. but has a highly non-uniform mass distribution in ............. . (Thomson’s model/Rutherford’s model) (E) The positively charged part of the atom possesses most of the mass in ............. . (Rutherford’s model/both the models)

25. The angular frequency of a cyclotron is



26. Tetra valent semiconductor is to be doped with



27. A 220 V A.C. supply is connected between points

independent of (1) Speed (2) Mass (3) Magnetic field (4) Charge

____ valent element to achieve _____ type extrinsic semiconductor. (1) penta, n (2) tri, p (3) penta, p (4) both (1) and (2) A and B in Figure. What will be the potential difference V across the capacitor?



(1) 220 V

(2) 110 V



(3) 0 V

(4) 220 2 V



28. Which one of the following diagrams depicts the

proper flow of electrons and holes in a forward biased p-n junction diode?

(1)

and B in the following circuit?

(2) 5 V (4) 20 V

23. When cell of e.m.f. E is connected with an external

resistance R, the p.d. across the cell becomes V. The expression for the internal resistance of the cell is E−V (1) R (2) 1023 V (3)

V−E R V

(4)

V−E R E

(2) 3 A (4) 1 A



22. What is the potential difference between points A

(1) 10 V (3) 2.5 V



part of the following circuit.

(1) 5 A (3) 7 A





24. Apply Kirchhoff ’s law to find the current I in the

19. Threshold wavelength of a photoelectric emission

from a material is 600 nm. Which of the following illuminating source will emit photoelectrons? (1) 400 W, infrared lamp (2) 10 W, ultraviolet lamp (3) 100 W, ultraviolet lamp (4) Both (2) & (3)





23 23

(2)

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

24 (3)





(3) two in parallel and one in series (4) two in series and one in parallel

35. A metal rod of length 10 cm and a rectangular

1 cm is connected to a 2 battery across opposite faces. The resistance will be (1) maximum when the battery is connected across 1 1 cm × cm faces. 2 (2) maximum when the battery is connected across 10 cm × 1 cm faces. (3) maximum when the battery is connected across 1 10 cm × cm faces. 2 (4) same irrespective of the three faces. cross-section of 1 cm ×

(4)





36. Kirchhoff ’s current law is based on the law of

conservation of (1) charge. (3) mass.

29. A capacitor plates are charged by a battery with



‘V’ volts. After charging, battery is disconnected and a dielectric slab with dielectric constant ‘K’ is inserted between its plates, the potential across the plates of a capacitor will become: (1) Zero (2) V/2 (3) V/K (4) KV



30. The best instrument for accurate measurement of



EMF of a cell is(1) Potentiometer (2) Metre bridge (3) Voltmeter (4) Ammeter and voltmeter



32. Three capacitors 2 µF, 3 µF and 6 µF are joined in



33. Which is not the unit of electric potential



34. The maximum capacitance of three identical

series with each other. The equivalent capacitance is: (1) 1/2 mF (2) 1 mF (3) 2 mF (4) 11 mF





37. One requires 11 eV of energy to dissociate a



38. Electromagnetic wave having frequency 5 × 1011

carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in (1) visible region (2) infrared region (3) ultraviolet region (4) microwave region Hz is (1) Ultraviolet wave (3) Microwave

(1) Joule/Coulomb (2) Newton/Coulomb (3) Newton-meter/Coulomb (4) Volt capacitors is obtained by connecting them (1) in parallel (2) in series

(2) Radio wave (4) X-rays

39. You are given the following 3 lenses. Two construct an astronomical telescope which one will you used as eyepiece and which one as objective?

31. A constant voltage is applied between the two ends

of a uniform metallic wire, heat ‘H’ is developed in it. If another wire of the same material, double the radius and twice the length as compared to original wire is used, then the heat developed in it will be: (1) H/2 (2) H (3) 2H (4) 4H

(2) energy. (4) (2) and (3)



Lens

Aperture (cm)

Power (4)

L1

8

3

L2

1

10

1 (2) L2, L1 (4) L3, L1

6

L3 (1) L1, L2 (3) L2, L3



40. When a monochromatic light is passed around



41. A proton, a neutron, an electron and an a-particle

a file wire a diffraction pattern is observed. How the fringe width will change by increasing the diameter? (1) Fringe width has no relation with the diameter of wire (2) Increases (3) Decreases (4) Fringe width changes with change of wavelength only



have same energy. Then, their de-Broglie wavelengths compare as (1) lp = ln > le > la (2) la < lp= ln< le (3) le < lp = ln > la (4) le= lp= ln= la

Sample Question Papers

42. In a hydrogen atom, the electron moves in an orbit



of radius 0.5 Ao making 10 revolutions per second, the magnetic moment associated with the orbital motion of the electron will be: (1) 2.512 × 10–38 Am2 (2) 1.256 × 10–38 Am2 (3) 0.628 × 10–38 Am2 (4) zero



43. A conducting square loop of side 'L' and resistance

'R' moves in its plane with the uniform velocity 'v' perpendicular to one of its sides. A magnetic induction 'B' constant in time and space pointing perpendicular and into the plane of the loop exists everywhere as shown in the figure. The current induced in the loop is:

I. Read the following text and answer the following questions on the basis of the same: Faraday Cage: A Faraday cage or Faraday shield is an enclosure made of conducting material. The fields within a conductor cancel out with any external fields, so the electric field within the encloser is zero. These Faraday cages act as big hollow conductors you can put things in to shield them from electrical fields. Any electrical shocks the cage receives, pass harmlessly around the outside of the cage.





(1) BLv/R Clockwise



(2) BLv/R Anticlockwise



(3) 2BLv/R Anticlockwise



(4) Zero



44. The magnetic flux linked with the coil (in Weber) is given by the equation: f = 5t2 + 3t + 16



The induced EMF in the coil at time, t = 4 will be:

(1) – 27 V

(2) – 43 V



(3) – 108 V

(4) 210 V



45. Assertion (A): Circuit containing capacitors should be handled very carefully even when the power is off.

Reason (R): The capacitors may break down at any time.

(1) Both A and R are true and R is the correct explanation of A



(2) Both A and R are true but R is NOT the correct explanation of A



(3) A is true but R is false



(4) A is false and R is true

25 25

46. Which of the following material can be used to make a Faraday cage ? (1) Plastic (3) Copper

(2) Glass (4) Wood



47. Example of a real-world Faraday cage is :



48. What is the electrical force inside a Faraday cage



49. If isolated point charge +q is placed inside the



50. A point charge of 2mC is placed at centre of Faraday

(1) car (3) lighting rod

(2) plastic box (4) metal rod

when it is struck by lightning ? (1) The same as the lightning (2) Half that of the lightning (3) Zero (4) A quarter of the lightning

Faraday cage. Its surface must have charge equal to : (1) Zero (2) +q (3) –q (4) +2q cage in the shape of cube with surface of 9 cm edge. The number of electric field lines passing through the cube normally will be : (1) 1.9 × 105 Nm2/C entering the surface. (2) 1.9 × 105 Nm2/C leaving the surface. (3) 2.0 × 105 Nm2/C leaving the surface. (4) 2.0 × 105 Nm2/C entering the surface. 

SAMPLE

Question Paper Maximum Marks : 200

6

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50. (ii) Correct answer or the most appropriate answer: Five marks (+5). (iii) Any incorrect option marked will be given minus one mark (– 1). (iv) Unanswered/Marked for Review will be given no mark (0). (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted.

Section - II



PHYSICS

1. Which of the following is NOT the property of equipotential surface? (1) They do not cross each other. (2) The rate of change of potential with distance on them is zero. (3) For a uniform electric field, they are concentric spheres. (4) They can be imaginary spheres.

2. Two point charges + 8q and –2q are located at x = 0 and x = L respectively.

The point on x axis at which net electric field is zero due to these charges is

(1) 8L

(2) 4L



(3) 2L

(4) L



smaller of the two. The null point shifts to 20 cm. What is the value of the bigger resistance? (1) 89 W (3) 100 W

(2) 144 W (4) None of the above



6. A straight conductor carries a current from south



7. A rod of length L, along east-west direction is

8. Ratio of total intensity of magnetic field at equator

to north. Point P and Q lie to the east and west of it at the same distance. The magnetic field at P is (1) equal to magnetic field at Q. (2) smaller than the magnetic field at Q. (3) greater than the magnetic field at Q. (4) cannot be predicted unless the value if I is known.



(1) 2pE

(2) pE

dropped from a height H. If B be the magnetic field due to Earth at that place and angle of dip is q, then the magnitude of the induced e.m.f. across two ends of the rod when the rod reachs the Earth is (1) BLH cos q (2) BL cos q × (2gH)1/2 1/2 (3) BL cos q / (2gH) (4) None of the above



(3) pE/2

(4) Zero





3. An electric dipole of moment p is placed parallel to the uniform electric field. The amount of work done in rotating the dipole by 90° is



4. Kirchhoff ’s voltage law is based on the law of

to poles is

(1) 1:1

(2) 1:2

(1) charge

(2) energy

(3) 2:1

(4) None of the above

(3) mass

(4) (2) and (3)



conservation of



5. Two resistances are connected in two gaps of Meter Bridge. The balance is 10 cm from the zero end. A resistance of 20 W is connected in series with the

9. To reduce the resonant frequency in an L-C-R series

circuit with a generator (1) the generator frequency should be reduced. (2) another capacitor should be added in parallel to the first.

Sample Question Papers

(3) the iron core of the inductor should be removed. (4) dielectric in the capacitor should be removed. 10. The output of a step-down transformer is measured to be 24 V when connected to a 12 W Light bulb. The value of the peak current is



27 27

18. The equivalent resistance between A and B is:

1/ 2 A (2) 2A (1) (3) 2 A (4) 2 2 A

11. From Maxwell’s hypothesis, a changing electric

field gives rise to (1) an electric field (3) a magnetic field

(2) an induced emf (4) a magnetic torque.

12. Proper arrangement of Gamma rays, Microwave,

IR wave and UV rays in ascending order of frequency is (1) Gamma rays > UV rays > IR rays > Microwave (2) Microwave > IR rays > UV rays > Gamma rays (3) UV rays > Gamma rays > Microwave > IR rays (4) IR rays > UV rays > Microwave > Gamma rays

13. In the combination of the following gates the

output Y can be written in terms of inputs A and B as :



Y





(1) A • B + A • B (3) A • B



(2) A • B + A • B

19. The SI unit of magnetic field intensity is:



20. The coil of a moving coil galvanometer is wound



(1) AmN–1 (3) NA–2m–2

two electrons accelerated through 25 V and 36 V is (1) 25/36 (2) 36/25 (3) 5/6 (4) 6/5

15. Which is true for p-type semiconductors?

(1) Electrons are majority carriers and trivalent atoms are the dopants. (2) Electrons are minority carriers and pentavalent atoms are dopants. (3) Holes are minority carriers and pentavalent atoms are dopants. (4) Holes are majority carriers and trivalent atoms are dopants.



21. The horizontal component of earth’s magnetic



field at a place is 3 times the vertical component. The angle of dip at that place is: (1) p/6 (2) p/3 (3) p/4 (4) 0



22. The maximum kinetic energy of the positive ion in the cyclotron is



(1)

qBR 2 2m

(2)

q 2B2 R 2 2m



(3)

q 2B2 R 2 m

(4)

qBR m



16. Two point charges placed in a medium of



17. Which statement is true for Gauss law?



dielectric constant 5 are at a distance r between them, experience an electrostatic force ‘F’. The electrostatic force between them in vacuum at the same distance r will be: (1) 5F (2) F (3) F/2 (4) F/5 (1) All the charges whether inside or outside the gaussian surface contribute to the electric flux. (2) Electric flux depends upon the geometry of the gaussian surface. (3) Gauss theorem can be applied to non-uniform electric field. (4) The electric field over the gaussian surface remains continuous and uniform at every point.

(2) NA–1m–1 (4) NA–1m–2

over a metal frame in order to: (1) reduce hysteresis (2) increase sensitivity (3) increase moment of inertia (4) provide electromagnetic damping

(4) A + B

14. The ratio of de-Broglie wavelength associated with

(2) 5.5 ohms (4) 9.5 ohms



A B

(1) 3 ohms (3) 7.5 ohms



23. Two coils are placed close to each other. The mutual

inductance of the pair of coils depends upon the: (1) rate at which current change in the two coils (2) relative position and orientation of the coils (3) rate at which voltage induced across two coils (4) currents in the two coils 24. A wire has resistance 15 W. It is bent in the form of equilateral triangle. The effective resistance between any two vertices will be: (1) 6.66 W (2) 15 W (3) 5 W (4) 3.33 W 25. An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of (1) Large focal length and large diameter (2) Large focal length and small diameter (3) Small focal length and large diameter (4) Small focal length and small diameter 26. A person can see clearly objects only when they lie between 50 cm and 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use, will be: (1) Convex, +2.25 diopter (2) Concave, – 0.25 diopter (3) Concave, – 0.2 diopter (4) Convex, + 0.15 diopter

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

28

27. An air bubble in a glass slab with refractive index



1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is : (1) 8 (2) 10 (3) 12 (4) 16



28. The interference pattern is obtained with two



coherent light sources of intensity ratio n. In the interference pattern, the ratio (Imax – Imin) / (Imax  +  Imin) will be : (1) √n/n+1 (2) 2√n/(n + 1) (3) √n/(n+1)2 (4) 2√n/(n + 1)2



29. The bending of beam of light around corners or



obstacles is called : (1) Reflection (3) Refraction

(2) Diffraction (4) Interference



(3) Energy is sometimes absorbed and sometimes released (4) Energy is neither absorbed nor released



36. The volume occupied by an atom is greater than





the volume of the nucleus by a factor of about : (1) 1015 (2) 101 5 (3) 10 (4) 1010

37. In a p-n junction diode, change in temperature due



to heating : (1) Does not affect resistance of p-n junction (2) Affects only forward resistance (3) Affects only reverse resistance (4) Affects the overall V-I characteristics of P-N junction



38. The input resistance of a silicon transistor is 100 Ω.



pattern from single slit. If blue light is used instead of red light, then diffraction pattern (1) Will be clearer (2) Will contract (3) Will expanded (4) Will not be visualized



Base current is changed by 40 μA which results in a change in collector current by 2 mA. This transistor is used as a common emitter amplifier with a load resistance of 4 KΩ. The voltage gain of the amplifier is : (1) 3000 (2) 4000 (3) 1000 (4) 2000



31. The de Broglie wavelength of a neutron in thermal



39. The electric potential at a point in free space due to



30. Red light is generally used to observe diffraction



equilibrium with heavy water at a temperature T (Kelvin) and mass m, is : (1) h/√3mkT (2) 2h/√3mkT (3) 2h/√mkT (4) h/√mkT



32. Electrons of mass m with de Broglie wavelength λ fall on the target in an X-ray tube. The cutoff wavelength λ0 of the emitted X-ray is :



(1) λ0 = (3) λ0 =

2mcλ2/h

2m2c2λ3/h2

(1)





p

(3)



p

(2)







p

(4)







34. The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is: (1) 1 (2) 4 (3) 0.5 (4) 2

35. When an electron jumps from L to K shell : (1) Energy is absorbed (2) Energy is released

40. A capacitor of capacity C1 is charged upto V volt





(4) λ0 = λ

variation of particle momentum and the associated de Broglie wavelength?







(2) λ0 = 2h/mc

33. Which of the following figures represent the

p



a charge Q coulomb is Q × 1011 volts. The electric field at that point is : (1) 12πε0 Q × 1022 Volt/m (2) 4πε0 Q × 1022 Volt/m (3) 12πε0 Q × 1020 Volt/m (4) 4πε0 Q × 1020 Volt/m



and then connected to capacity C2. Then final each will be : (1) C2V/(C1+C2) (3) C1V/(C1+C2)

an uncharged capacitor of potential difference across (2) (1+C2/C1) × V (4) (1−C2/C1) × V

41. If the distance between parallel plates of a capacitor



is halved and dielectric constant is doubled then the capacitance : (1) Decreases two times (2) Increases two times (3) Increases four times (4) Remain the same



42. A coil of self-inductance L is connected in series





with a bulb B and an AC source. Brightness of the bulb decreases when : (1) an iron rod is inserted in the coil. (2) frequency of the AC source is decreased. (3) number of turns in the coil is reduced. (4) A capacitance of reactance XC = XL is included in the same circuit.

43. A wire loop is rotated in magnetic field. The frequency of change of direction of the induced e.m.f. is : (1) Six times per revolution (2) Once per revolution (3) twice per revolution (4) four times per revolution

Sample Question Papers

44. I-V characteristics of four devices are shown in Figure.

I

I

V (i) I

V (ii) I

V V (ii) (iv) Identify devices that can be used for modulation : (1) ‘i’ and ‘iii’. (2) only ‘iii’. (3) ‘ii’ and some regions of ‘iv’. (4) All the devices can be used.



45. Assertion (A): In the α-particle scattering

experiment, most of the α-particles pass undeviated. Reason (R): Most of the space in the atom is empty. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is true I. Read the following text and answer the following questions on the basis of the same: Power factor corrector capacitor Power factor correction is a method to reduce the lagging power factor in inductive loads by fixing a high value capacitor across the phase and neutral line close to the load. When the Voltage and Current are in phase with each other in an AC circuit, the energy from the source is fully converted into another form to drive the load and in this case power factor is in unity. When the power factor drops, the system becomes less efficient. In inductive loads, current “lags” the voltage leading to “lagging power factor”. Power factor correction is the method to reduce the lagging power factor in inductive loads by fixing a high

29 29

value capacitor across the phase and neutral close to the load. These capacitors have leading power factor so that it will neutralize the lagging power factor of the load. Power capacitors are huge non-polarized metal film electrolytic type capacitors. Capacitors should be sufficiently rated to the load capacity. It should be connected to the lines, only when the loads are running and drawing current.

46. What is meant by power factor correction?



47. When the energy from source is fully converted

(1) The method to reduce the lagging power factor in inductive loads (2) The method to enhance the lagging power factor in inductive load (3) The method to reduce the lagging power factor in capacitive loads (4) The method to enhance the lagging power factor in capacitive loads into another form, the power factor is (1) 0.5 (2) 1.0 (3) 0 (4) ∞

48. Power capacitors for power factor correction are (1) polarized metal film electrolytic type. (2) non-polarized metal film electrolytic type. (3) non-polarized metal film non-electrolytic type. (4) polarized ceramic non- electrolytic type. 49. Power capacitors for power factor correction have (1) lagging power factor. (2) leading power factor. (3) leading or lagging power factor depending on the value of the capacitor. (4) leading or lagging power factor depending on the type of load.

50. Power factor corrector capacitors should be

connected (1) across the phase and ground near the inductive load. (2) across the phase and neutral away from the inductive load. (3) across the phase and neutral near the inductive load. (4) across the neutral and ground near the inductive load. 

SAMPLE

Question Paper Maximum Marks : 200

7

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50. (ii) Correct answer or the most appropriate answer: Five marks (+5). (iii) Any incorrect option marked will be given minus one mark (– 1). (iv) Unanswered/Marked for Review will be given no mark (0). (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted.

Section - II



PHYSICS

1. _______ increases in step-down transformer.   (1) Voltage (3) Power

(2) Current (4) Current density

2. The efficiency of transformer is very high because (1) There is no moving part (2) It uses AC only (3) It uses the copper wire for the coils (4) None of the above 3. An electric dipole is placed at an angle of 30° with an electric field intensity 2.0 × 105 N/C. It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is: (1) 8 mC (2) 2 mC (3) 5 mC (4) 7 μC 4. A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will: (1) increase four times (2) be reduced to half (3) remain the same (4) be doubled 5. Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities s, – s and s respectively. If VA, VB and VC denote the potentials of the three shells, then for c = a + b, we have: (1) VC = VB ≠ VA (2) VC ≠ VB ≠ VA (3) VC = VB = VA (4) VC = VA ≠ VB

6. A capacitor is charged by a battery. The battery

is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system:









(1) decreases by a factor of 2 (2) remains the same (3) increases by a factor of 2 (4) increases by a factor of 4

7. A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are: (1) Zero and Q/4peoR2 (2) Q/4peoR and Zero (3) Q/4peoR and Q/4peoR2 (4) Both are zero.

8. A carbon resistor of (47 ± 4.7) kW is to be marked



with rings of different colours for its identification. The colour code sequence will be: (1) Yellow – Green – Violet – Gold (2) Yellow – Violet – Orange – Silver (3) Violet – Yellow – Orange – Silver (4) Green – Orange – Violet – Gold



9. The resistance of a wire is ‘R’ ohm. If it is melted



and stretched to ’n’ times its original length, its new resistance will be: (1) R/n (2) n2R 2 (3) R/n (4) nR



10. The power dissipated in the circuit shown in the figure is 30 Watts. The value of R is:

Sample Question Papers R



5Ω

(1) 15 Ω (3) 30 Ω



(2) 10 Ω (4) 20 Ω

11. An electron is moving in a circular path under the



influence of a transverse magnetic field of 3.57 × 10-2 T. If the value of e/m is 1.76 × 1011 C/kg, frequency of revolution of the electron is: (1) 1 GHz (2) 100 MHz (3) 62.8 MHz (4) 6.28 MHz



12. A galvanometer has a coil of resistance 100 ohm

19. The frequency of a light wave in a material is 2 ×



1014 Hz and wavelength is 5000  Å. The refractive index of material will be: (1) 1.33 (2) 1.40 (3) 1.50 (4) 3.00



20. The penetration of light into the region of

10 V



31 31



geometrical shadow is called (1) Polarization (2) Interference (3) Diffraction (4) Refraction



21. Diffraction and interference of light suggest



22. A polaroid is placed at 45° to an incoming light

(1) Nature of light is electro-magnetic (2) Wave nature (3) Nature is quantum (4) Nature of light is transverse



and gives a full scale deflection for 30 mA current. If it is to work as a voltmeter of 30 volt range, the resistance required to be added will be: (1) 900 W (2) 1800 W (3) 500 W (4) 1000 W



of intensity I0. Now the intensity of light passing through polaroid after polarization would be: (1) I0 (2) I0/2 (3) I0/4 (4) Zero



13. A bar magnet is hung by a thin cotton thread in



23. A basic communication system consists of (A)

14. The given electrical network is equivalent to: A Y



24. A polariser is used to



25. Photoelectric emission occurs only when the

a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60o is W. Now the torque required to keep the magnet in this new position is: (1) W/√3 (2) √3 W (3) √3/2 W (4) 2/√3 W



B

(1) AND gate (3) NOR gate

(2) OR gate (4) NOT gate

15. A long solenoid has 1000 turns. When a current of

transmitter, (B) information source, (C) user of information, (D) channel and (E) receiver. Choose the correct sequence in which these are arranged in a basic communication system : (1) ABCDE. (2) BADEC. (3) BDACE. (4) BEADC. (1) Reduce intensity of light (2) Produce polarised light (3) Increase intensity of light (4) Produce unpolarised light incident light has more than a certain minimum: (1) power (2) wavelength (3) intensity (4) frequency





4A flows through it, the magnetic flux linked with each turn of the solenoid is 4 × 10–3 Wb. The self– inductance of the solenoid is: (1) 4 H (2) 3 H (3) 2 H (4) 1 H



16. If the current is halved in a coil, then the energy



stored is how much times the previous value? (1) 1/2 (2) 1/4 (3) 2 (4) 4



a frequency n (higher than the threshold frequency n0) is proportional to: (1) Frequency of light (n) (2) n – n0 (3) Threshold frequency (n0) (4) Intensity of light



17. The ratio of resolving powers of an optical



27. Ratio of longest wave lengths corresponding to



microscope for two wavelengths l1 = 4000 Å and l2 = 6000 Å is: (1) 9: 4 (2) 3: 2 (3) 16: 81 (4) 8: 27



18. For the angle of minimum deviation of a prism to



be equal to its refracting angle, the prism must be made of a material whose refractive index: (1) lies between 2 and √2 (2) is less than 1 (3) is greater than 2 (4) lies between √2 and 1



26. The number of photoelectrons emitted for light of



Lyman and Balmer series in hydrogen spectrum is : (1) 5/27 (2) 3/23 (3) 7/29 (4) 9/31



28. The first use of quantum theory to explain the



structure of atom was made by (1) Heisenberg (2) Bohr (3) Planck (4) Einstein

29. For a nuclear fusion process, the suitable nuclei are: (1) any nuclei (2) heavy nuclei

32

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS (3) light nuclei (4) nuclei lying in the middle of the periodic table

30. Solar energy is mainly caused due to:

(1) gravitational contraction (2) burning of hydrogen in the oxygen (3) fission of uranium present in the Sun (4) fusion of protons during synthesis of heavier elements 31. Cyclotron cannot accelerate (1) Electrons (2) Neutrons (3) Positive ions (4) Both (1) and (2)

32. In forward biasing of the p-n junction: (1) the positive terminal of the battery is connected to p-side and the depletion region becomes thick (2) the positive terminal of the battery is connected to n-side and the depletion region becomes thin (3) the positive terminal of the battery is connected to n-side and the depletion region becomes thick (4) the positive terminal of the battery is connected to p-side and the depletion region becomes thin



33. The barrier potential of a p-n junction depends on:



34. Zener diode is used for:



35. Electric field strength due to a point charge of 5μC



(A) type of semi conductor material (B) amount of doping (C) temperature Which one of the following is correct? (1) (B) and (C) only (2) (A), (B) and (C) (3) (A) and (B) only (4) (B) only (1) amplification (3) stabilisation in an oscillator

(2) rectification (4) producing oscillations

at a distance of 80 cm from the charge is: (1) 8 × 104 N/C (2) 7 × 104 N/C 4 (3) 5 × 10 N/C (4) 4 × 104 N/C

36. When one electron is taken towards the other



(1) Connecting them in parallel



(2) Connecting two in series and one in parallel



(3) Connecting two in parallel and one in series



(4) Connecting all of them in series





(4) Q/ε0

40. The capacity of a parallel plate condenser is C. Its (1) 4C

(2) 2C



(3) C/2

(4) C/4



41. The magnetic field at a distance r from a long wire carrying current i is 0.4 Tesla. The magnetic field at a distance 2r is:



(1) 0.2 Tesla

(2) 0.8 Tesla



(3) 0.1 Tesla

(4) 1.6 Tesla



42. A bar magnet of magnetic moment M is placed in



a magnetic field of induction B. The torque exerted on it is:      (1) MB (2) − MB     (3) M × B (4) − M × B



43. Faraday’s laws are consequence of conservation of  :



(1) Energy



(2) Energy and magnetic field



(3) Charge



(4) Magnetic field

44. The capacity of parallel plate condenser depends on:



(1) The type of metal used



(2) The thickness of plates



(3) The potential applied across the plates



(4) The separation between the plates



45. Assertion (1): In a non-uniform electric field, a dipole will have translatory as well as rotatory motion.

Reason (R): In a non-uniform electric field, a dipole experiences a force as well as torque.

r

B

(3) Q/8 ε0

(2) Q/6 ε0



37. ABC is an equilateral triangle. Charges +q are

+q

(1) Q/3 ε0

capacity when the separation between the plates is halved will be:



r

39. A charge Q is placed at the corner of a cube. The electric flux through all the six faces of the cube is  :





O

38. Three capacitors each of capacity 4μF are to





+q A

(2) 1/4πε0 × q/r (4) 1/4πε0 × 3q/r2

be connected in such a way that the effective capacitance is 6μF. This can be done by:

electron, then the electric potential energy of the system (1) Decreases (2) Increases (3) Remains unchanged (4) Becomes zero placed at each corner. The electric intensity at O will be:

(1) 1/4πε0 × q/r2 (3) Zero



(1) Both A and R are true and R is the correct explanation of A



(2) Both A and R are true but R is NOT the correct explanation of A



(3) A is true but R is false



(4) A is false and R is true

r

C

+q

33 33

Sample Question Papers I. Read the following text and answer the following questions on the basis of the same: Laser: Electromagnetic radiation is a natural phenomenon found in almost all areas of daily life, from radio waves to sunlight to x-rays. Laser radiation – like all light – is also a form of electromagnetic radiation. Electromagnetic radiation that has a wavelength between 380 nm and 780 nm is visible to the human eye and is commonly referred to as light. At wavelengths longer than 780 nm, optical radiation is termed infrared (IR) and is invisible to the eye. At wavelengths shorter than 380 nm, optical radiation is termed ultraviolet (UV) and is also invisible to the eye. The term “laser light” refers to a much broader range of the electromagnetic spectrum that just the visible spectrum, anything between 150 nm up to 11000 nm (i.e. from the UV up to the far IR). The term laser is an acronym which stands for “light amplification by stimulated emission of radiation”. Einstein explained the stimulated emission. In an atom, electron may move to higher energy level by absorbing a photon. When the electron comes back to the lower energy level it releases the same photon. This is called spontaneous emission. This may also so happen that the excited electron absorbs another photon , releases two photons and returns to the lower energy state. This is known as stimulated emission. Laser emission is therefore a light emission whose energy is used, in lithotripsy, for targeting and ablating the stone inside human body organ. Apart from medical usage, laser is used for optical disk drive, printer, barcode reader etc.



46. What is the full form of LASER ? (1) Light amplified by stimulated emission of radiation (2) Light amplification by stimulated emission of radiation (3) Light amplification by simultaneous emission of radiation (4) Light amplified by synchronous emission of radiation



47. The “stimulated emission” is the process of:



48. What is the range of amplitude of LASER?



49. Lithotripsy is:



50. LASER is used in:

(1) release of a photon when electron comes back from higher to lower energy level. (2) release of two photons by absorbing one photon when electron comes back from higher to lower energy level. (3) absorption of a photon when electron moves from lower to higher energy level. (4) None of the above (1) 150 nm – 400 nm (3) Both the above

(2) 700 nm – 11000 nm (4) None of the above

(1) an industrial application. (2) a medical application. (3) laboratory application. (4) process control application. (1) optical disk drive. (2) transmitting satellite signal. (3) radio communication. (4) ionization. 

SAMPLE

8

Question Paper Maximum Marks : 200

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50. (ii) Correct answer or the most appropriate answer: Five marks (+5). (iii) Any incorrect option marked will be given minus one mark (– 1). (iv) Unanswered/Marked for Review will be given no mark (0). (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted.

Section - II







PHYSICS

1. An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in a uniform electric field E. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively : (1) 2q⋅E and minimum (2) q⋅E and p⋅E (3) Zero and minimum (4) q⋅E and maximum 2. When a proton is accelerated through 1 V, then its kinetic energy will be : (1) 1840 eV (2) 13.6 eV (3) 1 eV (4) 0.54 eV 3. A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy stored in the capacitor is : (1) E2Ad/ε0 (2) (1/2) ε0E2Ad (3) ε0EAd (4) (1/2) ε0E2 4. In the figure, three capacitors each of capacitance 6 pF are connected in series. The total capacitance of the combination will be : C1

C2

C4





points A and B in the given figure is : 2Ω



+

3V

–

1Ω VB

I = 2A

A



(2) 6×10−12 F (4) 2×10−12 F

5. The potential difference (VA – VB) between the VA

B

(1) – 3 V (3) +6 V

(2) +3 V (4) +9 V

6. A cell can be balanced against 110 cm and 100 cm of



potentiometer wire, respectively with and without being short circuit through a resistance of 10 Ω. Its internal resistance is : (1) Zero (2) 1.0 ohm (3) 0.5 ohm (4) 2.0 ohm



7. What is the output Y in the following circuit, when all the three inputs A, B, C are first 0 and then 1? A B C



V

(1) 9×10−12 F (3) 3×10−12 F



(1) 0, 1 (3) 1, 0

P Q

Y

(2) 0, 0 (4) 1, 1

8. Two free parallel wires carrying currents in opposite directions (1) Attract each other (2) Repel each other (3) Neither attract nor repel (4) Get rotated to be perpendicular to each other

Sample Question Papers

9. Electromagnets are made of soft iron because soft iron has : (1) low retentivity and high coercive force (2) high retentivity and high coercive force (3) low retentivity and low coercive force (4) high retentivity and low coercive force

10. Unit of magnetic flux density (or magnetic induction) is : (1) Tesla (2) Weber/meter2 (3) Newton/ampere-metre (4) All of the above

11. A particle of mass m, charge Q and kinetic energy



T enters a transverse uniform magnetic field of induction B. After 3 seconds the kinetic energy of the particle will be : (1) 4 T (2) 3 T (3) 2 T (4) T



12. The magnetic flux linked with a coil, in weber's, is



given by the equations φ = 3t2 + 4t + 9. Then the magnitude of induced e.m.f. at t = 2 second will be  : (1) 2 volt (2) 4 volt (3) 8 volt (4) 16 volt



13. The magnetic potential energy stored in a certain



inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance : (1) 0.138 H (2) 138.88 H (3) 1.389 H (4) 13.89 H



14. In an ac circuit of capacitance, the current from



potential is : (1) Forward (2) Backward (3) Both are in the same phase (4) None of these

15. A wave travelling in the +ve x-direction having



displacement along y-direction as 1 m, wavelength 2πm and frequency of 1/π Hz is represented by : (1) y = sin (x – 2t) (2) y = sin (2πx – 2πt) (3) y = sin (10πx – 20πt) (4) y = sin (2πx + 2πt)



16. The ratio of amplitude of magnetic field to the



amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to : (1) reciprocal of speed of light in vacuum (2) the ratio of magnetic permeability to the electric susceptibility of vacuum (3) unity (4) the speed of light in vacuum



17. The refracting angle of a prism is A, and refractive





index of the material of the prism is cot (A/2). The angle of minimum deviation is : (1) 90° – A (2) 180° + 2A (3) 180° – 3A (4) 180° – 2A



35 35

18. If the speed of light in vacuum is c m/sec, then the



velocity of light in a medium of refractive index 1.5 is (1) 1.5 × c (2) c (3) c/1.5 (4) Can have any velocity



19. The idea of the quantum nature of light has



emerged in an attempt to explain (1) Interference (2) Diffraction (3) Radiation spectrum of a black body (4) Polarisation

20. A male voice after modulation-transmission



sounds like that of a female to the receiver. The problem is due to (1) poor selection of modulation index (selected 0 < m< 1). (2) poor bandwidth selection of amplifiers. (3) poor selection of carrier frequency. (4) loss of energy in transmission.



21. The photoelectric threshold wavelength for





potassium (work function being 2 eV) is : (1) 310 nm (2) 620 nm (3) 1200 nm (4) 2100 nm

22. The momentum of the photon of wavelength 5000 will be : (1) 1.3 × 10−27kg-m/s (3) 4 × 1029kg-m/s

(2) 1.3 × 10−28kg-m/s (4) 4 × 10−18kg-m/s

23. The energy of a hydrogen atom in the ground



state is – 13.6 eV. The energy of He+ ion in the first excited state will be : (1) – 13.6 eV (2) – 27.2 eV (3) – 54.4 eV (4) – 6.8 eV



24. Experimental evidence for the existence of the



atomic nucleus comes from : (1) Millikan’s oil drop experiment (2) Atomic emission spectroscopy (3) The magnetic bending of cathode rays (4) Alpha scattering by a thin metal foil

25. A sample has 4 × 1016 radioactive nuclei of half life



10 days. The number of atoms decaying in 30 days is : (1) 3.9 × 1016 (2) 5 × 1015 16 (3) 10 (4) 3.5 × 1016



26. A free neutron decays into a proton, an electron



and : (1) a beta particle (3) an anti-neutrino

(2) an alpha particle (4) a neutrino

27. If a small amount of antimony is added to germanium crystal: (1) it becomes a p-type semiconductor (2) the antimony become an acceptor atom (3) there will be more free electrons than holes in the semiconductor (4) its resistance is increased

36

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

28. Which of the following is used in the Davisson – Germer experiment? (1) Double slit (3) Electron gun

(2) Single slit (4) Electron microscope

29. An electric dipole is kept in non - uniform electric field. It experiences (1) A force and a torque (2) A force but not a torque (3) A torque but not a force (4) Neither a force nor a torque

30. If a unit positive charge is taken from one point to



another over an equipotential surface, then : (1) Work is done on the charge (2) Work is done by the charge (3) Work done is constant (4) No work is done



31. The capacity of parallel plate condenser depends



on : (1) The type of metal used (2) The thickness of plates (3) The potential applied across the plates (4) The separation between the plates



32. In a charged capacitor, the energy resides :



33. In semiconductors, at room temperature :



34. Green-house effect is the heating up of Earth’s





39. Through which character we can distinguish the





atmosphere due to : (1) green plants (3) X-rays

(2) infra-red rays (4) β-rays

35. Pick out the longest wavelength from the following types of radiation : (1) blue light (3) X-rays

(2) gamma rays (4) red light



36. Stars are twinkling due to :



37. Time taken by the sunlight to pass through a

(1) Diffraction (3) Refraction

(2) Reflection (4) Scattering



window of thickness 4 mm whose refractive index is 1.5, is (1) 2 × 10−8 sec (2) 2 × 108 sec −11 (3) 2 × 10 sec (4) 2 × 1011 sec



38. Unpolarised light is incident from air on a plane

surface of a material of refractive index ‘μ’. At a particular angle of incidence ‘i’, it is found that

light waves from sound waves (1) Interference (2) Refraction (3) Polarisation (4) Reflection

40. A simple pendulum of period T has a metal bob



which is negatively charged. If it is allowed to oscillate above a positively charged metal plate, its period will : (1) Remains equal to T (2) Less than T (3) Greater than T (4) Infinite



41. Figure shows three points A, B and C in a

(1) The positive charges (2) Both the positive and negative charges (3) The field between the plates (4) Around the edge of the capacitor plates (1) the conduction band is completely empty (2) the valence band is partially empty and conduction band is completely filled (3) the valence band is partially empty and the conduction band is partially filled (4) the valence band is completely filled





the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation? (1) i=sin-1(1/μ) (2) Reflected light is polarised with its electric vector perpendicular to the plane of incidence (3) Reflected light is polarised with its electric vector parallel to the plane of incidence (4) i = tan-1(1/μ)

region of uniform electric field E. The line AB is perpendicular and BC is parallel to the field lines. Then which of the following holds good. Where VA,VB and VC represent the electric potential at points A, B and C respectively : A

B



C

(1) VA = VB = VC (3) VA = VB < VC

(2) VA = VB > VC (4) VA > VB = VC

42. Three capacitors of capacitance 3 μF, 10 μF and 15 μF are connected in series to a voltage source of 100 V. The charge on 15 μF is :



(1) 50 μC

(2) 100 μC



(3) 200 μC

(4) 280 μC



43. Two thin dielectric slabs of dielectric constants K1

and K2 (K1 < K2) are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field ‘E’ between the plates with distance ‘d’ as measured from plate P is correctly shown by : P + + + + + + +

Q – – – – – – – K1

K2

Sample Question Papers

the pipe as it would take when dropped through the same height without the pipe. Now instead of PVC pipe use an aluminium pipe. Note the time it takes to come out of the pipe in each case. You will see that the magnet takes much longer time in the case of aluminium pipe. Why is it so ? It is due to the eddy currents that are generated in the aluminium pipe which oppose the change in magnetic flux, i.e., the motion of the magnet. The retarding force due to the eddy currents inhibits the motion of the magnet. Such phenomena are referred to as electromagnetic damping. Note that eddy currents are not generated in PVC pipe as its material is an insulator whereas aluminium is a conductor. This effect was discovered by physicist Foucault (1819-1868).

(1) E



0

d

0

d

(2) E



(3) E

0



d

(4) E



46. Eddy current is generated in a:



47. Eddy current was first observed by:



48. What is electromagnetic damping ?

0

d



44. Alpha-particles are :



45. Assertion (1): The resistance of superconductor is

(1) protons (3) neutrally charged

37 37

(2) positron (4) ionized helium atoms

zero. Reason (R): Super conductors are used for electrical power transmission. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is true I. Read the following text and answer the following questions on the basis of the same: Electromagnetic damping: Take two hollow thin cylindrical pipes of equal internal diameters made of aluminium and PVC, respectively. Fix them vertically with clamps on retort stands. Take a small cylindrical magnet having diameter slightly smaller than the inner diameter of the pipes and drop it through each pipe in such a way that the magnet does not touch the sides of the pipes during its fall. You will observe that the magnet dropped through the PVC pipe takes the same time to come out of





(1) metallic pipe. (3) glass pipe. (1) Helmholtz (3) D'Arsonval

(2) PVC pipe. (4) wooden pipe. (2) Foucault (4) Shock ley

(1) Generation of electromagnetic wave during the passage of a magnet through a metal pipe (2) Change of the direction of propagation of electromagnetic wave due to a variable magnetic flux (3) Change of the frequency of electromagnetic wave due to a variable magnetic flux (4) To slow down the motion of a magnet moving through a metal pipe due to electromagnetically induced current.

49. To observe electromagnetic damping a magnet

should be dropped through a metal pipe and: (1) the magnet should not touch inner wall of the pipe. (2) the magnet should touch the inner wall of the pipe. (3) it does not matter whether the magnet touches the inner wall of the pipe or not. (4) the magnet should be larger in size than the diameter of the pipe.

50. A piece of wood and a bar magnet of same

dimension is dropped through an aluminium pipe. Which of the following statements is true ? (1) The piece of wood will take more time to come out from the pipe. (2) The bar magnet will take more time to come out from the pipe. (3) Both will take same time to come out from the pipe. (4) The time required will depend on the mass of the wooden piece and the mass of the bar magnet. 

38

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

SAMPLE

Question Paper Maximum Marks : 200

9

Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50. (ii) Correct answer or the most appropriate answer: Five marks (+5). (iii) Any incorrect option marked will be given minus one mark (– 1). (iv) Unanswered/Marked for Review will be given no mark (0). (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted.

Section - II



PHYSICS

1. If a full wave rectifier circuit is operating from 50



(1) Hund

(2) Thomson



(3) Rutherford

(4) Sommerfield



Hz mains, the fundamental frequency in the ripple will be. (1) 100 Hz (2) 25 Hz (3) 50 Hz (4) 70.7Hz The depletion layer in the p-n junction region is caused by: (1) drift of holes (2) diffusion of charge carriers (3) migration of impurity ions (4) drift of electrons Atomic power station at Tarapore has a generating capacity of 200 MW. The energy generated in a day by this station is : (1) 200 MW (2) 200 J 6 (3) 4800 × 10 J (4) 1728 × 1010 J One requires energy En to remove a nucleon from a nucleus and an energy ‘Ee’ to remove an electron from the orbit of an atom. Then : (1) En = Ee (2) En < Ee (3) En > Ee (4) En ≥ Ee Rutherford’s experiment on scattering of particles showed for the first time that the atom has : (1) Electrons (2) Protons



(3) Nucleus











2.

3.

4.

5.

(4) Neutrons

6. Who modified Bohr’s theory by introducing elliptical orbits for electron path :



7. The momentum of a photon is 3.3×10−29 kg-m/sec. Its frequency will be : (1) 3 × 103Hz (3) 7.5 × 1012Hz

(2) 6 × 103Hz (4) 1.5 × 1013Hz

8. If the de Broglie wavelengths for a proton and for a

α particle are equal, then the ratio of their velocities will be : (1) 4 : 1 (2) 2 : 1 (3) 1 : 2 (4) 1 : 4



9. The transverse nature of light is shown by



10. Polarised glass is used in sun glasses because

(1) Interference of light (2) Refraction of light (3) Polarisation of light (4) Dispersion of light



(1) It reduces the light intensity to half on account of polarisation (2) It is fashionable (3) It has good colour (4) It is cheaper



11. Which of the following statement is false for the

properties of electromagnetic waves ? (1) both electric and magnetic field vectors attain the maxima and minima at the same place and same time. (2) The energy in electromagnetic wave is divided equally between electric and magnetic vectors.

Sample Question Papers





(3) Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of wave. (4) These waves do not require any material medium for propagation. medium of permittivity ε0 and permeability μ0 is given by (1)

µ0 / ε0

(3)

µ0ε0





(2) ε 0 / µ 0

14. An e.m.f. of 5 volt is produced by a self inductance, when the current changes at a steady rate from 3 A to 2 A in 1 millisecond. The value of self inductance is : (1) Zero (2) 5 H (3) 5000 H (4) 5 mH



15. In an A.C. circuit the current :



16. To get output 1 for the following circuit, the correct

(1) Always leads the voltage (2) Always lags behind the voltage (3) Is always in phase with the voltage (4) May lead or lag behind or be in phase with the voltage choice for the input is: A B C

17. A straight wire of diameter 0.5 mm carrying a



current of 1A is replaced by another wire of 1mm diameter carrying same current. The strength of magnetic field far away is : (1) twice the earlier value (2) same as earlier value (3) one-half of the earlier value (4) one-quarter of the earlier value 18. A charged particle moves through a magnetic field in a direction perpendicular to it. Then the (1) Speed of the particle remains unchanged (2) Direction of the particle remains unchanged (3) Acceleration remains unchanged (4) Velocity remains unchanged



19. Curie temperature is the temperature above





which  : (1) Ferromagnetic material becomes diamagnetic material (2) Ferromagnetic material becomes paramagnetic material

20. Magnetic field due to 0.l A current flowing through



21. From the graph between current i & voltage V shown, identify the portion corresponding to negative resistance : i C

E

B D

A



(1) DE (3) BC

V

(2) CD (4) AB

22. A wire of length L and resistance R is stretched to



get the radius of cross-section half. What is new resistance : (1) 5 R (2) 8 R (3) 4 R (4) 16 R



23. The electric intensity due to a dipole of length



10  cm and having a charge of 500 μC, at a point on the axis at a distance 20 cm from one of the charges in air, is : (1) 6.25×107 N/C (2) 9.28×107 N/C 11 (3) 13.1×11 N/C (4) 20.5×107 N/C



24. On rotating a point charge having a charge q

Y

(1) A = 0, B = 1, C = 0 (2) A = 1, B = 0, C = 0 (3) A = 1, B = 1, C = 0 (4) A = 1, B = 0, C = 1

(3) Paramagnetic material becomes diamagnetic material (4) Paramagnetic material becomes ferromagnetic material a circular coil of radius 0.l m and 1000 turns at the centre of the coil is : (1) 0.2 T (2) 2 × l0– 4 T –4 (3) 6.28 × 10 T (4) 9.8 × 10– 4 T

13. A coil of inductance 300 mH and resistance 2 Ω is







(4) 1 / µ 0 ε 0







12. The velocity of electromagnetic radiation in a

connected to a source of voltage 2 V. The current reaches half of its steady state value in : (1) 0.05 s (2) 0.1 s (3) 0.15 s (4) 0.2 s





39 39



around a charge Q in a circle of radius r. The work done will be : (1) q × 2πr (2) q × 2πQr (3) Zero (4) Q/2ε0r



25. A charge q is placed at the centre of the line joining



two equal charges Q. The system of the three charges will be in equilibrium, if q is equal to : Q Q (1) − (2) − 2 4 Q Q (3) (4) 4 2

26. A parallel plate condenser has a uniform electric



field E(V/m) in the space between the plates. If the distance between the plates is d(m) and area of each plate is A(m2) the energy (joules) stored in the condenser is : (1) E2Ad/ε0 (2) (1/2)ε0E2 (3) ε0EAd (4) (1/2)ε0E2Ad



27. Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The

40

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS



capacitance and breakdown voltage of the combination will be : (1) 3C, V/3 (2) C/3, 3V (3) 3C, 3V (4) C/3, V/3



28. Which crystal is used in the Davisson – Germer



experiment? (1) Aluminum (3) Cobalt

(2) Nickel (4) Zinc

29. Barrier potential of p-n junction diode does not depend on: (1) doping density (3) temperature

(2) diode design (4) forward bias



30. Solar energy is due to :



31. A slit of width a is illuminated by white light. For

(1) fusion reaction (2) fission reaction (3) combustion reaction (4) chemical reaction



red light (λ=6500Å), the first minima is obtained at θ  =  30°. Then the value of a will : (1) 3250 Å (2) 6.5 × 10−4mm (3) 1.24 microns (4) 2.6 × 10−4 cm



32. For a parallel beam of monochromatic light of



wavelength ‘λ’, diffraction is produced by a single slit whose width ‘a’ is of the order of the wavelength of the light. If ‘D’ is the distance of the screen from the slit, the width of the central maxima will be : (1) Da/λ (2) 2Da/λ (3) 2Dλ/a (4) Dλ/a



33. In an electromagnetic wave in free space the



root mean square value of the electric field is Erms = 6 V/m. The peak value of the magnetic field is : (1) 1.41 × 10–8 T (2) 2.83 × 10–8 T (3) 0.70 × 10–8 T (4) 4.23 × 10–8 T



34. The electric field part of an electromagnetic wave





in a medium is represented by Ex = 0; Ey = 2.5 N/C cos[(2π × 106 rad/c)t – (π × 10–2 rad/m)x], Ez = 0. The wave is : (1) Moving along – x direction with frequency 106 Hz and wavelength 200 m (2) Moving along y direction with frequency 2π × 106 Hz and wavelength 200 m (3) Moving along x-direction with frequency 106 Hz and wavelength 100 m (4) Moving along x direction with frequency 106 Hz and wavelength 200 m

35. Lenz’s law is consequence of the law of conservation of : (1) Charge (3) Mass





x-axis at points x = 0, x = a and x = 2a respectively, then (1) Only q is in stable equilibrium (2) None of the charges are in equilibrium (3) All the charges are in unstable equilibrium (4) All the charges are in stable equilibrium



38. Identify the mathematical expression for amplitude





henry is decreasing at the rate of 2 ampere/sec. The e.m.f. developing across the coil is : (1) 10 V (2) – 10 V (3) 2.5 V (4) – 2.5 V

modulated wave :

(1) Ac sin[ {ωc + k1vm ( t )} t+φ] (2) Ac sin {ωc t+φ+ k 2 vm ( t )}

(3) { Ac + k 2Vm ( t )} sin ( ωct+φ ) (4) Ac vm ( t ) sin ( ωc t+φ )

39. A cylindrical bar magnet is kept along the axis of a circular coil. If the magnet is rotated about its axis, then : (1) A current will be induced in a coil (2) No current will be induced in a coil (3) Only an e.m.f. will be induced in the coil (4) An e.m.f. and a current both will be induced in the coil

40. The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9.0× 10−13 cm is (1) 80 Volts (2) 8×106 Volts (3) 9 Volts (4) 9×105 Volts

41. A capacitor of 20 μF is charged to 500 volts and



connected in parallel with another capacitor of 10 μF and charged to 200 volts. The common potential is : (1) 200 Volts (2) 300 Volts (3) 400 Volts (4) 500 Volts



42. In the following circuit, the bulb will become suddenly bright if :

B



(1) Contact is made or broken (2) Contact is made (3) Contact is broken (4) Won’t become bright at all

43. An electric dipole when placed in a uniform electric



field E will have minimum potential energy, if the positive direction of dipole moment makes the following angle with E : (1) π (2) π/2 (3) Zero (4) 3π/2



44. A parallel plate condenser has a capacitance 50 μF

(2) Momentum (4) Energy

36. The current passing through a choke coil of 5

37. Point charges +4q,−q and +4q are kept on the

in air and 110 μF when immersed in an oil. The dielectric constant ′K′ of the oil is :

Sample Question Papers

(1) 0.45 (3) 1.10

(2) 0.55 (4) 2.20



(3) the diamond is transparent.



(4) rays always enter at angle greater than critical angle.



47. The critical angle for a diamond is 24.4°. Then its

45. Assertion (A): An electron and a proton moving with same velocity enters a magnetic field. The force experienced by the proton is more than the force experienced by the electron.

Reason (R): The mass of proton is more than the mass of the electron. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is true I. Read the following text and answer the following questions on the basis of the same: Sparking Brilliance of Diamond: The total internal reflection of the light is used in polishing diamonds to create a sparking brilliance. By polishing the diamond with specific cuts, it is adjusted the most of the light rays approaching the surface are incident with an angle of incidence more than critical angle. Hence, they suffer multiple reflections and ultimately come out of diamond from the top. This gives the diamond a sparking brilliance.

41 41



refractive index is: (1) 2.42 (3) 1

(2) 0.413 (4) 1.413

48. The basic reason for the extraordinary sparkle of suitably cut diamond is that: (1) it has low refractive index. (2) it has high transparency. (3) it has high refractive index. (4) it is very hard.

49. A diamond is immersed in a liquid with a refractive



index greater than water. Then the critical angle for total internal reflection will: (1) depend on the nature of the liquid. (2) decrease. (3) remains the same. (4) increase.



50. The following diagram shows same diamond cut in two different shapes.

Critical angle



Total reflection

Air

Diamond

The brilliance of diamond in the second diamond will be: (1) less than the first. (2) greater than first. (3) same as first. (4) will depend on the intensity of light.

46. Light cannot easily escape a diamond without multiple internal reflections. This is because:



(1) its critical angle with reference to air is too large.

(2) its critical angle with reference to air is too small. 

SAMPLE

Question Paper Maximum Marks : 200

10 Time : 45 Minutes

General Instructions: (i) This paper consists of 50 MCQs, attempt any 40 out of 50. (ii) Correct answer or the most appropriate answer: Five marks (+5). (iii) Any incorrect option marked will be given minus one mark (– 1). (iv) Unanswered/Marked for Review will be given no mark (0). (v) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted.

Section - II



PHYSICS

1. Which of the following statement does not form

part of Bohr’s model of the hydrogen atom : (1) Energy of the electrons in the orbit is quantized (2) The electron in the orbit nearest the nucleus has the lowest energy (3) Electrons revolve in different orbits around the nucleus (4) The position and velocity of the electrons in the orbit cannot be determined simultaneously

2. The energy of second Bohr orbit of the hydrogen



atom is – 328 kJ mol–1, hence the energy of fourth Bohr orbit would be : (1) – 41 kJ/mol (2) – 1312 kJ/mol (3) – 164 kJ/mol (4) – 82 kJ/mol



3. The de-Broglie wavelength λ associated with an



electron having kinetic energy E is given by the expression : (1) h/√2mE (2) 2h/mE (3) √2mhE (4) 2√2mE



4. Momentum of a photon of wavelength l is :



5. To observe diffraction the size of an obstacle

(1) h/λ (3) hλ/c2

(2) Zero (4) hλ/c

(1) Should be of the same order as wavelength (2) Should be much larger than the wavelength (3) Have no relation to wavelength (4) Should be exactly λ/2



6. Two slits are separated by a distance of 0.5 mm and



illuminated with light of λ=6000Å. If the screen is placed 2.5 m from the slits. The distance of the third bright image from the centre will be : (1) 1.5 mm (2) 3 mm (3) 6 mm (4) 9 mm



7. When a light wave goes from air into water, the



quality that remains unchanged is its (1) Speed (2) Amplitude (3) Frequency (4) Wavelength

8. Two identical thin plano-convex glass lenses



(refractive index 1.5) each having radius of curvature of 20 cm are laced with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is : (1) – 50 cm (2) 50 cm (3) – 20 cm (4) – 25 cm



9. The electric and the magnetic field, associated with



an e.m. wave, propagating along the +z-axis, can be represented by : (1) E = Eoî, B = Boĵ (2) E = Eoˆk, B = Boî (3) E = Eoĵ, B = Boî (4) E = Eoĵ, B = Boˆk



10. The decreasing order of wavelength of infrared,



microwave, ultraviolet and gamma rays is: (1) microwave, infrared, ultraviolet, gamma rays (2) gamma rays, ultraviolet, infrared, microwaves (3) microwaves, gamma rays, infrared, ultraviolet (4) infrared, microwave, ultraviolet, gamma rays

Sample Question Papers

11. The dimensions of magnetic flux are :



12. When the number of turns and the length of the

−2 −2

(1) [MLT A ] (3) [ML2T−1A−2]

2 −2 −2

(2) [ML T A ] (4) [ML2T−2A−1]



solenoid are doubled keeping the area of crosssection same, the inductance : (1) Remains the same (2) Is halved (3) Is doubled (4) Becomes four times



13. For a series LCR circuit, the power loss at resonance



is : (1) V2 / ωL – 1/ωC (3) I2R

2

(2) I Cω (4) V2/ωC



14. A choke coil should have :



15. Identify the expression for Bragg's law from the

(1) high resistance and low inductance (2) high resistance and high inductance (3) low resistance and high inductance (4) low resistance and low inductance



following. (1) 2d cosq = nl



(3) 2d sinq =



n λ

(2) 2d sinq = nl (4) 2d cosq =

n λ

16. Under the influence of a uniform magnetic field



a charged particle is moving in a circle of radius R with constant speed v. The time period of the motion : (1) depends on R and not on v (2) depends on v and not on R (3) depends on both R and v (4) is independent of both R and v



17. A charged particle (charge q) is moving in a circle



of radius R with uniform speed v. The associated magnetic moment μ is given by : (1) qvR (2) qvR/2 (3) qvR2 (4) qvR2/2



18. A diamagnetic substance is brought near a strong



magnet, then it is; (1) attracted by a magnet (2) repelled by a magnet (3) repelled by north pole and attracted by south pole (4) attracted by north pole and repelled by south pole





20. A hollow insulated conducting sphere is given a positive charge of 10μC. What will be the electric field at the centre of the sphere if its radius is 2 meters :

(2) 5μNC−1 (4) 8μNC−1

21. The acceleration of an electron in an electric field of



magnitude 50 V/cm, if e/m value of the electron is 1.76 × 1011 C/kg, is : (1) 8.8×1014m/sec2 (2) 6.2×1013 m/sec2 12 2 (3) 5.4×10 m/sec (4) Zero



22. Three condensers each of capacitance 2 F are put in series. The resultant capacitance is :



(1) 6 F

(2) 3/2 F



(3) 2/3 F

(4) 5 F



23. When air in a capacitor is replaced by a medium of



dielectric constant K, the capacity : (1) Decreases K times (2) Increases K times



(3) Increases K2 times



(4) Remains constant

24. The condensers of capacity C1 and C2 are connected in parallel, then the equivalent capacitance is:





(1) C1+C2

(2) C1C2/C1+C2

(3) C1/C2

(4) C2/C1

25. An audio signal of 15 kHz frequency cannot

be transmitted over long distances without modulation because (1) the size of the required antenna would be at least 50 km which is not convenient. (2) the audio signal can be transmitted through sky waves. (3) the size of the required antenna would be at least 20 km, which is not convenient. (4) effective power transmitted would be very low, if the size of the antenna is less than 5 km.

26. Light of two different frequencies whose photons



have energies 1   eV and 2.5   eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds emitted electrons will be : (1) 1 : 4 (2) 1 : 2 (3) 1 : 1 (4) 1 : 5



27. In the propagation of electromagnetic waves the



angle between the direction of propagation and plane of polarisation is: (1) 0° (2) 45° (3) 90° (4) 180°



28. In a double slit experiment, the two slits are 1

19. A point Q lies on the perpendicular bisector of an

electrical dipole of dipole moment p. If the distance of Q from the dipole is r (much larger than the size of the dipole), then electric field at Q is proportional to : (1) p−1 and r−2 (2) p and r−2 −1 −2 (3) p and r (4) p and r−3

(1) Zero (3) 20μNC−1

43 43



mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern? (1) 0.5 mm (2) 0.02 mm (3) 0.2 mm (4) 0.1 mm

29. Refractive index for a material for infrared light is (1) equal to that of ultraviolet light (2) less than for ultraviolet light (3) equal to that for red colour of light (4) greater than that for ultraviolet light

44

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS



30. Speed of light is maximum in



31. Two identical glass (μg = 3/2) equiconvex lenses

(1) Water (3) Glass

(2) Air (4) Diamond



of focal length f each are kept in contact. The space between the two lenses is filled with water (μw = 4/3). The focal length of the combination is : (1) f/3 (2) f (3) 4f/3 (4) 3f/4



32. Which of the following ray are not electromagnetic



waves : (1) X-rays (2) γ-rays (3) β-rays (4) Heat rays   33. If E and B represent electric and magnetic field vectors of the electromagnetic waves, then the direction of propagation of the waves will be along.    (1) B × E (2) E    (3) B (4) E × B



34. For a common emitter circuit if IC/IE = 0.98 then









current gain for common emitter circuit will be. (1) 49 (2) 98 (3) 4.9 (4) 25.5

35. In electromagnetic induction, the induced e.m.f. in a coil is independent of : (1) Change in the flux (2) Time (3) Resistance of the circuit (4) None of the above

36. The electric potential at a point on the axis of an

electric dipole depends on the distance r of the point from the dipole as : (1) ∝ 1/r (2) ∝ 1/r2 (3) ∝ r (4) ∝ 1/r3

37. The electric field inside a spherical shell of uniform



surface charge density is : (1) Zero (2) Constant, less than zero (3) Directly proportional to the distance from the centre (4) None of the above 38. A bullet of mass 2 gm is having a charge of 2μC. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of 10m/s. (1) 5 kV (2) 50 kV (3) 5 V (4) 50 V



39. Three capacitors of capacitances 3μF, 9μF and 18μF





are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases (Cs/Cp) will be : (1) 1 : 15 (2) 15 : 1 (3) 1 : 1 (4) 1 : 3



40. A condenser of capacity 50μF is charged to 10 volts. Its energy is equal to :



(1) 2.5×10−3 Joule (3) 5×10−2 Joule

(2) 2.5×10−4 Joule (4) 1.2×10−8 Joule

41. Half-lives of two radioactive substances A and



B are respectively 20 minutes and 40 minutes. Initially, the samples of A and B have equal number of nuclei. After 80 minutes the ratio of remaining numbers of A and B nuclei is : (1) 1 : 16 (2) 4 : 1 (3) 1 : 4 (4) 1 : 1



42. The stable nucleus that has a radius half that of



Fe56 is : (1) Li7 (3) S16

(2) Na21 (4) Ca40

43. When n-type semiconductor is heated:



(1) number of electrons increases while that of holes decreases (2) number of holes increases while that of electrons decreases (3) number of electrons and holes remain same (4) number of electrons and holes increases equally



44. When p-n junction diode is reverse biased the flow





of current across the junction is mainly due to: (1) diffusion of charges (2) drift of charges (3) depends on the nature of material (4) both drift and diffusion of charges

45. Assertion (1): Ferromagnetic substances become

paramagnetic beyond Curie temperature. Reason (R): Domains are destroyed at high temperature. (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is NOT the correct explanation of A (3) A is true but R is false (4) A is false and R is true I. Read the following text and answer the following questions on the basis of the same: Types of resistors Most common type of resistor is Carbon Composition Resistors. Carbon resistors are a cheap, general purpose resistor used in electrical and electronic circuits. Their resistive element is manufactured from a mixture of finely ground carbon dust or graphite and a non-conducting ceramic powder to bind it all together. The ratio of carbon dust to ceramic (conductor to insulator) determines the resistive value of the resistor. Higher the ratio of carbon, lower the overall resistance. Film Type Resistors consist of Metal Film, Carbon Film and Metal Oxide Film .Such resistors are generally made by depositing pure metals, such as nickel, or an oxide film, such as tin-oxide, on an insulating ceramic rod or substrate. The resistive value of the resistor is controlled by increasing the desired thickness of the deposited film giving them the names of either “thick-film

Sample Question Papers resistors” or “thin-film resistors”. Film type resistors can achieve much higher ohmic value compared to other types. Another type of resistor, called a Wirewound Resistor, is made by winding a thin metal alloy wire (Nichrome) or similar wire on an insulating ceramic former in the form of a spiral helix. These types of resistors are generally only available in very low ohmic value with high precision . They are able to handle much higher electrical currents than other resistors of the same ohmic value with much excessive power ratings. These high power resistors are moulded into an aluminium heat sink body with fins attached to increase their overall surface area to promote heat loss and cooling.

46. Carbon composition resistors are made from a

mixture of (1) finely ground metal dust and ceramic powder. (2) finely ground carbon dust or graphite and ceramic powder. (3) finely ground carbon dust or graphite and copper powder. (4) finely ground carbon dust or graphite.

45 45



47. In carbon composition resistors, ______ the ratio of



48. Metal Film Type Resistors are generally made by

carbon, _____ the overall resistance. (1) Higher, lower (2) Lower, higher (3) Lower, lower (4) Higher, higher



depositing pure ______, on ______ rod or substrate. (1) Ceramic, metal (2) Carbon, ceramic (3) Metal, ceramic (4) Carbon, metal



49. Wirewound Resistors are made by winding a thin



50. Wire wound resistors are available in very ____

_____ or similar wire on an ____ former in the form of a spiral helix. (1) Nichrome, copper (2) Nichrome, ceramic (3) Copper, ceramic (4) Copper, Nichrome

ohmic high precision values with ______ power rating. (1) High, high (2) Low, low (3) High, Low (4) Low, high



SOLUTIONS OF SAMPLE Question Paper Question Paper

01



Section - II



PHYSICS

1. Option (2) is correct.



Explanation: Inside the sphere, E = 0 Again E = – dV/dr So, dV/dr = 0 This is possible when V is constant.

Explanation: C = kε0A / d So, capacitance does not increase by increasing the distance between the plates (d) or decreasing the area of the plates (A). Thickness of plates has no connection with the capacitance of the capacitor.

2. Option (3) is correct. Explanation: Capacitance of a parallel plate capacitor filled with dielectric of constant k1 and thickness d1 is,



imilarly, for other capacitance of a parallel S plate capacitor filled with dielectric of constant k2 and thickness d2 is, k 2 ε0 A d2

C2 =

oth capacitors are in series so equivalent B capacitance C is related as : d1 d2 1 1 1 = + = + C C 1 C 2 k1 ε 0 A k 2 ε 0 A =

So,  

hich explains that electric potentials due to w point charge is same for all equidistant points. The locus of these equidistant points, which are at same potential, forms spherical surface.

1  k 2 d1 + k1d2   ε0 A  k1k 2 



C=

k1 k 2 ε 0 A     ( k1d2 + k 2 d1 )

…(i)

C′ =

kε0 A kε0 A = ( d1 + d2 ) d

…(ii)

   

where, d = (d1+d2) Comparing eqns. (i) and (ii), the dielectric constant of new capacitor is :



7. Option (2) is correct. Explanation: The triboelectric effect is a type of contact electrification on which certain materials become electrically charged after they come into frictional contact with a different material.

3. Option (1) is correct. Explanation: C’ = KC (where K is the dielectric constant). V = Q/C V’ = Q/C’ V’ = V/4 = Q/C’ = Q/KC = V/K \ K=4

6. Option (4) is correct. Explanation: Electric flux, through the closed surface (or space) depends only on the charge enclosed inside the surface. Here, charges inside all figures are same. So, electric flux will remain same.

k k (d + d ) k= 1 2 1 2 ( k1d2 + k 2 d1 )

5. Option (1) is correct. Explanation: For equipotential surface, these surfaces are perpendicular to the field lines. So there must be electric field, which cannot be without charge. So the algebraic sum of all charges must not be zero. Equipotential surface at a great distance means that space of charge is negligible as compared to distance. So the collection of charges is considered as a point charge. Electric potential due to point charge is, q V = ke r

k1ε0 A d1

C1 =

4. Option (2) is correct.



8. Option (1) is correct. Explanation: When a photon strikes a metal surface, the surface electrons come out with maximum speed and maximum kinetic energy. But if the electron emission takes place

Solutions from inner side of metal, then some energy of the electron is lost due to collision with other electrons and so their speed becomes less. So, ultimately the electrons come out with different speeds.

9. Option (1) is correct. Explanation: Photons are neural particles.

10. Option (1) is correct. Explanation: Tritium (31H) has 1 proton and 2 neutrons. If a neutron decays as, n→p+ē+ν then nucleus will have 2 protons and 1 neutron, i.e. triton atom converts in 2He3 (2 proton and 1 neutron). Binding energy of 1H3 is much smaller than 2He3, so transformation is not possible energetically. 11. Option (2) is correct. Explanation: Potentiometer is used to measure internal resistance of a cell, e.m.f. of a cell and to compare the e.m.f.'s of

17. Option (1) is correct. Explanation: To convert a galvanometer into a voltmeter, a high value resistance is to be connected in series with it. 18. Option (2) is correct. Explanation: When a charged particle perpendicularly enters a magnetic field to the direction, the path of the motion is circular. In circular motion, the direction of velocity changes at every point (the magnitude remains constant). Therefore, the linear momentum changes at every point. But kinetic energy remains constant since the magnitude of velocity does not change. 19. Option (2) is correct. Explanation: Current sensitivity of a galvanometer is the deflection produced when unit current passes through it. Current sensitivity = q/I = nBA/C 20. Option (1) is correct. Explanation:

12. Option (2) is correct. Explanation: Given that, emf of primary cells are 5 V and 10 V The potential drop across potentiometer wire must be slightly more than that larger emf 10 V. So, the battery should be of 15 V and about 4 V potential is dropped by using rheostat or resistances. So option (B) is correct. 13. Option (1) is correct. Explanation: End error of metre bridge is removed when the known and unknown resistances are interchanged.  14. Option (4) is correct. Explanation: The work done to rotate the loop in magnetic field, W = MB (cos q1 – cos q2). When current carrying coil is rotated then there will be no change in angle between magnetic moment and magnetic field. Here, q1 = q2 = a Þ W = MB (cos q - cos a) = 0. 15. Option (4) is correct. Explanation: As net current is zero, magnetic field at the empty space surrounded by toroid and outside the toroid is zero. 16. Option (1) is correct. Explanation: Shunt (S) is connected in parallel to the galvanometer (resistance G). So, the effective resistance is GS/(G+S).

47 47

M square M circular

=

NIA square NIA circular

= πR 2 / a 2

21. Option (1) is correct. Explanation: Magnetic dipole moment vector is directed from south pole to north pole. 22. Option (2) is correct. Explanation: On heating above Curie temperature, Ferromagnetic domains get randomly arranged and it transforms into paramagnetic substance. 23. Option (4) is correct. Explanation: BH = B cos q BV = B sin q \

B=

B2V + B2H

24. Option (2) is correct. Explanation: The radio waves emitted from a transmitter antenna can reach the receiver antenna by the following mode of operation : (i) Ground wave propagation (ii) Sky wave propagation (iii) Space wave propagation (iv) Mode of communication frequency range : (a) Ground wave propagation : 500–1,710 kHz (b) Sky wave propagation : 2–40 MHz (c) Space wave propagation : 54–42 MHz. As, A is transmitted via ground wave, B via sky wave and C via sky wave.

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

48

25. Option (3) is correct.

32. Option (3) is correct.

Explanation: Magnetic flux is defined as the total number of magnetic lines of force passing normally through an area placed in a magnetic field and is equal to the magnetic flux linked with that area.   Square lies in X-Y plane in B so A = L2 kˆ

Q = B.A = B ( 2ˆi + 3ˆj + 4kˆ).( L kˆ) = B [ 2 × ˆikˆ + 3 × ˆjkˆ + 4 × kˆ.kˆ] = B L [0 + 0 + 4]

Explanation: The voltmeter in AC circuit reads value and meter is calibrated to rms value which is multiplied by 2 to get Vrms. In other words, voltmeter connected to the AC main read root mean square value of AC voltage, i.e., < v 2 > . 33. Option (3) is correct. Explanation: Quality factor (Q) of an L-C-R circuit is given by, Q=

B L Wb. 26. Option (3) is correct. Explanation: According to Lenz’s law, the direction of an induced e.m.f. always opposes the change in magnetic flux that causes the e.m.f.

Tuning of an L-C-R circuit depends on quality factor of the circuit. Tuning will be better when quality factor of the circuit is high. For Q to be high, R should be low, L should be high and C should be low. Therefore, option (C) is most suitable. 34. Option (1) is correct.

27. Option (1) is correct. Explanation: NAND gates are one of the two basic logic gates (the other being NOR logic) from which any other logic gates can be built. Due to this property, NAND gate is sometimes called “universal gates”. However, modern integrated circuits are not constructed exclusively from a single type of gate. 28. Option (4) is correct. Explanation: If current I flows through a coil of self-inductance L, then magnetic field energy stored in it is ½ Li2 29. Option (1) is correct. Explanation: If two coils of self inductance L1 and L2 are coupled together, their mutual inductance becomes M = k L1L2 where k = coupling constant whose value lies between 0 and 1. 30. Option (3) is correct. Explanation: Eddy current is produced when a metal is kept in a time varying magnetic field. 31. Option (2) is correct. Explanation: Here, Irms = 5 A, n = 50 Hz and 1 t= s 300    I 0 = Peak value =

1 L R C

2Irms =

2 × 5= 5 2 A

Now, I = I 0 sin ωt =5 2 sin 2 πνt 1 3 = 5 2 sin 2 π × 50 = × 5 A 300 2

Explanation: Q factor of a series LCR circuit is 1 L given by Q = R C 35. Option (2) is correct. Explanation: The transformer is based on the principle of mutual induction which state that due to continuous change in current in the primary coil, an emf is induced across the secondary soil. 36. Option (2) is correct. Explanation: At resonance, LCR circuit behaves as purely resistive circuit. For purely resistive circuit, power factor is 1. 37. Option (2) is correct. Explanation: The phase of a wave changes by 180° or p radian after got reflected from a denser medium. But the type of waves remains identical. Therefore, for the reflected wave, we have z = − z , i = −i and additional phase of p in the incident wave. Incident electromagnetic wave. Then, E E ( −i )cos( kz − ωt ) = 0

Therefore, the reflected electromagnetic wave is given as: E= E ( −i )cos[ k ( − z ) − ωt + π] r

0

[∵ z = − z and i = −i ]

= − E0 i cos[ π − ( kz + ωt )] = − E i[ − cos{( kz + ωt )}] 0

= E0 i cos( kz + ωt )

49 49

Solutions 38. Option (1) is correct. Explanation: l = c/n =

3 × 10 8 5 × 1019

= 6 × 10–12 m

39. Option (2) is correct. Explanation: At 0 K temperature, all electrons of semiconductor are immovable from their shell as they do not have sufficient energy. So no free electron is available as charge carrier. This make the insulators to behave like insulators. 40. Option (2) is correct. Explanation: When p-n junction is in forward bias, it compresses or decreases the depletion layer, due to which potential barrier in forward bias decreases and in reverse bias potential barrier increases. 41. Option (2) is correct. Explanation: For refraction through prism, i1 + i2 = d + A and r1 + r2 = A For minimum deviation, i1 = i2 = i and r1 = r2 = r So, 2i = (A + dm)/2 d A + dm = 2i 42. Option (2) is correct. Explanation: l µ

1 V

43. Option (3) is correct. Explanation: According to Bohr’s model of an atom, radius of an atom in its ground state is r r = 0 Z where, r0 is Bohr’s radius and Z is atomic number. As given that, r0 = 53 pm and atomic number of Lithium atom is 3 53 so, = r = 17.67 pm ≈ 18 pm 3 44. Option (3) is correct.

Explanation: Radius of nucleus= R = R0A1/3. 4 So, Volume of nucleus, V = pR 30 A 3 Considering mass of proton = mass of neutron =m The mass of the nucleus = M = mA So, density = M/V =

mA m = 4 3 4 3 πR A πR 3 0 3 0

So, the mean density is independent of mass number. So, assertion and reason both are true and the reason properly explains the assertion. 46. Option (3) is correct. Explanation: The fact we hear sounds around corners and around barriers involves both diffraction and reflection of sound. 47. Option (1) is correct.

48. Option (1) is correct.

hc E = l

Now from the above relation, l µ

45. Option (1) is correct.

Explanation: In fact, diffraction is more pronounced with longer wavelengths

1 V

Since,

Explanation: Convex mirrors are preferred over plane mirrors as rear view mirror in automobile since these mirrors have larger field of view compared to plane and concave mirror.

Explanation: In fact, diffraction is more pronounced with longer wavelengths implies that you can hear low frequencies around obstacles better than high frequencies. 49. Option (3) is correct. Explanation: When the width of opening is comparatively less than the wavelength of sound wave, the sound spread out much better i.e. better diffraction occurs. When the width of the opening is larger than the wavelength, the wave passing through the opening does not spread out much on the other side. 50. Option (3) is correct. Explanation: Sound spreads out well through a gap whose width is slightly smaller than the wavelength of the sound wave as if it is a localised source of sound. 

02

SOLUTIONS OF Question Paper 

Section - II



1. Option (2) is correct.

PHYSICS

Explanation: Atoms of semiconductor are binding by covalent bonds between the atoms of same or different type. Due to thermal agitation when an electron leaves its position and become free, it leaves a vacancy of electron and this vacancy in the bond (covalent) is called hole.

2. Option (3) is correct. Explanation: When a forward bias is applied to a p-n junction, it lowers the value of potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced.



4. Option (1) is correct. Explanation: As the side or diameter of hemisphere is plane surface, and whole hemisphere is charged with positive charge so, the electric field line of forces emerging outward will be perpendicular to the plane surface or diameter.



5. Option (3) is correct. Explanation: From the given circuit diagram, we can see that this is AND Gate and their output will be multiple of its input. So the output of C = AB, and D = AB. The Resultant output of E is equal to the AND multiplication of C and D Output as inputs of E. We also know that NOT gate is also known as inverter that means if we apply ‘0’ as a input then output will be ‘1’ and vice versa. Resultant Truth Table

B

A

C

D

E

0

0

1

0

0

0

0

1

1

0

1

1

1

0

0

0

0

0

1

1

0

1

0

1

6. Option (3) is correct. Explanation: For an equipotential surface, VA = VB So, work done = 0



7. Option (4) is correct. Explanation: The electric potential at a point on the equatorial line of a electric dipole is zero.

3. Option (3) is correct. Explanation: Rubbing a rod with certain materials will cause the rod to become charged. A plastic rod when rubbed with fur becomes negatively charged and a glass rod when rubbed with silk becomes positively charged.





A



8. Option (3) is correct. Explanation: As the battery remains connected with the capacitor, the potential difference remain constant.



9. Option (4) is correct. Explanation: As capacitor offer infinite resistance for DC circuit. So current from cell will not flow across branch of 4 mF and 10 W. So current will flow across 2 ohm branch. So, Potential Difference (PD) across 2 W resistance V = RI = 2 × 1 = 2 Volts. As battery, capacitor and 2 branches are in parallel. So PD will remain same across all three branches. As current does not flow through capacitor branch, so no potential drop will be across 10 W. So PD across 4 mF capacitor = 2 Volt. Q = CV = 4 mF × 2 V = 8 mC

10. Option (1) is correct. Explanation: The magnitude of the electric force F is directly proportional to the amount of an electric charge, q1, multiplied by the other, q2, and inversely proportional to the square of the distance ‘r’ between their centres.

Solutions 11. Option (4) is correct.

17. Option (2) is correct.

Explanation: Either positive or negative charge will interact with the lines of electric field, so the electric field will become discontinuous. 12. Option (4) is correct.

13. Option (4) is correct. Explanation: Wheatstone bridge is suitable for measurement of medium value resistances because to ensure sensitivity, other resistors must be of comparable values. 14. Option (1) is correct. Explanation: For a balanced Meter Bridge  P/Q = L1 /(100 – L1). There is no parameter related to the radius of the wire. So, the null deflection of galvanometer does not depend on the radius of the wire. So, even if the radius of the wire is doubled, the null deflection of the galvanometer will not be changed. 15. Option (1) is correct. Explanation: Length of the building (L) is = 500 m And length of antenna = 100 m And we know wavelength of the wave which can be transmitted by L=

4

Explanation: Magnetic field at the end of a current carrying solenoid is half of the magnetic field inside it. 18. Option (2) is correct.

Explanation: Slope of I-V characteristic of an Ohmic conductor remains constant throughout.

λ

51 51

, So, l ~ 4l =4 × 100 = 400 m

Wavelength (l) is nearly equal to 400 m. 16. Option (1) is correct. Explanation: For a circular loop of radius R, carrying current I in x-y plane, the magnetic moment M = I × p R2. It acts perpendicular to the loop along z-direction. When half of the current loop is bent in y-z plane, then magnetic moment due to half current loop is x-y plane, M1 = I (pR2/2) acting along z-direction. Magnetic moment due to half current loop in y-z plane, M2 = I (pR2/2) along x-direction. Net magnetic moment due to entire bent current loop, M12

= M net

= 2 =

M 2

+ M 22 2

IπR 2

Therefore, Mnet < M or M diminishes.

Explanation: Magnetic dipole moment vector is directed from South pole to north pole. 19. Option (2) is correct. Explanation: As we know that, L =µr µ0

N2 A l

As L is constant for a coil, 1 L ∝ A and L ∝ l

As mr and N are constant here so, to increase L for a coil, area A must be increased and l must be decreased. So answer (B) is correct. Important point  : The self and mutual inductance of capacitance and resistance depend on the geometry of the devices as well as permittivity/permeability of the medium. 20. Option (2) is correct. Explanation: When a soft iron core is inserted in the inductor, the magnetic flux increases. According to Lenz’s law, it will be resisted by reducing the current. Since the current reduces, the intensity of the bulb decreases. 21. Option (4) is correct. Explanation: When coil A moves towards coil B with constant velocity, so rate of change of magnetic flux due to coil B in coil A will be constant that gives constant current in coil A in same direction as in coil B by Lenz’s law. 22. Option (1) is correct. Explanation: XC = 1/2pfC So, as f increases, XC becomes smaller and smaller. For very high value of f, XC will be too small which may be considered as SHORT. 23. Option (4) is correct. Explanation: Vrms = V0/√2 = 0.707V0 24. Option (3) is correct. Explanation: Laminated core means a layered core instead of a single solid core. Eddy currents are current loops generated by changing magnetic fields. They flow in a plane perpendicular to the magnetic field. Laminated magnetic core reduces eddy currents. For this reason, electrically isolated laminations are utilized to manufacture transformers.

52

OSWAAL CUET (UG) Sample Question Papers, PHYSICS

25. Option (3) is correct. Explanation: Transformer does not change the frequency of the applied AC. 26. Option (2) is correct. Explanation: Energy flux, f =20 W/cm2

Area A= 30 cm2, time t=30×60 sec U = Total energy falling in t sec= Energy flux × Area × time = fAt U= 20×30×30×60 J Momentum of the incident light u 20 × 30 × 30 × 60 = = = 36×10–4 kg-ms–1 c 3 × 10 8 s no reflection from the surface and for A complete absorption, momentum of reflected radiation is zero. Momentum delivered to surface = Change in −4 momentum =p f − pi =0 − 36 × 10 kgm/s = – 36 × 10–4 kg m/s (–) sign shows the direction of momentum. 27. Option (3) is correct. Explanation: A diode antenna radiates the electromagnetic waves outwards. The amplitude of electric field vector (E0) which transports significant energy from the source falls inversely as the distance (r) from the antenna. As we know that electromagnetic waves are radiated from dipole antenna and 1 radiated energy, so E ∝ . r 28. Option (1) is correct. Explanation: A Van de Graff generators produces large voltage and less current. A Van de Graff generator is an electrostatic generator which creates very high electric potentials. It produces very high voltage direct current electricity at low current levels. 29. Option (2) is correct. Explanation: Radiowaves are reflected by a layer of atmosphere called the Ionosphere, so they can reach distant parts of the Earth. The reflection of radiowaves by ionosphere is due to total internal reflection. It is the same as total internal reflection of light in air during a mirage, that is, angle of incidence is greater than critical angle. Ionosphere is transparent optical medium and radiowave is reflected back. Reflection through transparent surface is total internal reflection so that internal reflection of radio wave takes place.

30. Option (2) is correct. Explanation: u = f + x Using mirror formula, 1/v + 1/u = 1/f Or, 1/v – 1/(f+x) = – 1/f \ v = – f(f+x)/x So, the magnification = |m| = v/u = f/x 31. Option (3) is correct. Explanation: Magnifying power of compound microscope, M = m1 × m2 32. Option (3) is correct. Explanation: Wavefront is the locus of all points those are in same phase. 33. Option (1) is correct. Explanation: Wave nature of light cannot explain the photoelectric effect. Particle nature of light can only explain it. 34. Option (1) is correct. Explanation: β ∝ λ and λ ∝ 1/m So, β ∝ 1/m Since mWATER > mAIR \ βAIR > βWATER 35. Option (2) is correct. Explanation: Energy of the photon must be equal to the binding energy of proton So, energy of photon= 1 MeV = 106 × 1.6 × 10–19 J hc 6.63 × 10 −34 × 3 × 10 8 6.63 × 3 = = × 10 −26 +13 E 1.60 1.6 × 10 −13 19.89 = × 10 −13 =12.4 × 10 −13 =1.24 × 101 × 10 −13 1.60 = 1.24 × 10 −9 × 10 −3 = 1.24 × 10 −3 nm

λ=

36. Option (3) is correct. Explanation: KE = hn – f So, KE is independent of intensity of incident light. 37. Option (1) is correct. Explanation: A set of atoms in an excited state decays in general to any of the states with lower energy. 38. Option (1) is correct. Explanation: The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons because when we derive the formula for radius/energy levels, etc., we make the assumption that centripetal force is provided only by electrostatic force of attraction by the nucleus. So that, this will only work for single electron atoms. In multi-electron atoms, there will

Solutions also be repulsion due to other electrons. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. 39. Option (2) is correct. Explanation: Electrostatic force between proton-proton is repulsive which causes the instability of nucleus. So neutrons are more than the number of protons. 40. Option (2) is correct. Explanation: Nuclear force is the strongest short-range force which binds the neutrons and protons in a nucleus. 41. Option (1) is correct. Explanation: In β-emission, A remains same and Z increases by 1. In α-emission, A decreases by 4 and Z decreases by 2. In γ-emission, there is no change in A and Z. 42. Option (1) is correct.

45. Option (2) is correct. Explanation: Inertia is defined as the tendency of an object to resists its change of state of motion. Induced e.m.f. in a coil is changed by the change in current or magnetic flux. The property by which a coil opposes these parameters to incur any change in induced e.m.f. is known as self-inductance. Hence, self inductance may be called the inertia of electricity. So, the assertion and reason both are true but reason cannot explain why so happens. 46. Option (2) is correct. Explanation: Photocell is a technological application of the photoelectric effect 47. Option (1) is correct. Explanation: Photosensitive material used as emitter should be connected to -ve terminal of the battery so that the emitted electrons are repelled by emitter and collected by collector. 48. Option (2) is correct.

Explanation: In intrinsic semiconductor, conductivity increases with rise of temperature due to rupture of covalent bonds and thus charge carriers become available. In extrinsic semiconductor, conductivity increases due to doping and also due to rupture of covalent bonds with rise of temperature. 43. Option (1) is correct.

Explanation: A part of the bulb is left clean for the light to enter in it. 49. Option (3) is correct. Explanation: Photocurrent of the order of a few microampere can be normally obtained from a photocell. 50. Option (4) is correct.

Explanation: Induced e.m.f = – df/dt = (5t2 + 2t + 3) = – (10t + 2) = – 32 44. Option (2) is correct. Explanation: S =

53 53

igG i − ig

=

Explanation: A photocell converts a change in intensity of incident light into to change in photocurrent.

100 × 0.01 = 0.1 Ω 10 − 0.01 

SOLUTIONS OF

Question Paper

03



Section - II



1. Option (4) is correct.

PHYSICS

Explanation: In semiconductor, the density of charge carriers (electron, holes) are very small, so its resistance is high. When temperature increases, the charge carriers (density) increases which increases the conductivity. As temperature of semiconductor increases, the speed of free electrons increases which decreases the relaxation time. As the density of charge carrier is small, so there is small effect on decrease of relaxation time.





Explanation: Nuclear density is same for all nuclei. (Independent of mass number)



2. Option (3) is correct.

3. Option (2) is correct.



4. Option (3) is correct. Explanation: Positron is anti-particle of electron.



= 3.4 + 13.6 = –17.0 eV So that the loss in kinetic energy due to inelastic collision will be, DE = E2 – E1 = – 17.0 – (– 27.2) = – 10.2 eV = 27.2 – 17.0 = 10.2 eV

5. Option (3) is correct. Explanation: Half-life time for a radio-active substance is defined as the time in which a radio-active atomic substance remains half of its original value of radio-active atom. Given that, Half-life = 1 year. So, after 1 year means one half-life, that is, average atoms of radioactive substance remain after 1 year in each container is equal to half of 10,000 = 5,000 atoms (average).

8. Option (1) is correct. Explanation: Total energy of two H-atom in ground state = 2(−13.6) = −27.2 eV. The maximum amount by which their combined kinetic energy is reduced when any one H-atom goes into first excited state after the inelastic collision, that is, the total energy of two H-atom after inelastic collision : 13.6 E = - 2 + 13.6 n 13.6 = = – 2 + 13.61 [For excited state (n = 2)] 2

Explanation: In the middle right of the circuit the capacitor behaves like an open circuit for dc 0.2 mA current, so current will flow from A to B only. Let potential across A and B is V, so by Kirchhoff ’s loop law, VAB = (5,000 × 0.2 × 10−3) + 0.3 + 5,000 × 0.2 × 10−3 VAB = 1 V + 0.3 V + 1 V VAB = 2.3 V



7. Option (3) is correct. Explanation: In a hydrogen atom, electrons revolving around a fixed proton nucleus have some centripetal acceleration. So that, its frame of reference is non-inertial. In the frame of reference, where the electron is at rest, the given expression is not true as it forms the non-inertial frame of reference. As the mass of an electron is negligible as compared to proton, so the centripetal force cannot provide the electrostatic force, m v2 Fp = p r So the given expression is not true, as it forms non-inertial frame of reference due to me 0, it means converging nature of the lens. So, lens act as a convex lens irrespective of the side on which the object lies. 44. Option (1) is correct. Explanation: The current density (number of free electrons per m3) in metallic conductor is of the order of 1028. 45. Option (3) is correct. Explanation: Principle of operation of AC generator is electromagnetic induction. The assertion is true.

Resistance offered by inductor = 2pfL. For AC, f ¹ 0. So, 2pfL ¹ 0. So, the reason is false. 46. Option (2) is correct. Explanation: Core loss is the sum of hysteresis loss and eddy current loss. 47. Option (4) is correct. Explanation: Core loss is present even when no load is connected. So, these are also known as no-load loss. 48. Option (1) is correct. Explanation: If the currents in primary and secondary windings of the transformer are I1 and I2 respectively and the resistances of these windings are R1 and R2 then the copper losses that occurred in the windings are I12R1 and I22R2 respectively. So, the entire copper loss will be I12R1 + I22R2. This loss is also called variable or ohmic losses because this loss changes based on the load. 49. Option (1) is correct. Explanation: CRGO–Cold rolled grain oriented Si steal is used to reduce hysteresis loss. 50. Option (2) is correct. Explanation: The transformer is a highly efficient device which ranges between 95%98.5%. 

SOLUTIONS OF Question Paper

04



Section - II



1. Option (2) is correct.

PHYSICS

Explanation: As all charges are positive (or of same signs), so electric field lines on R.H.S. of Gaussian surface will be due to q2 and q4 only. On L.H.S. of Gaussian surface, the electric field lines on ‘E’ will be due to q1, q2, q3, q4 and q5. So, answer (2) is verified.

3. Option (3) is correct. Explanation: The relation between electric field intensity E and potential (V) is, dV E= − dr Where, Electric field intensity, E = 0 inside the sphere dV So that, =0 dr This means that V = constant. So, if E = 0 inside charged sphere, the potential is constant or V = 100 V everywhere inside the sphere and it verifies the shielding effect also. So, it verifies the option (3).



4. Option (3) is correct. Explanation: As we know that, an equipotential surface is always perpendicular to the direction of electric field. Positive charge experiences the force in the direction of electric field. When a positive charge is released from rest in uniform electric field, its velocity increases in the direction of electric field. So K.E. increases, and the P.E. decreases due to law of conservation of energy.

5. Option (1) is correct. Explanation: According to the law of conservation of charge, total charge on the two spheres is conserved.



6. Option (3) is correct. Explanation: All the capacitors are connected in parallel. So the equivalent capacitance will be 8 mF.

2. Option (3) is correct. Explanation: We know electric field emerges radially outward from positive point charge. In the figure given above, space between field lines is increasing (or density of electric field line is decreasing). In other words, the electric force is decreasing while moving from left to right. Thus, the force on charge – q is greater than the force on charge + q and in turn, dipole will experience a force towards left direction.







7. Option (1) is correct. Explanation: Since C1 and C2 are in parallel, Voltage across C1 = Voltage across C2



8. Option (1) is correct. Explanation: Coulomb force, also called electrostatic force or Coulomb interaction, attraction or repulsion of particles or objects because of their electric charge.



9. Option (1) is correct. Explanation: As we know that, I = AneVd So current, I µ Vd And, current (I) also depend on n, the number of free electrons which increases on increasing temperature which makes more collision between electrons and increases resistance or decrease current.

10. Option (2) is correct. Explanation: As we know that current density (J) depends on (i) conductivity [s = 1/ r = 1/ R.A] (ii) Electric field [J = s E] (iii) current, length and area of cross-section But in the given options only B, that is, electric field can be varied by the charges accumulated on the surface of wire. 11. Option (4) is correct. Explanation: Kirchhoff ’s laws are valid for both linear and non-linear circuits.

60

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

12. Option (4) is correct. Explanation: When charge/mass ratio of these two particles is same and charges on them are of opposite nature, then the charged particles will traverse identical helical paths in a completely opposite sense. Therefore, option (4) is correct. 13. Option (2) is correct. Explanation: F = qvB sin q When q = 1 C, v = 1 m/s, F = 1 N, q = 90°, then B = 1T 14. Option (4) is correct. Explanation: The direction of propagation of electromagnetic wave is perpendicular to   E and B both and by right thumb rule.

As it is a concave mirror so, ray cannot go behind the mirror so ray (4) is discarded. So, ray 2 is the reflected ray that verifies answer (B). 17. Option (2) is correct. Explanation: Radiowaves are reflected by a layer of atmosphere called the Ionosphere, so they can reach distant parts of the Earth. The reflection of radiowaves by ionosphere is due to total internal reflection. It is the same as total internal reflection of light in air during a mirage, that is, angle of incidence is greater than critical angle. Ionosphere is transparent optical medium and radiowave is reflected back. Reflection through transparent surface is total internal reflection so that internal reflection of radio wave takes place. 18. Option (1) is correct.

E

B

c

The direction of propagation of electromagnetic wave is perpendicular to both the electric field   vector E and B magnetic field vector B, i.e., in   the direction of E and B . Here, electromagnetic wave is along the z-direction   which is given by the cross product of E and B . 15. Option (3) is correct.

Explanation: The negative refractive index materials are those in which incident materials are those in which incident ray from air (medium 1) to them refract or bends differently or opposite and symmetric to normal to that of positive refractive index medium.

19. Option (4) is correct. Explanation: Magnifying power is inversely proportional to the focal length of eyepiece. So, if focal length of the eyepiece is halved, the magnification will be 2M.

Explanation: Average energy by electric field E0 is Uav 1 ε0 E02 U av= 20. Option (3) is correct. 2 Explanation: Traffic signals have two But E 0 = cB 0 conditions : ON and OFF, which is similar to 1 1 ε 0 ( cB 0 ) 2 = ε0 c 2 B02 (U av )electric field = that of digital conditions 1 and 0 which is the 2 2 basic of Boolean algebra. 1 1 1 = ε0 . (B0 )2 ∵ c 2 = µ0 ε0 2 µ0 ε0 21. Option (4) is correct. 1 2 Explanation: Light has dual nature. The (U av )electric field = B0 (U av )(magnetic field) 2µ 0 wave nature of light was confirmed by both (U av ) electric field Maxwell and Huygen but both are differed Ratio = (U av ) magnetic field is their views about requirement of medium. 1 For Maxwell, no medium is required while = , i.e., 1 : 1 1 Huygen was in favour that light requires medium to travel. 16. Option (2) is correct. 22. Option (4) is correct. Explanation: Incidence ray PQ is coming through principal focus F so it must be parallel to principal axis, that is, either 2 or 4.

Solutions Explanation: At P2 is minima due to two wave fronts in opposite phase coming from, two slits S1 and S2, but there is wave fronts from S1, S2. So P2 will act as a source of secondary wavelets. Wave front starting from P2 reaches at S3 and S4 slits which will again act as two monochromatic or coherent sources and will form pattern on second screen. 23. Option (3) is correct. Explanation: When a neutral body gains electrons, it becomes negatively charged. It means that the quantity of negative charges present is more than the quantity of positive charge present. 24. Option (3) is correct. Explanation: When a point positive charge is brought near an isolated conducting sphere, then there develops some negative charge on left side of the sphere and an equal positive charge on the right side of the sphere. Electric lines of force emanating from the point positive charge end normally on the left side of the sphere. And due to positive charge on the right side of the sphere, the electric lines of force emanate normally from the right side. So, the electric field is best given by Fig (ii). 25. Option (3) is correct. Explanation: Magnetic field at O due to ABCD µi straight conductor = 0 2 πr Magnetic field at O due to the BEC circular m 0i conductor = 2r The fields are in opposite direction. Hence the µi µi µi resultant field at O is 0 − 0 = 0 ( π − 1) 2r 2 πr 2 πr 26. Option (4) is correct. Explanation: Van de Graff generator was invented by Robert Jemison Van de Graff on November 28, 1933. Robert Jemison invented the Van de Graff generator, which is a kind of high-voltage electrostatic generator that accelerates particles, while he was doing his PhD in Princeton University. 27. Option (3) is correct. Explanation: Lenz’s law is a consequence of the law of conservation of energy. Lenz law says that induced current always tends to oppose the cause which produces it. So work is done against opposing force. This work is transformed into electrical energy. So it a consequence of law of conservation of energy.

61 61

28. Option (3) is correct. di Explanation: e = – L   dt 5=–L× \

0 −1 10 −3

L = 5 × 10–3 H

29. Option (4) is correct. Explanation: When a body is negatively charged by conduction, it gains electrons. Hence, its mass increases. When a body is positively charged by conduction, it loses electrons. Hence, its mass decreases. 30. Option (2) is correct. Explanation: Coulomb’s law states that : The magnitude of the electrostatic force of attraction or repulsion between two-point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distances between them. 31. Option (1) is correct. Explanation: The process of changing the amplitude of a carrier wave in accordance with the amplitude of the audio frequency (AF) signal is known as amplitude modulation (AM). In AM, frequency of the carrier wave remains unchanged. Side band frequencies : The AM wave contains three frequencies fC, (fC + fm) and (fC − fm), fC is called carrier frequency, (fC + fm) and (fC − fm) are called side band frequencies. (fC + fm) : Upper side band (USB) frequency (fC − fm) : Lower side band (LSB) frequency Side band frequencies are generally close to the carrier frequency. According to the problem, frequency of the carrier signal is fC = 1 MHz and frequency of speech signal = 3 kHz. = 3 × 10−3 MHz = 0.003 MHz We know that frequencies of side bands = (fC + fm)= (1 + 0.003) and (1 − 0.003) So, side band frequencies are 1.003 MHz and 0.997 MHz. 32. Option (3) is correct. Explanation: Above mentioned three given elements, the energy band gap of carbon is the maximum and that of germanium is the least. The energy band gaps of these elements are related as : (E g )C > (E g )Si > (E g )Ge

62

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

33. Option (2) is correct.

39. Option (1) is correct.

Explanation: In circuit, A is at −10 V and B is at 0 V. So B is positive than A. So D2 is in forward bias and D1 is in reverse bias, so no current flows from A to B or B to A. 34. Option (3) is correct. Explanation: When positive cycle is at A, diode will be in forward bias and resistance due to diode is approximately zero so potential across diode will be about zero. Similarly, when there is negative half cycle at A, diode will be in reverse bias and resistance will be maximum so potential difference across diode is Vm sin wt with negative at A. So we get only negative output at A, so it behaves like a half-wave rectifier with negative cycle at A in output, verifies the answer (C).

Explanation: According to Bohr’s second postulate of atomic model, angular momentum of revolving electron must be some integral h multiple of . So the Bohr’s model of 2p atom does not give correct value of angular momentum. 40. Option (4) is correct. Explanation: If a beam of electrons of having energy E0 is incident on a metal surface kept in an evacuated chamber.

35. Option (2) is correct. Explanation: During formation of H-atom, some mass of nucleons convert into energy by the equation E=mc2 This energy is used to bind the nucleons along with nucleus. So mass of atom becomes slightly less than sum of actual masses of nucleons and electrons. Actual mass of H-atom B.E. = M p+M e − 2 c

B   c 2 = Binding energy   

So, the binding energy of H atoms is 13.6 eV per atom. 36. Option (3) is correct. Explanation: Nuclear force is greater than Coulomb force. 37. Option (1) is correct. Explanation: Nuclear fusion involves fusion of two lighter nuclei. 38. Option (1) is correct. Explanation: As we know that the nuclear forces is too much stronger. Only attractive force as compared to electrostatic repulsive force and nuclear force decreases to zero on increasing distance. So in case of oxygen molecule, the distance between atoms of oxygen is larger as compared to the distances between nucleons in a nucleus. So that, the force between the nuclei of two oxygen atoms is not important as nuclear forces are short-ranged forces.

41. Option (1) is correct. Explanation: Photoelectric effect quantum nature of EM radiation.

shows

42. Option (2) is correct. Explanation: When an object is placed between f and 2f of a concave mirror, the image is formed beyond 2f. The image is real and magnified. 43. Option (3) is correct. Explanation: For sustained interference, the source must be coherent and should emit the light of same frequency. In this problem, one hole is covered with red and other with blue, which has different frequency, so no interference takes place. 44. Option (2) is correct. Explanation: According to Gauss’s law of electrostatic field, q ∫ E.ds = ε0 So it does not contradict for electrostatic field as the electric field lines do not form continuous path. According to Gauss’s law of magnetic field, ∫ B.ds = 0

Solutions I t is clear that it contradicts for magnetic field because there is magnetic field inside the solenoid, and no field outside the solenoid carrying current, but the magnetic field lines form the closed paths. 45. Option (4) is correct. Explanation: Speed of light is slower in glass compared to that in air. Hence the refractive index of glass is more than that of air. So the reason is true. When a double convex air bubble is formed within a glass slab, the refractive index of the medium of the bubble is less than the refractive index of the surrounding medium. Hence the lens will not behave like a converging lens. It will behave like a diverging lens. So, the assertion is false. 46. Option (3) is correct. Explanation: A toroid is a coil of insulated or enamelled wire wound on a donut-shaped form made of powdered iron. A toroid is used as an inductor in electronic circuits. 47. Option (2) is correct.

63 63

core of the same material and similar size. This makes it possible to construct high-inductance coils of reasonable physical size and mass. 48. Option (1) is correct. Explanation: In a toroid, all the magnetic flux is contained in the core material. This is because the core has no ends from which flux might leak off. 49. Option (4) is correct. Explanation: Audible vibration or hum in transformers is caused by vibration of the windings and core layers from the forces between the coil turns and core laminations. The toroidal transformer’s construction helps quiet this noise. For this reason, many sound system engineers prefer to use a toroidal transformer instead of a traditional laminated transformer. 50. Option (2) is correct. Explanation: Standard toroidal transformers typically offer a 95% efficiency, while standard laminated transformers typically offer less than a 90% rating.

Explanation: A toroid has more inductance , for a given number of turns, than a solenoid with a 

SOLUTIONS OF Question Paper

05



Section - II



1. Option (1) is correct.

PHYSICS

Explanation: Net force on charge q1, by other charges q2 and q3 is along the + x-direction, so nature of force between q1 and q2 and q1 and q3 is attractive. This is possible when charge q1 is negative. Now, if a positive charge Q is placed at (x, 0), then, the force on q1 will increase. The direction will be along positive x-axis.

2. Option (1) is correct. Explanation: When you place a positive charge near a conducting plane, then electric field lines from positive charges will enter into the conducting plane (from the side where positive charge is kept) and emerge from opposite side of the plane. In both cases, the direction of electric field lines will always be perpendicular to the surface of the plane.



Explanation: The magnetic field inside the long current carrying solenoid is uniform. Therefore, magnitude of force on the electron of charge (- e) is given by F = - evB sinq = 0 (q = 0°) as magnetic field and velocity are parallel. The electron will continue to move with uniform velocity along the axis of the solenoid.





Explanation: To calculate resistance, R   l1 R = S  (100 − l )  1  

= 2.98 W So to get balance point near to 50 cm (middle) we have to take S = 3 W, as here R : S =2.9  : 97.1 implies that S is nearly 33 times to R. In order to make ratio R and S = 1 : 1, we must take the resistance S = 3 W, which verifies options (C).

8. Option (2) is correct. Explanation: The process of changing the amplitude of carrier wave in accordance with the amplitude of the audio frequency (AF) signal is known as amplitude modulation (AM). In AM, frequency of the carrier wave remains unchanged or we can say that the frequency of modulated wave is equal to the frequency of carrier wave. Now, according to the problem, frequency of carrier wave is wc. Thus, the amplitude modulated wave also has frequency wc.

4. Option (3) is correct.

 2.9  = 100    97.1 

7. Option (4) is correct. Explanation: t = nBIA. So, torque is directly proportional to the magnetic field strength, area of the coil, number of turns and current flowing.

s o, it is clear that part ‘c’ and ‘d’ are incorrect by formula. According to this formula only option (A), is correct.

6. Option (1) is correct. Explanation: B = m0ni = 4p × 10–7 × 5 × 10 × 102 = 2p x 10–3 T

3. Option (1) is correct. Explanation: As we know that the equivalent emf in parallel combination of cells is : ( ε1r2 + ε 2 r1 ) eeq = ( r1 + r2 )

5. Option (4) is correct.



9. Option (1) is correct. Explanation: The magnetic field lines of the Earth resemble that of a hypothetical magnetic dipole located at the centre of the Earth. The axis of the dipole does not coincide with the axis of rotation of the Earth and it is tilted at some angle (angle of declination). In this situation, the angle of declination is approximately 11.3° with respect to the later. So, there is two possibilities arises as shown :

Solutions N 11.3˚

N 11.3˚

S

S

W

E

E

W

N S

N S

So that the declination varies between 11.3° W to 11.3° E. 10. Option (4) is correct. Explanation: Angle of inclination or dip is the angle between the direction of intensity of total magnetic field of the Earth and a horizontal line in the magnetic meridian. If the total magnetic field of the Earth is modelled by a point magnetic dipole at the centre, then it is in the same plane of geographical equator, thus the angle of dip on the geographical equator will be different at different points. It may be positive or negative or may be zero at some points. 11. Option (3) is correct. Explanation: Since in pure resistive circuit the current and voltage are in phase, the power dissipation is maximum. 12. Option (1) is correct. Explanation: When a capacitor C is charged to a certain potential and connected to an inductor L, energy stored in C oscillates between L and C. XL = XC 1 \ f= 2p LC 13. Option (4) is correct. Explanation: Ultraviolet rays are used for water purification and eye surgery. 14. Option (2) is correct. Explanation: Water molecules readily absorb infrared radiation and their thermal motion increases and therefore, they heat their surroundings. 15. Option (4) is correct. Explanation: When rear car approaches, initially it appears at rest as image is formed at focus. When car approaches nearer this speed will appear to increase. 16. Option (1) is correct. Explanation:

µ 2 µ1 µ 2 − µ1 − = v u R

65 65

Putting v = 3 cm m1 = 1 u=∞ R = 0.78 cm µ2 1 µ2 − 1 − = 3 ∞ 0.78 \ m2 = 1.35 17. Option (4) is correct. Explanation: From the transistor n-p-n circuit: Vi = 20 V, VBE = 0, VCE = 0, RB = 500 kW, RC = 4 kW, VCC = 20 V We need to find values of base and collector currents IB and IC and amplification factor β, so VCC = ICRC + VCE IC = VCC – VCE/RC = 20 V – 0 V/4000 = 5 mA Base current IB = Vi /RB = 20 V/500 k = 40 μA Now amplification factor β = IC/IB   = 5 mA/40 μA = 125 18. Option (4) is correct. Explanation: Angular width = 2sin–1l/d So, it is independent of D (distance between slit and screen). 19. Option (4) is correct. Explanation: The incident wavelength should be less than threshold wavelength for photoelectric emission. IR has wavelength more than 600 nm. UV has wavelength less than 600 nm. So, photoelectrons emitted when illuminated by UV lamp either 100 W or 10 W. 20. Option (3) is correct. Explanation: At stopping potential, photoelectrons have no kinetic energy.

the

21. Option (4) is correct. Explanation: (1)  The sizes of the atoms taken in Thomson’s model is not different from the atomic size in Rutherford’s model. (2) In the ground state of Thomson’s model, the electrons are in stable equilibrium, while in Rutherford’s model, electrons always experience a net force. 22. Option (2) is correct. Explanation: Equivalent resistance of the 1 1 1 = + =5W circuit = R ( 5 + 5) ( 5 + 5) 1 1 1 1 = + = R 10 10 5 R=5W Total circuit current = 10/5 = 2 A Current in each branch is 1A So, potential difference between points A and B i.e. across the 5 W resistor is 1 × 5 = 5 V.

66

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

23. Option (1) is correct.

Battery is disconnected.

Explanation: Current in the circuit = I = V/R If r = internal resistance of the cell, then V = E – Ir E−V E−V E−V \r= = = R I V/R V 24. Option (3) is correct. Explanation:

Q = Charge remains context C' = KC Q' = C'V' Q = C'V' Q = K C V' V' =

25. Option (1) is correct. Explanation: Angular frequency of cyclotron is not affected by any change in speed of the particle. 26. Option (4) is correct.

30. Option (1) is correct. Explanation: Potentiometer. 31. Option (3) is correct. Explanation:

Explanation: When a semiconductor having 4 valence electrons is doped with an element having 3 valence electrons, then an excess hole is generated and the semiconductor becomes p-type extrinsic semiconductor. When a semiconductor having 4 valence electrons is doped with an element having 5 valence electrons, then an excess electron becomes available and the semiconductor becomes n-type extrinsic semiconductor. 27. Option (4) is correct. Explanation: Potential difference across capacitors will be peak voltage when diode is in forward bias. Diode will be in forward bias when end A is at positive potential of cycle. So potential at C = peak value of V = Vrms = V

R=ρ

l A

R=ρ

l πr 2

R′ = ρ

2l π( 2r )2

2l π4 r 2

R′ = ρ H= 

V2 V2 t & H′ = t R R′ V = constant

H′ V2 R t = H R ′ V2 t R l 2 πr 2 =ρ 2 R′ ρl πr

=

28. Option (1) is correct. Explanation: In a forward biased p-n junction diode, the positive terminal of the battery is connected to the p-side and negative terminal of the battery is connected to the n-side of the diode. Holes flow from p-side to n-side and electrons flow from n-side to p-side.

Q V = KC K

H′ 2 = H 1 H' = 2H 32. Option (2) is correct. Explanation:

29. Option (3) is correct.

1 Cseries 1

Explanation:

Cseries



=

1 1 1 + + C1 C 2 C 3

=

1 1 1 + + 2 3 6

3+2+1 6 = 6 6

Cseries = 1 mF

Solutions 33. Option (2) is correct.

36. Option (1) is correct.

Explanation: Electric potential

V= So unit is

Work (W) Charge (q )

Joule Newton-meter or as Coulomb Coulomb

Work = Force × displacement. Additionally, the SI unit of electric potential is volt. So, Option (B) is not the unit of electric potential. 34. Option (1) is correct. Explanation: Maximum capacitance of capacitors is obtained when they are connected in parallel.

C p = C1 + C 2 + C 3 when, C1 = C 2 = C3 C p = 3C When these capacitors are connected in series, the effective capacitance is minimum.

1 1 1 1 = + + Cs C C C 1 3 = Cs C C Cs = 3 When two in parallel and one in series, the effective capacitance is

Ceff =

2C 3

When two in series and one in parallel, the effective capacitance is

Ceff =

67 67

3C 2

Explanation: Kirchhoff ’s current law is based on the law of conservation of charge. 37. Option (3) is correct. Explanation: E = 11 eV = 11 × 1.6 × 10–19 J 6.62 × 10 −−34 Js 6.62 × 10 34 Js hν hν + 34 E 11 × 1.6 × 10 −−19 8.8 × 10 −−19 19 8.8 × 10 19 +34 ν= E= 11 × 1.6 × 10 = ν= h= 6.62 × 10 −−34 = 3.31 h 3.31 6.62 × 10 34 880 880 × 1015 = 2.65 × 1015 15 Hz = 15 Hz = 2.65 × 10 Hz 331 × 10 Hz = 331

= h h = E= E=

his frequency radiation belongs to the T ultraviolet region. 38. Option (3) is correct. Explanation: Microwave frequency ranges from 1013 to 109 Hz. 39. Option (2) is correct. Explanation: The objective should have large aperture and large focal length while eyepiece should have small aperture and small focal length. 40. Option (3) is correct. Explanation: b = lD/d, where d is the diameter of the wire. So, if the diameter increases, fringe with decreases. 41. Option (2) is correct. Explanation: Matter waves (de-Broglie waves) According to de-Broglie, a moving material particle sometimes acts as a wave and sometimes as a particle. h De- Broglie wavelength: λd = p E= E= E= Ex p n e

Therefore, any combination of parallel and series capacitors, Ceff always be less than Cp.

K.E.= K= 2K = mv 2

35. Option (1) is correct. Explanation: As we know that,  l R=r    A The maximum resistance will be achieved l , is maximum, so that ‘A’ when the value of A must be minimum and it is minimum when 1 cm. area of cross section is 1 cm × 2

1 2 mv 2

2 Km = m 2 v 2 2mK = p 2 2mK = p h ∴ λd = p h λd = 2mK h or λ d = [as h and E (K.E.) is constt.] 2mK 1 ∴λ ∝ m ma > mp = mn < me ∴ λa < λ p = λn < λe

68

OSWAALh CUET (UG ) Sample Question Papers, PHYSICS

or λ d = ∴λ ∝

2mK

[as h and E (K.E.) is constt.]

So, assertion is true. Breakdown of capacitors requires high voltage. So, reason is false.

1

m ma > mp = mn < me

46. Option (3) is correct.

∴ λ < λ p = λn < λe a

42. Option (2) is correct. Explanation: R = 0.5 A° w = 10 rps = 10 × 2p rad/s n = 10 Hz M = IA = e n p r2 = 1.6 × 10–19 × 10 × 3.14 × 0.5 × 0.5 × 10–10 × 10–10 = 1.256 × 10–38 Am2



43. Option (4) is correct.

dφ dt

|e| R

But there is no change of flux with time, as

  B, A and θ all remain constant with time.

\ No current is induced. 44. Option (2) is correct.

f = 5t2 +3t + 16

Explanation:

|e| =

47. Option (1) is correct. Explanation: Cars are example of Faraday Cages in the real world. Cars can help keep us safe from lightning. Its metal body acts as a Faraday Cage. 48. Option (3) is correct.

Explanation: Current induced is I = Now |e| =

Explanation: A Faraday cage or Faraday shield is an enclosure made of a conducting material. Since copper is the only metal given in the list of options, copper is the correct answer.

=

dφ dt d  2 5t + 3t + 16  dt 

= 10t + 3 |e|t=4 = 10 (4) + 3 = 43 V e = – 43 Volts 45. Option (3) is correct.

Explanation: The field within a conductor cancel out with any external fields, so the electric field within the enclosure is zero. 49. Option (3) is correct. Explanation: If a charge is placed inside an ungrounded Faraday shield without touching the walls of the internal face of the shield becomes charged with – q, and + q accumulates on the outer face of the shield. If the cage is grounded, the excess charges will be neutralized by the ground connection. 50. Option (3) is correct. Explanation: The number of electric field lines passing through the cube normally and leaving the surface = Q/e0 Q = 2 mC = 2 × 10–6 C

e0 = 8.85 × 10–12 C2/Nm2

\ Q/e0 = 2.2 × 105 Nm2/C

Explanation: Even when power is off capacitor may have stored charge which may discharge through human body and thus one may get a shock. 

SOLUTIONS OF Question Paper

06



Section - II



PHYSICS

1. Option (3) is correct.

Explanation: As all other statements are correct. In uniform electric field, equipotential surfaces are never concentric spheres but are planes ^ to electric field lines.



2. Option (3) is correct. Explanation: Let P be the observation point at a distance r from –2q and at (L + r) from + 8q. Given  Now, Net EFI at P = 0 \ E1 = EFI (Electric Field Intensity) at P due to + 8q

\ \

  E1 = E2 k(8q ) (L + r ) 4

2

( L + r )2

= =





4. Option (2) is correct. Explanation: Kirchhoff ’s current law is based on the law of conservation of energy.



5. Option (2) is correct. Explanation: Case 1: P/Q = 10/90 = 1/9 ...(i) Case 2: (P+20)/Q = 20/80 = ¼ ...(ii) Dividing equation (i) by (ii) P/(P+20) = 4/9

8. Option (1) is correct. Explanation: BH = B cos q BV = B sin q At equator, q = 0°. So, BH = B, BV = 90° At poles, q = 90°. So, BH = B, BV = B So, the ratio of total intensity of magnetic field at equator to poles is 1 : 1.

( r )2

Explanation: W = pE (cos q1 – cos q2) q1 = 0° q2 = 90° W = pE (cos 0° – cos 90°) = pE (1 – 0) = PE



r2 1

3. Option (2) is correct.

7. Option (2) is correct. Explanation: Horizontal component of magnetic field = B cos q Velocity of the rod = (2 gH)1/2 Induced e.m.f. = BLv = BL cos q × (2 gH)1/2

k( 2q )

4r2 = (L + r)2 2r = L + r r = L \ P is at x = L + L = 2L from origin \ Correct option is (3) 2L.

6. Option (1) is correct. Explanation: B µ I, B µ 1/r So, if I and r remains constant, then magnetic field at P = Magnetic field at Q.

E2 = EFI (Electric Field Intensity) at P due to

– 2q

\ P = 16 W Putting in equation (i) Q = 144 W



9. Option (2) is correct. Explanation: The resonant frequency of L-C-R 1 series circuit is vo = 2 π LC o to reduce resonant frequency, we have S either to increase L or to increase C. To increase capacitance, another capacitor must be connected in parallel with the first.

10. Option (1) is correct. Explanation: Given, Power associated with secondary, Ps = 12 W Secondary voltage, Vs=24 V Ps 12 Is = = 0.5 A Current in the secondary, = Vs 24 Peak value of the current in the secondary, 1 = I 0 I= = 0.707 or A. s 2 (0.5)(1.414) 2

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

70

11. Option (3) is correct.

17. Option (4) is correct.

Explanation: A changing magnetic field induces an electromotive force (emf) and, hence, an electric field. The direction of the emf opposes the change. The third Maxwell’s equations is Faraday’s law of induction, and includes Lenz’s law.

Explanation: All other statements except (iv) are incorrect. The electric field over the Gaussian surface remains continuous and uniform at every point. 18. Option (3) is correct. Explanation: Redrawing the circuit, we get

12. Option (1) is correct. Explanation: Frequency range of Gamma rays: 1022 – 1019 Hz Frequency range of UV rays: 1017 – 1015 Hz Frequency range of IR rays: 1014 – 1012 Hz Frequency range of Microwave: 1013 – 109 Hz

3 W & 6 W are in parallel.

13. Option (2) is correct. Explanation: Now we see output of gate 1 and gate 2 A A G1

B B

Y A

\ R1 =

Now R1 and 8W in series \ R2 = R1 + 8 = 2 + 8 = 10 W Now R2 and 30 W in parallel Req =

G2 B

G1 output is A • B G2 output is A • B The output of G1 and G2 serve as input of OR gate, so now output Y is : A B Y

3 × 6 18 = =2Ω 3+6 9

=

R2 × 30 10 × 30 = R2 + 30 10 + 30

300 30 15 = = 40 4 2

= 7.5 W 19. Option (2) is correct. Explanation: We know

B=

A B

Output Y= A • B + A • B 14. Option (4) is correct. Explanation: l µ 1/√V \ l1/l2 = √(V2/V1 ) = 6/5 15. Option (4) is correct. Explanation: In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms. 16. Option (1) is correct.

1 Q1Q2 4 pe 0 r 2 (Force in medium) Explanation: K Hence, Fvacuum =

1 Q1Q2 ´K 4 pe 0 r 2

Fvacuum = 5K

F Il sin θ

N = NA −1m −1 Am

SI Unit of B =

20. Option (4) is correct. Explanation: The coil of a moving coil galvanometer is wound over metallic frame to provide electromagnetic damping so it becomes dead beat galvanometer. 21. Option (1) is correct. Explanation: Target law, BV = BH tan d tan d =

BV BH

Given BH = tan d =

d = 30° or

3 BV

BV 3 BV

=

1 3

π radians. 6

Solutions 22. Option (2) is correct. Explanation: K.E. =

71 71

28. Option (2) is correct. 1 q 2B2 R 2 2 M

23. Option (2) is correct. Explanation: Mutual inductance of a pair of two coils depends on the relative position and orientation of two coils.

29. Option (2) is correct.

24. Option (4) is correct. Explanation: A wire of resistance 15 W bent into an equilateral triangle. As resistance is directly proportional to length. So, resistance of each arm =

Explanation: Maximum Intensity is given as Imax = (√I1+√I2)2 Minimum intensity is given as Imin = (√I1 – √I2)2 Now I1 / I2 = n Imax = (√I1 + √I2)2 = (√n + 1)2 × I2 Imin = (√I1 – √I2)2 = (√n – 1)2 × I2 Now, (Imax – Imin)/(Imax + Imin) = 2√n/(n + 1)

15 = 5Ω. 3

Explanation: Diffraction is the phenomenon in which bending of light beans take place at obstacles or corners. 30. Option (2) is correct. Explanation: If λ Blue < λ Red, hence the fringe pattern will contract as fringe width ∝ λ

31. Option (1) is correct.

Since, RAC and RBC are in series, their combine resistance = 5 + 5 = 10 W. This 10 W resistance and RAB are in parallel. Therefore, equivalent resistance between A and B,

=

10 × 5 50 = = 3.33 Ω 10 + 5 15

25. Option (1) is correct. Explanation: In telescope, angular magnification fo/fe which shows that focal length of objective lens to be large. Angular resolution = D/1.22λ which is large, hence objective should have large focal length and larger diameter. 26. Option (2) is correct.

Explanation: Wavelength, λ = h/p λ = h/√2mE h λ= 2m × 3 / 2 kT     or, λ = h/√2m × (3/2)kT = h/√3mkT 32. Option (1) is correct. Explanation: It is observed that energy is given by, E = hn = hc/λ = p2/2m =

( h / λ )2 h2 = 2m 2mλ 2

For certain frequency, maximum wavelength that can be emitted is λ0 which is cut off wavelength obtained at cut off frequency, E0 = hc/l0 Since,    E = E0 h2

2mλ

Explanation: Given u = 400 cm = 4 m v = ∞ and f = ? Since,

1 1 1 1 1 = − f v u =∞−4

so, f = – 4 m Hence, lens will be concave Now P =

1 1 = = − 0.25 D f −4m

27. Option (3) is correct. Explanation: Seeing from one end, h1 = μ × (h – b)    = 3/2 × 5 = 15/2 cm From other end of the slab, h2 = μ × h    = 3/2 × 3 = 9/2 cm Now total height, (15/2 + 9/2) = 24/2 = 12 cm

2

=

hc 2mcλ 2 or, λ 0 = λ0 h

33. Option (4) is correct. Explanation: It is known that wavelength λ is, λ = h/p Also wavelength is proportional to momentum, so λ ∝ 1/p 34. Option (2) is correct. Explanation: In case of Balmer series : 4 n1 = ∞,  n2 = 2 Þ λ1 = R In case of Lyman series : n1 = ∞, n2 = 1 Þ λ2 = Now λ1/λ2 = 4

1 R

72

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

35. Option (2) is correct. Explanation: An electron jumps from L to K shell energy is released because EL > KK. 36. Option (1) is correct. Explanation: Now V µ r3 Vatom = r 3atom / r 3nucleus × Vnucleus But rnucleus / ratom = (1/105)3 So, V atom = 1015 × Vnucleus

44. Option (3) is correct.

37. Option (4) is correct.

Explanation: As a result of heating, temperature increases which generates large number of electron-hole pairs which lead to increase in conductivity. As current increases I = I0 (e– qv/ KT), overall resistance of diode changes which affects forward and reversed biasing. 38. Option (4) is correct. Explanation: We see that current gain (β) : β = ΔIC/ΔIB = 2 × 10–3/40 × 10–6 = 50 Further voltage gain = β (Rout/Rin) = 50 [4 × 103/100] = 2000 39. Option (2) is correct. Explanation: V = 1/4πε0 × Q/R = Q × 1011 volt   E = 1/4πε0 × Q/R2 = V/R = Q × 1011 × 4πε × 1011 0

…(i) from ...(i)

= 4πε0 × Q×1022 Volt/m

40. Option (3) is correct.

Explanation: By taking common potential:    C1V = C1V’ + C2V’ \

  V’ =

C1V C1 + C2

41. Option (3) is correct. Explanation: The relation of capacitance with dielectric constants is given as: K Cmedium ∝ d So, if the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance will be four times more. 42. Option (1) is correct. Explanation: If a bulb B and AC source are connected in series with self-inductance coil, then in such case the brightness of the bulb decreases when an iron rod is inserted in the coil which further increases impedance of the circuit. 43. Option (3) is correct.

Explanation: We know that flux of magnetic field through loop is ϕ = Bπ r2 cos ωt = Bπr2 cos θ Here θ is the angle which the normal makes with portion of induced loop generating emf ε = ωBπ r2 sin ωt It is zero when ωt=nπ, so if θ=0, π, 2π, then induced emf changes its direction in every half of rotation, hence frequency of change of direction of the induced e.m.f. is twice per revolution.

Explanation: A square law modulator is the device which can produce modulated waves by the application of the message signal and the carrier wave. Square law modulator is used for modulation purpose. Characteristics shown by (i) and (iii) corresponds to linear devices. And by (ii) corresponds to square law device which shows non-linear relations. Some part of (iv) also follows square law. 45. Option (1) is correct. Explanation: Most of the a-particles pass roughly in a straight line (within 10) without deviation. This shows that no force is acting on them. So assertion is true. Most of the space in the atom is empty. Only 0.14% of a-particles are scattered more than 1°. 46. Option (1) is correct. Explanation: Power factor correction is the method to reduce the lagging power factor in inductive loads by fixing a high value capacitor across the phase and neutral close to the load. 47. Option (2) is correct. Explanation: When the voltage and current are in phase with each other in an AC circuit, the energy from the source is fully converted into another form to drive the load and in this case, power factor is unity. When the power factor drops, the system becomes less efficient. 48. Option (2) is correct. Explanation: Power capacitors are huge non polarized metal film electrolytic type capacitors. 49. Option (2) is correct. Explanation: Power factor corrector capacitors have leading power factor so that they neutralize the lagging power factor of the inductive load. 50. Option (3) is correct. Explanation: Power capacitors are connected across the phase and neutral near the inductive load such as motor. 

07

SOLUTIONS OF Question Paper 

Section - II



PHYSICS

1. Option (2) is correct.

Explanation: Since VP/VS = IS/IP , so as as voltage reduces, the current increases in a step-down transformer.

2. Option (1) is correct. Explanation: Transformer is a static device which transforms power from one circuit to other through electromagnetic induction. In electrical transformer as there are no moving parts, no friction. Losses in the transformer are very less compared to any other rotating machine, hence efficiency of transformers will be very high which is about 95% to 98%.





Explanation: Total flux = Net Charge enclosed/ ε0 It depends only on net charge enclosed by the surface.



σ [ c − ( b − a )] ε0

6. Option (1) is correct.

7. Option (2) is correct.

+

+

+

Explanation: Given c = a + b

+ +

R +

b

+

+

Ein = Ecentre = 0 Vin = Vcentre = constant = KQ/R = Q/4peoR

a

A B C

σQ σb σc σ VA = − + = [ c − ( b − a )] ε0 ε0 ε0 ε0 VB =

=

Explanation: It is observed that in a conducting sphere, there is no charge

5. Option (4) is correct.

c

(b 2 − a2 )  σ  c −  c ε 0  

Explanation: E = q2/2C q’ = q C’ = 2C E’ = E/2 Hence, total electrostatic energy of resulting system decreases by a factor of 2

3. Option (2) is correct.

4. Option (3) is correct.

=

VA = VC ≠ VB

Explanation: Torque is given by τ = pE sin θ τ = pE sin θ = qlE sinθ or, q = τ /lE sinθ = 4/(2 × 10-2 × 0.5 × 2 × 105) =2 mC

(b 2 − a2 )  σ = c −  b ε  0  2 σ × 4 πa 2 σ × 4 πb σc 1 1 VC = − ⋅ + ⋅ ε0 4 π ε0 c c 4 πε 0

− σb σ × 4 πa 2 σc 1 + ⋅ + b ε0 ε0 4 πε 0



8. Option (2) is correct. Explanation: Given carbon resistor of (47 ± 4.7) k W R = (47 ± 4.7) × 103 = 47 × 103 ± 10% W As per colour code, 3 - Orange, 4 - Yellow, 7 Violet, tolerance 10% - Silver which shows the colour code Yellow – Violet – Orange – Silver

74

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

9. Option (2) is correct. Explanation: Resistance R = ρl/A A ρl ′ l’ = nl and A′ = , R ′ = n A′ ρnl ρl 2 R′ = = n A /n A

14. Option (3) is correct. Explanation: Consider the circuit NOR A B



R ′ = Rn 2 10. Option (2) is correct. Explanation: P = V2/Req Now putting values : 10 2 × ( 5 + R )   30 = 5R or, 5R/(R+5) = 10/3 or, 3R = 2R + 10 or R = 10 W 11. Option (1) is correct. Explanation: Given: B = 3.57 × 10–2 T e/m = 1.76 × 1011 C/kg Frequency of revolution of charge in magnetic field is given as f = eB/2πm Putting values (1.76 × 1011 C/kg ) × 3.57 × 10 −2 T = ( 2 × 3.14 ) = 109 Hz = 1 GHz 12. Option (1) is correct. Explanation: Resistance of galvanometer G = 100 Ω Current for full scale deflection, Ig = 30 mA = 30 × 10–3 A Voltmeter Range V = 30 V Required resistance V = R −G Ig 30 = − 100 30 × 10 − 3 = R 900 Ω Converting galvanometer to voltmeter for the given range, resistance R to be connected in series.

13. Option (2) is correct. Explanation: Torque on magnet is given by τ = MB sin θ Now, W = MB (cos θ1 – cos θ2) W = MB (cos 0° – cos 60°) MB = 2W τ = MB sin θ = 2W sin 60° = √3W

NOR y1

NOT y2

y

y= A+B 1

y= y1 + y 1 2

= A+B

y= y 2= A + B which is NOR-gate 15. Option (4) is correct. Explanation: Self inductance of coil: L = NΦ/I 1000 × 4 × 10 −3 = =1H 4 16. Option (2) is correct. Explanation: Energy stored is given as, U = (½) × Li2 Also, U2/U1 = (i2/i1) 2 = (1/2)2 = 1/4 U2 = (¼) × U1 17. Option (2) is correct. Explanation: Resolving powers of an optical microscope is: RP ∝ 1/λ Now, RP1 / RP2 = λ2/λ1 = 6000 Å/4000 Å= 3/2 18. Option (1) is correct. Explanation: Now refractive index, μ = sin [(δm+A)/2]/sin (A/2) μ = sin [(A+A)/2]/sin (A/2) = sin A/sin (A/2) = 2 sin (A/2) cos(A/2)/sin (A/2) = 2 cos (A/2) When δm = A, then the angle may vary for 0° to 90° For 0°, m = 260° =2 For 90°, m = 2 cos 45° = √2 So,refractive index lies between 2 and √2. 19. Option (4) is correct. Explanation: Refractive index, μ = c/v =

c νλ

= 3 × 108 / (2 × 1014 × 5000 × 10– 10) = 3 20. Option (3) is correct. Explanation: The penetration of light into the region of geometrical shadow results to turning of light around corners is known as Diffraction.

Solutions 21. Option (2) is correct.

75 75

22. Option (2) is correct.

Explanation: Diffraction, interference and polarization of light suggest wave nature of light.

Explanation: Now intensity I = I0 cos2 θ = I0 cos2 45° = I0/2

23. Option (2) is correct. Explanation: A basic communication system consists of an information source, a transmitter, a link (channel) and a receiver or a communication system is the set-up used in the transmission and reception of information from one place to another. The whole system consists of several elements in a sequence. It can be represented as the diagram given below: Information source

Message Signal

Message Received Use of Receiver Transmitter Transmitted Channel Signal information Signal Signal

Noise

24. Option (2) is correct. Explanation: Polarizer is used in producing polarized light. 25. Option (4) is correct. Explanation: For Photoelectric emission to occur, (1/2)mv2 = hn – hn0 So, required condition is n ≥ n0 26. Option (4) is correct. Explanation: Saturation current ∝ intensity It is noted that the number of photoelectrons emitted is independent of frequency while it depends on intensity of light. 27. Option (1) is correct. Explanation: We see that wavelength of Lyman series, λL = 1/R [1/1 – 1/4] = 4/3R We see that wavelength of Balmer series : λB = 1/R [1/4 – 1/9] λB = 1/R [1/5/36] = 36/5R Now ratio of longest wavelengths corresponds to Lyman and Balmer series: λL / λB = 4/3R × 5R/36 = 5/27 28. Option (2) is correct. Explanation: Bohr initially use quantum theory to explain the structure of atom and proposed that energy of electron in an atom is quantized. 29. Option (3) is correct. Explanation: In case of nuclear fusion, the nuclei having low mass is good. 30. Option (4) is correct. Explanation: With fusion, enormous amount of heat is liberated that appears to be the main cause of source of solar energy.

31. Option (4) is correct. Explanation: Cyclotron accelerates positive charged particles. 32. Option (4) is correct. Explanation: In forward biasing of the p-n junction, the positive terminal of the battery is connected to p-side and the negative terminal of the battery is connected to n-side. The depletion region becomes thin. 33. Option (2) is correct. Explanation: Barrier potential depends on difference in Fermi levels in n and p sides. If different materials are used, then there is an additional potential from intrinsic difference in electron affinity. It also depends upon temperature. 34. Option (3) is correct. Explanation: At a certain reverse bias voltage, zener diode allows current to flow through it and hence maintains the voltage supplied to any load. Hence, it is used for stabilisation. 35. Option (2) is correct. Explanation: Now the Electric field strength will be: E = 9×109 × Q/r2 = 9×109 × 5×10−6/(0.8)2=7 × 104 N/C 36. Option (2) is correct. Explanation: As seen, system potential energy will be: (−e) × (−e)/4pε0r=e2/4pε0r When radius r decreases, then potential energy of system increases.

76

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

37. Option (3) is correct.

44. Option (4) is correct.

Explanation: Since, all charges are equal in magnitude and equidistant from point O then magnitude of electric field intensity will be equal. Electric field intensity is vector quantity. So direction of vectors of all three will be 1200 apart from each other, so by adding these three vectors we get zero. 38. Option (2) is correct. Explanation: From the above options, two capacitors each of 4μF should be connected in series combination and one 4μF capacitor is to be connected in parallel. By doing this the Cseries will be 2μF and on parallel combination will give an effective capacitance of 6μF as 2μF + 4μF = 6μF 4F

4F

A

B 4F

Explanation: From the formula C = Kε0A/d, it is clear that the capacity of parallel plate condenser depends on the separation distance between the plates. 45. Option (1) is correct. Explanation: When an electric dipole is placed in a uniform electric field at an angle q with the field, the dipole experiences a torque. The torque produced by two parallel forces qE acting as couple = t ¬ t = qE (2l sin q) I n case of non-uniform field, force acting on both the ends of the dipole will not be equal. So, there will be a combination of couple and a net force. In this way, dipole will have both rotational as well as linear motion. So, both assertion and reason are true. Reason also explains the assertion. 46. Option (2) is correct. Explanation: The term laser is an acronym which stands for “light amplification by stimulated emission of radiation”.

39. Option (4) is correct. Explanation: According to the Gauss' theorem, the electric flux through a close surface in Q/e0 40. Option (2) is correct. Explanation: The capacity of a parallel plate condenser C = ε0A/d If the separation between the plates is halved, then ε A ε A C′= 1 = 2 × 0 = 2C d /2 d 41. Option (1) is correct. Explanation: B = μ0i/2πr Further, B ∝ 1/r when  r  is doubled,  B  is halved, hence the magnetic field will be 0.2 T.

47. Option (2) is correct. Explanation: Einstein explained the stimulated emission. In an atom, electron may move to higher energy level by absorbing a photon. When the electron comes back to the lower energy level, it releases the same photon. This is called spontaneous emission. This may also so happen that the excited electron absorbs another photon, releases two photons and returns to the lower energy state. This is known as stimulated emission. 48. Option (3) is correct. Explanation: The term “laser light” refers to a much broader range of the electromagnetic spectrum that just the visible spectrum, anything between 150 nm up to 11000 nm (i.e. from the UV up to the far IR). 49. Option (2) is correct.

42. Option (3) is correct. Explanation: When a bar magnet is placed



in an external magnetic field B , a magnetic torque τ acts on    it, which is given τ = M × B 43. Option (1) is correct. Explanation: As per Faraday’s laws, mechanical energy gets transformed into electric energy which is done in accordance with law of conservation of energy, hence Faraday’s laws are consequence of conservation of energy.

Explanation: Laser emission is therefore a light emission whose energy is used, in lithotripsy, for targeting and ablating the stone inside human body organ. 50. Option (1) is correct. Explanation: An optical disc drive (ODD) is a disc drive that uses laser light or electromagnetic waves within or near the visible light spectrum as part of the process of reading or writing data to form optical discs. 

SOLUTIONS OF Question Paper

08



Section - II



1. Option (3) is correct.

PHYSICS

Explanation: From the question, as dipole moment is along direction of field, in such case angle between p and E zero, so the potential energy will be U = − pE cos 0°   = − pE = minimum It is observed that in uniform electric field, net force Fnet = 0

3. Option (2) is correct. Explanation: Total energy = energy density × volume   = (1/2) ε0E2Ad



4. Option (4) is correct. Explanation: Total capacitance of the combination

C′ = C/n (here, n = 3)

= 6×10−12/3 =2×10−12F

5. Option (4) is correct. Explanation: On applying KVL in the circuit, the potential difference between the two points can be obtained. VA – VB = (2Ω × 2A)+3V+(2A × 1Ω)  = 4V+3V+2V=9 V



6. Option (2) is correct. Explanation: In potentiometer, the internal resistance: l  r =  1 − 1 R l  2

 110  = − 1 × 10  100 

   r = 1.0 Ω

7. Option (3) is correct. Explanation: Applying De Morgan’s law: Output Y = [(A × B) × C]’ = A’ + B’ + C’ When A, B, C are 0 Þ Y = 1 When A, B, C are 1 Þ Y = 0



8. Option (2) is correct. Explanation: If two free parallel wires carry current in opposite directions, they tend to repel each other.

2. Option (3) is correct. Explanation: Potential difference (V) = 1V Kinetic Energy acquired = qV So, K.E. = 1.6 × 10-19 × 1= 1.6 × 10–19 joules = 1 eV







9. Option (3) is correct. Explanation: Electromagnet are made of soft iron because soft iron has low retentivity and low coercive force or low coercivity.

10. Option (4) is correct. Explanation: Since, magnetic field density φ B= B A Weber = m2     = Tesla Volt × s = m2   Newton = A-m  

11. Option (4) is correct. Explanation: It is observed from the question that the work done by magnetic field on charge particle is zero, so the kinetic energy after 3 seconds of the particle is same as T. 12. Option (4) is correct. Explanation: According to Faraday’s Law, emf Em = −𝑑f/𝑑𝑡 Given φ=3t2+4t+9 Now differentiating above equation, w.r.t. t, emf = 𝑑(3t2 + 4t + 9)/𝑑𝑡 = 6𝑡 + 4 Now the induced emf at t = 2 s is, emf = 6 × 2 + 4 = 16 V

78

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

13. Option (4) is correct. Explanation: Magnetic potential energy (U) = 25 mJ = 25 × 10–3 J Now current flowing in the Inductor (I) = 60 mA = 60 × 10–3 A Magnetic potential energy stored in an inductor is given by U = (1/2)L × I² 25 ×10−3 = (1/2) × L × (60 × 10−3)2 L = (50 × 10–3)/(3600×10⁻⁶) = (50 × 10³)/3600 = 13.89 H 14. Option (1) is correct. Explanation: As observed in an ac circuit, the capacitance will have current from potential in forward biased direction. 15. Option (1) is correct. Explanation: Now y = a sin (ωt – kx) or y = a sin (kx – ωt) Further k = 2π/λ = 2π/2π Now angular frequency ω = 2π.f = 2π × 1/ π= 2 So equation of wave y = sin(x – 2t) 16. Option (1) is correct. Explanation: If the speed of light c then as E0/ B0 = c or, Bo/Eo = 1/c Hence ratio of amplitude of magnetic field to amplitude of electric field for electromagnetic wave in a vacuum is reciprocal of speed of light. 17. Option (4) is correct. Explanation: Formula for refractive index of material of prism, n = [sin (A + δ m)/2]/sin(A/2) δ m = angle of minimum deviation Now, n = cot(A/2) = cos (A/2)/sin (A/2) Applying in formula, cos (A/2) = sin (A + δm)/2 A + δm A  sin  90° −  = sin   2 2 \ 

δ m = 180° – 2A

18. Option (3) is correct. Explanation: Now refractive index, μ = c/cm It shows cm = c/1.5 19. Option (3) is correct. Explanation: A theory that explains blackbody radiation over the broad spectrum requires that light be emitted in energy quanta, which turns out to be the photon. A wave theory allows for energy to be emitted in a continuous range of values. Models that allows this to

happen are only valid for specific cases of the spectrum. So as per Planck’s hypothesis, black bodies emits radiations in form of photons. 20. Option (2) is correct. Explanation: In this problem, the frequency of modulated signal received becomes more, due to improper selection of bandwidth. This happens because bandwidth in amplitude modulation is equal to twice the frequency of modulating signal. But, the frequency of male voice is less than that of female. 21. Option (2) is correct. Explanation: Now frequency νo = hc/λ λ = hc/νo = 12400/2 = 6200 Å or   = 6200 × 10–10 m = 620 nm 22. Option (1) is correct. Explanation: Now momentum p = h/λ = 6.6 × 10–34/5000 × 10−10 = 1.3 × 10−27 kg-m/s 23. Option (1) is correct. Explanation: Energy En = – 13.6 [Z2/n2] = (– 13.6) × (4/4) = – 13.6 eV 24. Option (4) is correct. Explanation: Rutherford a-scattering experiment. 25. Option (4) is correct. Explanation: Number of atoms that are decayed : N = 4 × 1016(1/2)30/10 = (½) × 1016 Now atoms to be decayed: 4 × 1016 – (½) × 1016 = 3.5 × 1016 26. Option (3) is correct. Explanation: Consider a reaction: 0n1 → 1p1 + 0 –1e + X In this, X is zero charge and mass when the electrons emitted, so X need to be antineutrino (n). 27. Option (3) is correct. Explanation: If a small amount of antimony is added to germanium crystal, crystal becomes n-type semi-conductor. Hence, there will be more free electrons than holes. 28. Option (3) is correct. Explanation: Davisson – Germer experiment uses an electron gun to produce a fine beam of electrons which can be accelerated to any desired velocity by applying a suitable voltage across the gun. The others mentioned do not find an application here.

Solutions 29. Option (1) is correct. Explanation: An electric dipole if kept in nonuniform electric field will experiences two forces that are opposite and not equal. In this, a net force will exists and the forces will act at different points of a body along with a torque. 30. Option (4) is correct. Explanation: It is observed that on equipotential surfaces, electric field results as normal to the charged surface not allowing the work done. 31. Option (4) is correct. Explanation: From the formula C = Kε0A/d, it is clear that the capacity of parallel plate condenser depends on the separation distance between the plates. 32. Option (3) is correct. Explanation: In a charged capacitor, the energy resides in between the plates. 33. Option (3) is correct. Explanation: In semiconductors, conduction band is empty and valence band is completely filled at 0 K and there will be no electron that can cross from valence band to conduction band at 0 K. At room temperature, certain electrons in valence band will jump over to conduction band due to small forbidden gap as 1 eV. 34. Option (2) is correct. Explanation: Infrared ray is the cause of Green House effect. The glass transmits visible light while infrared rays are absorbed by plants. Then it emits long infrared rays, which are reflected back by glass. 35. Option (4) is correct. Explanation: The visible colors from shortest to longest wavelength are violet, blue, green, yellow, orange and red. 36. Option (3) is correct. Explanation: With the phenomenon of refraction, stars twinkle due to variation in refractive index in the atmosphere. 37. Option (3) is correct. Explanation: We see that velocity of light in window  = 3 × 108/1.5 m/s = 2 × 108m/s So time taken by the sunlight to pass through a window, t = 4 × 10−3/2 × 108 s =2 × 10−11 s

79 79

38. Option (2) is correct. Explanation: If reflected and refracted light rays are perpendicular, reflected light gets polarised with electric field vector perpendicular to the plane of incidence. 39. Option (3) is correct. Explanation: Interference, Refraction and Reflection are shown by both light and sound waves while polarisation is shown by light wave only. 40. Option (2) is correct. Explanation: When negative charged pendulum oscillates over a positively charged plate then effective value of acceleration g increases, so the time period: T = 2π√l/g, 41. Option (2) is correct. Explanation: In uniform electric field, in the direction of electric field potential decreases, so VA = VB > VC. 42. Option (3) is correct. Explanation: Equivalent capacitance will be: 1/Ceq = 1/3 + 1/10 + 1/15 or Ceq=2 μF Now charge on each capacitor ⇒

Q = Ceq× V 2×100 = 200 μC

43. Option (3) is correct. Explanation: E = σ/e0K = Q/Ae0K ∝ 1/K Eair > E1 > E2

E

0

d

44. Option (4) is correct. Explanation: It is known that alpha particles are the nucleus of helium atoms having two protons and two neutrons. 45. Option (2) is correct. Explanation: Resistance of superconductor falls to zero at critical temperature. This property is very useful for power transmission without any loss. Assertion and reason both are true but reason does not explain the assertion.

80

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

46. Option (1) is correct. Explanation: Eddy currents are not generated in non-conductor/insulator. Eddy currents are generated in conductor/metal. 47. Option (2) is correct. Explanation: The generation of eddy current was discovered by physicist Foucault (18191869). 48. Option (4) is correct. Explanation: The retarding force due to the eddy currents inhibits the motion of the magnet in a metal pipe. This phenomena is known as electromagnetic damping.

49. Option (1) is correct. Explanation: To observe electromagnetic damping, a magnet should be dropped through a metal pipe and the magnet should not touch the inner wall of the pipe. 50. Option (2) is correct. Explanation: When a piece of wood and a bar magnet of same dimension is dropped through an aluminium pipe, the bar magnet will take more time to come out from the pipe due to electromagnetic damping.



SOLUTIONS OF Question Paper

09



Section - II



1. Option (1) is correct.

PHYSICS

Explanation: In case of full wave rectifier, Fundamental frequency = 2 × main frequency = 2 × 50 =100 Hz

2. Option (2) is correct. Explanation: Diffusion of charge carriers.



3. Option (4) is correct. Explanation: Energy = Power × time = 200 × 106 × 24 × 3600 = 2 × 2.4 × 3.6 × 1012 = 1728 × 1010 J



4. Option (3) is correct. Explanation: Nuclear force is stronger than coulomb force, so En > Ee.



5. Option (3) is correct. Explanation: Rutherford results major mass of the atom is concentrated at nucleus.



6. Option (4) is correct. Explanation: Sommerfield extended Bohr Theory and gave his postulates who explained that stationary orbits where electrons are revolving around the nucleus in atom are not circular but elliptical in shape.



7. Option (4) is correct. Explanation: Now momentum p = hν/c or ν = pc/h = 3.3 × 10−29 × 3 × 108/6.6 × 10−34 = 1.5 × 1013 Hz



8. Option (1) is correct. Explanation: de-Broglie wavelength λ = h/m1v1 = h/m2v2 Now v1/v2 = m2/m1 = 4/1



9. Option (3) is correct. Explanation: Fact.

10. Option (1) is correct. Explanation: In sun glasses, polarised glasses are applied so as to reduces the light intensity to half during polarisation. 11. Option (3) is correct. Explanation: In an electromagnetic wave it is observed that electric and magnetic vectors are perpendicular to each other and are perpendicular to the direction of propagation of wave. 12. Option (4) is correct. Explanation: It is noted that velocity of electromagnetic radiation is the velocity of light (c) which is 1 / µ 0 ε 0 where permeability μ0 and permittivity ε0 of free space. 13. Option (2) is correct. R  − t Explanation: I = I 0  1 − e L   

0.693 = (R/Lt)   t = 0.3×0.693/2= 0.1 sec 14. Option (4) is correct. Explanation: E = – L × (dI/dt) 5 = – L × (2 – 3)/(1 × 10–3) = 1000 L or, L = 5 × 10– 3 H = 5 mH 15. Option (4) is correct. Explanation: In an ac circuit, when voltage across two terminals of resistance changes, current changes immediately. If voltage rises, current also rises and when voltage falls, current also falls, hence current and voltage are in phase. Further, as voltage rises and falls, current rises and falls, after a fraction of a second; so current flowing through inductor is always lagging behind voltage and current and voltage are out of phase.

82

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

16. Option (4) is correct. Explanation: Now, Y’= A + B Y = (A + B)C = 1 As seen, A or B can be zero while C should be 1 17. Option (2) is correct. Explanation: Magnetic field due to current carrying element is given by B=µ0i/2πr. Since, r is the distance between wire and measuring point. Thus, the magnetic field B is independent of the diameter of wire. Hence, the magnetic field will be same. 18. Option (1) is correct. Explanation: Here F ⊥ v ⇒ v = constant Hence, the speed of the particle remains unchanged. 19. Option (2) is correct.

Here, p = (500 × 10–6) × (10 × 10–2) = 5 × 10–5 Now r = 25 cm = 0.25 m l = 5 cm = 0.05 m Using the above expression and putting the values: E = 9×109×2×5×10−5×0.25/[(0.25)2−(0.05)2]2   = 6.25×107 N/C 24. Option (3) is correct. Explanation: On rotating a point charge q around a charge Q in a circle of radius r, work done will be zero as charge Q moves on equipotential surface where work done is zero. 25. Option (2) is correct. Explanation: Let the distance between given changes be 2x

B = μ0 Ni/2r



N = 1000, i = 0.1 A, r = 0.1 m

Substituting the values, we have B = 4π ×

l0-7

  = 4π × l0  

× l000 × 0.1/(2 × 0.1)

-4

= 6.28 × l0-4 Tesla

∆V , ∆I in the graph CD has negative slope, so in this, resistance R is negative.

Explanation: Since, the in resistance R =

22. Option (4) is correct. Explanation: To solve this, use relation 4

4

Now,

Hence,

B

| FAB | = − | FAC |

1 Q2 1 Qq =− 4 πε 0 4 x 2 4 πε 0 x 2 or,

q = – Q/4

26. Option (4) is correct. Explanation: U = (1/2) CV2 U = (1/2) (Aε0/d)Ed2= (1/2)Aε0E2d 27. Option (2) is correct.

21. Option (2) is correct.

R1  r2  = R2  r1 

Q

  In equilibrium, | FAB | + | FAC | = 0

Explanation: At the centre of circular coil carrying current, the magnetic field is,

C 2x

A

So,

20. Option (3) is correct.



q

Q

Explanation: It is observed that above the Curie temperature, domains breaks down, so ferromagnetic substances becomes paramagnetic.

4

 r  r  = 16 R R2 = R1  1  = R   r2   r / 2  R2 = 16 R

23. Option (1) is correct. Explanation: Considering electric field intensity E = 9 × 109 × 2pr/(r2 − l2)2

Explanation: Veff = V + V + V = 3 V 1/Ceff = 1/C + 1/C + 1/C C eff = C/3 28. Option (2) is correct. Explanation: The crystal used in the Davisson – Germer experiment is nickel. A fine beam of electrons is made to fall on the surface of the nickel crystal. As a result, the electrons are scattered in all directions by the atoms of the crystal. 29. Option (2) is correct. Explanation: Barrier potential does not depend in diode design while barrier potential depends upon temperature, doping density and forward biasing. 30. Option (1) is correct. Explanation: Solar energy is due to fusion reaction.

83 83

Solutions 31. Option (3) is correct.

37. Option (3) is correct.

Explanation: For first minima

Explanation: π     θ = = 30° 6

θ = λ/a ⇒ a = λ/θ  Now a = 6500 × 10−8 × 6/π = 1.24 × 10−4 cm = 1.24 microns

O

–q

4q

A

B

a 2a

Net force on – q FAO = FOB

32. Option (3) is correct. Explanation: θ = λ/a From the figure geometry

1 4q2 1 16 q 2 = 4 πε 0 a 2 4 πε 0 4 a 2

D Y a

4q

θ

Hence, the forces on charges at O and B are in equilibrium If any of the charge is displaced, the net force starts acting on all of them and the new system will be in unstable equilibrium. 38. Option (3) is correct.

sin θ = Y/D Y θ= D or, Y = λD/a Width of centre maxima 2Y = 2λD/a 33. Option (2) is correct. Explanation: The rms value of magnetic field Brms = Erms/c = 6/(3 × 108) = 2 × 10– 8 T Now peak value of magnetic field Bpeak = √2 Brms = 2.83 × 10– 8 T 34. Option (4) is correct. Explanation: Since coefficient of x is negative, but is moving along +ve x-axis, so equating equation as: Ey = 2.5 cos[(2π × 106)t – (π × 10–2)x] y = A cos(ωt – kx) ω = 2π × 106 So, f = ω/2π = 106 Hz k = π × 10–2 Also λ = 2π/k λ = 2π/(π × 10−2)= 200 m 35. Option (4) is correct. Explanation: Lenz’s Law is about conservation of energy that guarantees about energy in induced current from effect of change. The law describes that direction of induced emf tends to produce a current that oppose the change causing induction. Hence, Lenz’s law is a direct consequence of the principle of conservation of energy. 36. Option (2) is correct. Explanation: e = -L

di Þ – 5(2) = – 10V dt

Explanation: Let the change in phase angle of the modulating signal f. Now consider a sinusoidal modulating waveform signal m(t) is represented by m ( t ) = Amsinωmt (i) Where, Am = Amplitude of peak value of modulating signal. fm = Angular frequency = 2pvm= f(vm) Where vm is the amplitude of frequency. Also consider a sinusoidal carrier wave C(t) represented by C ( t )=Ac sinωc t (ii) Thus, modulated wave is given by Cm ( t ) = ( Ac + Amsinωmt ) sinωc t   A =Ac  1+ m sinωmt  sinωc t A c   Am =M Here, Ac      Cm ( t ) = ( Ac + Ac×µ sinωmt ) sinωc t (iii) Now, we know that Ac × μ = k  [Wave constant] And sinω mt = vm Thus Eq. (iii) becomes Cm (t )= ( Ac + k × vm )sin ωc t Now consider change in phase angle by f, then sinωc t → sin(ωc t+φ ) Thus, Cm (t )=(Ac +Kvm )(sinω c +φ ) 39. Option (2) is correct. Explanation: The magnetic field of cylindrical magnet is symmetrical about its axis and as it is coaxial with circular coil, the flux induced with the coil does not change, if the magnet is rotated about its axis. Hence, no current is induced in the coil.

84

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

40. Option (2) is correct. Explanation: The electric potential at the surface of an atomic nucleus will be: 1 q V= × 4 πε 0 r 9

−19

V = 9×10 × 50×1.6×10

/9×10

−15

6

=8×10 V

41. Option (3) is correct. Explanation: potential:

For

calculating

common

V = (V1C1 + V2C2)/(C1 + C2) Putting values =

500 × 20 + 200 × 10 20 + 10

= 400 V

42. Option (3) is correct. Explanation: From the circuit, if a circuit in broken, the induced e.m.f. is maximum. When contact is broken, current in the circuit decreases. In order to oppose, inductor releases the current as a result, emf is induced which has high value so bulb becomes suddenly bright. 43. Option (3) is correct. Explanation: In an electric dipole, when angle θ = 0, then potential energy = – pE cos θ 44. Option (4) is correct. Explanation: Now Cmedium = KCair Also, K = Cmedium/Cair = 110/50 = 2.20 45. Option (4) is correct. →

→ →

Explanation: F = q( v × B ) So, the force is mass independent. So, the assertion is false.

Proton is obviously heavier than electron. So, reason is true. 46. Option (2) is correct. Explanation: Light cannot easily escape a diamond, because its critical angle with air is too small. Most of the reflections are total and it is cut so that light can exit only in particular direction —thus light is concentrated inside and making the diamond sparkle. . 47. Option (1) is correct. Explanation: Refractive index = 1/ sin C C = critical angle = 24.4 \ µ = 1/sin 24.4 = 1/0.4131 = 2.42 48. Option (3) is correct. Explanation: The brilliance of diamond is due to its too small critical angle with air. As the critical angle become smaller, value of sine of critical angle also become small and hence refractive index increases (since μ= 1/sinC). So, the basic reason for the extraordinary sparkle of suitably cut diamond is its high refractive index. 49. Option (4) is correct. Explanation: A diamond is immersed in a liquid with a refractive index greater than water. Then the critical angle for total internal reflection will increase. This is because, as the refractive index of outer medium increases, the refracted ray bends less away from normal. So, angle of incidence should increase more to achieve 90° angle of refraction. 50. Option (1) is correct. Explanation: The brilliance of diamond in the second diamond will be less than the first since in the second case, no total internal reflection has taken place. 

SOLUTIONS OF Question Paper

10



Section - II



PHYSICS

1. Option (4) is correct. Explanation: It represents uncertainty principle.



 1 1 1 = ( µ − 1)  −  f R R  1 2

Heisenberg’s

1 1 1  = (1.5 − 1)  −  f1  ∞ − 20 

2. Option (4) is correct. Explanation: Now, E4/E2 = 22/42 = 4/16 = 1/4 E4 = E2/4 =−328/4 =−82 kJ/mol

⇒ f1 = 40 cm  1 1 1  = (1.7 − 1)  −  f2 − + 20 20  

3. Option (1) is correct. Explanation: Now E = 1/2mv2 mv = √2mE Hence, wavelength λ= h/mv = h/√2mE

⇒ and

4. Option (1) is correct.



6. Option (4) is correct. Explanation: Let the distance of nth bright fringe from centre, yn=nλD/d Distance of 3rd bright image from centre,    y3 = 3 × 6000 × 10−10 × 2.5/0.5 × 10−3 = 9 × 10−3 m = 9 mm





9. Option (1) is correct. Explanation: The e.m. wave is propagating along +z direction. Electric field and magnetic field are mutually perpendicular and also perpendicular to the direction of propagation.

  ˆ ∴ E E= = B0 ˆj 0 i and B 10. Option (1) is correct.

7. Option (3) is correct. Explanation: If ν′ and λ′ shows frequency and wavelength of light in a medium, ν′ = v/λ′ = (c/μ)/(λ/μ) = c/λ = ν



1 1 1  = (1.5 − 1)  −  ⇒ f3 = 40 cm f3  ∞ − 20 

∴ feq = – 50 cm Therefore, the focal length of the combination is – 50 cm

5. Option (1) is correct. Explanation: To have diffraction, it is observed that the size of an obstacle should be of similar order as that of wavelength.

100 cm 7

1 1 1 1 1 1 1 1 = + + ⇒ = + + feq f1 f2 f3 feq 40 −100 / 7 40

Explanation: As hν = mc2 So, p = mc = hν/c = h/λ

f2 = −

8. Option (1) is correct.

Explanation: From the electromagnetic wave spectrum, the correct decreasing order of wavelength is : microwave, infrared, ultraviolet, gamma rays. 11. Option (4) is correct.

Explanation:

Explanation: Magnetic flux is defined as the total number of magnetic field lines through a coil or given area. Magnetic flux has the dimension length2 × mass × time–2 × electriccurrent–1 which is [ML2T–2A–1]. n = 1.5

n = 1.7

n = 1.5

86

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

12. Option (3) is correct.

19. Option (4) is correct.

∝ N2/l

Explanation: As L so L2/L1 = (N2/N1)2(l1/l2) As N2 = 2N1 and l2 = 2l1 L2/L1 = 4(1/2) =2 L2 = 2L1

Explanation: In electric dipole, field Eequitorial = kp/r3 E ∝ p, so E ∝ r−3

20. Option (1) is correct.

Explanation: It is observed that charges are at the outer surface of a conducting hollow sphere of radius R. On considering spherical surface of radius r < R, then as per Gauss’s law

13. Option (3) is correct. Explanation: In series LCR circuit, resonance occurs when capacitive reactance becomes equal to inductive reactance. In series LCR circuit at resonance, capacitive reactance (XC) = inductive reactance (XL), so 1/ωC = ωL Total impedance in the circuit : 2 2 Z = R + ( X L − X C ) = R Now the power factor, cos f = R/Z = R/R = 1 Hence, the power loss at resonance, P = Erms × Irms × cos f   = Erms × Irms × 1 = (R × Irms) × Irms 2 2   = (Irms) × R = I × R

14. Option (3) is correct. Explanation: Choke is an inductor is used for blocking or decoupling higher frequencies and acts as an inductor in electronic filters or tuned circuits. Choke coil is a low-resistance inductor coil used to suppress or limit the flow of alternating current without affecting the flow of direct current. It has low resistance and high inductance, used in electrical circuits to pass direct current and attenuate alternating current. 15. Option (2) is correct. Explanation: Bragg's law is a special case of Laue diffraction, it gives the angles for coherent and incoherent scattering from a crystal lattice. The expression for Bragg's law is given as: 2d sinq = nl 16. Option (4) is correct. Explanation: T = 2πm/qB that is independent of both R and v. 17. Option (2) is correct. Explanation: Magnetic moment :  qv  × πR 2 M = IA =   2 πR  = (qvR)/2 18. Option (2) is correct. Explanation: When diamagnetic substances are placed in magnetic field of a strong magnet then it is feebly magnetised in the opposite direction of field or it is repelled by strong magnet.

∫

Eds = (1/εo) × charge inside

E × 4πr2 = 1/εo × 0 E = 0, hence intensity of electric field inside a hollow conducting sphere is zero. 21. Option (1) is correct. Explanation: E = 50 V/cm= 50/10-2 V/m Now e/m = 1.76 × 1011 C/kg As, eE = ma or a =eE/m   = 1.76 × 1011 × 5 × 103 = 88 × 1013 = 8.8 × 1014 m/s2 22. Option (3) is correct. Explanation: Capacitance in series combination will b 1/C = 1/2 + 1/2 + 1/2 2 or, C= F  3 23. Option (2) is correct. Explanation: In a dielectric : Cmedium = K × Cair Hence when air inside capacitor is replaced by a dielectric constant K, then the capacity will increase K times. 24. Option (1) is correct. Explanation: In parallel combination, the equivalent capacitance Ceq = C1+C2 25. Option (4) is correct. Explanation: vm = 15 kHz = 15 × 103 Hz c 3 × 108 1 Wavelength λ m= = = × 105 m ν m 15 × 103 5 λ 1 1  Size of the antenna required, L = = ×  × 105  4

103

4 5



=5× m = 5 km The audio signal cannot be transmitted through sky waves because in the sky wave propagation the frequency range is 2–40 MHz. Frequency of this wave is 15 kHz.

Solutions Effective power radiated by the antenna  L of length L. P ∝    λ

2

if L decreases, P also

decreases. 26. Option (2) is correct. Explanation: Kinetic energy will be, f – f0 So K .E1 = 1 eV – 0.5 eV = 0.5 eV Now K .E2 = 2.5 eV – 0.5 eV = 2 eV Now ratio of kinetic energies K.E1/K.E2 = 0.5 eV/2 eV = 1/4 Hence the ratio of maximum speed is, v1/v2 = √1/4 = 1/2 or 1 : 2 27. Option (1) is correct. Explanation: Plane containing electric field E and plane of polarisation where electromagnetic waves like light exhibit polarization is also the direction of propagation. Hence, the angle between them is 0°. 28. Option (3) is correct. Explanation: d = 1 mm D = 1 m width 2(λ/a)D λ = 500 × 10– 9 Putting formula, 10(Dλ/d) = 2λD/a a = d/5 = 0.2 mm 29. Option (2) is correct. Explanation: As μ ∝ 1/λ 30. Option (2) is correct. Explanation: As v ∝ 1/μ, the refractive index of material µ is smaller for air than water, glass and diamond. 31. Option (4) is correct. Explanation: Now 1/f = (μg – 1) × 2/R = 1/R, (μg = 3/2), R = f 1/f1 = – (μw – 1) × 2/R = – 2/3R = – 2/3f 1/feq = 1/f + 1/f + 1/f1 = 1/R + 1/R – 2/3R 1/feq = 1/f + 1/f – 2/3f = 2/f – 2/3f = 4/3f so feq = 3f/4 32. Option (3) is correct. Explanation: Electromagnetic waves are a form of energy waves that have both an electric and magnetic field such as Radio Waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, Gamma rays. So β-rays are not electromagnetic waves, but it is a beam of fast electrons. 33. Option (4) is correct. Explanation: Since, the direction of propagation of electro-magnetic waves is perpendicular to Electric field and Magnetic

87 87

field. Also electric and magnetic fields perpendicular so, direction of the wave will be   along E × B 34. Option (1) is correct. α Explanation: β = 1 − α

  ∆I C 0.98   ∝ = ∆I = 0.98 1 − 0.98  E  β = 49

β=

35. Option (3) is correct. Explanation: The induced e.m.f for a coil of wire depends on, magnetic strength of core in coil of wire, number of turns of wire in coil, cross-sectional area of coil, speed of magnet movement of the coil. 36. Option (2) is correct. Explanation: It is observed that electric potential due to dipole is normally be, V = k × p cos θ/r2 So, V ∝ 1/r2

37. Option (1) is correct.

Explanation: In spherical shell, the electric field is zero as in a shell every charge lies on outer surface, so from Gauss law, electric field inside a shell will be zero. 38. Option (2) is correct. Explanation: F = qE ma = qE a = qE/m [E=V/l]   a = qV/ml = qV/ml = (v2 – u2)/2l [ v2 = u2 + 2as] 2 V = mv /2q V = 2 × 10-3 × 100/(2 × 2 × 10-6) = (1/2) × 105 V = 50 kV 39. Option (1) is correct. Explanation: In series combination: 1/Cs = 1/3 + 1/9 + 1/18 = 1/2 or Cs = 2 μF In Parallel combination : Cp = 3 + 9 + 18 = 30 μF Ratio of equivalent capacitance Cs/Cp = 2/30 = 1/15 40. Option (1) is correct. Explanation: Energy stored U = (1/2) CV2 = (1/2) × 50 × 10−6 × (10)2 = 2.5 × 10−3 J

88

OSWAAL CUET (UG ) Sample Question Papers, PHYSICS

41. Option (3) is correct. Explanation: Given : Half-life of radioactive substance A = 20 minutes Half-life of radioactive substance B = 40 minutes Now, 80 = 20 × nA nA = 4 80 = 40 × nB nB = 2 Ratio of two substances: nA/nB = (1/2)4 / (1/2)2 1 1 1 = × = or 1 : 4 2 2 4 42. Option (1) is correct.

46. Option (2) is correct. Explanation: Carbon Composition Resistors are manufactured from a mixture of finely ground carbon dust or graphite and a nonconducting ceramic powder to bind it all together. 47. Option (1) is correct.

Explanation: Nuclear radius = R = R0A1/3 It shows that R µ A1/3 R3 µ A Further A µ R3 Now, A2/A1 = (R2/R1)3 = [(R1/2)/R1]3 = 1/8 Now A 2 = A1/8 = 56/8 = 7 Hence, stable nucleus having radius half of Fe56 is Li7. 43. Option (4) is correct. Explanation: Due to heating, when a free electron is produced then simultaneously a hole is also produced. 44. Option (2) is correct. Explanation: When p-n junction is reverse biased, the flow of current is due to drifting of thermally generated charge carriers across the junction. 45. Option (1) is correct. Explanation: From Curie Weiss law, c=

Paramagnetic substance has no magnetic domain. At a very high temperature, the domains of ferromagnetic substance get destroyed and the substance transforms into paramagnetic substance. So, the reason is also true and properly explains the assertion.

C T − TC

As temperature increases beyond Curie temperature, susceptibility decreases and the ferromagnetic substances become paramagnetic. So, the assertion is true.

Explanation: The ratio of carbon dust to ceramic (conductor to insulator) determines the resistive value of the resistor. Higher the ratio of carbon, lower the overall resistance. 48. Option (3) is correct. Explanation: Metal Film Type Resistors are generally made by depositing pure metals, such as nickel on an insulating ceramic rod or substrate. 49. Option (2) is correct. Explanation: Wirewound Resistor, is made by winding a thin metal alloy wire (Nichrome) or similar wire on an insulating ceramic former in the form of a spiral helix. 50. Option (4) is correct. Explanation: Wirewound Resistors are generally only available in very low ohmic high precision values. They are able to handle much higher electrical currents than other resistors of the same ohmic value with much excessive power ratings. These high power resistors are moulded into an aluminium heat sink body with fins attached to increase their overall surface area to promote heat loss and cooling.

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