PHYSICS

QUESTION BANK

PART-1
ADVANCED QUESTION BANK
WITH DETAILED INSTANT SOLUTIONS
PRACTICE ORIENTED CATEGORISATIONS
CHAPTERWISE AND TOPICWISE REFERENCE WITH EACH QUESTION

1
Table of Contents
PART 1. PHYSICS QUESTION BANK .......................................................................................... 3

PART 2. PHYSICS QUESTION BANK ........................................................................................ 11

PART 3. PHYSICS QUESTION BANK ........................................................................................ 18

PART 4. PHYSICS QUESTION BANK ........................................................................................ 27

PART 5. PHYSICS QUESTION BANK ........................................................................................ 35

PART 6. PHYSICS QUESTION BANK ........................................................................................ 44

PART 7. PHYSICS QUESTION BANK ........................................................................................ 53

PART 8. PHYSICS QUESTION BANK ........................................................................................ 61

PART 9. PHYSICS QUESTION BANK ........................................................................................ 69

PART 10. PHYSICS QUESTION BANK ...................................................................................... 77

PART 11. PHYSICS QUESTION BANK ...................................................................................... 84

PART 12. PHYSICS QUESTION BANK ...................................................................................... 92

PART 13. PHYSICS QUESTION BANK .................................................................................... 100

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 PART 1. PHYSICS
Chapter: Oscillation
[Topic: Time Period, Frequency, Simple Pendulum &
Spring Pendulum]
QUESTION BANK Q4. 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
Q1. The density of material in CGS system of units is (a) depends on both R and v
4g/cm3. In a system of units in which unit of length is 10 (b) is independent of both R and v
cm and unit of mass is 100 g, the value of density of (c) depends on R and not on v
material will be (d) depends on v and not on R
(a) 0.4 (b) 40 Ans: (b)
(c) 400 (d) 0.04
Ans: (b)
Solution: In CGS system,
The unit of mass is 100g and unit of length is 10 cm, so
. / . / ( )
density = =
. / . / ( )

( ) = 40 unit Solution:
( )
Chapter: Units and Measurement When a test charge q0 enters a magnetic field ⃗ directed
[Topic: Dimensions of Physical Quantities] along z-axis, with a velocity ⃗ making angles d with the
Q2. A rod PQ of mass M and length L is hinged at end P. z-axis. The time period of the motion is independent of R
The rod is kept horizontal by a massless string tied to and v.
point Q as shown in figure. When string is cut, the initial Chapter: Moving Charges and Magnetic Field
angular acceleration of the rod is [Topic: Motion of Charged Particle in Magnetic Field
& Moment]
Q5. An electron of mass m and a photon have same
energy E. The ratio of de-Broglie wavelengths associated
with them is :

(a) . /

(a) g /L (b) 2g/L (b) . /

(c) (d) (c) ( )
Ans: (d)
(d) . /
Solution: Weight of the rod will produce the torque
Ans: (a)
η = mg = I α = α0 1
Solution: For electron De-Broglie wavelength,
Hence, angular acceleration α =
Chapter: System of Particles and Rotational Motion √
[Topic: Torque, Couple and Angular Momentum] For photon E = pc
Q3. A simple pendulum performs simple harmonic ⇒ De-Broglie wavelength,
motion about x = 0 with an amplitude a and time period
T. The speed of the pendulum at x = will be: ∴ . /
√
(a) Chapter - Dual Nature of Radiation and Matter
[Topic: Matter Waves, Cathode & Positive Rays]
(b) Q6. Which logic gate is represented by the following
(c)
√ combination of logic gate ?
√
(d)
Ans: (c)
Solution: Speed √
√ √
v=ω√ √ = (a) NAND

3
(b) AND k = = = 1[∵ λ = 2 π]
(c) NOR
∴ Y = 1 sin (2t – x + θ)[∵ A = 1 m]
(d) OR
Ans: (b) Chapter: Waves
Solution: First two gates are NOT gates and the last gate [Topic: Basic of Waves]
is NOR gate. Q10. A long solenoid carrying a current produces a
magnetic field B along its axis. If the current is double
Thus, y1 = y2 = and y = ̅̅̅̅̅̅̅̅̅̅
and the number of turns per cm is halved, the new value
The truth table corresponding to this is as follows:
of the magnetic field is
(a) 4B
(b) B/2
(c) B (d) 2B
Ans: (c)
Solution:
( ). /( )
Thus the combination of gate represents AND gate. ⇒ B1 = B
Chapter: Semiconductor Electronics Materials, Devices Chapter: Moving Charges and Magnetic Field
[Topic: Digital Electronics and Logic Gates] [Topic: Magnetic Field, Biot-Savart's Law & Ampere's
Q7. The time dependence of a physical quantity p is Circuital Law]
given by p = p0 exp (– α t2), where α is a constant and t is Q11. Monochromatic radiation emitted when electron on
the time. The constant α hydrogen atom jumps from first excited to the ground
(a) is dimensionless (b) has dimensions T–2 state irradiates a photosensitive material. The stopping
2
(c) has dimensions T (d) has dimensions of p potential is measured to be 3.57 V. The threshold
Ans: (b) frequency of the materials is :
Solution: In p = p0 exp (– αt2), αt2 dimensionless (a) 4 × 1015 Hz (b) 5 × 1015 Hz
∴ , - (c) 1.6 × 1015 Hz (d) 2.5 × 1015Hz
Ans: (c)
Chapter: Units and Measurement Solution: n → 2 – 1
[Topic: Dimensions of Physical Quantities] E = 10.2 eV
Q8. A solid homogeneous sphere of mass M and radius kE = E – θ
R is moving on a rough horizontal surface, partly rolling Q = 10.20 – 3.57
and partly sliding. During this kind of motion of the h υ0 = 6.63 eV
sphere
= 1.6 × 1015 Hz
(a) total kinetic energy is conserved
(b) the angular momentum of the sphere about the point Chapter - Dual Nature of Radiation and Matter
of contact with the plane is conserved [Topic: Electron Emission, Photon Photoelectric Effect
(c) only the rotational kinetic energy about the centre of & X-ray]
mass is conserved Q12. If the velocity of a particle is v = At + Bt2, where A
(d) angular momentum about the centre of mass is and B are constants, then the distance travelled by it
conserved. between 1s and 2s is :
Ans: (b) (a) (b) 3A + 7B
Solution: Angular momentum about the point of contact
with the surface includes the angular momentum about (c) (d)
the centre. Due to friction linear momentum is conserved. Ans: (c)
Chapter: System of Particles and Rotational Motion Solution: Given : Velocity
[Topic: Moment of Inertia, Rotational K.E. and Power] V = At + Bt2 ⇒ = At + Bt2
Q9. A wave travelling in the +ve x-direction having By integrating we get distance travelled
displacement along y-direction as 1m, wavelength 2π m
⇒∫ ∫ ( )
and frequency Hz is represented by
Distance travelled by the particle between 1s and 2s
(a) y = sin (2πx – 2πt)
(b) y = sin (10πx – 20πt) x= ( ) ( )
(c) y = sin (2πx + 2πt) Chapter: Kinematics Motion in a Straight Line
(d) y = sin (x – 2t) [Topic: Non-uniform motion]
Ans: (d) Q13. A small object of uniform density rolls up a curved
Solution: As Y = A sin (ωt – kx + θ) surface with an initial velocity „ν‟. It reaches upto a
ω = 2πf = = 2[∵ f = ]

4
maximum height of with respect to the initial [Topic: Force & Torque on a Current Carrying
Conductor]
position. The object is a Q16. The photoelectric work function for a metal surface
[2013] is 4.125 eV. The cut off wavelength for this surface is
(a) solid sphere (a) 4125 Å (b) 3000 Å
(b) hollow sphere (c) 6000 Å (d) 2062.5 Å
(c) disc
Ans: (b)
(d) ring
Solution: Let λ0 be cut off wavelength.
Ans: (c)
Work function =

Chapter - Dual Nature of Radiation and Matter

[Topic: Electron Emission, Photon Photoelectric Effect
& X-ray]
Solution:
From law of conservation of mechanical energy Q17. A bus is moving with a speed of 10 ms–1 on a
straight road. A scooterist wishes to overtake the bus in
Iω2 + 0 + mv2 = mg × 100 s. If the bus is at a distance of 1 km from the
⇒ Iω2 = mv2 – mv2 scooterist, with what speed should the scooterist chase
the bus?
= . / (a) 40 ms–1 (b) 25 ms–1
–1
(c) 10 ms (d) 20 ms–1
or, I = or, I = mR2 Ans: (d)
Hence, object is a disc. Solution: Let v be the relative velocity of scooter w.r.t
Chapter: System of Particles and Rotational Motion bus as v = vS– vB
[Topic: Rolling Motion] v=
Q14. A standing wave is represented by y = Asin (100t) ∴vS = v + vB,
cos (0.01x), where y and A are in millimetre, t in seconds
and x is in metre. Velocity of wave is
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(a) 10 m/s
(b) 1 m/s = 10 + 10 = 20 ms–1
(c) 10–4 m/s ∴velocity of scooter = 20 ms–1
(d) not derivable from above data Chapter: Kinematics Motion in a Straight Line
Ans: (a) [Topic: Relative Velocity]
Solution: The wave equation is y = Asin (ωt)cos (kx); Q18. The density of a newly discovered planet is twice
c = ω/k = 100/0.01 = 104 m/s. that of earth. The acceleration due to gravity at the
Chapter: Waves surface of the planet is equal to that at the surface of the
[Topic: Basic of Waves] earth. If the radius of the earth is R, the radius of the
Q15. When a charged particle moving with velocity ⃗ is planet would be
subjected to a magnetic field of induction⃗ , the force on (a) ½ R (b) 2 R
it is non-zero. This implies that (c) 4 R (d) 1/4 R
(a) angle between ⃗ and ⃗ can have any value other than Ans: (a)
90° Solution: . Also,
(b) angle between ⃗ and ⃗ can have any value other than .
zero and 180°
(c) angle between ⃗ and ⃗ is either zero or 180° At the surface of planet, ( ) ,
(d) angle between ⃗ and ⃗ is necessarily 90° At the surface of the earth
Ans: (b) ge = gp ⇒ dR = 2d R' ⇒ R' = R/2
Solution: Force on a particle moving with velocity v in a Chapter: Gravitation
magnetic field B is (⃗ ⃗ ) [Topic: Acceleration due to Gravity]
⃗
If angle between ⃗ & is either zero or 180º, then value Q19. A closed organ pipe (closed at one end) is excited
of F will be zero as cross product of ⃗ and ⃗ will be zero. to support the third overtone. It is found that air in the
So option (b) is correct. pipe has
Chapter: Moving Charges and Magnetic Field (a) three nodes and three antinodes
(b) three nodes and four antinodes
(c) four nodes and three antinodes

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(d) four nodes and four antinodes Hence, correct option is (c) .
Ans: (d) Chapter: Atoms
Solution: Third overtone has a frequency 7 n, which [Topic: Atomic Structure, Rutherford's Nuclear Model
means = three full loops + one half loop, which of Atom]
would make four nodes and four antinodes. Q22. If the angle between the vectors ⃗ and ⃗ is θ, the
Chapter: Waves value of the product (⃗ ⃗ ) ⃗ is equal to
[Topic: Beats, Interference & Superposition of Waves] [2005]
Q20. A coil in the shape of an equilateral triangle of side (a) BA2 sinθ
l is suspended between the pole pieces of a permanent (b) BA2 cosθ
magnet such that ⃗ is in the plane of the coil. If due to a (c) BA2 sinθ cosθ
current i in the triangle a torque η acts on it, the side l of (d) zero
the triangle is Ans: (d)
Solution: (⃗ ⃗ ) ⃗ ⃗ ⃗ = CA cos90º = 0.
(a) . / (b) . /
√ √
(c) . / (d)
√ √
Ans: (b)
Solution: η = MB sinθ
η = iAB sin90º
Chapter: Kinematics Motion in a Plane
[Topic: Vectors]
Q23. A particle of mass m is thrown upwards from the
surface of the earth, with a velocity u. The mass and the
radius of the earth are, respectively, M and R. G is
∴ gravitational constant and g is acceleration due to gravity
on the surface of the earth. The minimum value of u so
Also, A = 1/2 (BC) (AD)
that the particle does not return back to earth, is
But ( )( ) ( )√ . / (a) √
√
(b) √
√
() (c) √

∴ . / (d) √
√
Chapter: Magnetism and Matter Ans: (a)
[Topic: Magnetism, Gauss’s Law] Solution: The velocity u should be equal to the escape
Q21. An alpha nucleus of energy bombards a velocity. That is,
heavy nuclear target of charge Ze. Then the distance of u=√
closest approach for the alpha nucleus will be But g =
proportional to
(a) √ ⇒√
(b)
(c) Chapter: Gravitation
[Topic: Motion of Satellites, Escape Speed and Orbital
(d) Velocity]
Ans: (c) Q24. A vehicle, with a horn of frequency n is moving
Solution: Kinetic energy of alpha nucleus is equall to with a velocity of 30 m/s in a direction perpendicular to
electrostatic potential energy of the system of the alpha the straight line joining the observer and the vehicle. The
particle and the heavy nucleus. That is, observer perceives the sound to have a frequency n + n1.
= Then (if the sound velocity in air is 300 m/s)
(a) n1 = 10n
where is the distance of closest approach (b) n1 = 0
(c) n1 = 0.1n
(d) n1 = – 0.1n
Ans: (b)

6
Solution: As the source is not moving towards or away As we know, compressibility,
from the observer in a straight line, so the Doppler‟s . /
effect will not be observed by the observer. K= ( )
Chapter: Waves So,(∆V/V) = Kρgh
[Topic: Musical Sound & Doppler's Effect] = 45.4 × 10–11 × 103 × 10 × 2700 = 1.2258 × 10–2
Q25. A coil of resistance 400Ω is placed in a magnetic Chapter: Mechanical Properties of Fluids
field. If the magnetic flux θ (wb) linked with the coil [Topic: Pressure, Density Pascal's Law & Archimedes'
varies with time t (sec) as θ = 50t2 + 4. The current in the Principle]
coil at t = 2 sec is : Q29. Intensity of an electric field (E) depends on
(a) 0.5 A (b) 0.1 A distance r, due to a dipole, is related as
(c) 2 A (d) 1 A (a)
Ans: (a)
Solution: According, to Faraday‟s law of induction (b)
Induced e.m.f. ( ) (c)
Induced current i at t = 2 sec. (d)
=| | Ans: (c)
Chapter: Electromagnetic Solution: Intensity of electric field due to a Dipole
[Topic: Magnetic Flux, Faraday's & Lenz's Law] E= √ ⇒
Q26. The Bohr model of atoms Chapter: Electrostatic Potential and capacitance
(a) predicts the same emission spectra for all types of [Topic: Electric Flux & Gauss's Law]
atoms
Q30. An inductor may store energy in
(b) assumes that the angular momentum of electrons is
(a) its electric field
quantised
(b) its coils
(c) uses Einstein‟s photoelectric equation
(c) its magnetic field
(d) predicts continuous emission spectra for atoms
(d) both in electric and magnetic fields
Ans: (b)
Ans: (c)
Solution: In Bohr‟s model, angular momentum is
Solution: An inductor stores energy in its magnetic field.
quantised i.e . / Chapter: Electromagnetic
Chapter: Atoms [Topic: Motional and Static EMI & Applications]
[Topic: Bohr Model & The Spectra of the Hydrogen Q31. The mass number of a nucleus is
Atom] (a) sometimes less than and sometimes more than its
Q27. A missile is fired for maximum range with an atomic number
initial velocity of 20 m/s. If g = 10 m/s2, the range of the (b) always less than its atomic number
missile is (c) always more than its atomic number
(a) 40 m (b) 50 m (d) sometimes equal to its atomic number
(c) 60 m (d) 20 m Ans: (d)
Ans: (a) Solution: In case of hydrogen atom,
Solution: For maximum range, the angle of projection, θ Mass number = atomic number
= 45°. Chapter: Nuclei
( ) ( ) [Topic: Composition and Size of the Nucleus]
∴R= = = = 40 m.
Q32. When a body moves with a constant speed along a
Chapter: Kinematics Motion in a Plane circle
[Topic: Projectile Motion] (a) its velocity remains constant
Q28. The approximate depth of an ocean is 2700 m. The (b) no force acts on it
compressibility of water is 45.4 × 10–11 Pa–1 and density of (c) no work is done on it
water is 103 kg/m3.What fractional compression of water (d) no acceleration is produced in it
will be obtained at the bottom of the ocean ? Ans: (c)
(a) 1.0 × 10–2 (b) 1.2 × 10–2 Solution: On circular motion, the force acts along the
(c) 1.4 × 10–2 (d) 0.8 × 10–2 radius and displacement at a location is perpendicular to
Ans: (b) the radius i.e., θ = 90°
Solution: Compressibility of water, As work done = ⃗ ⃗
K = 45.4 × 10–11 Pa–1 Chapter: Kinematics Motion in a Plane
density of water P = 103 kg/m3 [Topic: Relative Velocity in2D & Circular Motion]
depth of ocean, h = 2700 m
We have to find ?

7
Q33. Steam at 100°C is passed into 20 g of water at [Topic: A.C. Circuit, LCR Circuit, Quality & Power
10°C. When water acquires a temperature of 80°C, the Factor]
mass of water present will be: Q36. The binding energy of deuteron is 2.2 MeV and
[Take specific heat of water = 1 cal g–1 °C–1 and latent that of is 28 MeV. If two deuterons are fused to form
heat of steam = 540 cal g–1] one , then the energy released is
(a) 24 g (b) 31.5 g (a) 23.6 MeV (b) 19.2 MeV
(c) 42.5 g (d) 22.5 g (c) 30.2 MeV (d) 25.8 MeV
Ans: (d) Ans: (a)
Solution: According to the principle of calorimetry. Solution:
Heat lost = Heat gained Energy released = 28 – 2 × 2.2 = 23.6 MeV
mLv + msw∆θ = mwsw (Binding energy is energy released on formation of
⇒m × 540 + m × 1 × (100 – 80) Nucleus)
= 20 × 1 × (80 – 10) Chapter: Nuclei
⇒m = 2.5 g [Topic: Mass-Energy & Nuclear Reactions]
Therefore total mass of water at 80°C Q37. Three blocks with masses m, 2 m and 3 m are
= (20 + 2.5) g = 22.5 g connected by strings as shown in the figure. After an
Chapter: Thermal Properties upward force F is applied on block m, the masses move
[Topic: Calorimetry & Heat Transfer] upward at constant speed v. What is the net force on the
Q34. A hollow metal sphere of radius 10 cm is charged block of mass 2m? (g is the acceleration due to gravity)
such that the potential on its surface is 80 V. The
potential at the centre of the sphere is
(a) zero (b) 80 V
(c) 800 V (d) 8 V
Ans: (b)
Solution: Potential at the centre of the sphere
= potential on the surface = 80 V.
Chapter: Electrostatic Potential and capacitance
[Topic: Electric Potential Energy & Work Done in (a) 2 mg (b) 3 mg
Carrying a Charge] (c) 6 mg (d) zero
Q35. In a circuit, L, C and R are connected in series with Ans: (d)
an alternating voltage source of frequency f. The current
leads the voltage by 45°. The value of C is
(a) ( )
(b) ( )
(c) ( )
(d) ( )
Ans: (d)
Solution: From figure,
Solution:
From figure
F = 6 mg,
As speed is constant, acceleration a = 0
∴ 6 mg = 6ma = 0, F = 6 mg
∴ T = 5 mg , T′ = 3 mg
T″ = 0
Fnet on block of mass 2 m
= T – T' – 2 mg = 0
tan 45º = Alternate :
⇒ =R v = constant
so, a = 0, Hence, Fnet = ma = 0
⇒ ( ) Chapter: Dynamics Laws of Motion
[Topic: Motion of Connected Bodies, Pulleys]
C= ( )
= ( ) Q38. A black body is at temperature of 500 K. It emits
Chapter: Alternating Current energy at rate which is proportional to

8
(a) (500)4 (b) (500)3 between the tyres of the car and the road is µs. The
(c) (500)2 (d) 500 maximum safe velocity on this road is :
Ans: (a)
(a) √
Solution: According to Stefan's Law E =
; so, ( )
Chapter: Thermal Properties (b) √
[Topic: Calorimetry & Heat Transfer]
Q39. The capacity of a parallel plate condenser is 10 µF (c) √
when the distance between its plates is 8 cm. If the
distance between the plates is reduced to 4 cm then the (d) √
capacity of this parallel plate condenser will be Ans: (b)
(a) 5 µF (b) 10 µF Solution: On a banked road,
(c) 20 µF (d) 40 µF
Ans: (c) ( )
Solution: C = 10 µFd = 8 cm
C ' = ?d ' = 4 cm Maximum safe velocity of a car on the banked road
C= ⇒ Vmax = √ 0 1
If d is halved then C will be doubled.
Chapter: Dynamics Laws of Motion
Hence, C ' = 2C = 2 × 10 µF = 20 µF
[Topic: Circular Motion, Banking of Road]
Chapter: Electrostatic Potential and capacitance
[Topic: Capacitors, Capacitance, Grouping of Q43. A mass of diatomic gas (γ = 1.4) at a pressure of 2
Capacitors & Energy Stored in a Capacitor.] atmospheres is compressed adiabatically so that its
temperature rises from 27°C to 927°C. The pressure of
Q40. The electric and the magnetic field associated with the gas in final state is
an E.M. wave, propagating along the +z-axis, can be
(a) 28 atm (b) 68.7 atm
represented by
(c) 256 atm (d) 8 atm
(a) [⃗ ̂⃗ ]̂ Ans: (c)
(b) [ ⃗ ⃗ ⃗ ̂] Solution: T1 = 273 + 27 = 300K
(c) [⃗ ̂⃗ ̂] T2 = 273 + 927 = 1200K
̂] For adiabatic process,
(d) [⃗ ̂⃗ P1–γ Tγ= constant
Ans: (a) ⇒ P11–γT1γ= P21–γ T2γ
Solution: E.M. wave always propagates in a direction
perpendicular to both electric and magnetic fields. So, . / =. /
electric and magnetic fields should be along + X– and +
Y–directions respectively. Therefore, option (a) is the ⇒. / =. /
correct option.
Chapter - Electromagnetic Waves ( ) ( )
[Topic: Electromagnetic Waves, Conduction &
Displacement Current] ( ) ( )
Q41. A radio isotope „X‟ with a half life 1.4 × 109 years
decays to „Y‟ which is stable. A sample of the rock from
a cave was found to contain „X‟ and „Y‟ in the ratio 1 : 7. ( )
The age of the rock is : . / . /
(a) 1.96 × 109 years (b) 3.92 × 109 years 7
9
(c) 4.20 × 10 years (d) 8.40 × 109 years = P1 (2 ) = 2 × 128 = 256 atm
Ans: (c) Chapter: Heat & Thermodynamics
[Topic: Specific Heat Capacity & Thermodynamic
Solution: As (Given)
Processes]
⇒ . / Q44. A, B and C are voltmeters of resistance R, 1.5 R
and 3R respectively as shown in the figure. When some
Therefore, age of the rock potential difference is applied between X and Y, the
t = 3T1/2 = 3 × 1.4 × 109 yrs = 4.2 × 109 yrs. voltmeter readings are VA, VB and VC respectively. Then
Chapter: Nuclei
[Topic: Radioactivity]
Q42. A car is negotiating a curved road of radius R. The
road is banked at an angle θ. the coefficient of friction

9
(a) VA ≠ VB = VC acquired by the person and the average force exerted on
(b) VA = VB ≠ VC the person are
(c) VA ≠ VB ≠ VC (a) –1.6 ms–1; 8 N (b) –0.08 ms–1; 16 N
–1
(d) VA = VB = VC (c) – 0.8 ms ; 8 N (d) –1.6 ms–1; 16 N
Ans: (d) Ans: (c)
Solution: Effective resistance of B and C Solution: According to law of conservation of
momentum
MV + mnv = 0
i.e., equal to resistance of voltmeter A. ⇒
In parallel potential difference is same so, VB = VCand in ⇒ – 0.8 m/s
series current is same According to work energy theorem,
So, VA = VB = VC Average work done = Change in average kinetic energy
Chapter: Current Electricity i.e.,
[Topic: Combination of Resistances]
Q45. Match the corresponding entries of column-1 with ⇒
column-2 (Where m is the magnification produced by the ⇒ Fav = 8 N
mirror): Chapter: Work, Energy and Power
Column-1 Column-2 [Topic: Energy]
(P) m = –2(A) Convex mirror Q48. The temperature of source and sink of a heat
(Q) m = - ½ (B) Concave mirror engine are 127ºC and 27ºC respectively. An inventor
(R) m = +2 (C) Real image claims its efficiency to be 26%, then:
(S) m = + ½ (D) Virtual image (a) it is impossible
(a) P → B and C, Q → B and C, R → B and D,S → A (b) it is possible with high probability
and D. (c) it is possible with low probability
(b) P → A and C, Q → A and D, R → A and B, S → C (d) data are insufficient.
and D Ans: (a)
(c) P → A and D, Q → B and C, R → B and D, S → B
and C Solution:
(d) P → C and D, Q → B and D, R → B and C, S → A
and D
Ans: (a) Hence, it is not possible to have efficiency more than
Solution: Magnitude m = +ve ⇒ virtual image 25%.
m = –ve ⇒ real image Chapter: Heat & Thermodynamics
magnitude of magnification, Q49. Consider the following two statements:
| m | > 1 ⇒ magnified image (A) Kirchhoff's junction law follows from the
| m | < 1 ⇒ diminished image conservation of charge.
Chapter - Ray Optics and Optical (B) Kirchhoff's loop law follows from the conservation of
[Topic: Plane, Spherical Mirror & Reflection of Light] energy.
Q46. Half-lives of two radioactive substances A and B Which of the following is correct?
are respectively 20 minutes and 40 minutes. Initially, the (a) Both (A) and (B) are wrong
samples of A and B have equal number of nuclei. After (b) (a) is correct and (B) is wrong
80 minutes the ratio of remaining numbers of A and B (c) (a) is wrong and (B) is correct
nuclei is (d) Both (A) and (B) are correct
(a) 1 : 16 (b) 4 : 1 Ans: (d)
(c) 1 : 4 (d) 1 : 1 Solution: Junction law follows from conservation of
Ans: (c) charge and loop law is the conservation of energy
Solution: 80 = 20 × nA = nA = 4 Chapter: Current Electricity
80 = 40 × nB = nB = 2 [Topic: Kirchhoff's Laws, Cells, Thermo emf &
Electrolysis]
. /
= Q50. When a biconvex lens of glass having refractive
. / index 1.47 is dipped in a liquid, it acts as a plane sheet of
Chapter: Nuclei glass. This implies that the liquid must have refractive
[Topic: Radioactivity] index.
Q47. A person holding a rifle (mass of person and rifle (a) equal to that of glass
together is 100 kg) stands on a smooth surface and fires (b) less than one
10 shots horizontally, in 5 s. Each bullet has a mass of 10 (c) greater than that of glass
g with a muzzle velocity of 800 ms–1. The final velocity (d) less than that of glass

10
 Ans: (a) (a) 7.5 JK–1 mol–1 (b) 7.0 JK–1 mol–1
Solution: . /. / (c) 8.5 JK–1 mol–1 (d) 8.0 JK–1 mol–1
Ans: (d)
If µg = µm, then ( ). / Solution: Molar mass of the gas = 4g/mol
Speed of sound
⇒ = 0  infinity
This implies that the liquid must have refractive index V=√ ⇒ 952 = √
equal to glass.
⇒ γ = 1.6 =
Chapter - Ray Optics and Optical
[Topic: Refraction at Curved Surface, Lenses & Power Also, γ =
of Lens]
So, CP = = 8JK–1mol–1
PART 2. PHYSICS –1
[CV = 5.0 JK given]
Chapter: Kinetic Theory
QUESTION BANK [Topic: Degree of Freedom, Specific Heat Capacity &
Mean Free Path]
Q54. The total power dissipated in watts in the circuit
Q51. In forward biasing of the p–n junction shown here is
(a) the positive terminal of the battery is connected to p–
side and the depletion region becomes thick
(b) the positive terminal of the battery is connected to n–
side and the depletion region becomes thin
(c) the positive terminal of the battery is connected to n–
side and the depletion region becomes thick (a) 40 (b) 54
(d) the positive terminal of the battery is connected to p– (c) 4 (d) 16
side and the depletion region becomes thin
Ans: (b)
Ans: (d)
Solution: Power dissipiated = P
Solution: 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. Chapter: Current Electricity
Chapter: Semiconductor Electronics Materials, Devices [Topic: Heating Effects of Current]
[Topic: Solids, Semiconductors and P-N Junction Q55. The angle of a prism is „A‟. One of its refracting
Diode] surfaces is silvered. Light rays falling at an angle of
Q52. An engine pumps water through a hose pipe. Water incidence 2A on the first surface returns back through the
passes through the pipe and leaves it with a velocity of 2 same path after suffering reflection at the silvered
m/s. The mass per unit length of water in the pipe is 100 surface. The refractive index µ, of the prism is :
kg/m. What is the power of the engine? (a) 2 sin A (b) 2 cos A
(a) 400 W (b) 200 W (c) (d) tan A
(c) 100 W (d) 800 W Ans: (b)
Ans: (d)
Solution: Amount of water flowing per second from the
pipe
= = . /
Power = K.E. of water flowing per second
= . / Solution:
= . / According to Snell‟s law µ =
= = 400 W ⇒ (1) sin 2A = (µ) sin A ⇒ µ = 2 cos A
Chapter - Ray Optics and Optical
Chapter: Work, Energy and Power [Topic: Prism & Dispersion of Light]
[Topic: Power]
Q56. An alternating current can be converted into direct
Q53. 0 g of a gas occupies 22.4 litres at NTP. The current by a
specific heat capacity of the gas at constant volume is (a) transformer
5.0JK–1. If the speed of sound in this gas at NTP is 952 (b) dynamo
ms–1, then the heat capacity at constant pressure is (Take (c) motor
gas constant R = 8.3 JK–1 mol–1)

11
(d) rectifier monochromatic light wavelength 500 nm is used. What
Ans: (d) will be the width of each slit for obtaining ten maxima of
Solution: Chapter: Semiconductor Electronics double slit within the central maxima of single slit pattern
Materials, Devices ?
[Topic: Solids, Semiconductors and P-N Junction (a) 0.1 mm (b) 0.5 mm
Diode] (c) 0.02 mm (d) 0.2 mm
Q57. The co-efficient of restitution e for a perfectly Ans: (d)
elastic collision is Solution: Here, distance between two slits,
(a) 1 (b) 0 d = 1mm = 10–3m
(c) ∞ (d) –1 distance of screen from slits, D = 1 m
Ans: (a) wavelength of monochromatic light used,
Solution: e = | v1 – v2 |/ | u1 – u2 | which is 1 for a perfectly λ = 500nm = 500 × 10–9m
elastic collision. width of each slit a = ?
Chapter: Work, Energy and Power Width of central maxima in single slit pattern =
[Topic: Centre of Mass, Centre of Gravity & Principle
Fringe width in double slit experiment
of Moments]
Q58. Which one of the following statements is true for So,required condition
the speed v and the acceleration a of a particle executing
⇒ m = 0.2 mm
simple harmonic motion ?
(a) When v is maximum, a is zero Chapter - Wave Optics
(b) When v is maximum, a is maximum [Topic: Young's Double Slit Experiment]
(c) Value of a is zero, whatever may be the value of v Q61. In a CE transistor amplifier, the audio signal
(d) When v is zero, a is zero Solution: (d) In S.H.M., vmax voltage across the collector resistance of 2kΩ is 2V. If
= A ω = A(2πf) the base resistance is 1kΩ and the current amplification
persec. of the transistor is 100, the input signal voltage is :
( ) (a) 0.1 V (b) 1.0 V
Chapter: Oscillation (c) 1 mV (d) 10 mV
[Topic: Displacement, Phase, Velocity & Acceleration Ans: (d)
of SHM] Solution:
Q59. The resistance of the four arms P, Q, R and S in a
Wheatstone‟s bridge are 10 ohm, 30 ohm, 30 ohm and 90
ohm, respectively. The e.m.f. and internal resistance of
the cell are 7 volt and 5 ohm respectively. If the
galvanometer resistance is 50 ohm, the current drawn
from the cell will be
(a) 0.2 A (b) 0.1 A
(c) 2. 0 A (d) 1. 0 A
Ans: (a)
Solution: Given : V = 7 V The output voltage, across the load RC
r = 5Ω V 0 = IC R C = 2
The collector current (IC)
Amp
Current gain (β)
(β) current gain =
Amp
Input voltage (Vi)
Req = Ω Vi = RB IB = 1 × 103 × 10–5 = 10–2 Volt
Vi = 10 mV
I= =
Chapter: Semiconductor Electronics Materials, Devices
= = = 0.2 A. [Topic: Junction Transistor]
Q62. The unit of permittivity of free space, εo is
Chapter: Current Electricity
(a) Coulomb2/(Newton-metre)2
[Topic: Wheatstone Bridge & Different Measuring
(b) Coulomb/Newton-metre
Instruments] (c) Newton-meter2/Coulomb2
Q60. In a double slit experiment, the two slits are 1 mm (d) Coulomb2/Newton-meter2
apart and the screen is placed 1 m away. A Ans: (d)

12
Solution: [Topic: Motion of Charged Particle in Magnetic Field
( ) & Moment]
⇒ unit of εo is (coulomb) / newton-metre
2 2
Q66. The output from a NAND gate is divided into two
Chapter: Units and Measurement in parallel and fed to another NAND gate. The resulting
[Topic: Dimensions of Physical Quantities] gate is a
Q63. Two discs are rotating about their axes, normal to
the discs and passing through the centres of the discs.
Disc D1 has 2 kg mass and 0.2 m radius and initial
angular velocity of 50 rad s–1. Disc D2 has 4kg mass, 0.1
m radius and initial angular velocity of 200 rad s–1. The
two discs are brought in contact face to face, with their (a) NOT gate
axes of rotation coincident. The final angular velocity (in (b) AND gate
rad s–1) of the system is (c) NOR gate
(a) 40 (b) 60 (d) OR gate
(c) 100 (d) 120 Ans: (b)
Ans: (c) Solution: ̿̿̿̿̿̿
Solution: Given:m1 = 2 kgm2= 4 kg Hence the resultant gate is AND gate.
r1= 0.2 mr2= 0.1 m Chapter: Semiconductor Electronics Materials, Devices
w1= 50 rad s–1w2= 200 rad s–1 [Topic: Digital Electronics and Logic Gates]
As, angular momentum, I1W1 = I2W2 = Constant Q67. Turpentine oil is flowing through a tube of length
∴ = and radius r.
The pressure difference between the two ends of the tube
By putting the value of m1, m2, r1, r2 and solving we get = is p. The viscosity of oil is given by
100 rad s–1 ( )
Chapter: System of Particles and Rotational Motion
[Topic: Torque, Couple and Angular Momentum]
where v is the velocity of oil at a distance x from the axis
Q64. A block of mass M is attached to the lower end of a of the tube. The dimensions of η are
vertical spring. The spring is hung from a ceiling and has (a) [M0L0T0] (b) [MLT–1]
force constant value k. The mass is released from rest 2 –2
(c) [ML T ] (d) [ML–1T–1]
with the spring initially unstretched. The maximum Ans: (d)
extension produced in the length of the spring will be: ( ) , -, -
(a) 2 Mg/k (b) 4 Mg/k Solution: , -, -
(c) Mg/2k (d) Mg/k , -
Ans: (d) Chapter: Units and Measurement
Solution: Restoring force, f ′ = – kx [Topic: Dimensions of Physical Quantities]
where x is the extension produced in the spring. Q68. Two discs of same moment of inertia rotating
Weight of the mass acting downward = Mg. about their regular axis passing through centre and
In equilibrium perpendicular to the plane of disc with angular velocities
kx = Mg or x = ω1 and ω2 . They are brought into contact face to face
Chapter: Oscillation coinciding the axis of rotation. The expression for loss of
[Topic: Time Period, Frequency, Simple Pendulum & energy during this process is:-
Spring Pendulum] (a) ( )
Q65. An electron moves in a circular orbit with a (b) ( )
uniform speed v. It produces a magnetic field B at the
(c) ( )
centre of the circle. The radius of the circle is
proportional to (d) ( )
(a) √ Ans: (a)
Solution: Here,
(b) ⇒ =

(c) √ (K.E.)i =

(d) (K.E.)f = = . /
Ans: (d) Loss in K.E. ( ) ( ) = ( )
Solution: ⇒r Chapter: System of Particles and Rotational Motion
Chapter: Moving Charges and Magnetic Field [Topic: Moment of Inertia, Rotational K.E. and Power]

13
Q69. Two waves are represented by the equations y1 = a [Topic: Electron Emission, Photon Photoelectric Effect
sin (ωt + kx + 0.57) m and y2= a cos (ωt + kx) m, where x & X-ray]
is in meter and t in sec. The phase difference between Q72. A particle of unit mass undergoes one-dimensional
them is motion such that its velocity varies according to v(x) =
(a) 1.0 radian (b) 1.25 radian bx–2n where b and n are constants and x is the position of
(c) 1.57 radian (d) 0.57 radian the particle. The acceleration of the particle as d function
Ans: (a) of x, is given by:
Solution: Here, y1 = a sin (ωt + kx + 0.57) (a) –2nb2x–4n–1 (b) –2b2x–2n+1
2 –4n+1
and y2 = a cos (ωt + kx) (c) –2nb e (d) –2nb2x–2n–1
= a sin 0 ( )1 Ans: (a)
Solution: According to question,
Phase difference, ∆θ = θ2– θ1 V (x) = bx–2n
=
So, = – 2 nb x–2n–1
= Acceleration of the particle as function of x,
= 1.57 – 0.57 a = v = bx * (
–2n
) +
= 1 radian 2 –4n–1
= – 2nb x
Chapter: Waves
Chapter: Kinematics Motion in a Straight Line
[Topic: Basic of Waves]
[Topic: Non-uniform motion]
Q70. A wire carries a current. Maintaining the same
current it is bent first to form a circular plane coil of one
Q73. A drum of radius R and mass M, rolls down
turn which produces a magnetic field B at the centre of without slipping along an inclined plane of angle θ. The
the coil. The same length is now bent more sharply to frictional force
give a double loop of smaller radius. The magnetic field (a) dissipates energy as heat
at the centre of the double loop, caused by the same (b) decreases the rotational motion
current is (c) decreases the rotational and translational motion
(a) 4B (b) B/4 (d) converts translational energy to rotational energy
(c) B/2 (d) 2B Ans: (d)
Ans: (a) Solution: Net work done by frictional force when drum
Solution: Let I be current and l be the length of the wire. rolls down without slipping is zero.
Wnet = 0
For Ist case : where and n = 1
For IInd Case : ( )⇒

Wtrans. + Wrot. = 0; ∆Ktrans. + ∆Krot. = 0

Chapter: Moving Charges and Magnetic Field
∆Ktrans = –∆Krot.
[Topic: Magnetic Field, Biot-Savart's Law & Ampere's
i.e., converts translation energy to rotational energy.
Circuital Law]
Chapter: System of Particles and Rotational Motion
Q71. Two radiations of photons energies 1 eV and 2.5 [Topic: Rolling Motion]
eV, successively illuminate a photosensitive metallic
surface of work function 0.5 eV. The ratio of the
Q74. Which of the following equations represent a
wave?
maximum speeds of the emitted electrons is :
(a) 1 : 4 (b) 1 : 2 (a) y = A sin ωt
(c) 1 : 1 (d) 1 : 5 (b) y = Acos kx
(c) y = A sin (at – bx + c)
Ans: (b)
(d) y = A (ωt – kx)
Solution: According to Einsten‟s photoelectric effect, the
K.E. of the radiated electrons Ans: (c)
K.Emax = E – W Solution: y = A sin (at – bx + c) represents a wave,
where a may correspond to ω and b may correspond to K.
mv12 = (1 – 0.5) eV = 0.5 eV Chapter: Waves
mv22 = (2.5 – 0.5) eV = 2 eV [Topic: Basic of Waves]
Q75. A very long straight wire carries a current I. At the
√ instant when a charge + Q at point P has velocity ⃗ , as
√ shown, the force on the charge is
Chapter - Dual Nature of Radiation and Matter

14
 Chapter: Kinematics Motion in a Straight Line
[Topic: Motion Under Gravity]
Q78. The acceleration due to gravity on the planet A is 9
times the acceleration due to gravity on planet B. A man
jumps to a height of 2m on the surface of A. What is the
height of jump by the same person on the planet B?
(a) (b)
(c) 18 m (d) 6 m
Ans: (c)
Solution: Applying conservation of total mechanical
(a) along oy energy principle
(b) opposite to oy
(c) along ox
(d) opposite to ox =
Ans: (a) . / = 9 × 2 = 18 m
Solution: The direction of ⃗ is along ( ̂ ) Chapter: Gravitation
∴ The magnetic force [Topic: Acceleration due to Gravity]
⃗ (⃗ ⃗ ) ( ) ( ̂) ̂ Q79. A source of unknown frequency gives 4 beats/s,
⇒ ⃗ is along OY. when sounded with a source of known frequency 250 Hz.
Chapter: Moving Charges and Magnetic Field The second harmonic of the source of unknown
[Topic: Force & Torque on a Current Carrying frequency gives five beats per second, when sounded
Conductor] with a source of frequency 513 Hz. The unknown
frequency is
Q76. In a photo-emissive cell, with exciting wavelength (a) 246 Hz (b) 240 Hz
λ, the fastest electron has speed v. If the exciting
(c) 260 Hz (d) 254 Hz
wavelength is changed to , the speed of the fastest Ans: (d)
emitted electron will be Solution: When sounded with a source of known
(a) (3/4)1/2. v frequency fundamental frequency
(b) (4/3)1/2. v = 250 ± 4 Hz = 254 Hz or 246 Hz
(c) less than (4/3)1/2. v 2nd harmonic if unknown frequency (suppose) 254 Hz = 2
(d) greater than (4/3)1/2. v × 254 = 508 Hz
Ans: (d) As it gives 5 beats
Solution: or and ∴ 508 + 5 = 513 Hz
Hence, unknown frequency is 254 Hz
Chapter: Waves
. / [Topic: Beats, Interference & Superposition of Waves]
Q80. Current i is flowing in a coil of area A and number
( ) of turns N, then magnetic moment of the coil, M is
(a) NiA
So, v1 is greater than . / . (b)
Chapter - Dual Nature of Radiation and Matter (c)
[Topic: Electron Emission, Photon Photoelectric Effect √
& X-ray] (d) N2Ai
Ans: (a)
Q77. A train of 150 metre long is going towards north Solution: Magnetic moment linked with one turn = iA
direction at a speed of 10 m/s . A parrot flies at the speed
Magnetic moment linked with N turns = iNA amp-m2.
of 5 m/s towards south direction parallel to the railway
Here, A = Area of current loop.
track. The time taken by the parrot to cross the train is
Chapter: Magnetism and Matter
(a) 12 sec (b) 8 sec
[Topic: Magnetism, Gauss’s Law]
(c) 15 sec (d) 10 sec
Ans: (d) Q81. In a Rutherford scattering experiment when a
Solution: Relative velocity of parrot w.r.t the train projectile of charge Z1 and mass M1approaches a target
= 10 – (–5) = 15 ms–1. nucleus of charge Z2 and mass M2, the distance of closest
Time taken by parrot to cross the train approach is r0. The energy of the projectile is
(a) directly proportional to Z1 Z
(b) inversely proportional to Z1

15
(c) directly proportional to mass M1 velocity of star (v) = 1.5 × 106 m/s.
(d) directly proportional to M1 × M2 We know that wavelength of the approaching star (λ') =
Ans: (a)
Solution: The kinetic energy of the projectile is given by
( ) or,
= =
Thus energy of the projectile is directly proportional to or, . Therefore,
Z1 , Z2 A.
Chapter: Atoms
[Topic: Atomic Structure, Rutherford's Nuclear Model [where ∆λ = Change in the wavelength]
of Atom] Chapter: Waves
[Topic: Musical Sound & Doppler's Effect]
Q82. If |⃗ ⃗ | √ ⃗ ⃗ then the value of |⃗ ⃗ | is
Q85. In a coil of resistance 10 Ω, the induced current
(a) ( √ ) developed by changing magnetic flux through it, is
shown in figure as a function of time. The magnitude of
(b) ( ) change in flux through the coil in Weber is :
(c) . /
√
(d) A + B
Ans: (b)
Solution: |⃗ ⃗ |
⃗ ⃗ =ABcosθ
|⃗ ⃗ | √ ⃗ ⃗ ⇒ AB sin θ = √3 AB cos θ
or, tan θ = √3,∴ θ = 60º
|⃗ ⃗| √ (a) 8 (b) 2
(c) 6 (d) 4
√
Ans: (b)
Chapter: Kinematics Motion in a Plane Solution: The charge through the coil = area of current-
[Topic: Vectors] time(i – t) graph
Q83. The radii of circular orbits of two satellites A and = 0.2 C
B of the earth, are 4R and R, respectively. If the speed of
satellite A is 3 V, then the speed of satellite B will be: ∵ Change in flux (∆θ) = q × R
(a) 3 V/4 (b) 6 V
(c) 12 V (d) 3 V/2
Ans: (b) ∆θ = 2 Weber
Solution: Orbital velocity of a satellite in a circular orbit Chapter: Electromagnetic
of radius a is given by [Topic: Magnetic Flux, Faraday's & Lenz's Law]
Q86. Energy E of a hydrogen atom with principal
√ quantum number n is given by E = – 13.6/n2 eV. The
energy of photon ejected when the electron jumps from n
v √ = 3 state to n = 2 state of hydrogen is approximately
(a) 1.9 eV. (b) 1.5 eV
(c) 0.85 eV (d) 3.4 eV
=√
Ans: (a)
Solution: ∆E = E3 – E2
v2 = v1√ = 2 v1= 6V
Chapter: Gravitation
[Topic: Motion of Satellites, Escape Speed and Orbital 2 3 = 1.9 eV
Velocity] Chapter: Atoms
Q84. A star, which is emitting radiation at a wavelength [Topic: Bohr Model & The Spectra of the Hydrogen
of 5000 Å, is approaching the earth with a velocity of Atom]
1.50 × 106 m/s. The change in wavelength of the radiation Q87. A projectile is fired at an angle of 45° with the
as received on the earth is horizontal. Elevation angle of the projectile at its highest
(a) 0.25 Å (b) 2.5 Å point as seen from the point of projection is
(c) 25 Å (d) 250 Å (a) 60°
Ans: (c)
Solution: Given : Wavelength (λ) = 5000 Å (b) . /

16
 √
(c) . / (d) 45° ∫ ( )
Ans: (b)
Chapter: Electromagnetic
Solution: H = ... (1)
[Topic: Motional and Static EMI & Applications]
R= Q91. The volume occupied by an atom is greater than
the volume of the nucleus by a factor of about
... (2) [2003]
(a) 1015 (b) 101
5
(c) 10 (d) 1010
Ans: (a)
Solution:
= . /
But . /
15
Hence, Vatom = 10 × Vnucleus
Chapter: Nuclei
[Topic: Composition and Size of the Nucleus]
Q92. A boat is sent across a river with a velocity of 8 km
h–1. If the resultant velocity of boat is 10 km h–1, then the
Chapter: Kinematics Motion in a Plane velocity of the river is
[Topic: Projectile Motion] (a) 12.8 km h–1 (b) 6 km h–1
Q88. In rising from the bottom of a lake, to the top, the (c) 8 km h –1
(d) 10 km h–1
temperature of an air bubble remains unchanged, but its Ans: (b)
diameter gets doubled. If h is the barometric height
Solution: √ √
(expressed in m of mercury of relative density ρ) at the
Chapter: Kinematics Motion in a Plane
surface of the lake, the depth of the lake is
[Topic: Relative Velocity in2D & Circular Motion]
(a) 8 ρh m (b) 7ρh m
(c) 9 ρh m (d) 12 ρh m Q93. A piece of iron is heated in a flame. It first
Ans: (b) becomes dull red then becomes reddish yellow and
finally turns to white hot. The correct explanation for the
Solution: ( ) ( ) above observation is possible by using
This gives H = 7hρ [2013]
Chapter: Mechanical Properties of Fluids (a) Wien‟s displacement law
[Topic: Pressure, Density Pascal's Law & Archimedes' (b) Kirchoff‟s law
Principle] (c) Newton‟s law of cooling
Q89. The formation of a dipole is due to two equal and (d) Stefan‟s law
dissimilar point charges placed at a Ans: (a)
(a) short distance Solution: Wein‟s displacement law
(b) long distance According to this law
(c) above each other λmax
(d) none of these or, λmax × T = constant
Ans: (a) So, as the temperature increases λ decreases.
Solution: Dipole is formed when two equal and unlike Chapter: Thermal Properties
charges are placed at a short distance. [Topic: Calorimetry & Heat Transfer]
Chapter: Electrostatic Potential and capacitance
Q94. The potential energy of particle in a force field is
[Topic: Electric Flux & Gauss's Law]
Q90. The total charge induced in a conducting loop , where A and B are positive constants and r is
when it is moved in a magnetic field depends on the distance of particle from the centre of the field. For
(a) the rate of change of magnetic flux stable equilibrium, the distance of the particle is :
(b) initial magnetic flux only (a) B / 2A (b) 2A / B
(c) the total change in magnetic flux (c) A / B (d) B / A
(d) final magnetic flux only Ans: (b)
Ans: (c) Solution: for equilibrium
Solution: ⇒
Total charge induced = ∫ ∫ r=
for stable equilibrium

17
 should be positive for the value of r. Chapter: Thermal Properties
[Topic: Calorimetry & Heat Transfer]
here is +ve value for Q99. Energy stored in a capacitor is
So (a)
Chapter: Electrostatic Potential and capacitance (b) QV
[Topic: Electric Potential Energy & Work Done in (c)
Carrying a Charge]
(d)
Q95. A coil of 40 henry inductance is connected in
series with a resistance of 8 ohm and the combination is Ans: (a)
joined to the terminals of a 2 volt battery. The time Solution: Energy stored in capacitor
constant of the circuit is = ( )
[2004]
(a) 20 seconds (b) 5 seconds =
(c) 1/5 seconds (d) 40 seconds Chapter: Electrostatic Potential and capacitance
Ans: (b) [Topic: Capacitors, Capacitance, Grouping of
Solution: Time constant is L/R Capacitors & Energy Stored in a Capacitor.]
Given, L = 40H & R = 8Ω Q100. Which of the following statement is false for the
∴ η = 40/8 = 5 sec. properties of electromagnetic waves?
Chapter: Alternating Current (a) Both electric and magnetic field vectors attain the
[Topic: A.C. Circuit, LCR Circuit, Quality & Power maxima and minima at the same place and same time.
Factor] (b) The energy in electromagnetic wave is divided
Q96. In the reaction, , if the equally between electric and magnetic vectors
binding energies of and are respectively, a, b (c) Both electric and magnetic field vectors are parallel to
and c (in MeV), then the energy (in MeV) released in this each other and perpendicular to the direction of
reaction is propagation of wave
(a) a + b + c (d) These waves do not require any material medium for
(b) a + b – c propagation.
(c) c – a – b Ans: (c)
(d) c + a – b Solution: Electromagnetic waves are the combination of
Ans: (c) mutually perpendicular electric and magnetic fields. So,
Solution: and requires a and b amount of energies option (c) is false.
for their nucleons to be separated. Chapter - Electromagnetic Waves
releases c amount of energy in its formation i.e., in [Topic: Electromagnetic Waves, Conduction &
assembling the nucleons as nucleus. Displacement Current]
Hence, Energy released =c – (a + b) = c – a – b
Chapter: Nuclei PART 3. PHYSICS
QUESTION BANK
[Topic: Mass-Energy & Nuclear Reactions]
Q97. A person of mass 60 kg is inside a lift of mass 940
kg and presses the button on control panel. The lift starts
moving upwards with an acceleration 1.0 m/s2. If g = 10
ms–2, the tension in the supporting cable is Q1. The half life of a radioactive isotope „X‟ is 20 years.
(a) 8600 N (b) 9680 N It decays to another element „Y‟ which is stable. The two
(c) 11000 N (d) 1200 N elements „X‟ and „Y‟ were found to be in the ratio of 1 :
Ans: (c) 7 in a sample of a the given rock. The age of the rock is
Solution: Total mass = (60 + 940) kg = 1000 kg estimated to be
Let T be the tension in the supporting cable, then (a) 60 years (b) 80 years
T – 1000g = 1000 × 1 (c) 100 years (d) 40 years
⇒ T = 1000 × 11 = 11000 N Ans: (a)
Chapter: Dynamics Laws of Motion Solution: The value of x is
[Topic: Motion of Connected Bodies, Pulleys]
= = ⇒ t = 3T = 3 × 20 = 60 years
Q98. If the temperature of the sun is doubled, the rate of
energy received on earth will be increased by a factor of Hence the estimated age of the rock is 60 years
(a) 2 (b) 4 Alternate :X→Y0at t = 0N00at t = tNN0 – N
(c) 8 (d) 16 = = =
Ans: (d) t = 3T
Solution: Amount of energy radiated T4

18
= 3 × 20 = 60 years ∴ ζeq =
Chapter: Nuclei
[Topic: Radioactivity] Chapter: Current Electricity
[Topic: Combination of Resistances]
Q2. What is the minimum velocity with which a body of
mass m must enter a vertical loop of radius R so that it Q5. Two plane mirrors are inclined at 70°. A ray
can complete the loop ? incident on one mirror at angle θ after reflection falls on
second mirror and is reflected from there parallel to first
(a) √
mirror. The value of θ is
(b) √ [NEET Kar. 2013]
(c) √ (a) 50° (b) 45°
(d) √ (c) 30° (d) 55°
Ans: (a)
Ans: (d)
Solution: To complete the loop a body must enter a
vertical loop of radius R with the minimum velocity v =
√ .
Chapter: Dynamics Laws of Motion
[Topic: Circular Motion, Banking of Road]
Q3. If and represent the increase in internal
energy and work done by the system respectively in a
thermodynamical process, which of the following is true?
(a) in an adiabatic process Solution:
(b) in an isothermal process From fig. 40° + θ = 90° ∴ θ = 90° – 40° = 50°
(c) in an adiabatic process Chapter - Ray Optics and Optical
(d) in an isothermal process [Topic: Plane, Spherical Mirror & Reflection of Light]
Ans: (a) Q6. The activity of a radioactive sample is measured as
Solution: By first law of thermodynamics, 9750 counts per minute at t = 0 and as 975 counts per
minute at t = 5 minutes. The decay constant is
In adiabatic process, =0 approximately
(a) 0.922 per minute (b) 0.691 per minute
In isothermal process, =0 (c) 0.461 per minute (d) 0.230 per minute
Ans: (c)
Chapter: Heat & Thermodynamics Solution:
[Topic: Specific Heat Capacity & Thermodynamic 9750 = KN0............. (1)
Processes] 975 = KN............. (2)
Q4. Two metal wires of identical dimension are Dividing (1) by (2)
connected in series. If ζ1 and ζ2 are the conductivities of
the metal wires respectively, the effective conductivity of
the combination is : K=
(a)
= 0.4606 = 0.461 per minute
(b) Chapter: Nuclei
[Topic: Radioactivity]
(c)
Q7. A particle with total energy E is moving in a
(d) potential energy region U(x) . Motion of the particle is
restricted to the region when
Ans: (d) [NEET Kar. 2013]
Solution: In figure, two metal wires of identical (a) U(x) > E
dimension are connected in series (b) U(x) < E
(c) U(x) = O
(d) U(x) ≤ E
Ans: (d)
Solution: As the particle is moving in a potential energy
region.
Req = ∴ Kinetic energy ≥ 0
= . / And, total energy E = K.E. + P.E.
⇒ U(x) ≤ E

19
 Chapter: Work, Energy and Power
[Topic: Energy]
Q8. A reversible engine converts one-sixth of the heat
input into work. When the temperature of the sink is
reduced by 62ºC, the efficiency of the engine is doubled.
The temperatures of the source and sink are
(a) 99ºC, 37ºC (b) 80ºC, 37ºC
(c) 95ºC, 37ºC (d) 90ºC, 37ºC Solution:
Ans: (a) d = f2 + 2f1
Solution: Initially the efficiency of the engine was Chapter - Ray Optics and Optical
[Topic: Refraction at Curved Surface, Lenses & Power
which increases to when the sink temperature reduces of Lens]
by 62º C. Q11. If a small amount of antimony is added to
, when T2 = sink temperature germanium crystal
(a) it becomes a p–type semiconductor
T1 = source temperature
(b) the antimony becomes an acceptor atom
(c) there will be more free electrons than holes in the
semiconductor
Secondly,
(d) its resistance is increased
Ans: (c)
or, T1= 62 × 6 = 372K =372– 273 = 99ºC Solution: When small amount of antimony (pentavalent)
& T2 = is added to germanium crystal then crystal becomes n-
type semi conductor. Therefore, there will be more free
Chapter: Heat & Thermodynamics electrons than holes in the semiconductor.
[Topic: Kinetic Theory of an Ideal Gas & Gas Laws] Chapter: Semiconductor Electronics Materials, Devices
Q9. The thermo e.m.f E in volts of a certain [Topic: Solids, Semiconductors and P-N Junction
thermocouple is found to vary with temperature Diode]
difference θ in °C between the two junctions according to Q12. Water falls from a height of 60 m at the rate of 15
the relation kg/s to operate a turbine. The losses due to frictional
force are 10% of energy. How much power is generated
by the turbine?(g = 10 m/s2)
The neutral temperature for the thermocouple will be (a) 8.1 kW (b) 10.2 kW
(a) 30° C (b) 450° C (c) 12.3 kW (d) 7.0 kW
(c) 400 ° C (d) 225° C Ans: (a)
Ans: (d) Solution: Given, h = 60m, g = 10 ms–2,
Solution: Rate of flow of water = 15 kg/s
∴ Power of the falling water
For neutral temperature, = 15 kgs–1 × 10 ms–2 × 60 m = 900 watt.
0= Loss in energy due to friction
∴
= 225° C
Hence, neutral temperature is 225°C. ∴ Power generated by the turbine
= (9000 – 900) watt = 8100 watt = 8.1 kW
Chapter: Current Electricity
Chapter: Work, Energy and Power
[Topic: Kirchhoff's Laws, Cells, Thermo emf &
[Topic: Power]
Electrolysis]
Q10. A concave mirror of focal length „f1‟ is placed at a Q13. The mean free path of molecules of a gas, (radius
„r‟) is inversely proportional to :
distance of 'd‟ from a convex lens of focal length „f2‟. A
(a) r3 (b) r2
beam of light coming from infinity and falling on this
(c) r
convex lens-concave mirror combination returns to
infinity. (d) √
The distance „d‟ must equal : Ans: (b)
(a) f1 + f2 (b) –f1 + f2 Solution: Mean free path λm =
√
(c) 2f1 + f2 (d) –2f1 + f2 where d = diameter of molecule and d = 2r
Ans: (c)
∴λm
Chapter: Kinetic Theory

20
 [Topic: Degree of Freedom, Specific Heat Capacity & (C) A couple on a body produce both translational and
Mean Free Path] rotation motion in a body
Q14. Power dissipated across the 8Ω resistor in the (D) Mechanical advantage greater than one means that
circuit shown here is 2 watt. The power dissipated in watt small effort can be used to lift a large load
units across the 3Ω resistor is (a) (A) and (B)
(b) (B) and (C)
(c) (C) and (D)
(d) (B) and (D)
Ans: (d)
Solution: Centre of mass may or may not coincide with
centre of gravity. Net torque of gravitational pull is zero
about centre of mass.
(a) 1.0 (b) 0.5
(c) 3.0 (d) 2.0
Ans: (c) Mechanical advantage , M. A.=
Solution: Power = V . I = I2R If M.A. > 1 ⇒ Load > Effort
Chapter: System of Particles and Rotational Motion
√ √ √ A [Topic: Centre of Mass, Centre of Gravity & Principle
Potential over 8Ω = of Moments]
Q18. Two simple harmonic motions act on a particle.
This is the potential over parallel branch. So,
These harmonic motions are x = A cos (ωt + δ), y = A
A cos (ωt + α) when , the resulting motion is
Power of 3Ω = i12R = 1 × 1 × 3 = 3W (a) a circle and the actual motion is clockwise
Chapter: Current Electricity (b) an ellipse and the actual motion is counterclockwise
[Topic: Heating Effects of Current] (c) an elllipse and the actual motion is clockwise
Q15. Rainbow is formed due to a combination of (d) a circle and the actual motion is counter clockwise
[2000] Ans: (d)
(a) dispersion and total internal reflection Solution: ( )
(b) refraction and absorption ( ) ....(1)
(c) dispersion and focussing
When
(d) refraction and scattering
Ans: (a) . /
Solution: Rainbow is formed due to combination of total
( ) ....(2)
internal reflection and dispersion.
Squaring (1) and (2) and then adding
Chapter - Ray Optics and Optical
x2 + y2 = A2 [cos2 (ωt + α) + sin2 (ωt + α)]
[Topic: Prism & Dispersion of Light]
or x2 + y2 = A2, which is the equation of a circle. The
Q16. In forward bias, the width of potential barrier in a present motion is anticlockwise.
p-n junction diode Chapter: Oscillation
(a) increases
[Topic: Displacement, Phase, Velocity & Acceleration
(b) decreases
of SHM]
(c) remains constant
(d) first „1‟ then „2‟ Q19. In the circuit shown the cells A and B have
Ans: (b) negligible resistances. For VA = 12V, R1 = 500Ω and R =
Solution: We know that in forward bias of p-n junction 100Ω the galvanometer (G) shows no deflection. The
diode, when positive terminal is connected to p-type value of VB is :
diode, the repulsion of holes takes place which decreases
the width of potential barrier by striking the combination
of holes and electrons.
Chapter: Semiconductor Electronics Materials, Devices
[Topic: Solids, Semiconductors and P-N Junction
Diode]
Q17. Which of the following statements are correct ? (a) 4 V (b) 2 V
[2017] (c) 12 V (d) 6 V
(A) Centre of mass of a body always coincides with the Ans: (b)
centre of gravity of the body Solution: Since deflection in galvanometer is zero so
(B) Centre of mass of a body is the point at which the current will flow as shown in the above diagram.
total gravitational torque on the body is zero current =

21
So VB = IR = [Topic: Dimensions of Physical Quantities]
Chapter: Current Electricity Q23. When a mass is rotating in a plane about a fixed
[Topic: Wheatstone Bridge & Different Measuring point, its angular momentum is directed along :
Instruments] [2012]
(a) a line perpendicular to the plane of rotation
Q20. Two slits in Young‟s experiment have widths in (b) the line making an angle of 45° to the plane of
the ratio 1 : 25. The ratio of intensity at the maxima and rotation
minima in the interference pattern, is: (c) the radius
(d) the tangent to the orbit
(a)
Ans: (a)
(b) Solution: ⃗ ⃗
(c)
(d)
Ans: (d)
Solution: The ratio of slits width = (given)
∴ =

I A2 ⇒ or
By right hand screw, rule, the direction of ⃗ is ⊥ to the
plane containing & ⃗ .
The mass is rotating in the plane, about a fixed point, thus
∴ . / this plane will contain & ⃗ and the direction of ⃗ , will
be ⊥ to the this plane.
Chapter - Wave Optics
Chapter: System of Particles and Rotational Motion
[Topic: Young's Double Slit Experiment]
[Topic: Torque, Couple and Angular Momentum]
Q21. Transfer characteristics [output voltage (V0) vs
Q24. A point performs simple harmonic oscillation of
input voltage (V1)] for a base biased transistor in CE
period T and the equation of motion is given by x = a sin
configuration is as shown in the figure. For using
(ωt + π/6). After the elapse of what fraction of the time
transistor as a switch, it is used :
period the velocity of the point will be equal to half of its
[2012]
maximum velocity?
(a) T/8 (b) T/6
(c) T/3 (d) T/12
Ans: (d)
Solution: We have . /
∴ Velocity, . /
(a) in region III
Maximum velocity = aω
(b) both in region (I) and (III)
According to question,
(c) in region II
(d) in region (I) . /
Ans: (b)
Solution: I → ON or, . / = cos 60° or
II → OFF ⇒
nd
In II state it is used as a amplifier it is active region.
or,
Chapter: Semiconductor Electronics Materials, Devices
[Topic: Junction Transistor] or, ⇒
Q22. The unit of the Stefan-Boltzmann's constant is Chapter: Oscillation
(a) W/m2K4 (b) W/m2 [Topic: Time Period, Frequency, Simple Pendulum &
2
(c) W/m K (d) W/m2K2 Spring Pendulum]
Ans: (a) Q25. A charged particle moves through a magnetic field
Solution: E = ζAT4 in a direction perpendicular to it. Then the
E is energy dissipated per second. (a) velocity remains unchanged
(b) speed of the particle remains unchanged
(c) direction of the particle remains unchanged
Chapter: Units and Measurement (d) acceleration remains unchanged

22
 Ans: (b) , -, -
Solution: Magnetic force acts perpendicular to the , -
velocity. Hence speed remains constant. [ ]
Chapter: Moving Charges and Magnetic Field Chapter: Units and Measurement
[Topic: Motion of Charged Particle in Magnetic Field [Topic: Dimensions of Physical Quantities]
& Moment] Q29. From a disc of radius R and mass M, a circular
Q26. If the kinetic energy of the particle is increased to hole of diameter R, whose rim passes through the centre
16 times its previous value, the percentage change in the is cut. What is the moment of inertia of the remaining
de-Broglie wavelength of the particle is : part of the disc about a perpendicular axis, passing
(a) 25 (b) 75 through the centre ?
(c) 60 (d) 50 (a) 15 MR2/32 (b) 13 MR2/32
Ans: (b) 2
(c) 11 MR /32 (d) 9 MR2/32
Solution: As we know Ans: (b)
λ= = (∵ √ ) Solution: Moment of inertia of complete disc about point
√
'O'.
or √ √ ITotaldisc =
Therefore the percentage change in de-Broglie Mass of removed disc
wavelength = MRemoved = (Mass area)
Chapter - Dual Nature of Radiation and Matter Moment of inertia of removed disc about point 'O'.
[Topic: Matter Waves, Cathode & Positive Rays] IRemoved (about same perpendicular axis)
Q27. The output(X) of the logic circuit shown in figure = Icm + mx2
will be . /
= +
. /
Therefore the moment of inertia of the remaining part of
the disc about a perpendicular axis passing through the
(a) X =
centre,
(b) X = A.B
IRemaingdisc = ITotal – IRemoved
(c) X =
(d) X = ̿ ̿ =
Ans: (b) Chapter: System of Particles and Rotational Motion
Solution: [Topic: Moment of Inertia, Rotational K.E. and Power]
Q30. Sound waves travel at 350 m/s through a warm air
and at 3500 m/s through brass. The wavelength of a 700
i.e., output X = A.B Hz acoustic wave as it enters brass from warm air
Truth Table (a) decreases by a factor 10 (b) increases by a factor 20
(c) increases by a factor 10 (d) decreases by a factor 20
Ans: (c)
Solution: We have, v = nλ
⇒ v λ (as n remains constant)
Thus, as v increases 10 times, λ also increases 10 times.
Chapter: Waves
[Topic: Basic of Waves]
Chapter: Semiconductor Electronics Materials, Devices Q31. Two long parallel wires P and Q are both
[Topic: Digital Electronics and Logic Gates] perpendicular to the plane of the paper with distance of 5
Q28. P represents radiation pressure, c represents speed m between them. If P and Q carry currents of 2.5 amp
of light and S represents radiation energy striking unit and 5 amp respectively in the same direction, then the
area per sec. The non zero integers x, y, z such that Px Sy magnetic field at a point half-way between the wires is
cz is dimensionless are (a)
(a) x = 1, y = 1, z = 1 (b) x = – 1, y = 1, z = 1
(b)
(c) x = 1, y = – 1, z = 1 (d) x = 1, y = 1, z = – 1
√
Ans: (c) (c)
Solution: Try out the given alternatives.
(d)
When x = 1, y = – 1, z = 1
Ans: (d)

23
Solution: When the current flows in both wires in the ∴ x = (3 – 3)2
same direction then magnetic field at half way due to the ⇒ x = 0.
wire P, Chapter: Kinematics Motion in a Straight Line
⃗ (where I1= 2.5 amp) [Topic: Non-uniform motion]
Q34. A solid cylinder of mass m and radius R rolls down
The direction of ⃗ is downward an inclined plane of height h without slipping. The speed
of its centre of mass when it reaches the bottom is
(a) √( )

(b) √

(c) √

(d) √
Magnetic field at half way due to wire Q
⃗ [upward] Ans: (b)
Solution:
[where ]
Net magnetic field at half way ( )
⃗ ⃗ ⃗
= (upward)
Hence, net magnetic field at midpoint Now, gain in K.E. = Loss in P.E.
Chapter: Moving Charges and Magnetic Field ⇒ √. /
[Topic: Magnetic Field, Biot-Savart's Law & Ampere's
Chapter: System of Particles and Rotational Motion
Circuital Law]
[Topic: Rolling Motion]
Q32. Photoelectric emmision occurs only when the
incident light has more than a certain minimum Q35. A wave of frequency 100 Hz is sent along a string
(a) power towards a fixed end. When this wave travels back after
(b) wavelength reflection, a node is formed at a distance of 10 cm from
(c) intensity the fixed end of the string. The speeds of incident (and
(d) frequency reflected) waves are
(a) 5 m/s (b) 10 m/s
Ans: (d)
(c) 20 m/s (d) 40 m/s
Solution: For occurence of photoelectric effect, the
Ans: (c)
incident light should have frequency more than a certain
minimum which is called the threshold frequency (v0). Solution: As fixed end is a node, therefore, distance
between two consecutive nodes
We have,
cm.
For photoelectric effect emission ν > ν0
where ν is the frequency of the incident light. λ = 20 cm = 0.2 m
Chapter - Dual Nature of Radiation and Matter As v=nλ ∴v=100×02=20m/s
[Topic: Electron Emission, Photon Photoelectric Effect Chapter: Waves
& X-ray] [Topic: Basic of Waves]
Q33. The displacement „x‟ (in meter) of a particle of Q36. A particle having charge q moves with a velocity
mass „m‟ (in kg) moving in one dimension under the ⃗ through a region in which both an electric field ⃗ and a
action of a force, is related to time „t‟ (in sec) by t = magnetic field⃗ are present .The force on the particle is
√ . The displacement of the particle when its (a) ⃗ (⃗ ⃗ )
velocity is zero, will be (b) ⃗ ( ⃗ ⃗)
[NEET Kar. 2013]
(a) 2 m (b) 4 m (c) ⃗ (⃗ ⃗ )
(c) zero (d) 6 m (d) ⃗ (⃗ ⃗ )
Ans: (c) Ans: (d)
Solution: ∵ t = √ Solution: Force due to electric field = ⃗
⇒√ = t – 3 ⇒ x = (t – 3)2 Force due to magnetic field = (⃗ ⃗ )
v = = 2(t – 3) = 0 Net force experienced = ⃗ (⃗ ⃗ )
⇒t=3 Chapter: Moving Charges and Magnetic Field

24
 [Topic: Force & Torque on a Current Carrying (a) zero (b) 4
Conductor] (c) 8 (d) 2
Q37. The 21 cm radio wave emitted by hydrogen in Ans: (b)
interstellar space is due to the interaction called the Solution:
hyperfine interaction in atomic hydrogen. The energy of
the emitted wave is nearly ⇒
[1998]
(a) 10–17 J (b) 1 J ⇒ Beats =
(c) 7 × 10–8 J (d) 10–24 J Chapter: Waves
Ans: (d) [Topic: Beats, Interference & Superposition of Waves]
Solution: E = hν = Q41. A bar magnet, of magnetic moment ⃗⃗⃗ , is placed in
a magnetic field of induction ⃗ . The torque exerted on it
=
is
Chapter - Dual Nature of Radiation and Matter (a) ⃗⃗⃗ ⃗ (b) –⃗⃗⃗ ⃗
[Topic: Electron Emission, Photon Photoelectric Effect ⃗⃗⃗ ⃗
(c) (d) ⃗ ⃗⃗⃗
& X-ray]
Ans: (c)
Q38. A stone falls freely under gravity. It covers Solution: We know that when a bar magnet is placed in
distances h1, h2 and h3 in the first 5 seconds, the next 5
the magnetic field at an angle θ, then torque acting on the
seconds and the next 5 seconds respectively. The relation
bar magnet (η)
between h1, h2 and h3 is
= ⃗⃗⃗ ⃗ .
(a) h1 = = Note : This torque η has a tendency to make the axis of
(b) h2 = 3h1 and h3 = 3h2 the magnet parallel to the direction of the magnetic field.
(c) h1 = h2 = h3 (d) h1 = 2h2 = 3h3 Chapter: Magnetism and Matter
Ans: (a) [Topic: Magnetism, Gauss’s Law]
Solution: ∵h = gt2 Q42. J.J. Thomson‟s experiment demonstrated that
∴h1 = g(5) 2 = 125 [2003]
(a) the e/m ratio of the cathode-ray particles changes
h1 + h2 = g(10)2 = 500 when a different gas is placed in the discharge tube
⇒h2 = 375 (b) cathode rays are streams of negatively charged ions
(c) all the mass of an atom is essentially in the nucleus
h1 + h2 + h3 = g(15)2 = 1125
(d) the e/m of electrons is much greater than the e/m of
⇒h3 = 625 protons
h2 = 3h1 , h3 = 5h1 Ans: (b)
orh1 = = Solution: Cathode rays are streams of negatively charged
Chapter: Kinematics Motion in a Straight Line ions
[Topic: Motion Under Gravity] Chapter: Atoms
Q39. Assuming earth to be a sphere of uniform density, [Topic: Atomic Structure, Rutherford's Nuclear Model
what is the value of „g‟ in a mine 100 km below the of Atom]
earth‟s surface? (Given, R = 6400 km) Q43. The vector sum of two forces is perpendicular to
[2001] their vector differences. In that case, the forces
(a) 9.65 m/s2 (b) 7.65 m/s2 (a) cannot be predicted
(c) 5.06 m/s2 (d) 3.10 m/s2 (b) are equal to each other
Ans: (a) (c) are equal to each other in magnitude
Solution: We know that effective gravity g' at depth (d) are not equal to each other in magnitude
below earth surface is given by Ans: (c)
Solution: ⃗ ⃗ ⃗
g' = . /
⃗⃗ ⃗ ⃗
Here, d = 100 km, R = 6400 km,
⃗ ⃗
Since and are perpendicular
∴ . / 9.65m/
∴⃗ ⃗ ⇒ (⃗ ⃗ )(⃗ ⃗)
Chapter: Gravitation ⃗ ⃗
⇒A =B⇒| | | |
2 2
[Topic: Acceleration due to Gravity]
Q40. Two sources P and Q produce notes of frequency Chapter: Kinematics Motion in a Plane
660 Hz each. A listener moves from P to Q with a speed [Topic: Vectors]
of 1 ms–1. If the speed of sound is 330 m/s, then the Q44. Two satellites of earth, S1 and S2 are moving in the
number of beats heard by the listener per second will be same orbit. The mass of S1 is four times the mass of S2.
Which one of the following statements is true?

25
(a) The potential energies of earth satellites in the two Solution: The magnitude of the resultant velocity at the
cases are equal. point of projection and the landing point is same.
(b) S1 and S2 are moving with the same speed.
(c) The kinetic energies of the two satellites are equal.
(d) The time period of S1 is four times that of S2.
Ans: (b)
Solution: Since orbital velocity of satellite is
√ , it does not depend upon the mass of the
satellite.
Therefore, both satellites will move with same speed.
Chapter: Gravitation
[Topic: Motion of Satellites, Escape Speed and Orbital
Clearly, change in momentum along horizontal (i.e along
Velocity] x-axis)
Q45. Two trains move towards each other with the same = mv cos θ – mv cos θ = 0
speed. The speed of sound is 340 m/s. If the height of the Change in momentum along vertical (i.e. along y–axis) =
tone of the whistle of one of them heard on the other mv sinθ – (–mv sinθ)
changes 9/8 times, then the speed of each train should be = 2 mvsinθ = 2mv × sin 45°
(a) 20 m/s (b) 2 m/s
(c) 200 m/s (d) 2000 m/s = √
√
Ans: (a) Hence, resultant change in momentum = √
Solution: Here, Chapter: Kinematics Motion in a Plane
Source and observer are moving in opposite direction, [Topic: Projectile Motion]
therefore, apparent frequency Q48. The compressibility of water is 4 × 10–5 per unit
( ) atmospheric pressure. The decrease in volume of 100
( ) cm3of water under a pressure of 100 atmosphere will be
(a) 0.4 cm3 (b) 4 × 10–5 cm3
3
(c) 0.025 cm (d) 0.004 cm3
Ans: (a)
⇒ =20m/sec. Solution: . Here, P = 100 atm,
Chapter: Waves K = 4 × 10–5 and V = 100 cm3.
[Topic: Musical Sound & Doppler's Effect] Hence, ∆V = 0.4 cm3
Q46. In which of the following systems will the radius Chapter: Mechanical Properties of Fluids
of the first orbit (n = 1) be minimum ? [Topic: Fluid Flow, Reyonld's Number & Bernoulli's
(a) Hydrogen atom Principle]
(b) Doubly ionized lithium Q49. An electric dipole, consisting of two opposite
(c) Singly ionized helium charges of C each separated by a distance 3 cm
(d) Deuterium atom is placed in an electric field of N/C. Torque
Ans: (b) acting on the dipole is
Solution: ; Z(=3) is maximum for Li2+. (a) (b)
Chapter: Atoms (c) (d)
[Topic: Bohr Model & The Spectra of the Hydrogen Ans: (c)
Atom] Solution: Charges (q) = 2 × 10–6 C, Distance (d) = 3 cm =
Q47. A particle of mass m is projected with velocity v 3 × 10–2 m and electric field (E) = 2 × 105 N/C. Torque (η)
making an angle of 45° with the horizontal. When the = q.d.
particle lands on the level ground the magnitude of the E =(2 × 10–6) × (3 × 10–2) × (2 × 105)
change in its momentum will be: = 12 × 10–3N–m .
[2008] Chapter: Electrostatic Potential and capacitance
(a) 2mv [Topic: Electric Flux & Gauss's Law]
(b) Q50. The current in self inductance L = 40 mH is to be
√ increased uniformly from 1 amp to 11 amp in 4
(c) √ milliseconds. The e.m.f. induced in the inductor during
(d) zero the process is
Ans: (c) (a) 100 volt (b) 0.4 volt
(c) 4.0 volt (d) 440 volt

26
 Ans: (a) [Topic: Calorimetry & Heat Transfer]
Solution: Q54. Three charges, each +q, are placed at the corners of
Given that L = 40 × 10–3 H, an isosceles triangle ABC of sides BC and AC, 2a. D and
di = 11 A – 1 A = 10 A E are the mid points of BC and CA. The work done in
and dt = 4 × 10–3 s taking a charge Q from D to E is
∴ e = 40 × 10–3 × . /
Chapter: Electromagnetic
[Topic: Alternating Current, Voltage & Power]

PART 4. PHYSICS
QUESTION BANK
(a)
Q51. Mn and Mp represent mass of neutron and proton (b)
respectively. If an element having atomic mass M has N-
neutron and Z-proton, then the correct relation will be (c) zero
(a) M < [NMn + ZMp] (d)
(b) M > [NMn + ZMp]
Ans: (c)
(c) M = [NMn + ZMp]
Solution: AC = BC
(d) M = N[Mn + Mp]
VD = VE
Ans: (a)
We have,
Solution: Given : Mass of neutron = Mn
W = Q (VE – VD)
Mass of proton = Mp; Atomic mass of the element = M ;
⇒W=0
Number of neutrons in the element = N and number of
protons in the element = Z. We know that the atomic Chapter: Electrostatic Potential and capacitance
mass (M) of any stable nucleus is always less than the [Topic: Electric Potential Energy & Work Done in
sum of the masses of the constituent particles. Carrying a Charge]
Therefore, M < [NMn + ZMp]. Q55. In a series resonant circuit, having L, C and R as its
X is a neutrino, when β-particle is emitted. elements, the resonant current is i. The power dissipated
Chapter: Nuclei in the circuit at resonance is
[Topic: Composition and Size of the Nucleus] [2002]
Q52. An electric fan has blades of length 30 cm (a)
. /
measured from the axis of rotation. If the fan is rotating
at 120 rpm, the acceleration of a point on the tip of the (b) zero
blade is (c)
(a) 1600 ms–2 (b) 47.4 ms–2 (d)
(c) 23.7 ms –2
(d) 50.55 ms–2 where ω is the angular resonance frequency.
Ans: (c) Ans: (d)
Solution: Centripetal acc. = ω2r = 4π2v2r Solution: At resonance Lω = ,ω=
√
= ( ) Current through circuit i =
[ ∵ ω = 2πv] Power dissipated at Resonance = i2R
Chapter: Kinematics Motion in a Plane Chapter: Alternating Current
[Topic: Ist, IInd & IIIrd Laws of Motion] [Topic: A.C. Circuit, LCR Circuit, Quality & Power
Q53. Two metal rods 1 and 2 of same lengths have same Factor]
temperature difference between their ends. Their thermal Q56. In any fission process, the ratio
conductivities are K1 and K2 and cross sectional areas A1
and A2, respectively. If the rate of heat conduction in rod is [2005]
1 is four times that in rod 2, then (a) equal to 1 (b) greater than 1
(a) K1A1 = K2A2 (b) K1A1 = 4K2A2 (c) less than 1 (d) depends on the mass of
(c) K1A1 = 2K2A2 (d) 4K1A1 = K2A2 the parent nucleus
Ans: (b) Ans: (c)
Solution: Q1 = 4Q2 (Given) Solution: Binding energy per nucleon for fission
⇒ ⇒ K1A1 = 4K2A2. products is higher relative to Binding energy per nucleon
for parent nucleus, i.e., more masses are lost and are
Chapter: Thermal Properties

27
obtained as kinetic energy of fission products. So, the now connected to the first capacitor in parallel. The
given ratio < 1. energy in each of the capacitor is
Chapter: Nuclei (a) U / 2 (b) 3U / 2
[Topic: Mass-Energy & Nuclear Reactions] (c) U
Q57. (d) U / 4
Ans: (a)
Solution: In Ist case when capacitor C connected with
battery charged with the energy,
U1= U (stored energy on capacitor).
In IInd case, after disconnection of battery similar
capacitor is connected in parallel with Ist capacitor then
Ceq.=C' = 2C.

Now,
Three forces acting on a body are shown in the figure. To
have the resultant force only along the y- direction, the ( )
magnitude of the minimum additional force needed is:
(a) 0.5 N (b) 1.5 N
√
(c) (d) √
Chapter: Electrostatic Potential and capacitance
Ans: (a) [Topic: Capacitors, Capacitance, Grouping of
Solution: The components of 1 N and 2N forces along + Capacitors & Energy Stored in a Capacitor.]
x axis = 1 cos 60° + 2 sin 30° Q60. The electric field part of an electromagnetic wave
= in a medium is represented by Ex=0;
0. / . / 1;
Ez= 0. The wave is :
(a) moving along x direction with frequency 106 Hz and
wave length 100 m.
(b) moving along x direction with frequency 106 Hz and
wave length 200 m.
(c) moving along – x direction with frequency 106 Hz and
wave length 200 m.
(d) moving along y direction with frequency 2π × 106 Hz
and wave length 200 m.
Ans: (b)
Solution: Comparing with the equation of wave.
Ey = E0 cos (ωt – kx)
ω = 2 πf = 2π × 106∴ f = 106 Hz
The component of 4 N force along –x-axis = 4 sin 30° = = k = π × 10–2 m–1, λ = 200 m
. Chapter - Electromagnetic Waves
[Topic: Electromagnetic Waves, Conduction &
Therefore, if a force of 0.5N is applied along + x-axis, the
Displacement Current]
resultant force along x-axis will become zero and the
resultant force will be obtained only along y-axis. Q61. α-particles, β-particles and γ-rays are all having
Chapter: Dynamics Laws of Motion same energy. Their penetrating power in a given medium
[Topic: Motion of Connected Bodies, Pulleys] in increasing order will be
[NEET Kar. 2013]
Q58. Thermal capacity of 40 g of aluminium (s = 0.2 cal (a) β, γ, α
/g K) is
(b) γ, α, β
(a) 168 joule /°C (b) 672 joule/°C
(c) α, β, γ
(c) 840 joule/°C (d) 33.6 joule/°C
(d) β, α, γ
Ans: (d)
Ans: (c)
Solution: Thermal capacity = ms = 40 × 0.2 = 8 cal/°C
= 4.2 × 8 J = 33.6 joules/°C Solution: Increasing order of penetrating power :
α < β < γ.
Chapter: Thermal Properties
For same energy, lighter particle has higher penentrating
[Topic: Calorimetry & Heat Transfer]
power.
Q59. A capacitor is charged to store an energy U. The Chapter: Nuclei
charging battery is disconnected. An identical capacitor is

28
 [Topic: Radioactivity] ( )
RAC = ( )
Q62. Two stones of masses m and 2 m are whirled in
∴ RAB : RBC : RAC = 27 : 32 : 35
horizontal circles, the heavier one in radius and the Chapter: Current Electricity
lighter one in radius r. The tangential speed of lighter [Topic: Combination of Resistances]
stone is n times that of the value of heavier stone when Q65. A rod of length 10 cm lies along the principal axis
they experience same centripetal forces. The value of n is of a concave mirror of focal length 10 cm in such a way
: that its end closer to the pole is 20 cm away from the
(a) 3 (b) 4 mirror. The length of the image is :
(c) 1 (d) 2 (a) 10 cm (b) 15 cm
Ans: (d) (c) 2.5 cm (d) 5 cm
Solution: According to question, two stones experience Ans: (d)
same centripetal force
i.e.
or, or,
. /
So, V1 = 2V2 i.e., n = 2
Chapter: Dynamics Laws of Motion
[Topic: Circular Motion, Banking of Road]
Q63. In thermodynamic processes which of the
following statements is not true? Solution:
(a) In an isochoric process pressure remains constant The focal length of the mirror
(b) In an isothermal process the temperature remains –
constant For A end of the rod the image distance
(c) In an adiabatic process PVγ = constant When u1 = – 20 cm
(d) In an adiabatic process the system is insulated from ⇒
the surroundings
Ans: (a) =
Solution: In an isochoric process volume remains v1 = – 20 cm
constant whereas pressure remains constant in isobaric For when u2 = – 30 cm
process.
Chapter: Heat & Thermodynamics
[Topic: Specific Heat Capacity & Thermodynamic
Processes] =
Q64. A 12 cm wire is given a shape of a right angled v2 = – 15 cm
triangle ABC having sides 3 cm, 4 cm and 5 cm as shown L = v2 – v1 = – 15 – (– 20)
in the figure. The resistance between two ends (AB, BC, L = 5 cm
CA) of the respective sides are measured one by one by a Chapter - Ray Optics and Optical
multi-meter. The resistances will be in the ratio of [Topic: Plane, Spherical Mirror & Reflection of Light]
[NEET Kar. 2013] Q66. A free neutron decays into a proton, an electron
and
(a) a beta particle
(b) an alpha particle
(c) an anti-neutrino
(d) a neutrino
Ans: (c)
(a) 3 : 4 : 5 (b) 9 : 16 : 25 Solution: +X
(c) 27 : 32 : 35 (d) 21 : 24 : 25 X must have zero charge and almost zero mass as
Ans: (c) electron is emitted. Hence X must be anti-neutrino.
Solution: Resistance is directly proportional to length Chapter: Nuclei
( ) [Topic: Radioactivity]
= ( )( )
Q67. The potential energy of a system increases if work
( )
RAB = is done
( )
(a) upon the system by a non conservative force
Similarly, (b) by the system against a conservative force
( )
RBC = ( ) (c) by the system against a non conservative force
(d) upon the system by a conservative force

29
 Ans: (d) best describe the image formed of an object of height 2
Solution: When work is done upon a system by a cm placed 30 cm from the lens?
conservative force then its potential energy increases. (a) Virtual, upright, height = 1 cm[2011]
Chapter: Work, Energy and Power (b) Virtual, upright, height = 0.5 cm
[Topic: Energy] (c) Real, inverted, height = 4 cm
Q68. The efficiency of a Carnot engine operating (d) Real, inverted, height = 1cm
between the temperatures of 100ºC and –23ºC will be Ans: (c)
(a) Solution: R = 20 cm
h0 = 2
(b) u = –30 cm
(c) We have, ( ). /
(d) =. /0 . /1
Ans: (d)
⇒ . /
Solution:
∴ f = 20 cm
T1 = –23°C = 250 K, T2 = 100°C = 373K
⇒

Chapter: Heat & Thermodynamics =

[Topic: Kinetic Theory of an Ideal Gas & Gas Laws] v = 60 cm
Q69. See the electric circuit shown in the figure. m=
= cm
So, image is inverted.
Chapter - Ray Optics and Optical
[Topic: Refraction at Curved Surface, Lenses & Power
of Lens]
Q71. How much water, a pump of 2 kW can raise in one
2
minute to a height of 10 m, take g = 10 m/s ?
(a) 1000 litres (b) 1200 litres
(c) 100 litres (d) 2000 litres
Ans: (b)
Which of the following equations is a correct equation for Solution: . Here, P = 2kW = 2000 W.
it?
W = Mgh = M × 10 × 10 = 100 M
(a) ε2 – i2 r2 – ε1 – i1 r1 = 0 (b) − ε2 – (i1 + i2) R+ i2 r2 =
and t = 60 s.
0
This gives, M = 1200 kg
(c) ε1 – (i1 + i2) R + i1 r1 = 0 (d) ε1 – (i1 + i2) R– i1 r1 = 0
Its volume = 1200 litre as 1 litre of water contains 1 kg of
Ans: (d) its mass.
Chapter: Work, Energy and Power
[Topic: Collisions]
Q72. The amount of heat energy required to raise the
temperature of 1g of Helium at NTP, from T1K to T2K is
(a) NakB(T2 – T1)
(b) NakB(T2 – T1)
(c) NakB
Solution:
Applying Kirchhoff ‟s rule in loop abcfa (d) NakB(T2 – T1)
ε1 – (i1 + i2) R – i1 r1 = 0. Ans: (d)
Chapter: Current Electricity Solution: From first law of thermodynamics
[Topic: Kirchhoff's Laws, Cells, Thermo emf & ∆Q = ∆U + ∆W = . R (T2 – T1) + 0
Electrolysis]
Q70. A biconvex lens has a radius of curvature of = NaKB (T2 – T1) [∵ K = ]
magnitude 20 cm. Which one of the following options Chapter: Kinetic Theory

30
 [Topic: Degree of Freedom, Specific Heat Capacity & Solution: There is no external force so centre of mass of
Mean Free Path] the system will not shift
Q73. A 5–ampere fuse wire can withstand a maximum 3. (a)
power of 1 watt in the circuit. The resistance of the fuse ( ) ( )
wire is
(a) 0.04 ohm (b) 0.2 ohm
(c) 5 ohm (d) 0.4 ohm
Ans: (a)
Solution: = 40 cm
Chapter: System of Particles and Rotational Motion
Chapter: Current Electricity
[Topic: Heating Effects of Current] [Topic: Centre of Mass, Centre of Gravity & Principle
of Moments]
Q74. A ray of light is incident at an angle of incidence, i,
on one face of prism of angle A (assumed to be small) Q77. A particle executing S.H.M. has amplitude 0.01m
and emerges normally from the opposite face. If the and frequency 60 Hz. The maximum acceleration of the
refractive index of the prism is µ, the angle of incidence i, particle is
is nearly equal to: (a) 144 π2 m/s2 (b) 120 π2 m/s2
2 2
(a) µA (c) 80 π m/s (d) 60 π2 m/s2
Ans: (a)
(b) Solution: Amplitude (A) = 0.01 m, Frequency = 60 Hz
(c) Maximum acceleration
= = ( )
(d) = m/sec2
Ans: (a) Chapter: Oscillation
Solution: For normally emerge e = 0 [Topic: Displacement, Phase, Velocity & Acceleration
Therefore r2 = 0 and r1= A of SHM]
Snell‟s Law for Incident ray‟s Q78. A potentiometer circuit is set up as shown. The
1sin i = µ sin r1 = µsin A potential gradient, across the potentiometer wire, is k
For small angle volt/cm and the ammeter, present in the circuit, reads 1.0
i = µA A when two way key is switched off. The balance points,
Chapter - Ray Optics and Optical when the key between the terminals (i) 1 and 2 (ii) 1 and
[Topic: Prism & Dispersion of Light] 3, is plugged in, are found to be at lengths l1 cm and l2 cm
Q75. A depletion layer consists of respectively. The magnitudes, of the resistors R and X, in
(a) electrons ohms, are then, equal, respectively, to
(b) protons
(c) mobile ions
(d) immobile ions
Ans: (d)
Solution: Depletion layer is formed by diffusion of holes
and electrons from p-type semiconductor to n-type
semiconductor and vice-versa. Hence, donor and acceptor
atom get positive and negative charge leading to
formation of p-n junction. Thus, donor and acceptor are
immobile.
Chapter: Semiconductor Electronics Materials, Devices
[Topic: Solids, Semiconductors and P-N Junction
Diode]
Q76. Two persons of masses 55 kg and 65 kg (a) k (l2– l1) and k l2 (b) k l1 and k (l2 – l1)
respectively, are at the opposite ends of a boat. The (c) k (l2 – l1) and k l1 (d) k l1 and k l2
length of the boat is 3.0 m and weighs 100 kg. The 55 kg Ans: (b)
man walks up to the 65 kg man and sits with him. If the Solution: (i) When key between the terminals 1 and 2 is
boat is in still water the centre of mass of the system plugged in,
shifts by : P.D. across R = IR = k l1
(a) 3.0 m (b) 2.3 m ⇒ R = k l1 as I = 1A
(c) zero (d) 0.75 m (ii)When key between terminals 1 and 3 is plugged in,
Ans: (c) P.D. across (X + R) = I(X + R) = k l2
⇒ X + R = k l2
∴ X = k (l2 – l1)

31
∴ R = kl1 and X = k (l2 – l1)
Chapter: Current Electricity
[Topic: Wheatstone Bridge & Different Measuring
Instruments]
Q79. In the Young‟s double-slit experiment, the
intensity of light at a point on the screen where the path
difference is λ is K, (λ being the wave length of light
used). The intensity at a point where the path difference
is λ/4, will be:
(a) K (b) K/4 (a) F1 + F2 (b) F1 – F2
(c) K/2 (d) Zero (c) (d) 2(F1 + F2)
Ans: (c) Ans: (a)
Solution: For path difference λ, phase difference= 2π rad.
Solution: F1 x + F2 x = F3 x
For path difference , phase difference= rad. F3 = F1 + F2
As K = 4I0 so intensity at given point where path Chapter: System of Particles and Rotational Motion
difference is [Topic: Torque, Couple and Angular Momentum]
Q83. A mass of 2.0 kg is put on a flat pan attached to a
K′ = . /. / vertical spring fixed on the ground as shown in the figure.
= The mass of the spring and the pan is negligible. When
pressed slightly and released the mass executes a simple
Chapter - Wave Optics harmonic motion. The spring constant is 200 N/m. What
[Topic: Young's Double Slit Experiment] should be the minimum amplitude of the motion so that
Q80. The input resistance of a silicon transistor is 100 the mass gets detached from the pan (take g = 10 m/s2)?
W. 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 :
(a) 2000 (b) 3000
(c) 4000 (d) 1000
Ans: (a) (a) 10.0 cm
(b) any value less than 12.0 cm
Solution: Voltage gain (AV) =
(c) 4.0 cm
AV = 2 × 100 = 2000 (d) 8.0 cm
Ans: (a)
Chapter: Semiconductor Electronics Materials, Devices
Solution: Mass gets detached at the upper extreme
[Topic: Junction Transistor] position when pan returns to its mean position.
Q81. In a particular system, the unit of length, mass and At that point, R = mg – mω2a = 0
time are chosen to be 10 cm, 10 g and 0.1 s respectively. i.e. g = ω2a
The unit of force in this system will be equivalent to ⇒a=g/ =mg/k
(a) 0.1 N (b) 1 N
(c) 10 N (d) 100 N [ ]
Ans: (a)
Solution: [F] = MLT–2 = (10g) (10 cm) (0.1s)–2
= (10–2 kg) (10–1m) (10–1s)–2 = 10–1N.
Chapter: Oscillation
Chapter: Units and Measurement
[Topic: Time Period, Frequency, Simple Pendulum &
[Topic: Dimensions of Physical Quantities]
Spring Pendulum]
Q82. ABC is an equilateral triangle with O as its centre.
⃗ ⃗ and ⃗ represent three forces acting along the sides
Q84. In a certain region of space electric field ⃗ and
AB, BC and AC respectively. If the total torque about O magnetic field ⃗ are perpendicular to each other and an
electron enters in region perpendicular to the direction of
is zero the magnitude of ⃗ is : ⃗ and ⃗ both and moves undeflected, then velocity of
electron is
|⃗ |
(a)
|⃗⃗ |
(b) ⃗ ⃗
|⃗⃗ |
(c)
|⃗ |

32
(d) ⃗ ⃗ (a) MLT–2A–2 (b) M0L1T
Ans: (a) (c) M0L2T–1A2 (d) None of the above
Solution: Electron moves undeflected if force exerted Ans: (a)
due to electric field is equal to force due to magnetic Solution: ; [B] = MT–2A–1,
field. [n] = L–1,[I] = A
|⃗ |
|⃗ ||⃗ | |⃗ | ⇒ |⃗ | [ ] , -
|⃗⃗ |
Chapter: Moving Charges and Magnetic Field Chapter: Units and Measurement
[Topic: Motion of Charged Particle in Magnetic Field [Topic: Dimensions of Physical Quantities]
& Moment] Q88. Three identical spherical shells, each of mass m
Q85. The wavelength λe of an electron and λp of a photon and radius r are placed as shown in figure. Consider an
are of same energy E are related by axis XX' which is touching to two shells and passing
[2013] through diameter of third shell. Moment of inertia of the
(a) system consisting of these three spherical shells about
(b) √ XX' axis is
(c) (a) 3mr2
√
(d)
Ans: (d)
Solution: As P =
λp = ... (i)
⇒ ... (ii) (b)
√
From equations (i) and (ii) (c) 4mr2 (d)
λp λe 2
Ans: (c)
Chapter - Dual Nature of Radiation and Matter
Solution: Moment of inertia of shell 1 along diameter
[Topic: Matter Waves, Cathode & Positive Rays]
Q86. The figure shows a logic circuit with two inputs A Idiameter =
and B and the output C. The voltage wave forms across Moment of inertia of shell 2 = m. i of shell 3
A, B and C are as given. The logic circuit gate is : = Itangential =

So,I of the system along x x1

= Idiameter + (Itangential) × 2
or,Itotal = . /
=
(a) OR gate
(b) NOR gate Chapter: System of Particles and Rotational Motion
(c) AND gate [Topic: Moment of Inertia, Rotational K.E. and Power]
(d) NAND gate Q89. A transverse wave is represented by y = A sin (ω t–
Ans: (a) kx). For what value of the wavelength is the wave
Solution: velocity equal to the maximum particle velocity?
(a)
(b) π A
(c) 2πA
Chapter: Semiconductor Electronics Materials, Devices (d) A
[Topic: Digital Electronics and Logic Gates] Ans: (c)
Solution: y = A sin (ω t–kx)
Q87. The dimensional formula for permeability µ is Particle velocity,
given by

33
vp = = A ωcos (ωt – kx)
∴ vpmax= A ω
wave velocity =
∴ Aω = Solution:
i. e., A = But k = Torque, Iα = f. R.
∴ γ = 2π A Using Newton's IInd law, mg sin θ – f = ma
Chapter: Waves ∵pure rolling is there, a = Rα
[Topic: Basic of Waves] mg sin θ
Q90. Magnetic field intensity at the centre of a coil of 50 mg sin θ .∵ /
turns, radius 0.5 m and carrying a current of 2 A is
(a) 0.5 × 10–5T (b) 1.25 × 10–4 T or, acceleration, a =
. /
(c) 3 × 10–5 T (d) 4 × 10–5 T
Ans: (b) Using, s = ut + at2
Solution: We know that magnetic field at the centre of or, s = at2 t α
circular coil, √
t minimum means a should be more. This is possible
B= when I is minimum which is the case for solid cylinder.
Chapter: Moving Charges and Magnetic Field Therefore, solid cylinder will reach the bottom first.
[Topic: Magnetic Field, Biot-Savart's Law & Ampere's Chapter: System of Particles and Rotational Motion
Circuital Law] [Topic: Rolling Motion]
Q91. The momentum of a photon of energy hν will be Q94. The temperature at which the speed of sound
[2011] becomes double as was at 27°C is
(a) hν/c (a) 273° C (b) 0° C
(b) c/hν (c) 927° C (d) 1027°C
(c) hν Ans: (c)
(d) hν/c2 Solution: We have √
Ans: (a)
Solution: Chapter - Dual Nature of Radiation and √
Matter
[Topic: Electron Emission, Photon Photoelectric Effect
& X-ray] ( )
⇒ √
Q92. A particle has initial velocity ( ) and
acceleration ( ) . The magnitude of velocity ⇒ T2 = 1200k = 927°C
after 10 seconds will be : Chapter: Waves
(a) √ units (b) √ units [Topic: Basic of Waves]
(c) 5 units (d) 9 units Q95. Two long parallel wires are at a distance of 1
Ans: (b) metre. Both of them carry one ampere of current. The
Solution: ⃗ ⃗ ⃗ force of attraction per unit length between the two wires
( )̂ ( )̂ = ̂ is
|⃗ | = √ (a) 2 × 10–7 N/m (b) 2 × 10–8 N/m
–8
|⃗ | √ (c) 5 × 10 N/m (d) 10–7 N/m
Chapter: Kinematics Motion in a Straight Line Ans: (a)
[Topic: Non-uniform motion] Solution: =
Q93. A solid cylinder and a hollow cylinder both of the = 2 × 10–7 N/m.
same mass and same external diameter are released from [This relates to the definition of ampere]
the same height at the same time on an inclined plane. Chapter: Moving Charges and Magnetic Field
Both roll down without slipping. Which one will reach [Topic: Force & Torque on a Current Carrying
the bottom first? Conductor]
(a) Both together Q96. Light of wavelength 5000 Å falls on a sensitive
(b) Solid cylinder plate with photo-electric work function of 1.9 eV. The
(c) One with higher density kinetic energy of the photo-electrons emitted will be
(d) Hollow cylinder (a) 0.58 eV (b) 2.48 eV
Ans: (b) (c) 1.24 eV (d) 1.16 eV
Ans: (a)

34
Solution: K.E. = hν – hν0 ( ) ( )
,
( ) ( )
= where A1, A2 are amplitudes of given two sound wave.
= 2.48 – 1.9 = 0.58 eV Chapter: Waves
Chapter - Dual Nature of Radiation and Matter [Topic: Beats, Interference & Superposition of Waves]
[Topic: Electron Emission, Photon Photoelectric Effect Q100. The work done in turning a magnet of magnetic
& X-ray] moment M by an angle of 90° from the meridian, is n
Q97. A boy standing at the top of a tower of 20m height times the corresponding work done to turn it through an
drops a stone. Assuming g = 10 ms–2, the velocity with angle of 60°. The value of n is given by
which it hits the ground is (a) 2 (b) 1
(a) 10.0 m/s (b) 20.0 m/s (c) 0.5 (d) 0.25
(c) 40.0 m/s (d) 5.0 m/s Ans: (a)
Ans: (b) Solution: Magnetic moment = M; Initial angle through
Solution: Here, u = 0 which magnet is turned (θ1) = 90º and final angle through
We have, v2 = u2 + 2gh which magnet is turned (θ2)= 60º. Work done in turning
⇒v=√ =√ = 20 m/s the magnet through 90º(W1) = MB (cos 0º – cos 90º)
Chapter: Kinematics Motion in a Straight Line = MB (1–0) = MB.
[Topic: Motion Under Gravity] Similarly, W2 = MB (cos 0º – cos 60º)
Q98. A body weighs 72 N on the surface of the earth. ( )
What is the gravitational force on it due to earth at a
height equal to half the radius of the earth from the ∴ W1 = 2W2 or n = 2.
surface? Chapter: Magnetism and Matter
(a) 32 N (b) 28 N [Topic: The Earth's Magnetism, Magnetic Materials
(c) 16 N (d) 72 N and their Properties]

Solution: mg = 72 N (body weight on the surface)

Ans: (a)
PART 5. PHYSICS
QUESTION BANK
At a height

Q1. Who indirectly determined the mass of the electron

. / by measuring the charge of the electron?
(a) Millikan
Body weight at height , (b) Rutherford
(c) Einstein
(d) Thomson
= Ans: (d)
Solution: Chapter: Atoms
= [Topic: Atomic Structure, Rutherford's Nuclear Model
Chapter: Gravitation of Atom]
[Topic: Acceleration due to Gravity] Q2. The angle between the two vectors
Q99. Two sources of sound placed close to each other ⃗ ̂ ̂ ̂ and ⃗ ̂ ̂ ̂ will be
are emitting progressive waves given by y1 = 4 sin 600 πt [2001, 1994]
and y2 = 5 sin 608 πt. An observer located near these two (a) zero (b) 45º
sources of sound will hear : (c) 90º (d) 180º
(a) 4 beats per second with intensity ratio 25 : 16 between Ans: (c)
waxing and waning. (b) 8 beats per second with ̂,⃗
Solution: ⃗ ̂ ̂ ̂
intensity ratio 25 : 16 between waxing and waning
⃗ ⃗ ( ̂ )( ̂ ̂)
(c) 8 beats per second with intensity ratio 81 : 1 between
waxing and waning (d) 4 beats per second with |⃗ ||⃗ |
intensity ratio 81 : 1 between waxing and waning |⃗ | , |⃗ | , hence, cos θ = 0, θ = 90°
Ans: (d) Chapter: Kinematics Motion in a Plane
Solution: 2π f1 = 600 π [Topic: Vectors]
f1 = 300... (1)
Q3. For a satellite moving in an orbit around the earth,
2π f2 = 608 π
the ratio of kinetic energy to potential energy is
f2 = 304... (2)
|f1 – f2| = 4 beats (a)

35
(b) Q6. An electron changes its position from orbit n = 2 to
√
the orbit n = 4 of an atom. The wavelength of the emitted
(c) 2 (d) √ radiations is (R = Rydberg‟s constant)
Ans: (a)
Solution: K.E. of satellite moving in an orbit around the (a)
earth is (b)

K= 4√ 5 (c)
(d)
P.E. of satellite and earth system is
Ans: (b)
⇒ Solution: . /, where n1 = 2, n2 = 4
Chapter: Gravitation
[Topic: Motion of Satellites, Escape Speed and Orbital ( )
Velocity]
Q4. Suppose the charge of a proton and an electron ( )⇒
differ slightly. One of them is – e, the other is (e + ∆e). If Chapter: Atoms
the net of electrostatic force and gravitational force [Topic: Bohr Model & The Spectra of the Hydrogen
between two hydrogen atoms placed at a distance d Atom]
(much greater than atomic size) apart is zero, then ∆e is Q7. For angles of projection of a projectile (45° – θ) and
of the order of [Given mass of hydrogen mh = 1.67 × 10–27 (45° + θ), the horizontal ranges described by the
kg] projectile are in the ratio of
(a) 10–23C (b) 10–37 C (a) 1: 3 (b) 1 : 2
–47
(c) 10 C (d) 10–20 C (c) 2 : 1 (d) 1 : 1
Ans: (b) Ans: (d)
Solution: According to question, the net electrostatic Solution: (45º – θ) & (45º + θ) are complementary angles
force (FE) = gravitational force (FG) as 45º – θ + 45º + θ = 90º. We know that if angle of
FE = FG projection of two projectiles make complementary
or angles, their ranges are equal. In this case also, the range
will be same. So the ratio is 1 : 1.
⇒ ∆e = √ . / Chapter: Kinematics Motion in a Plane
[Topic: Projectile Motion]
= 1.67 × 10–27 √ Q8. A wind with speed 40 m/s blows parallel to the roof
of a house. The area of the roof is 250 m2. Assuming that
∆e ≈ 1.436 × 10–37C the pressure inside the house is atmosphere pressure, the
Chapter: Electrostatic Potential and capacitance force exterted by the wind on the roof and the direction of
[Topic: Electric Field, Electric Field Lines & Dipole] the force will be (ρair = 1.2 kg/m3)
Q5. A conducting circular loop is placed in a uniform (a) 4.8 × 105 N, upwards (b) 2.4 × 105 N, upwards
magnetic field, B = 0.025 T with its plane perpendicular (c) 2.4 × 105 N, downwards (d) 4.8 × 105 N, downwards
to the loop. Ans: (b)
The radius of the loop is made to shrink at a constant rate Solution: According to Bernoulli‟s theorem,
of 1 mm s–1. The induced e.m.f. when the radius is 2 cm,
is
(a) (b) So,∆P = ρv2
(c) (d)
Ans: (b)
Solution: Magnetic flux linked with the loop is
| | =
When r = 2 cm, = 1 mm s–1
e = 0.025× π ×2 ×2 ×10–2×10–3 F = ∆PA = ρv2A
= 0.100 × π × 10–5 = π × 10–6 V = πµV = × 1.2 × 40 × 40 × 250
Chapter: Electromagnetic
= 2.4 × 105 N (upwards)
[Topic: Magnetic Flux, Faraday's & Lenz's Law]
Chapter: Mechanical Properties of Fluids
[Topic: Fluid Flow, Reyonld's Number & Bernoulli's
Principle]

36
Q9. There is an electric field E in x-direction. If the work
done on moving a charge of 0.2 C through a distance of 2
m along a line making an angle 60° with x-axis is 4J,
then what is the value of E?
(a) 3 N/C (b) 4 N/C
(c) 5 N/C (d) 20 N/C
Ans: (d)
Solution: Charge (q) = 0.2 C; Distance (d) = 2m; Angle θ (a) 24 Ns (b) 20 Ns
= 60º and work done (W) = 4J. (c) 12 Ns (d) 6 Ns
Work done in moving the charge (W) Ans: (c)
= F.d cos θ = qEd cos θ Solution: Change in momentum,
or, ∆p= ∫
= Area of F-t graph
= 20 N/C.
= = 12 N-s
Chapter: Electrostatic Potential and capacitance
[Topic: Electric Flux & Gauss's Law] Chapter: Dynamics Laws of Motion
Q10. A small signal voltage V(t) = V0 sin ωt is applied [Topic: Ist, IInd & IIIrd Laws of Motion]
across an ideal capacitor C : Q13. If the radius of a star is R and it acts as a black
(a) Current I (t) , lags voltage V(t) by 90°. body, what would be the temperature of the star, in which
(b) Over a full cycle the capacitor C does not consume the rate of energy production is Q ?
any energy from the voltage source. [2012]
(c) Current I (t) is in phase with voltage V(t) . (a) Q/4πR2ζ
(d) Current I (t) leads voltage V(t) by 180°. (b) (Q/4πR2ζ)–1/2
Ans: (b) (c) (4πR2Q/ζ)1/4 (d) (Q/4πR2ζ)1/4
Solution: As we know, power P = Vrms · Irms cosθ (ζ stands for Stefan‟S constant)
as cosθ = 0(∵ θ = 90°) Ans: (d)
∴Power consumed = 0 Solution: Stefan‟s law for black body radiation
(in one complete cycle) Q = ζe AT4
Chapter: Alternating Current
[Topic: Alternating Current, Voltage & Power] [ ]
( )
Q11. Atomic weight of Boron is 10.81 and it has two Here e = 1
isotopes 5B10and 5B11 .Then the ratio in nature A = 4πR2
would be Chapter: Thermal Properties
(a) 19 : 81 (b) 10 : 11 [Topic: Calorimetry & Heat Transfer]
(c) 15 : 16 (d) 81 : 19 Q14. Charges +q and –q are placed at points A and B
Ans: (a) respectively which are a distance 2L apart, C is the
Solution: Suppose that, midpoint between A and B. The work done in moving a
The number of 10B type atoms = x charge +Q along the semicircle CRD is
and the number of 11B type atoms = y
Weight of 10B type atoms = 10x
Weight of 11B type atoms = 11y
Total number of atoms = x + y
∴ Atomic weight =
⇒ 10x + 11y = 10.81x + 10.81 y (a)
⇒ 0.81x = 0.19 y (b)
Chapter: Nuclei (c)
[Topic: Composition and Size of the Nucleus]
Q12. The force „F‟ acting on a particle of mass „m‟ is (d)
indicated by the force-time graph shown below. The Ans: (c)
change in momentum of the particle over the time
interval from zero to 8 s is :

Solution:
Potential at C = VC = 0

37
Potential at D = VD heat is taken from the surroundings, the temperature of
water in the tumbler becomes nearly (L = 80 cal/g)
. / (a) 31°C (b) 22°C
Potential difference (c) 19°C (d) 15°C
VD – VC = . / Ans: (b)
Solution: Let the final temperature be T
⇒ Work done = Q (VD – VC) Heat gained by ice = mL + m × s × (T – 0)
= 10 × 80 + 10 × 1 × T
Heat lost by water = 55 × 1× (40 – T)
Chapter: Electrostatic Potential and capacitance By using law of calorimetery,
[Topic: Electric Potential Energy & Work Done in 800 + 10 T = 55 × (40 – T)
Carrying a Charge] ⇒ T = 21.54°C = 22°C
Q15. A capacitor has capacity C and reactance X. If Chapter: Thermal Properties
capacitance and frequency become double, then reactance [Topic: Newton's Law of Cooling]
will be Q19. A parallel plate condenser with oil between the
(a) 4X (b) X/2 plates (dielectric constant of oil K = 2) has a capacitance
(c) X/4 (d) 2X C. If the oil is removed, then capacitance of the capacitor
Ans: (c) becomes
Solution: Capacitive reactance, (a) √ (b) 2C
(c)
⇒ √
(d)
Ans: (d)
Solution: When oil is placed between space of plates
...(1) 0 1
Chapter: Alternating Current
[Topic: A.C. Circuit, LCR Circuit, Quality & Power When oil is removed ...(2)
Factor] On comparing both equations, we get C ' = C/2
Q16. Fission of nuclei is possible because the binding Chapter: Electrostatic Potential and capacitance
energy per nucleon in them [Topic: Capacitors, Capacitance, Grouping of
(a) increases with mass number at low mass numbers. Capacitors & Energy Stored in a Capacitor.]
(b) decreases with mass number at low mass numbers. Q20. The velocity of electromagnetic radiation in a
(c) increases with mass number at high mass numbers. medium of permittivity ε0 and permeability µ0 is given by
(d) decreases with mass number at high mass numbers.
Ans: (d) (a) √
Solution: B.E. per nucleon is smaller for lighter as well (b) √
as heavier nucleus. But fusion reaction occurs for small
mass number nuclei and fission reaction occurs for larger (c)
√
mass number nuclei to attain reaction binding energy per (d) √
nucleon.
Ans: (c)
Chapter: Nuclei
Solution: The velocity of electromagnetic radiation in a
[Topic: Mass-Energy & Nuclear Reactions]
Q17. The mass of a lift is 2000 kg. When the tension in medium of permittivity and permeability µ0 is is equal
0

the supporting cable is 28000 N, then its acceleration is: to .

√
(a) 4 ms–2 upwards (b) 4 ms–2 downwards
–2
(c) 14 ms upwards (d) 30 ms–2 downwards Chapter - Electromagnetic Waves
Ans: (a) [Topic: Electromagnetic Waves, Conduction &
Solution: Net force, F = T – mg Displacement Current]
ma = T – mg Q21. A mixture consists of two radioactive materials A1
2000 a = 28000 – 20000 = 8000 and A2 with half lives of 20 s and 10 s respectively.
Initially the mixture has 40 g of A1 and 160 g of A2. The
amount of the two in the mixture will become equal after
Chapter: Dynamics Laws of Motion :
[Topic: Motion of Connected Bodies, Pulleys] (a) 60 s (b) 80 s
(c) 20 s (d) 40 s
Q18. 10 gm of ice cubes at 0°C are released in a tumbler
(water equivalent 55 g) at 40°C. Assuming that negligible Ans: (d)

38
Solution: Let, the amount of the two in the mixture will Q24. A ring is made of a wire having a resistance R0 =
become equal after t years. 12 Ω. Find the points A and B as shown in the figure, at
The amount of A1, which remains after t years which a current carrying conductor should be connected
so that the resistance R of the sub-circuit between these
( ) points is equal to .
The amount of A2, which remains, after t years

( )
According to the problem
N1 = N2

( ) ( )
. /
(a)
(b)
(c)
(d)
t = 40 s Ans: (d)
Chapter: Nuclei Solution: Let x is the resistance per unit length then
[Topic: Radioactivity]
Q22. A car is moving in a circular horizontal track of
radius 10 m with a constant speed of 10 m/s. A bob is
suspended from the roof of the car by a light wire of
length 1.0 m. The angle made by the wire with the
vertical is equivalent resistance
(a) 0° ( )( )
(b)
(c) ⇒
(d) = ... (i)
Ans: (d)
Solution: Given; speed = 10 m/s; radius r = 10 m alsoR0 = xl1 + x l2
Angle made by the wire with the vertical 12 = x(l1 + l2)
12 = . / ... (ii)

⇒
() . /
Chapter: Dynamics Laws of Motion ⇒
[Topic: Circular Motion, Banking of Road] ( )
. / . /
Q23. If Q, E and W denote respectively the heat added,
change in internal energy and the work done in a closed
cyclic process, then: ( )
(a) W = 0 (b) Q = W = 0
)
2
(c) E = 0 (d) Q = 0 (y + 1 + 2y) × y (where
2
Ans: (c) 8y + 8 + 16y = 36y
Solution: In a cyclic process, the initial state coincides ⇒8y2 – 20y + 8 = 0
with the final state. Hence, the change in internal energy ⇒2y2 – 5y + 2 = 0
is zero, as it depends only on the initial and final states. ⇒2y2 – 4y – y + 2 = 0
But Q & W are non-zero during a cycle process. ⇒2y (y – 2) – 1(y – 2) = 0
Chapter: Heat & Thermodynamics ⇒(2y – 1) (y – 2) = 0
[Topic: Specific Heat Capacity & Thermodynamic ⇒ or 2
Processes]
Chapter: Current Electricity

39
 [Topic: Combination of Resistances] Solution: Efficiency of carnot engine (η1) = 40%= 0.4;
Q25. A person is six feet tall. How tall must a vertical Initial intake temperature (T1) = 500K and new efficiency
mirror be if he is able to see his entire length? (η2) = 50% = 0.5.
[2000] Efficiency ( ) or .
(a) 3 ft (b) 4.5 ft
(c) 7.5 ft (d) 6 ft Therefore in first case, .
Ans: (a) ⇒ T2 = 0.6×500=300K
Solution: To see his full image in a plane mirror a person And in second case,
requires a mirror of at least half of his height.
⇒
Chapter: Heat & Thermodynamics
[Topic: Kinetic Theory of an Ideal Gas & Gas Laws]
Q29. A student measures the terminal potential
difference (V) of a cell (of emf E and internal resistance
r) as a function of the current (I) flowing through it. The
slope and intercept, of the graph between V and I, then,
Chapter - Ray Optics and Optical respectively, equal:
[Topic: Plane, Spherical Mirror & Reflection of Light] (a) – r and E
Q26. The most penetrating radiation of the following is (b) r and – E
(a) gamma-rays (c) – E and r
(b) alpha particles (d) E and – r
(c) beta-rays 47.A cell can be balanced against 110 cm and 100 cm of
(d) X-rays potentiometer wire, respectively with and without being
Ans: (a) short circuited through a resistance of 10Ω. Its internal
Solution: The penetrating power of radiation is directly resistance is
proportional to the energy of its photon. [2008]
Energy of a photon Ans: (a)
Solution: The terminal potential difference of a cell is
∴ Penetrating power given by V + Ir = E
λ is minimum for γ-rays, so penetrating power is V = VA – VB
maximum of γ-rays. or V = E – Ir
Chapter: Nuclei ⇒ = – r, Also for, i = 0 then V = E
[Topic: Radioactivity]
∴slope = – r, intercept = E
Q27. An engine pumps water continuously through a Chapter: Current Electricity
hose. Water leaves the hose with a velocity v and m is the
[Topic: Kirchhoff's Laws, Cells, Thermo emf &
mass per unit length of the water jet. What is the rate at
Electrolysis]
which kinetic energy is imparted to water?
Q30. A converging beam of rays is incident on a
(a) mv2 (b) diverging lens. Having passed through the lens the rays
(c) intersect at a point 15 cm from the lens on the opposite
side. If the lens is removed the point where the rays meet
(d) will move 5 cm closer to the lens. The focal length of the
Ans: (d) lens is
Solution: m : mass per unit length (a) – 10 cm (b) 20 cm
∴ rate of mass leaving the the hose per sec (c) –30 cm (d) 5 cm
= . Ans: (c)
Rate of K.E.= (mv) v2 = mv3
Chapter: Work, Energy and Power
[Topic: Energy]
Q28. An ideal carnot engine, whose efficiency is 40%
receives heat at 500 K. If its efficiency is 50%, then the
intake temperature for the same exhaust temperature is
(a) 600 K (b) 700 K
(c) 800 K (d) 900 K Solution:
Ans: (a) By lens formula,

40
 respectively. If γ = and R is the universal gas constant,
u = 10 cm then Cv is equal to
v = 15 cm [2013]
f=? (a) ( )
Putting the values, we get ( )
(b)
(c) γR
= – 30 cm (d)
Chapter - Ray Optics and Optical Ans: (a)
[Topic: Refraction at Curved Surface, Lenses & Power Solution: Cp – Cv = R ⇒ Cp = Cv + R
of Lens] ∵γ= = = +
Q31. Pure Si at 500K has equal number of electron (ne)
and hole (nh) concentrations of 1.5 × 1016 m–3. Doping by ⇒γ=1+ ⇒ =γ–1
indium increases nh to 4.5 × 1022 m–3. The doped
semiconductor is of ⇒ Cv =
(a) n–type with electron concentration ne = 5 × 1022 m–3 Chapter: Kinetic Theory
(b) p–type with electron concentration ne = 2.5 ×1010 m–3 [Topic: Degree of Freedom, Specific Heat Capacity &
(c) n–type with electron concentration ne = 2.5 × 1023 m–3 Mean Free Path]
d) p–type having electron concentration ne = 5 × 109 m–3 Q34. When three identical bulbs of 60 watt, 200 volt
Ans: (d) rating are connected in series to a 200 volt supply, the
Solution: ni2 = nenh power drawn by them will be
(1.5 × 1016)2 = ne (4.5 × 1022) (a) 20 watt (b) 60 watt
⇒ ne = 0.5 × 1010 (c) 180 watt (d) 10 watt
or ne = 5 × 109 Ans: (a)
Given nh = 4.5 × 1022 Solution: or
⇒ nh >> ne
∴ Semiconductor is p-type and ⇒ Peq = 20 watt.
ne = 5 × 109 m–3. Chapter: Current Electricity
Chapter: Semiconductor Electronics Materials, Devices [Topic: Heating Effects of Current]
[Topic: Solids, Semiconductors and P-N Junction Q35. For the angle of minimum deviation of a prism to
Diode] be equal to its refracting angle, the prism must be made
Q32. Two particles A and B, move with constant of a material whose refractive index
velocities ⃗ and ⃗ . At the initial moment their position (a) lies between√ and 1 (b) lies between 2 and√
vectors are and respectively. The condition for (c) is less than 1 (d) is greater than 2
particles A and B for their collision is: Ans: (b)
(a) ⃗⃗⃗ ⃗ ⃗
(b) ⃗⃗⃗ ⃗ ⃗
(c) ⃗⃗⃗ ⃗⃗⃗ ⃗ ⃗
⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗ ⃗
(d) |⃗⃗⃗⃗ ⃗⃗⃗⃗ | |⃗ ⃗ |
Ans: (d)
Solution: For collision ⃗ should be along

( ) Solution:
The angle of minimum deviation is given as
⃗⃗ ⃗⃗ ⃗⃗⃗⃗ ⃗
So, | = i + e–A
| | |
for minimum deviation
= A then
2A = i + e
in case of i=e
Chapter: Work, Energy and Power
[Topic: Collisions] 2A = 2i r1 = r2 =
Q33. The molar specific heats of an ideal gas at constant i = A = 90°
pressure and volume are denoted by Cp and Cv, from smell‟s law
1 sin i = n sin r1

41
sin A = n sin connected in parallel to S the bridge gets balanced. What
is the value of S?
(a) 3Ω (b) 6Ω
(c) 1Ω (d) 2Ω
Ans: (a)
Solution: A balanced wheatstone bridge simply requires
when A = 90° = imin
then nmin = √ ⇒
i = A = 0 nmax = 2 Therefore, S should be 2Ω.
Chapter - Ray Optics and Optical A resistance of 6Ω is connected in parallel.
[Topic: Prism & Dispersion of Light] In parallel combination,
Q36. Which of the following when added acts as an
impurity into silicon produced n-type semi-conductor?
(a) P
(b) Al
(c) B
(d) Mg Chapter: Current Electricity
Ans: (a) [Topic: Wheatstone Bridge & Different Measuring
Solution: n-type of silicon semiconductor is formed Instruments]
when impurity is mixed with pentavalent atom. Out of Q40. In Young‟s double slit experiment, the slits are 2
given choices only phosphorus is pentavalent. mm apart and are illuminated by photons of two
Chapter: Semiconductor Electronics Materials, Devices wavelengths λ1 = 12000Å and λ2 = 10000Å. At what
[Topic: Solids, Semiconductors and P-N Junction minimum distance from the common central bright fringe
Diode] on the screen 2 m from the slit will a bright fringe from
Q37. Three masses are placed on the x-axis : 300 g at one interference pattern coincide with a bright fringe
origin, 500 g at x = 40 cm and 400 g at x = 70 cm. The from the other ?
distance of the centre of mass from the origin is : [2013]
(a) 40 cm (b) 45 cm (a) 6 mm (b) 4 mm
(c) 50 cm (d) 30 cm Solution: (c) 3 mm (d) 8 mm
Chapter: System of Particles and Rotational Motion Ans: (a)
[Topic: Centre of Mass, Centre of Gravity & Principle Solution: ∵ y =
of Moments]
∴ n1 λ1 = n2λ2
Q38. Two simple harmonic motions with the same
frequency act on a particle at right angles i.e., along x and ⇒ n1 × 12000 × 10–10 = n2 × 10000 × 10–10
y axis. If the two amplitudes are equal and the phase or, n (12000 × 10–10) = (n + 1) (10000 × 10–10)
difference is π/2, the resultant motion will be ⇒n=5
(a) a circle ( )
(b) an ellipse with the major axis along y-axis
(c) an ellipse with the major axis along x-axis
(d) a straight line inclined at 45º to the x-axis
Ans: (a)
Solution: Equation of two simple harmonic motions
( ) ....(1)
. /
⇒x=Acos( ) ....(2) Hence, ycommon
On squaring and adding equations (1) and (2) ,
x2 + y2 = A2 ( )
This is an equation of a circle. Hence, resulting motion
will be a circular motion. (∵ d=2mm and D=2m)
Chapter: Oscillation = 5 × 12 × 10–4 m
[Topic: Displacement, Phase, Velocity & Acceleration = 60 × 10–4 m
of SHM] = 6 × 10–3m = 6 mm
Q39. Three resistances P, Q, R each of 2Ω and an Chapter - Wave Optics
unknown resistance S form the four arms of a [Topic: Young's Double Slit Experiment]
Wheatstone bridge circuit. When a resistance of 6Ω is Q41. A transistor is operated in common emitter
configuration at VC = 2V such that a change in the base

42
current from 100 µA to 300 µA produces a change in the
collector current from 10mA to 20 mA. The current gain ( )
is According to problem y = a/2
(a) 50 (b) 75
(c) 100 (d) 25 ( )
Ans: (a)
Solution: The current gain ( )
β= This is the minimum time taken by the particle to travel
half of the amplitude from the equilibrium position.
Chapter: Semiconductor Electronics Materials, Devices Chapter: Oscillation
[Topic: Junction Transistor] [Topic: Time Period, Frequency, Simple Pendulum &
Q42. If x = at + bt2, where x is the distance travelled by Spring Pendulum]
the body in kilometers while t is the time in seconds, then Q45. A charged particle of charge q and mass m enters
the unit of b is
perpendicularly in a magnetic field ⃗ . Kinetic energy of
(a) km/s (b) kms
the particle is E; then frequency of rotation is
(c) km/s2 (d) kms2
Ans: (c) (a)
Solution: [x] = [bt2]. Hence [b] = [x /t2] = km/s2. (b)
Chapter: Units and Measurement
[Topic: Dimensions of Physical Quantities] (c)
Q43. A circular platform is mounted on a frictionless (d)
vertical axle. Its radius R = 2 m and its moment of inertia
Ans: (b)
about the axle is 200 kgm2. It is initially at rest. A 50 kg
Solution: For circular path in magnetic field,
man stands on the edge of the platform and begins to
mrω2= qvB
walk along the edge at the speed of 1 ms–1 relative to the
ground. Time taken by the man to complete one ⇒ As v = rω
revolution is (
⇒
)
(a) πs
∴ If ν is frequency of roatation, then
(b)
⇒
(c) 2πs
Chapter: Moving Charges and Magnetic Field
(d)
[Topic: Motion of Charged Particle in Magnetic Field
Ans: (c) & Moment]
Solution: Using conservation Q46. The de-Broglie wavelength of neutron in thermal
Li = 0 (Initial moment) equilibrium at temperature T is
Lf = mvR – Iω (Final moment)
According to the conservation of momentum (a)
√
Li = Lf (b)
√
⇒mvR – I.ω = 0
mvR = I.ω (c)
√

( ) (d)
√

(v + ωR)t = 2πR Ans: (a)

Solution: From formula
t. /
λ=
t = 2π sec. √

Chapter: System of Particles and Rotational Motion =

√
[Topic: Torque, Couple and Angular Momentum] [By placing value of h, m and k)
Q44. A particle executes simple harmonic oscillation =
with an amplitude a. The period of oscillation is T. The √
minimum time taken by the particle to travel half of the Chapter - Dual Nature of Radiation and Matter
amplitude from the equilibrium position is [Topic: Matter Waves, Cathode & Positive Rays]
(a) T/8 (b) T/12 Q47. According to Newton, the viscous force acting
(c) T/2 (d) T/4 between liquid layers of area A and velocity gradient
Ans: (b) ∆V/∆Z is given by where η is constant
Solution: Displacement from the mean position called coefficient of viscosity. The dimensional formula
of η is

43
(a) ML–2T–2 (b) M0L0T0 Thus, the magnetic field is produced outside the pipe
(c) ML2T–2 (d) ML–1T–1 only.
Ans: (d) Chapter: Moving Charges and Magnetic Field
Solution: Substitute the dimensional formula of F, A, ∆V [Topic: Magnetic Field, Biot-Savart's Law & Ampere's
and ∆Z on both sides and find that for η. Circuital Law]

PART 6. PHYSICS
Chapter: Units and Measurement
[Topic: Dimensions of Physical Quantities]
Q48. The ratio of radii of gyration of a circular ring and
a circular disc, of the same mass and radius, about an axis QUESTION BANK
passing through their centres and perpendicular to their
planes are
[NEET Kar. 2013, 2008] Q51. Light of two different frequencies whose photons
(a) √ (b) 1:√ have energies 1 eV and 2.5 eV respectively illuminate a
(c) 3 : 2 (d) 2 : 1 metallic surface whose work function is 0.5 eV
Ans: (a) successively. Ratio of maximum speeds of emitted
electrons will be
Solution: ∵ I = MK2 ∴ K = √ (a) 1 : 4 (b) 1 : 2
Iring = MR2 and Idisc = (c) 1 : 1 (d) 1 : 5
Ans: (b)
Solution: The maximum kinetic energy of emitted
√ √ √ electrons is given by
( ) K.E = θ – θ0
K.E1 = 1 eV – 0.5 eV = 0.5 eV
Chapter: System of Particles and Rotational Motion K.E2 = 2.5 eV – 0.5 eV = 2 eV
[Topic: Moment of Inertia, Rotational K.E. and Power]
Q49. A wave in a string has an amplitude of 2 cm. The
wave travels in the + ve direction of x axis with a speed
of 128 m/sec and it is noted that 5 complete waves fit in 4 √
m length of the string. The equation describing the wave
is Chapter - Dual Nature of Radiation and Matter
(a) y = (0.02) m sin (15.7x – 2010t)
[Topic: Electron Emission, Photon Photoelectric Effect
(b) y = (0.02) m sin (15.7x + 2010t)
& X-ray]
(c) y = (0.02) m sin (7.85x – 1005t)
(d) y = (0.02) m sin (7.85x + 1005t) Q52. The motion of a particle along a straight line is
Ans: (c) described by equation :
x = 8 + 12t – t3
Solution: = 128 ms–1, 5λ = 4, λ = m where x is in metre and t in second. The retardation of the
y = A sin (kx – ωt), particle when its velocity becomes zero, is :
k= (a) 24 ms–2 (b) zero
(c) 6 ms–2 (d) 12 ms–2
y = 0.02 m sin (7.85x – 1005 t)
Ans: (d)
ω = 128 × 7.85 = 1005
Solution: x = 8 + 12t – t3
Chapter: Waves
The final velocity of the particle will be zero, because it
[Topic: Basic of Waves]
retarded.
Q50. If a long hollow copper pipe carries a current, then V = 0 + 12 – 3t2 = 0
magnetic field is produced 3t2 = 12
(a) inside the pipe only t = 2 sec
(b) outside the pipe only Now the retardation
(c) both inside and outside the pipe
(b) no where = 0 – 6t
Ans: (b) a [t = 2] = – 12 m/s2
Solution: Inside a hollow pipe carrying current, the retardation = 12 m/s2
magnetic field is zero, since according to Ampere‟s law, Chapter: Kinematics Motion in a Straight Line
Bi. 2πr = µ0 × 0 ⇒ Bi = 0. [Topic: Non-uniform motion]
But for external points, the current behaves as if it was Q53. A spherical ball rolls on a table without slipping.
concentrated at the axis only; so, outside, . Then the fraction of its total energy associated with
rotation is
(a) 2/5 (b) 2/7

44
(c) 3/5 (d) 3/7 Q57. A ball is dropped from a high rise platform at t = 0
Ans: (b) starting from rest. After 6 seconds another ball is thrown
downwards from the same platform with a speed v. The
Solution: two balls meet at t = 18s. What is the value of v?
, -
(take g = 10 m/s2)
= (a) 75 m/s (b) 55 m/s
(c) 40 m/s (d) 60 m/s
Here, Ans: (a)
Chapter: System of Particles and Rotational Motion Solution: Clearly distance moved by 1st ball in 18s =
[Topic: Rolling Motion] distance moved by 2nd ball in 12s.
Q54. With the propagation of a longitudinal wave Now, distance moved in 18 s by 1st ball = ×10×182 = 90
through a material medium, the quantities transmitted in ×18 = 1620 m
the propagation direction are Distance moved in 12 s by 2nd ball
(a) Energy, momentum and mass
(b) Energy = ut + gt2
(c) Energy and mass 1620 = 12 v + 5 × 144
(d) Energy and linear momentum v = 135 – 60 = 75 ms –1
Ans: (b) Chapter: Kinematics Motion in a Straight Line
Solution: With the propagation of a longitudinal wave, [Topic: Motion Under Gravity]
energy alone is propagated. Q58. In a rocket a seconds pendulum is mounted. Its
Chapter: Waves period of oscillation decreases when the rocket
[Topic: Basic of Waves] [1991]
Q55. A coil carrying electric current is placed in uniform (a) comes down with uniform acceleration
magnetic field, then (b) moves round the earth in a geostationary orbit
(a) torque is formed (c) moves up with a uniform velocity
(b) e.m.f is induced (d) moves up with uniform acceleration
(c) both (a) and (b) are correct Ans: (d)
(d) none of the above
Solution: √ . When the rocket accelerates
Ans: (a)
Solution: A current carrying coil has magnetic dipole upwards g increases to (g + a).
moment. Hence, a torque ⃗ ⃗ acts on it in magnetic Chapter: Gravitation
field. [Topic: Gravitational Field, Potential and Energy]
Chapter: Moving Charges and Magnetic Field Q59. Two identical piano wires kept under the same
[Topic: Force & Torque on a Current Carrying tension T have a fundamental frequency of 600 Hz. The
Conductor] fractional increase in the tension of one of the wires
Q56. Which of the following statement is correct? which will lead to occurrence of 6 beats/s when both the
(a) Photocurrent increases with intensity of light wires oscillate together would be
(b) Photocurrent is proportional to the applied voltage (a) 0.02 (b) 0.03
(c) Current in photocell increases with increasing (c) 0.04 (d) 0.01
frequency Ans: (a)
(d) Stopping potential increases with increase of incident Solution: For fundamental mode,
light
f= √
Ans: (a)
Solution: According to photoelectric effect, speed of Taking logarithm on both sides, we get
electron (kinetic energy) emitted depends upon frequency
of incident light while number of photoelectrons emitted log f = . / (√ )
depends upon intensity of incident light. Hence, as the = . / . /
intensity of light increases, the photocurrent increases. In
a photo-cell, the photocurrent has no relation with the or log f = log . / , -
applied voltage. Differentiating both sides, we get
Stopping potential is the (negative) potential at which the
current is just reduced to zero. It is independent of (as l and µ are constants)
intensity of light but depends on the frequency of light
similar to K.E. ⇒
Chapter - Dual Nature of Radiation and Matter Here df = 6
[Topic: Electron Emission, Photon Photoelectric Effect f = 600 Hz
& X-ray]

45
 = 0.02 Chapter: Kinematics Motion in a Plane
[Topic: Vectors]
Chapter: Waves
[Topic: Beats, Interference & Superposition of Waves]
Q63. The escape velocity on the surface of earth is 11.2
km/s. What would be the escape velocity on the surface
Q60. If θ1 and θ2 be the apparent angles of dip observed of another planet of the same mass but 1/4 times the
in two vertical planes at right angles to each other, then radius of the earth?
the true angle of dip θ is given by :- (a) 22.4 km/s (b) 44.8 km/s
(a) tan2θ = tan2θ1 + tan2θ2 (b) cot2θ = cot2θ1 – cot2θ2 (c) 5.6 km/s (d) 11.2 km/s
2 2 2
(c) tan θ = tan θ1 – tan θ2 (d) cot2θ = cot2θ1 + cot2θ2 Ans: (a)
Ans: (d)
Solution: If θ1 and θ2 are opparent angles of dip Solution: √
Let α be the angle which one of the plane make with the
magnetic meridian. √ √ =√

√ √ =2
i.e., …(i)
∴ = 22.4 km/s
Chapter: Gravitation
i.e., …(ii) [Topic: Motion of Satellites, Escape Speed and Orbital
Velocity]
Squaring and adding (i) and (ii), we get Q64. Two identical charged spheres suspended from a
( ) ( ) common point by two massless strings of lengths l, are
initially at a distance d (d << l) apart because of their
i.e., , - mutual repulsion. The charges begin to leak from both
the spheres at a constant rate. As a result, the spheres
or approach each other with a velocity v. Then v varies as a
i.e., function of the distance x between the spheres, as :
Chapter: Magnetism and Matter (a)
[Topic: The Earth's Magnetism, Magnetic Materials (b)
and their Properties]
(c)
Q61. In Rutherford scattering experiment, what will be (d)
the correct angle for α-scattering for an impact parameter, Ans: (c)
b=0? Solution: From figure tan θ =
(a) 90° (b) 270°
(c) 0° (d) 180°
Ans: (d)
Solution: Impact parameter for Rutherford scattering
experiment, orx3 q2 …(1)
orx3/2 q …(2)
. /
⇒ ( ) Differentiating eq. (1) w.r.t. time
3x2 2q but is constant
or so x2(v) q Replace q from eq.
Chapter: Atoms (2)
[Topic: Atomic Structure, Rutherford's Nuclear Model x2(v) x3/2 or v x–1/2
of Atom] Chapter: Electrostatic Potential
Q62. A particle moves with a velocity ⃗ ̂ ̂ and capacitance
m/s under the influence of a constant force ⃗ [Topic: Electric Field, Electric Field Lines & Dipole]
̂ ̂ N. The instantaneous power applied to the Q65. A rectangular, a square, a circular and an elliptical
particle is loop, all in the (x – y) plane, are moving out of a uniform
[2000] magnetic field with a constant velocity, ⃗ ̂ . The
(a) 45 J/s (b) 35 J/s magnetic field is directed along the negative z axis
(c) 25 J/s (d) 195 J/s direction. The induced emf, during the passage of these
Ans: (a) loops, out of the field region, will not remain constant for
Solution: ⃗ ⃗ ( ̂ ̂) ( ̂ ̂) [2009]
= 6 × 20 – 4 × 15 –3 ×5 = 45 J/s (a) the circular and the elliptical loops.
(b) only the elliptical loop.

46
(c) any of the four loops. (d)
(d) the rectangular, circular and elliptical loops.
Ans: (a) Ans: (a)
Solution: The induced emf will remain constant only in Solution: Inflow rate of volume of the liquid = Outflow
the case of rectangular and square loops. In case of the rate of volume of the liquid
circular and the elliptical loops, the rate of change of area πR2V = nπr2(v) ⇒ v =
of the loops during their passage out of the field is not Chapter: Mechanical Properties of Fluids
constant, hence induced emf will not remain constant for [Topic: Fluid Flow, Reyonld's Number & Bernoulli's
them. Principle]
Chapter: Electromagnetic
Q69. The electric field in a certain region is acting
[Topic: Magnetic Flux, Faraday's & Lenz's Law]
radially outward and is given by E = Aa. A charge
Q66. The energy of hydrogen atom in nth orbit is En, contained in a sphere of radius 'a' centred at the origin of
then the energy in nth orbit of single ionised helium atom the field, will be given by
will be (a) A ε0 a2 (b) 4 πε0 Aa3
(a) 4En (c) ε0 Aa 3
(d) 4 πε0 Aa2
(b) En/4 Ans: (b)
(c) 2En Solution: Net flux emmited from a spherical surface of
(d) En/2
radius a according to Gauss‟s theorem
Ans: (a)
Solution: We have . For helium Z = 2.
Hence requisite answer is 4En or,(Aa) (4πa2) =
Chapter: Atoms So,qin = 4πε0 A a3
[Topic: Bohr Model & The Spectra of the Hydrogen Chapter: Electrostatic Potential and capacitance
Atom] [Topic: Electric Flux & Gauss's Law]
Q67. Two projectiles are fired from the same point with Q70. The stable nucleus that has a radius half that of Fe56
the same speed at angles of projection 60º and 30º is
respectively. Which one of the following is true? (a) Li7 (b) Na21
(a) Their maximum height will be same (c) S16
(d) Ca40
(b) Their range will be same Ans: (a)
(c) Their landing velocity will be same
(d) Their time of flight will be same Solution: The nuclear radius
Ans: (b) or
Solution: Given, u1 = u2 = u, θ1 = 60º, θ2 = 30º or,
In 1st case, we know that range
( ) ∴ . / 4 5

( )
Chapter: Nuclei
( ) √ [Topic: Composition and Size of the Nucleus]
Q71. A stone is dropped from a height h. It hits the
In 2nd case, when , then ground with a certain momentum P. If the same stone is
√ dropped from a height 100% more than the previous
= R1 = R2
height, the momentum when it hits the ground will
[we get same value of ranges]. change by :
Chapter: Kinematics Motion in a Plane (a) 68% (b) 41%
[Topic: Projectile Motion] (c) 200% (d) 100%
Q68. The cylindrical tube of a spray pump has radius, R, Ans: (b)
one end of which has n fine holes, each of radius r. If the Solution: Momentum P = mv = √
speed of the liquid in the tube is V, the speed of the (∵ v2 = u2 + 2gh; Here u = 0)
ejection of the liquid through the holes is : When stone hits the ground momentum
(a) P= √
(b) when same stone dropped from 2h (100% of initial) then
momentum
(c) P′= √ ( ) √
Which is changed by 41% of initial.

47
 Chapter: Dynamics Laws of Motion (c) m3 = | m1 – m2|
[Topic: Ist, IInd & IIIrd Laws of Motion] (d) m3 < (m1 + m2)
Q72. As per the diagram, a point charge +q is placed at Ans: (d)
the origin O. Work done in taking another point charge – Solution: m3 < (m1 + m2)(∵ m1+ m2 = m3 + E ]
Q from the point A [coordinates (0, a)] to another point B as E = [m1+ m2– m3] C2
[coordinates (a, 0)] along the straight path AB is: Chapter: Nuclei
[Topic: Mass-Energy & Nuclear Reactions]
Q75. The coefficient of static friction, µs, between block
A of mass 2 kg and the table as shown in the figure is 0.2.
What would be the maximum mass value of block B so
that the two blocks do not move? The string and the
pulley are assumed to be smooth and massless.
(a) zero (g = 10 m/s2)
(b) . /√
(c) . /
√
(d) . / √
Ans: (a) (a) 0.4 kg (b) 2.0 kg
(c) 4.0 kg (d) 0.2 kg
Solution: We know that potential energy of two charge
system is given by Ans: (a)
Solution: mBg = µs mAg{
mAg = µs mAg}
⇒ mB = µs mA
According to question, or, mB= 0.2 × 2 = 0.4 kg
( )( )
UA = Chapter: Dynamics Laws of Motion
( )( ) [Topic: Motion of Connected Bodies, Pulleys]
and UB =
Q76. Certain quantity of water cools from 70°C to 60°C
∆U = UB–UA = 0 in the first 5 minutes and to 54°C in the next 5 minutes.
We know that for conservative force, The temperature of the surroundings is:
W = –∆U = 0 (a) 45°C (b) 20°C
Chapter: Electrostatic Potential and capacitance (c) 42°C (d) 10°C
[Topic: Electric Potential Energy & Work Done in Ans: (a)
Carrying a Charge] Solution: Let the temperature of surroundings be θ0
Q73. An inductance L having a resistance R is connected By Newton's law of cooling
to an alternating source of angular frequency ω. The
[ ]
quality factor Q of the inductance is
(a) ⇒ 0 1
⇒2 = k [65 – θ0]...(i)
(b) . /
Similarly, 0 1
(c) . / ⇒ = k [57 – θ0]...(ii)
(d) By dividing (i) by (ii) we have
Ans: (d) ⇒ θ0 = 45º
Solution: Quality factor Chapter: Thermal Properties
= [Topic: Newton's Law of Cooling]
= Q77. The electric and magnetic field of an
electromagnetic wave are
Chapter: Alternating Current (a) in opposite phase and perpendicular to each other
[Topic: A.C. Circuit, LCR Circuit, Quality & Power (b) in opposite phase and parallel to each other
Factor] (c) in phase and perpendicular to each other
Q74. If in nuclear fusion process the masses of the (d) in phase and parallel to each other.
fusing nuclei be m1 and m2 and the mass of the resultant Ans: (c)
nucleus be m3, then Solution: Variation in magnetic field causes electric field
(a) m3 > (m1 + m2) and vice-versa.
(b) m3 = m1 + m2

48
In electromagnetic waves, ⃗ ⊥ ⃗ Both ⃗ ⃗ are in the Ans: (a)
same phase. Solution: T1 = T, W = 6R joules,
In electromagnetic waves
( )
( ) ( )
The electromagnetic waves travel in the direction of
(⃗ ⃗ ) ( )
Chapter - Electromagnetic Waves n = 1, T1 = T
[Topic: Electromagnetic Waves, Conduction & ⇒ T2 = (T–4)K
Displacement Current] Chapter: Heat & Thermodynamics
Q78. The half life of a radioactive nucleus is 50 days. [Topic: Specific Heat Capacity & Thermodynamic
The time interval (t2 – t1) between the time t2when of it Processes]
has decayed and the time t1when of it had decayed is : Q81. A wire of resistance 12 ohms per meter is bent to
form a complete circle of radius 10 cm. The resistance
(a) 30 days (b) 50 days between its two diametrically opposite points, A and B as
(c) 60 days (d) 15 days shown in the figure, is
Ans: (b)
–λt
Solution: N1 = N0 e

... (i)

(a) (b)
(c) (d) 0.
Ans: (a)
... (ii)
Dividing equation (i) by equation (ii)
( )

( )
= T1/2 = 50 days
Chapter: Nuclei
Solution:
[Topic: Radioactivity]
Q79. A car of mass 1000 kg negotiates a banked curve
of radius 90 m on a frictionless road. If the banking angle
is 45°, the speed of the car is :
(a) 20 ms–1 (b) 30 ms–1
(c) 5 ms–1 (d) 10 ms–1
Ans: (c) The resistance of length 2πR is 12Ω. Hence the resistance
Solution: For banking tan of length πR is 6Ω. Thus two resistances of 6Ω can be
represented as shown in fig. 2.
tan 45 = ∴ Equivalent resistance R =
V = 30 m/s Chapter: Current Electricity
Chapter: Dynamics Laws of Motion [Topic: Combination of Resistances]
[Topic: Circular Motion, Banking of Road] Q82. If two mirrors are kept inclined at 60° to each other
Q80. One mole of an ideal gas at an initial temperature and a body is placed at the middle, then total number of
of T K does 6R joules of work adiabatically. If the ratio images formed is
of specific heats of this gas at constant pressure and at (a) six
constant volume is 5/3, the final temperature of gas will (b) five
be (c) four
(a) (T – 4) K (d) three
(b) (T + 2.4) K Ans: (b)
(c) (T – 2.4) K
(d) (T + 4) K

49
Solution: Angle between two mirrors (θ) = 60º. Number [Topic: Kinetic Theory of an Ideal Gas & Gas Laws]
of images formed by the inclined mirror ( ) Q86.
(a) 1.0 ohm (b) 0.5 ohm
(c) 2.0 ohm (d) zero
Chapter - Ray Optics and Optical Ans: (a)
[Topic: Plane, Spherical Mirror & Reflection of Light] Solution: Here hence the lengths 110 cm and
Q83. What is the respective number of α and β-particles 100 cm are interchanged.
emitted in the following radioactive decay Without being short-circuited through R, only the battery
E is balanced.
(a) 6 and 8 (b) 6 and 6
()
(c) 8 and 8 (d) 8 and 6
Ans: (d) When R is connected across E,
Solution: . We know that
or, . / ( )
.
Therefore, in this process, Dividing (i) by (ii), we get
200 = 4n + 168 or . =
Also, 90 = 2n – m + 80 or, 100 R + 100 r = 110 R
or, m = 2n + 80 – 90 = (2 × 8 + 80 – 90) = 6. or, 10 R = 100 r
Thus, respective number of α and β-particles will be 8 ∴ (∴ )
and 6. ⇒ r = 1Ω.
Chapter: Nuclei Chapter: Current Electricity
[Topic: Radioactivity] [Topic: Kirchhoff's Laws, Cells, Thermo emf &
Q84. A body of mass 1 kg is thrown upwards with a Electrolysis]
velocity 20 m/s. It momentarily comes to rest after Q87. A lens having focal length f and aperture of
attaining a height of 18 m. How much energy is lost due diameter d forms an image of intensity I. Aperture of
to air friction? (g = 10 m/s2)
(a) 30 J (b) 40 J diameter in central region of lens is covered by a black
(c) 10 J (d) 20 J paper. Focal length of lens and intensity of image now
Ans: (d) will be respectively:
Solution: When the body is thrown upwards. its K.E is (a) f and
converted into P.E. The loss of energy due to air friction
is the difference of K.E and P.E. (b) and
(c) f and
= 200 – 180 = 20 J (d) and
Chapter: Work, Energy and Power Ans: (c)
[Topic: Energy] Solution: By covering aperture, focal length does not
Q85. Two vessels separately contain two ideal gases A change. But intensity is reduced by times, as aperture
and B at the same temperature. The pressure of A being
twice that of B. Under such conditions, the density of A diameter is covered.
is found to be 1.5 times the density of B. The ratio of ∴ I' =
molecular weight of A and B is :
(a) (b) 2 ∴ New focal length = f and intensity = .
Chapter - Ray Optics and Optical
(c) [Topic: Refraction at Curved Surface, Lenses & Power
(d) of Lens]
Ans: (a) Q88. A zener diode, having breakdown voltage equal to
Solution: From PV = nRT 15V, is used in a voltage regulator circuit shown in
figure. The current through the diode is
PA = and
From question,

So,
Chapter: Kinetic Theory (a) 10 mA (b) 15 mA

50
(c) 20 mA (d) 5 mA =
Ans: (d)
√
Solution: Voltage across zener diode is constant. or = ⇒ V2 = V
Chapter: Work, Energy and Power
[Topic: Collisions]
Q90. The molar specific heat at constant pressure of an
ideal gas is (7/2) R. The ratio of specific heat at constant
pressure to that at constant volume is
(a) 8/7 (b) 5/7
Current in 1kΩ resistor, (c) 9/7 (d) 7/5
(i) 1kΩ= = 15 mA Ans: (d)
Current in 250Ω resistor, Solution:
( )
(i) 250Ω=

() ( ) Chapter: Kinetic Theory

Chapter: Semiconductor Electronics Materials, Devices [Topic: Degree of Freedom, Specific Heat Capacity &
[Topic: Solids, Semiconductors and P-N Junction Mean Free Path]
Diode] Q91. In India electricity is supplied for domestic use at
Q89. On a frictionless surface a block of mass M 220 V. It is supplied at 110 V in USA. If the resistance of
moving at speed v collides elastically with another block a 60 W bulb for use in India is R, the resistance of a 60
of same mass M which is initially at rest. After collision W bulb for use in USA will be
the first block moves at an angle θ to its initial direction (a) R/2 (b) R
and has a speed . The second block's speed after the (c) 2R
collision is : (d) R/4
Ans: (d)
(a) ( ) ( )
Solution:
(b)
√ ( )
√
(c)
√ Chapter: Current Electricity
(d) [Topic: Heating Effects of Current]
Ans: (d) Q92. A thin prism of angle 15º made of glass of
Solution: Here, M1 = M2 and u2 = 0 refractive index µ1 = 1.5 is combined with another prism
u1 = V, V1 = ; V2 = ? of glass of refractive index µ2 = 1.75. The combination of
the prism produces dispersion without deviation. The
angle of the second prism should be
(a) 7° (b) 10°
(c) 12° (d) 5°
Ans: (b)
Solution: Deviation = zero
So, δ = δ1 + δ2 = 0
From figure, along x-axis, ⇒ (µ1 – 1)A1 + (µ2 – 1) A2 = 0
M1u1 + M2u2 = M1V1 cosθ + M2V2 cosθ ...(i)
⇒ A2 (1.75 – 1) = – (1.5 – 1) 15°
Along y-axis
0 = M1V1 sinθ – M2Vs sinθ ...(ii) ⇒ A2 =
By law of conservation of kinetic energy or A2 = – 10°.
...(iii) Negative sign shows that the second prism is inverted
with respect to the first.
Putting M1 = M2 and u2 = 0 in equation (i) , (ii) and (iii)
Chapter - Ray Optics and Optical
we get
[Topic: Prism & Dispersion of Light]
θ + θ = = 90°
Q93. In a junction diode, the holes are due to
and (a) protons
V2 = . / 0 1 (b) extra electrons
(c) neutrons
or, V – . / =
2
(d) missing electrons

51
 Ans: (d) Solution: Current will flow from B to A
Solution: Holes are produced due to missing of electrons. Potential drop over the resistance CA will be more due to
Chapter: Semiconductor Electronics Materials, Devices higher value of resistance. So potential at A will be less
[Topic: Solids, Semiconductors and P-N Junction as compared with at B. Hence, current will flow from B
Diode] to A.
Q94. Two particles which are initially at rest, move Chapter: Current Electricity
towards each other under the action of their internal [Topic: Wheatstone Bridge & Different Measuring
attraction. If their speeds are v and 2v at any instant, then Instruments]
the speed of centre of mass of the system will be: Q97. In Young‟s double slit experiment the distance
(a) 2v between the slits and the screen is doubled. The
(b) zero separation between the slits is reduced to half. As a result
(c) 1.5 (d) v the fringe width
Ans: (b) (a) is doubled
Solution: If no external force actson a system of (b) is halved
particles, the centre of mass remains at rest. So, speed of (c) becomes four times
centre of mass is zero. (d) remains unchanged
Chapter: System of Particles and Rotational Motion Ans: (c)
[Topic: Centre of Mass, Centre of Gravity & Principle Solution: Fringe width ;
of Moments]
Q95. A particle starts simple harmonic motion from the From question D′ = 2D and
mean position. Its amplitude is A and time period is
T.What is its displacement when its speed is half of its ∴
maximum speed Chapter - Wave Optics
(a)
√ [Topic: Young's Double Slit Experiment]
√
Q98. A common emitter amplifier has a voltage gain of
(b) 50, an input impedance of 100Ω and an output impedance
(c) of 200Ω The power gain of the amplifier is
√ (a) 500 (b) 1000
(d) (c) 1250 (d) 50
√
Ans: (b) Ans: (c)
Solution: Power gain = voltage gain × current gain
Solution: vmax = Aω when
=
√
√ = =
√
=
[Alt : The displacement at which the speed is n times the
Chapter: Semiconductor Electronics Materials, Devices
maximum speed is given by √ ]
[Topic: Junction Transistor]
Chapter: Oscillation
[Topic: Displacement, Phase, Velocity & Acceleration
Q99. A physical quantity of the dimensions of length
of SHM] that can be formed out of c, G and is [c is velocity of
Q96. In the circuit shown, if a conducting wire is light, G is universal constant of gravitation and e is
connected between points A and B, the current in this charge]
wire will
(a) 0 1

(b) 0 1
(c)

(d) | |
(a) flow in the direction which will be decided by the Ans: (d)
value of V Solution: Let dimensions of length is related as,
(b) be zero
(c) flow from B to A L , - , - 0 1
(d) flow from A to B
= ML3T–2
Ans: (c)

52
L = [LT–1]x [M–1L3T–2]y[ML3T–2]z Solution: =
[L] = [Lx +3y+3z M–y+z T–x–2y–2z]
Comparing both sides
– y + z = 0 ⇒ y = z...(i)
x + 3y + 3z = 1...(ii) Chapter: Moving Charges and Magnetic Field
– x – 4z = 0 (∵ y = z)...(iii) [Topic: Motion of Charged Particle in Magnetic Field
From (i) , (ii) & (iii) & Moment]
z=y= x=–2 Q3. If the momentum of electron is changed by P, then
the de Broglie wavelength associated with it changes by
Hence, L = 0 1 0.5%. The initial momentum of electron will be :
Chapter: Units and Measurement (a) 200 P (b) 400 P
[Topic: Dimensions of Physical Quantities] (c) (d) 100 P
Q100. The instantaneous angular position of a point on a Ans: (a)
rotating wheel is given by the equation θ(t) = 2t3 – 6t2. Solution: The de-Broglie‟s wavelength associated with
The torque on the wheel becomes zero at the moving electron
(a) t = 1s
(b) t = 0.5 s Now, according to problem
(c) t = 0.25 s
(d) t = 2s
Ans: (a)
Solution: When angular acceleration (α) is zero then
torque on the wheel becomes zero. P′ = 200 P
θ(t) = 2t3 – 6t2 Chapter - Dual Nature of Radiation and Matter
[Topic: Matter Waves, Cathode & Positive Rays]
Q4. The device that can act as a complete electronic
∴ t = 1 sec. circuit is
Chapter: System of Particles and Rotational Motion (a) junction diode
[Topic: Torque, Couple and Angular Momentum] (b) integrated circuit
(c) junction transistor
PART 7. PHYSICS (d) zener diode
Ans: (b)
QUESTION BANK Solution: Integrated circuit can act as a complete
electronic circuit.
Chapter: Semiconductor Electronics Materials, Devices
[Topic: Digital Electronics and Logic Gates]
Q1. Two springs of spring constants k1 and k2 are joined Q5. The frequency of vibration f of a mass m suspended
in series. The effective spring constant of the
from a spring of spring constant k is given by a relation
combination is given by
of the type f = c mx ky, where c is a dimensionless
(a) k1k2 /(k1+ k2)
constant. The values of x and y are
(b) k1k2
(c) (k1+ k2) /2 (d) k1+ k2 (a)
Ans: (a) (b)
Solution: ⇒ (c)
⇒ (d)
Ans: (d)
Chapter: Oscillation
Solution: f = c mx ky;
[Topic: Time Period, Frequency, Simple Pendulum &
Spring constant k = force/length.
Spring Pendulum]
[M0L0T–1] = [Mx (MT–2)y]=[ Mx+y T–2y]
Q2. A proton moving with a velocity 3 × 105 m/s enters a
magnetic field of 0.3 tesla at an angle of 30º with the ⇒x + y=0,-2y = -1 or y =
field. The radius of curvature of its path will be (e/m for Therefore,
8
proton = 10 C/kg) Chapter: Units and Measurement
(a) 2 cm (b) 0.5 cm [Topic: Dimensions of Physical Quantities]
(c) 0.02 cm (d) 1.25 cm
Ans: (b)

53
Q6. The moment of inertia of a uniform circular disc is . / ,
maximum about an axis perpendicular to the disc and
passing through : using (1) (where n = 2)
Chapter: Moving Charges and Magnetic Field
[Topic: Magnetic Field, Biot-Savart's Law & Ampere's
Circuital Law]
Q9. In photoelectric emission process from a metal of
work function 1.8 eV, the kinetic energy of most
energetic electrons is 0.5 eV. The corresponding stopping
potential is
(a) B (a) 1.8 V (b) 1.2 V
(b) C (c) 0.5 V (d) 2.3 V
(c) D Ans: (c)
(d) A
Solution: The stopping potential is equal to maximum
Ans: (a) kinetic energy.
Solution: According to parallel axis theorem of the Chapter - Dual Nature of Radiation and Matter
moment of Inertia
[Topic: Electron Emission, Photon Photoelectric Effect
I = Icm + md2
& X-ray]
d is maximum for point B so Imax about B.
Chapter: System of Particles and Rotational Motion Q10. A body is moving with velocity 30 m/s towards
[Topic: Moment of Inertia, Rotational K.E. and Power] east. After 10 seconds its velocity becomes 40 m/s
towards north. The average acceleration of the body is
Q7. The wave described by y = 0.25 sin (10πx – (a) 1 m/s2 (b) 7 m/s2
2πt),where x and y are in meters and t in seconds, is a (c) 7 m/s2
(d) 5 m/s2
wave travelling along the: Ans: (d)
(a) –ve x direction with frequency 1 Hz.
Solution: Average acceleration
(b) +ve x direction with frequency π Hz and wavelength
λ = 0. 2 m. < a> =
(c) +ve x direction with frequency 1 Hz and wavelength λ | ̂ ̂|
<a>=
= 0.2 m (d) –ve x direction with
amplitude 0.25 m and wavelength λ = 0.2 m √ ( )
Ans: (c) < a > = 5 m/sec2
Solution: y = 0.25 sin (10 πx – 2πt) Chapter: Kinematics Motion in a Straight Line
Comparing this equation with the standard wave equation [Topic: Non-uniform motion]
y = asin (kx – ωt) Q11. A thin uniform circular ring is rolling down an
We get, k = 10π inclined plane of inclination 30° without slipping. Its
⇒ ⇒ λ = 0.2 m linear acceleration along the inclination plane will be
(a)
And ω = 2π or, 2πv = 2π ⇒ v = 1Hz.
The sign inside the bracket is negative, hence the wave (b)
travels in + ve x- direction. (c)
Chapter: Waves
[Topic: Basic of Waves] (d)
Q8. A coil of one turn is made of a wire of certain length Ans: (c)
and then from the same length a coil of two turns is Solution:
made. If the same current is passed in both the cases, then
the ratio of the magnetic inductions at their centres will Chapter: System of Particles and Rotational Motion
be [Topic: Rolling Motion]
(a) 2 : 1 (b) 1 : 4 Q12. The frequency of sinusoidal wave n = 0.40 cos
(c) 4 : 1 (d) 1 : 2 [2000 t + 0.80] would be
Ans: (b) (a) 1000 π Hz (b) 2000 Hz
Solution: Let (c) 20 Hz
be length of wire. (d)
Ist case :
Ans: (d)
=2πr ⇒ Solution: Comparing with the equation
[ ∵ n = 1]…(1) ( )
2nd Case : ( )⇒ We get 2πν = 2000

54
 Chapter: Waves (c) 1400 km (d) 2000 km
[Topic: Basic of Waves] Ans: (a)
Q13. A straight wire of length 0.5 metre and carrying a Solution: As we know, gravitational potential (v) and
current of 1.2 ampere is placed in uniform magnetic field acceleration due to gravity (g) with height
of induction 2 tesla. The magnetic field is perpendicular V= = –5.4 × 107…(1)
to the length of the wire. The force on the wire is
(a) 2.4 N (b) 1.2 N and g = ( )
…(2)
(c) 3.0 N (d) 2.0 N Dividing (1) by (2)
Ans: (b)
Solution: F = Bi
= 2 ×1.2 × 0.5 = 1.2 N
Chapter: Moving Charges and Magnetic Field ( )
[Topic: Galvanometer and Its Conversion into Ammeter ⇒ ( )
& Voltmeter] ⇒R + h = 9000 km so, h = 2600 km
Q14. The X-rays cannot be diffracted by means of an Chapter: Gravitation
ordinary grating because of [Topic: Gravitational Field, Potential and Energy]
(a) high speed
Q17. A tuning fork of frequency 512 Hz makes 4 beats
(b) short wavelength
per second with the vibrating string of a piano. The beat
(c) large wavelength
frequency decreases to 2 beats per sec when the tension
(d) none of these
in the piano string is slightly increased. The frequency of
Ans: (b)
the piano string before increasing the tension was
Solution: We know that the X-rays are of short (a) 510 Hz (b) 514 Hz
wavelength as compared to grating constant of optical
(c) 516 Hz (d) 508 Hz
grating. As a result of this, it makes difficult to observe
Ans: (d)
X-rays diffraction with ordinary grating.
Solution: The frequency of the piano string = 512 + 4 =
Chapter - Dual Nature of Radiation and Matter
516 or 508. When the tension is increased, beat frequency
[Topic: Electron Emission, Photon Photoelectric Effect decreases to 2, it means that frequency of the string is
& X-ray] 508 as frequency of string increases with tension.
Q15. A man of 50 kg mass is standing in a gravity free Chapter: Waves
space at a height of 10 m above the floor. He throws a [Topic: Beats, Interference & Superposition of Waves]
stone of 0.5 kg mass downwards with a speed 2 m/s.
Q18. The magnetic susceptibility is negative for :
When the stone reaches the floor, the distance of the man
(a) diamagnetic material only
above the floor will be:
(b) paramagnetic material only
(a) 9.9 m (b) 10.1 m
(c) ferromagnetic material only
(c) 10 m (d) 20 m
(d) paramagnetic and ferromagnetic materials
Ans: (b)
Ans: (a)
Solution: No external force is acting, therefore,
Solution: Magnetic susceptibility χ for dia-magnetic
momentum is conserved.
materials only is negative and low |χ| = –1; for
By momentum conservation,
paramagnetic substances low but positive |χ| = 1 and for
50 u + 0.5 × 2 = 0
where u is the velocity of man. ferromagnetic substances positive and high |χ| = 102.
Chapter: Magnetism and Matter
[Topic: The Earth's Magnetism, Magnetic Materials
Negative sign of u shows that man moves upward. and their Properties]
Time taken by the stone to reach the ground Q19. What is the radius of iodine atom (At. no. 53, mass
no. 126)
=
(a) 2.5 × 10–11 m (b) 2.5 × 10–9 m
–9
Distance moved by the man (c) 7 × 10 m (d) 7 × 10–6m
∴ when the stone reaches the floor, the distance of the Ans: (a)
man above floor = 10.1 m Solution: 53 electrons in iodine atom are distributed as 2,
Chapter: Kinematics Motion in a Straight Line 8, 18, 18, 7
[Topic: Motion Under Gravity] ∴n=5
Q16. At what height from the surface of earth the ( )
gravitational potential and the value of g are –5.4 × 107 J
kg–1 and 6.0 ms–2 respectively ?
Take the radius of earth as 6400 km :
(a) 2600 km (b) 1600 km Chapter: Atoms

55
 [Topic: Bohr Model & The Spectra of the Hydrogen
Atom] Solution: From figure, tan θ = ⇒ =
Q20. What is the linear velocity if angular velocity [∵ F = from coulomb‟s law]
vector⃗⃗ ̂ ̂ ̂ and position vector ̂ ̂
̂ ⇒r 3
y ⇒ r'3 ⇒ =
(a) ̂ ̂
⇒ r' =
(b) ̂ ̂ √
̂ ̂ Chapter: Electrostatic Potential and capacitance
(c) ̂ (d) ̂ [Topic: Electric Field, Electric Field Lines & Dipole]
Ans: (b)
Q23. A conducting circular loop is placed in a uniform
Solution: As we know that
magnetic field of 0.04 T with its plane perpendicular to
⃗ ⃗⃗ =( ̂ ̂) ( ̂ ̂)
the magnetic field. The radius of the loop starts shrinking
= ̂ ̂ at 2 mm/s. The induced emf in the loop when the radius
Chapter: Kinematics Motion in a Plane is 2 cm is
[Topic: Vectors] (a) 4.8 π µV (b) 0.8 π µV
Q21. The escape velocity of a sphere of mass m is given (c) 1.6 π µV (d) 3.2 π µV
by (G = Universal gravitational constant; M = Mass of Ans: (d)
the earth and Re = Radius of the earth) Solution: Induced emf in the loop is given by e = –
(a) √ where A is the area of the loop.
e=– (π r2) = – B π 2r
(b) √ r = 2cm = 2 × 10–2 m
dr = 2 mm = 2 × 10–3 m
(c) √ dt = 1s
e = – 0.04 × 3.14 × 2 × 2 ×10–2 ×
(d) √
= 0.32 π × 10–5V
Ans: (b) =3.2 π ×10–6V
Solution: Escape velocity is the minimum velocity with =3.2 π µV
which a body is projected to escape from earth's Chapter: Electromagnetic
gravitational field [Topic: Magnetic Flux, Faraday's & Lenz's Law]
⇒ √
Q24. When an electron jumps from the fourth orbit to
the second orbit, one gets the
Chapter: Gravitation (a) second line of Lyman series
[Topic: Motion of Satellites, Escape Speed and Orbital (b) second line of Paschen series
Velocity] (c) second line of Balmer series
Q22. Two pith balls carrying equal charges are (d) first line of Pfund series
suspended from a common point by strings of equal Ans: (c)
length. The equilibrium separation between them is r. Solution: When the electron drops from any orbit to
Now the strings are rigidly clamped at half the height. second orbit, then wavelength of line obtained belongs to
The equilibrium separation between the balls now Balmer series.
become Chapter: Atoms
[Topic: Bohr Model & The Spectra of the Hydrogen
Atom]
Q25. If a body A of mass M is thrown with velocity v at
an angle of 30° to the horizontal and another body B of
the same mass is thrown with the same speed at an angle
of 60° to the horizontal, the ratio of horizontal range of A
to B will be
(a) . / (a) 1 : 3 (b) 1 : 1
√
(b) . / (c) 1:√ (d) √
√
Ans: (b)
(c) . / Solution: Horizontal range is same when angle of
projection with the horizonatal is θ and (90° – θ).
(d) . /
√ Chapter: Kinematics Motion in a Plane
Ans: (a) [Topic: Projectile Motion]

56
Q26. A fluid is in streamline flow across a horizontal
pipe of variable area of cross section. For this which of Irms = √
. /
√
the following statements is correct? cos θ =
[NEET Kar. 2013]
(a) The velocity is minimum at the narrowest part of the <P>=
pipe and the pressure is minimum at the widest part of the Chapter: Alternating Current
pipe [Topic: Alternating Current, Voltage & Power]
(b) The velocity is maximum at the narrowest part of the Q29. A nucleus ruptures into two nuclear parts, which
pipe and pressure is maximum at the widest part of the have their velocity ratio equal to 2:1. What will be the
pipe ratio of their nuclear size (nuclear radius)?
(c) Velocity and pressure both are maximum at the [1996]
narrowest part of the pipe (a) 21/3 : 1 (b) 1 : 21/3
(d) Velocity and pressure both are maximum at the 1/2
(c) 3 : 1 (d) 1 : 31/2
widest part of the pipe Ans: (b)
Ans: (b) Solution: Applying law of conservation of momentum,
Solution: According to Bernoulli‟s theorem, m1v1 = m2v2
= constant and Av = constant
If A is minimum, v is maximum, P is minimum.
Chapter: Mechanical Properties of Fluids As m = ⇒
[Topic: Viscosity & Terminal Velocity]
Hence,
Q27. What is the flux through a cube of side 'a' if a point
charge of q is at one of its corner :
(a) ⇒ ( )

(b) Chapter: Nuclei

[Topic: Composition and Size of the Nucleus]
(c)
Q30. A body of mass M hits normally a rigid wall with
(d) velocity V and bounces back with the same velocity. The
impulse experienced by the body is
Ans: (b)
(a) MV (b) 1.5 MV
Solution: Eight identical cubes are required to arrange so
(c) 2 MV (d) zero
that this charge is at centre of the cube formed so flux.
Ans: (c)
Solution: Impulse experienced by the body
= change in momentum
= MV – (–MV)
Chapter: Electrostatic Potential and capacitance
= 2MV.
[Topic: Electric Flux & Gauss's Law]
Chapter: Dynamics Laws of Motion
Q28. The instantaneous values of alternating current and [Topic: Ist, IInd & IIIrd Laws of Motion]
voltages in a circuit are given as
Q31. A slab of stone of area 0.36 m2 and thickness 0.1 m
( ) amper is exposed on the lower surface to steam at 100°C. A
√
. / Volt block of ice at 0°C rests on the upper surface of the slab.
√ In one hour 4.8 kg of ice is melted. The thermal
The average power in Watts consumed in the circuit is : conductivity of slab is :
(a) (Given latent heat of fusion of ice = 3.36 × 105 J kg–1.) :
√ (a) 1.24 J/m/s/°C (b) 1.29 J/m/s/°C
(b) (c) 2.05 J/m/s/°C (d) 1.02 J/m/s/°C
(c) Ans: (a)
(d)
Ans: (d)
Solution: The average power in the circuit where cos θ =
power factory
< P > = Vrms × Irmscos θ Solution:
Rate of heat given by steam = Rate of heat taken by ice
θ = π/3 = phase difference = where K = Thermal conductivity of the slab
√ m = Mass of the ice
Vrms = volt L = Latent heat of melting/fusion
√

57
A = Area of the slab Q34. Mp denotes the mass of a proton and Mn that of a
( ) neutron. A given nucleus, of binding energy B, contains Z
protons and N neutrons. The mass M(N, Z) of the nucleus
is given by (c is the velocity of light)
(a) M(N, Z) = NMn + ZMp + B/c2
K =1.24 J/m/s/°C (b) M(N, Z) = NMn + ZMp – Bc2
2
Chapter: Thermal Properties (c) M(N, Z) = NMn + ZMp + Bc
[Topic: Calorimetry & Heat Transfer] (d) M(N, Z) = NMn + ZMp – B/c2
Q32. Two charges q1 and q2 are placed 30 cm apart, as Ans: (d)
shown in the figure. A third charge q3 is moved along the Solution: Mass defect
arc of a circle of radius 40 cm from C to D. The change Mass of nucleus = Mass of proton + mass of neutron –
in the potential energy of the system is where k is mass defect
Chapter: Nuclei
[Topic: Mass-Energy & Nuclear Reactions]
Q35. A block of mass m is placed on a smooth wedge of
inclination θ. The whole system is accelerated
horizontally so that the block does not slip on the wedge.
The force exerted by the wedge on the block (g is
(a) 8q1 (b) 6q1 acceleration due to gravity) will be
(c) 8q2 (d) 6q2 (a) mg/cos θ
Ans: (c) (b) mg cos θ
Solution: We know that potential energy of discrete (c) mg sin θ
system of charges is given by (d) mg
Ans: (a)
( )
According to question,
Uinitial = . /
Ufinal = . /
Ufinal – Uinitial = . / Solution:
N = m a sin θ + mg cos θ.....(1)
= , -= ( ) Also, m g sin θ = m a cos θ ....(2)
Chapter: Electrostatic Potential and capacitance From (1) & (2) , a = g tan θ
[Topic: Electric Potential Energy & Work Done in .
Carrying a Charge]
Q33. In an experiment, 200 V A.C. is applied at the ends ( )
of an LCR circuit. The circuit consists of an inductive or,
reactance (XL) = 50 Ω, capacitive reactance (XC) = 50 Ω
and ohmic resistance (R) = 10 Ω. The impedance of the Chapter: Dynamics Laws of Motion
circuit is [Topic: Motion of Connected Bodies, Pulleys]
[1996] Q36. A beaker full of hot water is kept in a room. If it
(a) 10Ω (b) 20Ω cools from 80°C to 75°C in t1 minutes, from 75° C to
(c) 30Ω (d) 40Ω 70°C in t2 minutes and from 70°C to 65°C in t3 minutes,
Ans: (a) then
Solution: Given : Supply voltage (Vac) = 200 V (a)
Inductive reactance (XL) = 50 Ω (b)
Capacitive reactance (XC) = 50 Ω (c)
Ohmic resistance (R) = 10 Ω. (d)
We know that impedance of the LCR circuit (Z) Ans: (c)
Solution: Let θ0 be the temperature of the surrounding.
=√*( ) +
Then
√*( ) ( ) +
Chapter: Alternating Current ( )
[Topic: A.C. Circuit, LCR Circuit, Quality & Power
or, ( )
Factor]

58
or, ( )
… (1) ∴ Part of Nx = ( )
Similarly, …(2) = ( )
( )
So, total 4 half lives are passed, so, age of rock is 4 × 50
and ( )
… (3) = 200 years
From (1) , (2) & (3) , it is obvious that Chapter: Nuclei
t1 < t2 < t3 [Topic: Radioactivity]
Chapter: Thermal Properties Q40. A car of mass m is moving on a level circular track
[Topic: Newton's Law of Cooling] of radius R. If µs represents the static friction between the
Q37. If the potential of a capacitor having capacity 6 µF road and tyres of the car, the maximum speed of the car
is increased from 10 V to 20 V, then increase in its in circular motion is given by :
energy will be (a) √
(a) (b)
(c) (d) (b) √
Ans: (c)
Solution: Capacitance of capacitor (C) = 6 µF = 6 ×10–6 (c) √
F; Initial potential (V1) = 10 V and final potential (V2) =
20 V. (d) √
The increase in energy (∆U) Ans: (d)
Solution: For smooth driving maximum speed of car v
( ) then
( ) ,( ) ( ) -
( ) . √
Chapter: Electrostatic Potential and capacitance Chapter: Dynamics Laws of Motion
[Topic: Electric Current, Drift of Electrons, Ohm's [Topic: Circular Motion, Banking of Road]
Law, Resistance & Resistivity] Q41. An ideal gas at 27ºC is compressed adiabatically to
Q38. If ⃗ and ⃗ represent electric and magnetic field of its original volume. The rise in temperature is
vectors of the electromagnetic waves, then the direction
of propagation of the waves will be along . /
(a) ⃗ ⃗ (a) 475ºC (b) 402ºC
(b) ⃗ (c) 275ºC (d) 175ºC
(c) ⃗ Ans: (b)
Solution: T = 27°C = 300 K
(d) ⃗ ⃗
Ans: (d) ; ;
Solution: Direction of propagation of electro-magnetic From adiabatic process we know that
waves is perpendicular to Electric field and Magnetic
field. Hence, direction is given by vector⃗ ⃗ ⃗⃗
⃗ ⃗⃗
. ( ) ( )
Chapter - Electromagnetic Waves
[Topic: Electromagnetic Waves, Conduction &
Displacement Current] T2 = 675 – 273°C = 402°C
Q39. The half life of a radioactive isotope 'X' is 50 years. Chapter: Heat & Thermodynamics
It decays to another element 'Y' which is stable. The two [Topic: Specific Heat Capacity & Thermodynamic
elements 'X' and 'Y' were found to be in the ratio of 1 : 15 Processes]
in a sample of a given rock. The age of the rock was Q42. When a wire of uniform cross–section a, length l
estimated to be and resistance R is bent into a complete circle, resistance
(a) 150 years (b) 200 years between any two of diametrically opposite points will be
(c) 250 years (d) 100 years
(a) (b) 4R
Ans: (b)
Solution: Let number of atoms in X = Nx (c) (d)
Number of atoms in Y = Ny Ans: (a)
By question . /
Solution: ∴

59
 Chapter: Current Electricity
[Topic: Combination of Resistances]
Q43. Ray optics is valid, when characteristic dimensions
are
(a) of the same order as the wavelength of light
(b) much smaller than the wavelength of light
(c) of the order of one millimetre
(d) much larger than the wavelength of light (a) P2 > P1 (b) P2 < P1
Ans: (d) (c) Cannot be predicted
Solution: Characteristic dimensions must be much larger (d) P2 = P1
than the wavelength of light. Ans: (b)
Chapter - Ray Optics and Optical Solution: P1 > P2
[Topic: Refraction of Light at Plane Surface & Total As V = constant ⇒ P T
Internal Reflection] Hence from V–T graph P1 > P2
Chapter: Kinetic Theory
Q44. The count rate of a Geiger Muller counter for the
[Topic: Kinetic Theory of an Ideal Gas & Gas Laws]
radiation of a radioactive material of half-life 30 minutes
decreases to 5 sec–1 after 2 hours. The initial count rate Q47. A steady current of 1.5 amp flows through a copper
was voltameter for 10 minutes. If the electrochemical
(a) 20 sec–1 (b) 25 sec–1 equivalent of copper is 30 × 10–5 g coulomb–1, the mass of
(c) 80 sec –1
(d) 625 sec–1 copper deposited on the electrode will be
Ans: (c) (a) 0.50 g (b) 0.67 g
Solution: Half-life = 30 minutes; Rate of decrease (N) = (c) 0.27 g (d) 0.40 g.
5 per second and total time = 2 hours = 120 minutes. Ans: (c)
Relation for initial and final count rate Solution: We have, m = ZIt
where, Z is the electrochemical equivalent of copper.
. / . / . / . Therefore, N0 = 16
= 0.27 gm.
× N = 16 × 5 = 80 s–1.
Chapter: Current Electricity
Chapter: Nuclei
[Topic: Kirchhoff's Laws, Cells, Thermo emf &
[Topic: Radioactivity]
Electrolysis]
Q45. A particle of mass m1 is moving with a velocity
v1and another particle of mass m2 is moving with a
Q48. Two thin lenses of focal lengths f1 and f2 are in
contact and coaxial. The power of the combination is:
velocity v2. Both of them have the same momentum but
their different kinetic energies are E1 and E2 respectively. (a) √
If m1 > m2 then
(a) E1 = E2 (b) E1 < E2 (b) √
(c) (d) E1 > E2
(c)
Ans: (b)
Solution: (d)
Ans: (d)
or,
Solution: The focal length of the combination
or,

⇒ ∴ Power of the combinations,

, - (∴ )
or, E2 > E1 Chapter - Ray Optics and Optical
Chapter: Work, Energy and Power [Topic: Refraction at Curved Surface, Lenses & Power
[Topic: Energy] of Lens]
Q46. In the given (V – T) diagram, what is the relation Q49. Which one of the following statement is FALSE ?
between pressure P1 and P2 ? (a) Pure Si doped with trivalent impurities gives a p-type
semiconductor
(b) Majority carriers in a n-type semiconductor are holes
(c) Minority carriers in a p-type semiconductor are
electrons

60
(d) The resistance of intrinsic semiconductor decreases Ans: (c)
with increase of temperature Solution: We know that ratio of specific heats,
Ans: (b) or
Solution: Majority carriers in an n-type semiconductor
are electrons. [where n = Degree of freedom]
Chapter: Semiconductor Electronics Materials, Devices Chapter: Kinetic Theory
[Topic: Solids, Semiconductors and P-N Junction [Topic: Degree of Freedom, Specific Heat Capacity &
Diode] Mean Free Path]
Q50. A ball is thrown vertically downwards from a Q52. An electric kettle has two heating coils. When one
height of 20 m with an initial velocity v0. It collides with of the coils is connected to an a.c. source, the water in the
the ground loses 50 percent of its energy in collision and kettle boils in 10 minutes. When the other coil is used,
rebounds to the same height. The initial velocity v0 is : the water boils in 40 minutes. If both the coils are
(Take g = 10 ms–2) connected in parallel, the time taken by the same quantity
(a) 20 ms–1 (b) 28 ms–1 of water to boil will be
(c) 10 ms –1
(d) 14 ms–1 (a) 15 min (b) 8 min
Ans: (a) (c) 4 min (d) 25 min
Solution: When ball collides with the ground it loses its Ans: (b)
50% of energy Solution: Time
Chapter: Current Electricity
∴ ⇒
[Topic: Heating Effects of Current]
Q53. The refractive index of the material of a prism is
√2 and its refracting angle is 30º. One of the refracting
surfaces of the prism is made a mirror inwards. A beam
of monochromatic light enters the prism from the mirror
surface if its angle of incidence of the prism is
(a) 30° (b) 45°
(c) 60° (d) 0°
Ans: (b)
Solution: ∠r = 30° (using law of triangle)
⇒ µ=

or
√
√
or,
√ √
or, 4gh =
∴ V0 = 20ms–1
Chapter: Work, Energy and Power √
[Topic: Collisions]
⇒
PART 8. PHYSICS √
Chapter - Ray Optics and Optical

QUESTION BANK [Topic: Prism & Dispersion of Light]

Q54. A semi-conducting device is connected in a series
circuit with a battery and a resistance. A current is found
to pass through the circuit. If the polarity of the battery is
Q51. If γ be the ratio of specific heats of a perfect gas, reversed, the current drops to almost zero. The device
the number of degrees of freedom of a molecule of the may be
gas is (a) a p-n junction
(a) ( ) (b) an intrinsic semi-conductor
(c) a p-type semi-conductor
(b) (d) an n-type semi-conductor
(c) Ans: (a)
Solution: In reverse bias, the current through a p-n
(d) ( )
junction is almost zero.

61
 Chapter: Semiconductor Electronics Materials, Devices Chapter: Current Electricity
[Topic: Solids, Semiconductors and P-N Junction [Topic: Wheatstone Bridge & Different Measuring
Diode] Instruments]
Q55. Two bodies of mass 1 kg and 3 kg have position Q58. In Young‟s double slit experiment carried out with
vectors ̂ ̂ and ̂ ̂ respectively. The light of wavelength (λ) = 5000Å, the distance between
centre of mass of this system has a position vector: the slits is 0.2 mm and the screen is at 200 cm from the
(a) ̂ ̂ (b) ̂ ̂ slits. The central maximum is at x = 0. The third
(c) ̂ ̂ (d) ̂ maximum (taking the central maximum as zeroth
maximum) will be at x equal to
Ans: (a)
(a) 1.67 cm (b) 1.5 cm
Solution: The position vector of the centre of mass of
(c) 0.5 cm (d) 5.0 cm
two particle system is given by
Ans: (b)
⃗ ⃗
⃗ Solution: ( )
( )
m = 1.5 cm
[ ̂] ̂ ̂ Chapter - Wave Optics
Chapter: System of Particles and Rotational Motion [Topic: Young's Double Slit Experiment]
[Topic: Centre of Mass, Centre of Gravity & Principle Q59. For transistor action :
of Moments] (1) Base, emitter and collector regions should have
Q56. Which one of the following is a simple harmonic similar size and doping concentrations.
motion? (2) The base region must be very thin and lightly doped.
(a) Ball bouncing between two rigid vertical walls (3) The eimtter-base junction is forward biased and base-
(b) Particle moving in a circle with uniform speed collector junction is reverse based.
(c) Wave moving through a string fixed at both ends (4) Both the emitter-base junction as well as the base-
(d) Earth spinning about its own axis. collector junction are forward biased.
Ans: (c) (a) (3) , (4)
Solution: A wave moving through a string fixed at both (b) (4) , (1)
ends, is a transverse wave formed as a result of simple (c) (1) , (2)
harmonic motion of particles of the string. (d) (2) , (3)
Chapter: Oscillation Ans: (d)
[Topic: Energy in Simple Harmonic Motion] Solution: For transistor action, the base region must be
very thin and lightly doped. Also, the emitter-base
Q57. For the network shown in the Fig. the value of the junction is forward biased and base-collector junction is
current i is reverse biased.
Chapter: Semiconductor Electronics Materials, Devices
[Topic: Junction Transistor]
Q60. If energy (E) , velocity (V) and time (T) are chosen
as the fundamental quantities, the dimensional formula of
surface tension will be :
(a) [EV–1T–2]
(b) [EV–2T–2]
(c) [E–2V–1T–3]
(d) [EV–2T–1]
Ans: (b)
Solution: Let surface tension
(a) s = Ea Vb Tc
(b)
( ) ( ) ( )
(c) Equating the dimension of LHS and RHS
(d) ML0T–2 = MaL2a+b T–2a–b+c
Ans: (d) ⇒a = 1, 2a + b = 0, – 2a – b + c = – 2
Solution: It is a balanced Wheatstone bridge. Hence ⇒a = 1, b = – 2, c = – 2
resistance 4Ω can be eliminated. Hence, the dimensions of surface tension are [E V–2 T–2]
Chapter: Units and Measurement
∴ [Topic: Dimensions of Physical Quantities]
∴ Q61. A circular disk of moment of inertia It is rotating in
a horizontal plane, its symmetry axis, with a constant

62
angular speed . Another disk of moment of inertia Ib is [Topic: Motion of Charged Particle in Magnetic Field
dropped coaxially onto the rotating disk. Initially the & Moment]
second disk has zero angular speed. Eventually both the Q64. Electrons used in an electron microscope are
disks rotate with a constant angular speed . The accelerated by a voltage of 25 kV. If the voltage is
energy lost by the initially rotating disk to friction is: increased to 100kV then the de–Broglie wavelength
(a) associated with the electrons would
( ) (a) increase by 2 times
(b) ( (b) decrease by 2 times
)
(c) decrease by 4 times
(c) ( ) (d) increase by 4 times
(d) Ans: (b)
( )
Solution:
Ans: (d) √
Solution: By conservation of angular momentum,
It =(It+Ib) √ √
where is the final angular velocity of disks
=. / ⇒
Loss in K.E., = Initial K.E. – Final K.E. Chapter - Dual Nature of Radiation and Matter
= – ( ) [Topic: Matter Waves, Cathode & Positive Rays]
= Q65. The following Figure shows a logic gate circuit
with two inputs A and B and the output Y. The voltage
( )( waveforms of A, B and Y are given :
)
= ( )=
( )
Chapter: System of Particles and Rotational Motion
[Topic: Torque, Couple and Angular Momentum]
Q62. The time period of a mass suspended from a spring
is T. If the spring is cut into four equal parts and the same
mass is suspended from one of the parts, then the new
time period will be
(a) 2T
(b)
(c) 2 (d)
Ans: (d)
Solution: √
When a spring is cut into n parts
The logic gate is :
Spring constant for each part = nk
(a) NAND gate
Here, n = 4
(b) NOR gate
√ (c) OR gate
(d) AND gate
Chapter: Oscillation Ans: (a)
[Topic: Time Period, Frequency, Simple Pendulum & Solution: From the given waveforms, the truth table is as
Spring Pendulum] follows.
Q63. When a proton is accelerated through 1 V, then its
kinetic energy will be
(a) 1840 eV (b) 13.6 eV
(c) 1 eV (d) 0.54 eV
Ans: (c)
Solution: Potential difference (V) = 1V,
K.E. acquired = qV
= 1.6 × 10–19 × 1 The above truth table is for NAND gate.
= 1.6 × 10–19joules= 1 eV Therefore, the logic gate is NAND gate.
Chapter: Moving Charges and Magnetic Field Chapter: Semiconductor Electronics Materials, Devices
[Topic: Digital Electronics and Logic Gates]

63
Q66. The dimensional formula of pressure is (a) 2 V (b) 3 V
(a) [MLT–2] (c) 5 V (d) 1 V
(b) [ML–1T2] Ans: (a)
(c) [ML–1T–2] Solution: K.E. = hν – hνth = eV0 (V0 = cut off voltage)
(d) [MLT2]
( )
Ans: (c)
Solution: [Pressure] = [Force] / [Area] =
= ML–1T–2 Chapter - Dual Nature of Radiation and Matter
Chapter: Units and Measurement [Topic: Electron Emission, Photon Photoelectric Effect
[Topic: Dimensions of Physical Quantities] & X-ray]
Q67. The moment of inertia of a thin uniform rod of Q71. A particle has initial velocity( ̂ )̂ and has
mass M and length L about an axis passing through its acceleration ( ̂ )̂ . It's speed after 10 s is:
midpoint and perpendicular to its length is I0. Its moment (a) 7 units (b) √ units
of inertia about an axis passing through one of its ends (c) 8.5 units (d) 10 units
and perpendicular to its length is Ans: (b)
(a) I0 + ML2/2 (b) I0 + ML2/4 Solution: ⃗ ̂ ̂⃗ ̂ ̂
2
(c) I0 + 2ML (d) I0 + ML2 ⇒ux = 3units, uy= 4 units
Ans: (b) ax = 0.4 units, ay = 0.3 units
Solution: By theorem of parallel axes, ∴ = 3 + 4 = 7 ms–1
I = Icm + Md2 and = 4 + 3 = 7 ms–1
I = I0 + M (L/2)2 = I0 + ML2/4
Chapter: System of Particles and Rotational Motion ∴v=√ = √ ms–1
[Topic: Moment of Inertia, Rotational K.E. and Power]
Chapter: Kinematics Motion in a Straight Line
Q68. Which one of the following statements is true ?
[Topic: Non-uniform motion]
(a) The sound waves in air are longitudinal while the
light waves are transverse Q72. A solid sphere, disc and solid cylinder all of the
(b) Both light and sound waves in air are longitudinal same mass and made of the same material are allowed to
(c) Both light and sound waves can travel in vacuum roll down (from rest) on the inclined plane, then
(d) Both light and sound waves in air are transverse (a) solid sphere reaches the bottom first
Ans: (a) (b) solid sphere reaches the bottom last
Solution: Sound waves in air are longitudinal and the (c) disc will reach the bottom first
light waves are transverse. (d) all reach the bottom at the same time
Chapter: Waves Ans: (a)
[Topic: Basic of Waves] Solution: For solid sphere,
Q69. The magnetic field ( ⃗ ) due to a small element For disc and solid cylinder,
( ) at a distance ( ) and element carrying current i is
⃗ As for solid sphere is smallest, it takes minimum time
(a) ⃗ ( )
to reach the bottom of the incline
⃗ ⃗
(b) ⃗ ( ) Chapter: System of Particles and Rotational Motion
⃗ ⃗
[Topic: Rolling Motion]
(c) ⃗ ( ) Q73. Velocity of sound waves in air is 330 m/s. For a
⃗
particular sound wave in air, a path difference of 40 cm is
(d) ⃗ ( ) equivalent to phase difference of 1.6 π. The frequency of
Ans: (d) this wave is
Solution: According to Biot Savart law, (a) 165 Hz (b) 150 Hz
(c) 660 Hz (d) 330 Hz
( ) Ans: (c)
⃗
Chapter: Moving Charges and Magnetic Field Solution: From
[Topic: Magnetic Field, Biot-Savart's Law & Ampere's ( )
Circuital Law]
Q70. The threshold frequency for a photosensitive metal Hz
is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is
Chapter: Waves
incident on this metal, the cut-off voltage for the
[Topic: Basic of Waves]
photoelectric emission is nearly
[2011M]

64
Q74. A circuit contains an ammeter, a battery of 30V Solution: Let t1 & t2 be the time taken by A and B
and a resistance 40.8Ω all connected in series. If the respectively to reach the ground then from the formula,
ammeter has a coil of resistance 480Ω and a shunt of
20Ω, the reading in the ammeter will be:
[2015 RS] For first body,
(a) 0.25 A (b) 2A
(c) 1 A (d) 0.5 A For second body,
Ans: (d) ∴ ⇒ .
Solution: From circuit diagram
Chapter: Kinematics Motion in a Straight Line
[Topic: Motion Under Gravity]
Q77. Two sound waves with wavelengths 5.0 m and
5.5m respectively, each propagate in a gas with velocity
330 m/s. We expect the following number of beats per
second
(a) 0 (b) 1
(c) 6 (d) 12
Ans: (c)
Solution: Frequencies of sound waves are &
i.e., 66 Hz and 60 Hz
Resistance of ammeter = = 19.2Ω. Frequencies of beat = 66 – 60 = 6 per second
Total resistance R = 40.8 + 19.2 = 60Ω Chapter: Waves
[Topic: Beats, Interference & Superposition of Waves]
Reading in the ammeter i =
Q78. There are four light–weight–rod samples A,B,C,D
= = 0.5A separately suspended by threads. A bar magnet is slowly
Chapter: Moving Charges and Magnetic Field brought near each sample and the following observations
[Topic: Galvanometer and Its Conversion into Ammeter are noted
& Voltmeter] (i) A is feebly repelled
Q75. An electron of mass m and charge e is accelerated (ii)B is feebly attracted
from rest through a potential difference of V volt in (iii)C is strongly attracted
vacuum. Its final speed will be (iv)D remains unaffected
Which one of the following is true ?
(a) (a) B is of a paramagnetic material
(b) (b) C is of a diamagnetic material
(c) D is of a ferromagnetic material
(c) √ (d) A is of a non–magnetic material
Ans: (a)
(d) √ Solution: A → diamagnetic
B → paramagnetic
Ans: (c) C → Ferromagnetic
Solution: Kinetic energy of electron accelerated through D → Non magnetic
a potential V= eV Chapter: Magnetism and Matter
⇒ [Topic: The Earth's Magnetism, Magnetic Materials
and their Properties]
⇒ √ Q79. The ratio of wavelengths of the last line of Balmer
Chapter - Dual Nature of Radiation and Matter series and the last line of Lyman series is :-
[Topic: Electron Emission, Photon Photoelectric Effect (a) 1 (b) 4
& X-ray] (c) 0.5 (d) 2
Q76. Two bodies, A (of mass 1 kg) and B (of mass 3 Ans: (b)
kg), are dropped from heights of 16m and 25m, Solution: For last line of Balmer series : n1 = 2 and n2 =
respectively. The ratio of the time taken by them to reach ∞
the ground is
6 7 [ ]
(a) 12/5 (b) 5/12
(c) 4/5 (d) 5/4 ... (i)
Ans: (c)
For last line of Lyman series : n1 = 1 and n2 = ∞

65
 For this, force between charge at A and B + force
6 7 [ ] between charge at point O and either at A or B is zero.
... (ii) i.e.,
. /
Dividing equation (i) by (ii) By solving we get,
q=
Chapter: Electrostatic Potential and capacitance
[Topic: Electric Field, Electric Field Lines & Dipole]
Ratio of wavelengths is Q83. A circular disc of radius 0.2 meter is placed in a
Chapter: Atoms uniform magnetic field of induction (Wb/ ) in such a
[Topic: Bohr Model & The Spectra of the Hydrogen
way that its axis makes an angle of 60° with ⃗ .The
Atom] magnetic flux linked with the disc is:
Q80. The angle between two vectors of magnitude 12 (a) 0.02 Wb (b) 0.06 Wb
and 18 units when their resultant is 24 units, is (c) 0.08 Wb (d) 0.01 Wb
(a) 63º 51´ (b) 75º 52´ Ans: (a)
(c) 82º 31´ (d) 89º 16´
Ans: (b) Solution: Here, (Wb/m2)
Solution: We know that, θ = 60°
( ) ( ) ( ) ( )( ) Area normal to the plane of the disc
⇒ ‟ =
Chapter: Kinematics Motion in a Plane Flux = B × normal area
[Topic: Vectors] =
Q81. The escape velocity of a body on the surface of the Chapter: Electromagnetic
earth is 11.2 km/s. If the earth‟s mass increases to twice [Topic: Magnetic Flux, Faraday's & Lenz's Law]
its present value and the radius of the earth becomes half, Q84. Which of the following transitions in a hydrogen
the escape velocity would become atom emits the photon of highest frequency?
(a) 44.8 km/s (b) 22.4 km/s [2000]
(c) 11.2 km/s (remains unchanged) (a) n = 2 to n = 1 (b) n = 2 to n = 6
(d) 5.6 km/s (c) n = 6 to n = 2 (d) n = 1 to n = 2
Ans: (b) Ans: (a)
Solution: Escape velcocity
Solution: 0 1 or 0 1
√ √ Frequency, 0 1
Note : See the greatest energy difference and also see that
∴ √ the transition is from higher to lower energy level. Hence,
it is highest in case of n = 2 to n = 1.
Given and Chapter: Atoms
[Topic: Bohr Model & The Spectra of the Hydrogen
∴ √ √
Atom]
= 2×11.2 = 22.4 km/s Q85. The maximum range of a gun of horizontal terrain
Chapter: Gravitation is 16 km. If g = 10 ms–2, then muzzle velocity of a shell
[Topic: Motion of Satellites, Escape Speed and Orbital must be
Velocity] (a) 160 ms–1 (b) 200√
Q82. A charge „q‟ is placed at the centre of the line (c) 400 ms–1 (d) 800 ms–1
joining two equal charges „Q‟. The system of the three Ans: (c)
charges will be in equilibrium if „q‟ is equal to Solution: =16000[16km=16000m]
(a) Q/2 (b) – Q/4
(c) Q/4 (d) – Q/2 or ( ) ( )
Ans: (b) = 400 ms –1
Solution: The system of three charges will be in Chapter: Kinematics Motion in a Plane
equilibrium. [Topic: Projectile Motion]
Q86. The terminal velocity vr of a small steel ball of
radius r falling under gravity through a column of a
viscous liquid of coefficient of viscosity η depends on

66
mass of the ball m, acceleration due to gravity g, (c) 10√ kg (d) 2√ kg
coefficient of viscosity η and radius r. Which of the Ans: (c)
following relations is dimensionally correct ? Solution: ⃗ ̂ ̂
(a) | | √ √
(b)
(∵ √ )
(c)
–2
a = 1 ms
(d)
∵F=ma
Ans: (c) √
Solution: Note that according to Stoke's law ∴ √ kg
Chapter: Dynamics Laws of Motion
Hence, the valid relation is [Topic: Ist, IInd & IIIrd Laws of Motion]
Q91. A cylindrical metallic rod in therrnal contact with
Chapter: Mechanical Properties of Fluids two reservoirs of heat at its two ends conducts an amount
[Topic: Surface Tension, Surface Energy & Capillarity] of heat Q in time t. The metallic rod is melted and the
Q87. A charge Q is enclosed by a Gaussian spherical material is formed into a rod of half the radius of the
surface of radius R. If the radius is doubled, then the original rod. What is the amount of heat conducted by the
outward electric flux will new rod, when placed in thermal contact with the two
(a) increase four times reservoirs in time t?
(b) be reduced to half
(a)
(c) remain the same
(d) be doubled (b)
Ans: (c) (c) 2 Q
Solution: By Gauss‟s theorem,
(d)
θ=
Thus, the net flux depends only on the charge enclosed Ans: (b)
by the surface. Hence, there will be no effect on the net Solution: The rate of heat flow is given by
flux if the radius of the surface is doubled. = K. A.
Chapter: Electrostatic Potential and capacitance Area of Original rod A = ;
[Topic: Electric Flux & Gauss's Law]
Q88. In an ac circuit an alternating voltage e = 200 Areal of new rod A′ = .
√ sin 100 t volts is connected to a capacitor of capacity Volume of original rod will be equal to the volume of
1 µF. The r.m.s. value of the current in the circuit is new rod.
(a) 10 mA (b) 100 mA ∴ = . / ‟
(c) 200 mA (d) 20 mA
Ans: (d) ⇒ =
. /
√
Solution: Vrms =
√ = =
Irms = Q' =
= 2 × 10–2 = 20mA Chapter: Thermal Properties
Chapter: Alternating Current [Topic: Calorimetry & Heat Transfer]
[Topic: Alternating Current, Voltage & Power] Q92. An electric dipole has the magnitude of its charge
Q89. The mass number of a nucleus is equal to the as q and its dipole moment is p. It is placed in uniform
number of electric field E. If its dipole moment is along the direction
(a) protons it contains of the field, the force on it and its potential energy are
(b) nucleons it contains respectively
(c) neutrons it contains (a) q.E and max. (b) 2 q.E and min.
(d) electron it contains (c) q.E and p.E (d) zero and min.
Ans: (b) Ans: (d)
Solution: Chapter: Nuclei Solution: When the dipole is in the direction of field then
[Topic: Composition and Size of the Nucleus] net force is qE + (–qE) = 0
Q90. A body under the action of a force⃗ ̂
̂ acquires an acceleration of 1 m/s2. The mass of this
body must be
(a) 10 kg (b) 20 kg and its potential energy is minimum
= – P.E. = –qaE

67
 Chapter: Electrostatic Potential and capacitance between system and surrounding.
[Topic: Capacitors, Capacitance, Grouping of As the temperature difference is halved, so the rate of
Capacitors & Energy Stored in a Capacitor.] cooling will also be halved.
Q93. An LCR series circuit is connected to a source of So time taken will be doubled
alternating current. At resonance, the applied voltage and t = 2 × 5 sec. = 10 sec.
the current flowing through the circuit will have a phase Chapter: Thermal Properties
difference of [Topic: First Law of Thermodynamics]
(a) π (b) Q97. The four capacitors, each of 25µ F are connected as
shown in fig. The dc voltmeter reads 200 V. The charge
(c) (d) 0 on each plate of capacitor is
Ans: (d)
Solution: At resonance, . The circuit behaves as
if it contains R only. So, phase difference = 0
At resonance, impedance is minimum Zmin = R and
current is maximum, given by

(a)
It is interesting to note that before resonance the current
leads the applied emf, at resonance it is in phase, and (b)
after resonance it lags behind the emf. LCR series circuit (c)
is also called as acceptor circuit and parallel LCR circuit (d)
is called rejector circuit. Ans: (b)
Chapter: Alternating Current Solution: Charge on each plate of each capacitor
[Topic: A.C. Circuit, LCR Circuit, Quality & Power
Factor]
Q94. Solar energy is mainly caused due to Chapter: Electrostatic Potential and capacitance
(a) gravitational contraction [Topic: Electric Current, Drift of Electrons, Ohm's
(b) burning of hydrogen in the oxygen Law, Resistance & Resistivity]
(c) fission of uranium present in the Sun Q98. The electromagnetic radiations are caused by
(d) fusion of protons during synthesis of heavier elements (a) a stationary charge
Ans: (d) (b) uniformly moving charges
Solution: As a result of fusion, enormous amount of heat (c) accelerated charges
is liberated which is the main cause of source of solar (d) all of above
energy. Ans: (c)
Chapter: Nuclei Solution: A stationary charge produces electric field
[Topic: Mass-Energy & Nuclear Reactions] only; an uniformly moving charge produces localised
Q95. A man weighing 80 kg, stands on a weighing scale electromagnetic field; an accelerated charge produces
in a lift which is moving upwards with a uniform electromagnetic radiations.
acceleration of 5m/s2. What would be the reading on the Chapter - Electromagnetic Waves
scale ? (g = 10 m/s2) [Topic: Electromagnetic Waves, Conduction &
(a) 1200 N (b) zero Displacement Current]
(c) 400 N (d) 800 N Q99. A nucleus emits one α-particle and two β-
Ans: (a) particles. The resulting nucleus is
Solution: Reading of the scale , -
(a) , -
= Apparent wt. of the man = m(g + a)
(b)
= 80 (10 + 5) = 1200 N
(c)
Chapter: Dynamics Laws of Motion , -
[Topic: Motion of Connected Bodies, Pulleys] (d) , -
Q96. A body cools from 50.0°C to 48°C in 5s. How long Ans: (c)
will it take to cool from 40.0°C to 39°C? Assume the Solution: When emits one α-particle then its atomic
temperature of surroundings to be 30.0°C and Newton's mass decreases by 4 units and atomic number by 2.
, -
law of cooling to be valid. Therefore, the new nucleus becomes , - . But as it
(a) 2.5 s (b) 10 s emits two β– particles, its atomic number increases by 2.
(c) 20 s (d) 5 s Thus the resulting nucleus is
Ans: (b) Chapter: Nuclei
Solution: Rate of cooling temperature difference [Topic: Radioactivity]

68
Q100. A 500 kg car takes a round turn of radius 50 m Q3. Which of the following is not due to total internal
with a velocity of 36 km/h. The centripetal force is reflection?
(a) 250 N (b) 750 N (a) Working of optical fibre
(c) 1000 N (d) 1200 N (b) Difference between apparent and real depth of pond
Ans: (c) (c) Mirage on hot summer days
Solution: Centripetal force =
( ) (d) Brilliance of diamond
Ans: (b)
= 1000 N[ Solution: Difference between apparent and real depth of
36 km/hr = 10 m/s] a pond is due to the refraction of light, not due to the total
Chapter: Dynamics Laws of Motion internal reflection. Other three phenomena are due to the
[Topic: Circular Motion, Banking of Road] total internal reflection.
PART 9. PHYSICS Chapter - Ray Optics and Optical
[Topic: Refraction of Light at Plane Surface & Total

QUESTION BANK Q4. In a given reaction

Internal Reflection]

, - , -
Radioactive radiations are emitted in the sequence of
Q1. If the ratio of specific heat of a gas at constant
(a) α, β, γ
pressure to that at constant volume is γ, the change in
(b) γ, α, β
internal energy of a mass of gas, when the volume
(c) β, α, γ
changes from V to 2V at constant pressure P, is
(d) γ, β, α
(a) ( ) Ans: (c)
(b) PV Solution: Increase of charge number by 1 indicates β
(c) ( emission. Decrease of mass number by 4 and charge
) number by 2 indicates α emission. No change of mass
(d) ( )
number and charge number indicates γ-emission.
Ans: (c) Chapter: Nuclei
Solution: Change in internal energy is equal to work [Topic: Radioactivity]
done in adiabatic system Q5. A ball of mass 2 kg and another of mass 4 kg are
∆W = –∆U (Expansion in the system) dropped together from a 60 feet tall building. After a fall
= ( )( ) of 30 feet each towards earth, their respective kinetic
energies will be in the ratio of
( ) (a) 1 : √2 (b) √2 : 1
( ) (c) 1 : 4 (d) 1 : 2
Here, Ans: (d)
Solution: Since height is same for both balls, their
, -
velocities on reaching the ground will be same

Chapter: Heat & Thermodynamics

[Topic: Specific Heat Capacity & Thermodynamic Chapter: Work, Energy and Power
Processes] [Topic: Energy]
Q2. Resistances n, each of r ohm, when connected in Q6. At 10° C the value of the density of a fixed mass of
parallel give an equivalent resistance of R ohm. If these an ideal gas divided by its pressure is x. At 110°C this
resistances were connected in series, the combination ratio is:
would have a resistance in ohms, equal to (a) x
(a) nR
(b)
(b) n2R
(c) R/n2 (d) R/n (c)
Ans: (b)
(d)
Solution: ⇒
Ans: (d)
When connected in series, Req = nr Solution: Let the mass of the gas be m.
= n (nR) = n2R At a fixed temperature and pressure, volume is fixed.
Chapter: Current Electricity
Density of the gas,
[Topic: Combination of Resistances]

69
Now Chapter: Semiconductor Electronics Materials, Devices
[Topic: Solids, Semiconductors and P-N Junction
⇒ = x (By question)
Diode]
⇒ xT = constant ⇒ x1T1 = x2T2 Q10. A body of mass (4m) is lying in x-y plane at rest. It
⇒ x2 ⇒ = { } suddenly explodes into three pieces. Two pieces, each of
mass (m) move perpendicular to each other with equal
Chapter: Kinetic Theory speeds (v) . The total kinetic energy generated due to
[Topic: Kinetic Theory of an Ideal Gas & Gas Laws] explosion is :
Q7. In producing chlorine through electrolysis, 100 watt (a) mv2 (b)
power at 125 V is being consumed. How much chlorine 2
per minute is liberated? E.C.E. of chlorine is 0.367 × 10– (c) 2 mv (d) 4 mv2
6
kg/ coulomb. Ans: (b)
(a) 21.3 mg (b) 24.3 mg Solution: By conservation of linear momentum
(c) 13.6 mg (d) 17.6 mg 2mv1 = √ ⇒
√
Ans: (d) As two masses of each of mass m move perpendicular to
Solution: Power = V × I each other.
Total KE generated
= ( )
E.C.E. of chlorine is 0.367 × 10–6 kg/coulomb
Charge passing in one minute =
= 48 coulomb Chapter: Work, Energy and Power
Chlorine precipitated = 0.367 × 10–6 × 48 [Topic: Collisions]
= 17.6 × 10–6 kg Q11. The degree of freedom of a molecule of a triatomic
= 17.6 mg gas is
Chapter: Current Electricity (a) 2 (b) 4
[Topic: Kirchhoff's Laws, Cells, Thermo emf & (c) 6 (d) 8
Electrolysis] Ans: (c)
Q8. A boy is trying to start a fire by focusing sunlight on Solution: No. of degree of freedom = 3 K – N
a piece of paper using an equiconvex lens of focal length where K is no. of atom and N is the number of relations
10 cm. The diameter of the Sun is 1.39 ×109 m and its between atoms. For triatomic gas,
mean distance from the earth is 1.5 × 1011 m. What is the K = 3, N =
diameter of the Sun‟s image on the paper? No. of degree of freedom = 3 (3) – 3 = 6
(a) 9.2 × 10–4 m (b) 6.5 × 10–4m Chapter: Kinetic Theory
(c) 6.5 × 10–5 m (d) 12.4 × 10–4 m [Topic: Degree of Freedom, Specific Heat Capacity &
Ans: (a) Mean Free Path]
Solution: We have, | | Q12. Two 220 volt, 100 watt bulbs are connected first in
series and then in parallel. Each time the combination is
or, Size of image = | | Size of object connected to a 220 volt a.c. supply line. The power drawn
by the combination in each case respectively will be
=. / ( )
[2003]
= 0.92 ×10–3 m = 9.2 ×10–4 m (a) 50 watt, 200 watt (b) 50 watt, 100 watt
∴ Diameter of the sun‟s image = 9.2 × 10–4 m. (c) 100 watt, 50 watt (d) 200 watt, 150 watt
Chapter - Ray Optics and Optical Ans: (a)
[Topic: Refraction at Curved Surface, Lenses & Power Solution: Power
of Lens]
In series combination, resistance doubles.
Q9. Which one of the following bonds produces a solid Hence, power will be halved.
that reflects light in the visible region and whose In parallel combination, resistance halves.
electrical conductivity decreases with temperature and Hence, power will be doubled.
has high melting point?
Chapter: Current Electricity
(a) metallic bonding
[Topic: Heating Effects of Current]
(b) van der Waal‟s bonding
(c) ionic bonding Q13. The refractive index of the material of the prism is
(d) covalent bonding √ ; then the angle of minimum deviation of the prism is
Ans: (a) (a) 30º (b) 45º
Solution: For a metal, conductivity decreases with (c) 60º (d) 75º
increase in temperature. Ans: (c)
Also, metal has high melting point. Solution: Angle of minimum deviation

70
 ( ) Q16. A particle is subjected to two mutually per-
. /
√ pendicular simple harmonic motions such that its x and y
. / . / coordinates are given by
x = 2 sin ωt; . /
√
( ) ⇒ The path of the particle will be
⇒ δm = 60°. (a) a straight line
(b) a circle
Chapter - Ray Optics and Optical
(c) an ellipse
[Topic: Prism & Dispersion of Light]
(d) a parabola
Q14. The cause of the potential barrier in a p-n diode is Ans: (c)
(a) depletion of positive charges near the junction
(b) concentration of positive charges near the junction Solution: As phase difference the resultant path of
(c) depletion of negative charges near the junction particle is an ellipse.
(d) concentration of positive and negative charges near Chapter: Oscillation
the junction [Topic: Energy in Simple Harmonic Motion]
Ans: (d) Q17. Five equal resistances each of resistance R are
Solution: During the formation of a junction diode, holes connected as shown in the figure. A battery of V volts is
from p-region diffuse into n-region and electrons from n- connected between A and B. The current flowing in
region diffuse into p-region. In both cases, when an AFCEB will be
electron meets a hole, they cancel the effect of each other
and as a result, a thin layer at the junction becomes free
from any of charge carriers. This is called depletion layer.
There is a potential gradient in the depletion layer,
negative on the p-side and positive on the n-side. The
potential difference thus developed across the junction is
called potential barrier.
Chapter: Semiconductor Electronics Materials, Devices
[Topic: Solids, Semiconductors and P-N Junction
Diode] (a)
Q15. If the linear density (mass per unit length) of a rod
of length 3m is proportional to x, where x is the distance (b)
from one end of the rod, the distance of the centre of (c)
gravity of the rod from this end is
(a) 2.5 m (b) 1 m (d)
(c) 1.5 m (d) 2 m Ans: (d)
Ans: (d) Solution: A balanced Wheststone‟s bridge exists between
Solution: Consider an element of length dx at a distance A & B.
x from end A. ∴ Req = R
Here, mass per unit length λ of rod Current through circuit = V/R
Current through AFCEB = V/2R
∴ dm = λdx = kx dx Chapter: Current Electricity
[Topic: Wheatstone Bridge & Different Measuring
Instruments]
Q18. If yellow light emitted by sodium lamp in Young‟s
Position of centre of gravity of rod from end A. double slit experiment is replaced by a monochromatic
∫ blue light of the same intensity
[1992]
∫ (a) fringe width will decrease
( ) (b) finge width will increase
( ) [ ]
∫ (c) fringe width will remain unchanged
∴ (d) fringes will become less intense
∫ ( )
[ ] Ans: (a)
Chapter: System of Particles and Rotational Motion Solution: As and ,
[Topic: Centre of Mass, Centre of Gravity & Principle ∴ Fringe width β will decrease
of Moments] Chapter - Wave Optics
[Topic: Young's Double Slit Experiment]

71
Q19. A transistor is operated in common-emitter (Moment of inertia of a thin circular ring about an axis
configuration at Vc = 2 V such that a change in the base vertical to its plane = MR2)
current from 100 µA to 200 µA produces a change in the ∴ .
collector current from 5 mA to 10 mA. The current gain
is Chapter: System of Particles and Rotational Motion
(a) 100 (b) 150 [Topic: Torque, Couple and Angular Momentum]
(c) 50 (d) 75 Q22. A body of mass M, executes vertical SHM with
Ans: (c) periods t1 and t2, when separately attached to spring A and
Solution: ∆ IE = ∆IB + ∆IC spring B respectively. The period of SHM, when the
body executes SHM, as shown in the figure is t0. Then

( )

( )
(a) t0–1 = t1–1 + t2–1 (b) t0 = t1 + t2
(c) t02 = t12 + t22 (d) t0–2 = t1–2 + t2–2
Ans: (d)
Chapter: Semiconductor Electronics Materials, Devices
[Topic: Junction Transistor] Solution: √
Q20. If dimensions of critical velocity υc of a liquid ⇒k=Const.
flowing through a tube are expressed as , - , where Here the springs are joined in parallel. So
η, ρ and r are the coefficient of viscosity of liquid,
density of liquid and radius of the tube respectively, then where k0 is resultant force constant
the values of x, y and z are given by : ∴ Const. Const. +Const.
(a) – 1, – 1, 1 (b) – 1, – 1, – 1 or,
(c) 1, 1, 1 (d) 1, – 1, – 1 Chapter: Oscillation
Ans: (d) [Topic: Time Period, Frequency, Simple Pendulum &
Solution: Applying dimensional method : Spring Pendulum]
vc = ηxρyrz Q23. A positively charged particle moving due east
[M0LT–1] = [ML–1T–1]x [ML–3T0]y [M0LT0]z enters a region of uniform magnetic field directed
Equating powers both sides vertically upwards. The particle will
x + y = 0; – x = – 1 ∴ x = 1 (a) continue to move due east
1+y=0∴y=–1 (b) move in a circular orbit with its speed unchanged
– x – 3y + z = 1 (c) move in a circular orbit with its speed increased
– 1 – 3(– 1) + z = 1 (d) gets deflected vertically upwards.
–1+3+z=1 Ans: (b)
∴z=–1 Solution: In a perpendicular magnetic field, the path of a
Chapter: Units and Measurement charged particle is a circle, and the magnetic field does
[Topic: Dimensions of Physical Quantities] not cause any change in energy.
Q21. A thin circular ring of mass M and radius R is Chapter: Moving Charges and Magnetic Field
rotating in a horizontal plane about an axis vertical to its [Topic: Motion of Charged Particle in Magnetic Field
plane with a constant angular velocity ω. If two objects & Moment]
each of mass m be attached gently to the opposite ends of Q24. In the Davisson and Germer experiment, the
a diameter of the ring, the ring will then rotate with an velocity of electrons emitted from the electron gun can be
angular velocity: increased by
(a) (a) increasing the potential difference between the anode
( ) and filament
(b) (b) increasing the filament current
(c) (c) decreasing the filament current
( ) (d) decreasing the potential difference between the anode
(d) and filament
Ans: (a) Ans: (a)
Solution: In absence of external torque, L = Iω = Solution: In the Davisson and Germer experiment, the
constant velocity of electrons emitted from the electron gun can be
increased by increasing the potential difference between
the anode and filament.

72
 Chapter - Dual Nature of Radiation and Matter In 2nd case, K.E. =
[Topic: Matter Waves, Cathode & Positive Rays] ( )

Q25. The circuit But r′ =

= ⇒ K.E.′ = 4 K.E.

∴ K.E. is increased by a factor of 4.

Chapter: System of Particles and Rotational Motion
is equivalent to [Topic: Moment of Inertia, Rotational K.E. and Power]
(a) AND gate
Q28. A transverse wave propagating along x-axis is
(b) NAND gate
(c) NOR gate represented by ( ) . / where
(d) OR gate x is in metres and t is in seconds. The speed of the wave
Ans: (c) is
Solution: Let A and B be inputs and Y the output. (a) 0.5 π m/s (b) m/s
(c) 8 m/s (d) 4π m/s
Ans: (c)
Solution: Speed of a wave represented by the equation
Then ( )
y(x, t) = A sin (kx – ωt + θ) is
(By De-Morgan‟s theorem)
By comparison, ω = 4π; k = 0.5π
(̿̿̿̿̿̿) ̿ ̿
(̅̅̅̅̅̅̅) 8m/sec
∴
Hence, the given circuit is equivalent to a NOR gate. Chapter: Waves
Chapter: Semiconductor Electronics Materials, Devices [Topic: Basic of Waves]
[Topic: Digital Electronics and Logic Gates] Q29. Two equal electric currents are flowing
Q26. The dimensional formula of torque is perpendicular to each other as shown in the figure. AB
(a) [ML2T–2] and CD are perpendicular to each other and
(b) [MLT–2] symmetrically placed with respect to the current flow.
(c) [ML–1T–2] Where do we expect the resultant magnetic field to be
(d) [ML–2T–2] zero?
Ans: (a)
Solution: [Force distance]
= [MLT–2] [L] = ML2T–2
Chapter: Units and Measurement
[Topic: Dimensions of Physical Quantities]
Q27. A small mass attached to a string rotates on
frictionless table top as shown. If the tension in the string
is increased by pulling the string causing the radius of the
circular motion to decrease by a factor of 2, the kinetic
energy of the mass will (a) on AB
(b) on CD
(c) on both AB and CD
(d) on both OD and BO
Ans: (a)
Solution: Net magnetic field on AB is zero because
(a) remain constant
magnetic field due to both current carrying wires is equal
(b) increase by a factor of 2
in magnitude but opposite in direction.
(c) increase by a factor of 4 (d) decrease by a factor of
Chapter: Moving Charges and Magnetic Field
2
[Topic: Magnetic Field, Biot-Savart's Law & Ampere's
Ans: (c)
Circuital Law]
Solution: K.E. = Q30. A source S1 is producing, 1015 photons per second
The angular momentum L remains conserved about the of wavelength 5000 Å. Another source S2 is producing
centre. 15
1.02×10 photons per second of wavelength 5100Å
That is, L = constant. Then, (power of S2) (power of S1) is equal to :
I = mr2 (a) 1.00 (b) 1.02
∴ K.E. = (c) 1.04 (d) 0.98
Ans: (a)

73
Solution: Energy emitted/sec by
Energy emitted/sec by Speed of propagation,
∴

Chapter: Waves
Chapter - Dual Nature of Radiation and Matter [Topic: Vibration of String & Organ Pipe]
[Topic: Electron Emission, Photon Photoelectric Effect
Q34. In an ammeter 0.2% of main current passes
& X-ray]
through the galvanometer. If resistance of galvanometer
Q31. A particle moves a distance x in time t according to is G, the resistance of ammeter will be :
equation x = (t + 5)–1. The acceleration of particle is
proportional to: (a)
(a) (velocity) 3/2 (b) (distance)2 (b)
(c) (distance)–2 (d) (velocity)2/3
Ans: (a) (c)
Solution: (d)
∴v = Ans: (c)
( ) Solution: As 0.2% of main current passes through the
∴a = =( )
= 2x3 galvanometer hence current through the shunt.
Now ( )

∴( )
Chapter: Kinematics Motion in a Straight Line
[Topic: Non-uniform motion]
Q32. The speed of a homogenous solid sphere after
rolling down an inclined plane of vertical height h from
rest without sliding is . / . / ⇒S=
(a) √ Total resistance of Ammeter
. /
(b) √ R=
. /

(c) √ Chapter: Moving Charges and Magnetic Field

[Topic: Galvanometer and Its Conversion into Ammeter
& Voltmeter]
(d) √ Solution: (a) P.E. = total K.E.
Q35. The nature of ions knocked out from hot surfaces is
(see sol. of Q.9) (a) Protons
(b) Neutrons
(c) Electrons
√ (d) Nuclei
Ans: (c)
Chapter: System of Particles and Rotational Motion Solution: Chapter - Dual Nature of Radiation and
[Topic: Rolling Motion] Matter
Q33. The transverse wave represented by the equation [Topic: Electron Emission, Photon Photoelectric Effect
. / ( ) has & X-ray]
(a) amplitude = 4 (b) wavelength = Q36. A ball is thrown vertically upward. It has a speed
of 10 m/sec when it has reached one half of its maximum
(c) speed of propagation = 5 (d) period height. How high does the ball rise?
Ans: (c) Take g = 10 m/s2.
Solution: Compare the given equation with standard (a) 10 m (b) 5 m
form (c) 15 m (d) 20 m
Ans: (a)
[ ] Solution: For part AB
and From 3rd equation of motion
v2 = u2 – 2gH

74
 (c) low retentivity and low coercive force
(d) high retentivity and low coercive force
Ans: (d)
Solution: Soft iron has high retentivity and low coercive
force.
Chapter: Magnetism and Matter
[Topic: The Earth's Magnetism, Magnetic Materials
and their Properties]
Q40. Given the value of Rydberg constant is 107m–1, the
wave number of the last line of the Balmer series in
0 = u2 – 2g(H/2) = u2 – gH hydrogen spectrum will be :
H= (a) 0.025 × 104 m–1 (b) 0.5 × 107 m–1
7 –1
(c) 0.25 × 10 m (d) 2.5 × 107 m–1
Chapter: Kinematics Motion in a Straight Line Ans: (c)
[Topic: Motion Under Gravity] Solution: According to Bohr's theory, the wave number
Q37. A body of mass „m‟ is taken from the earth‟s of the last line of the Balmer series in hydrogen spectrum,
surface to the height equal to twice the radius (R) of the For hydrogen atom z = 1
earth. The change in potential energy of body will be
4 5
(a) mgR (b) 3 mgR
(c) mgR (d) mg2R = 107 × 12 . /
Ans: (a) ⇒wave number = 0.25 × 107 m–1
Solution: Initial P. E., Ui = , Chapter: Atoms
Final P.E., Uf = [∵ R' = R + 2R = 3R] [Topic: Bohr Model & The Spectra of the Hydrogen
Atom]
∴ Change in potential energy, ̂,
Q41. If a unit vector is represented by ̂
∆U = + the value of c is
= . /= = mgR. / (a) 1 (b) √
(c) √ (d) 0.39
Alternate : ∆U =
Ans: (b)
By placing the value of h = 2R we get Solution: ̂ ̂ ̂
∆U = mgR. |̂| √( ) ( )
Chapter: Gravitation ( ) ( )
[Topic: Gravitational Field, Potential and Energy] ⇒ √
Q38. Two vibrating tuning forks produce progressive Chapter: Kinematics Motion in a Plane
waves given by y1 = 4 sin 500 πt and y2 = 2 sin 506 πt. [Topic: Vectors]
Number of beats produced per minute is Q42. A ball is dropped from a satellite revolving around
(a) 360 (b) 180 the earth at a height of 120 km. The ball will
(c) 60 (d) 3 [1996]
Ans: (b) (a) continue to move with same speed along a straight
Solution: Equation of progressive wave is given by line tangentially to the satellite at that time
y = A sin2πf t (b) continue to move with the same speed along the
Given y1 = 4sin500 πt and y2 = 2sin506πt. original orbit of satellite
Comparing the given equations with equation of (c) fall down to earth gradually
progressive wave, we get (d) go far away in space
2f1 = 500 f1 = 250 Ans: (b)
2f2 = 506 f2 = 253 Solution: The orbital speed of satellite is independent of
Beats = f2– f1 = 253 – 250 = 3 beats/sec mass of satellite, so the ball will behave as a satellite and
= 3 × 60 = 180 beats/minute. will continue to move with the same speed in the original
Chapter: Waves orbit.
[Topic: Beats, Interference & Superposition of Waves] Chapter: Gravitation
Q39. Electromagnets are made of soft iron because soft [Topic: Motion of Satellites, Escape Speed and Orbital
iron has Velocity]
(a) low retentivity and high coercive force Q43. Two metallic spheres of radii 1 cm and 3 cm are
(b) high retentivity and high coercive force given charges of –1×10–2 C and 5×10–2 C, respectively. If

75
these are connected by a conducting wire, the final (a) 1 : 1 (b) 1 : 2
charge on the bigger sphere is : (c) 1 : 3 (d) 2:√
(a) 2 × 10–2 C (b) 3 × 10–2 C Ans: (a)
–2
(c) 4 × 10 C (d) 1 × 10–2 C Solution: Horizontal range is same when angle of
Ans: (b) projection is θ or (90° – θ).
Solution: At equilibrium potential of both sphere Chapter: Kinematics Motion in a Plane
becomes same if charge of sphere one x and other sphere [Topic: Relative Velocity in2D & Circular Motion]
Q – x then Q47. Water rises to a height 'h' in a capillary tube. If the
where Q = 4 × 10–2 C length of capaillary tube above the surface of water is
v1 = v2 made less than 'h' then :
( )
(a) water rises upto the top of capillary tube and stays
there without overflowing
3x = Q – x ⇒ 4x = Q (b) water rises upto a point a little below the top and stays
x= there
Q′ = Q – x = 3 × 10–2C (c) water does not rise at all.
Chapter: Electrostatic Potential and capacitance (d) Water rises upto the tip of capillary tube and then
starts overflowing like fountain.
[Topic: Electric Field, Electric Field Lines & Dipole]
Ans: (a)
Q44. As a result of change in the magnetic flux linked to Solution: Water rises upto the top of capillary tube and
the closed loop shown in the Fig,
stays there without overflowing.
Chapter: Mechanical Properties of Fluids
[Topic: Surface Tension, Surface Energy & Capillarity]
Q48. A square surface of side L meter in the plane of the
paper is placed in a uniform electric field E (volt/m)
acting along the same plane at an angle θ with the
an e.m.f. V volt is induced in the loop. The work done horizontal side of the square as shown in Figure. The
(joules) in taking a charge Q coulomb once along the electric flux linked to the surface, in units of volt. m, is
loop is
(a) QV (b) 2QV
(c) QV/2 (d) Zero
Ans: (a)
Solution:
Chapter: Electromagnetic
[Topic: Magnetic Flux, Faraday's & Lenz's Law]
Q45. In the Bohr model of a hydrogen atom, the
centripetal force is furnished by the coulomb attraction
between the proton and the electron. If a0 is the radius of
the ground state orbit, m is the mass, e is the charge on
the electron and ε0 is the vacuum permittivity, the speed
of the electron is (a) EL2 (b) EL2 cosθ
(c) EL2 sinθ (d) zero
(a) 0 (b)
√ Ans: (d)
(c) (d)
√ Solution: Electric flux, ϕ= EA cos θ , where θ = angle
√ between E and normal to the surface.
Ans: (c) Here
Solution: Centripetal force = Coulombian force
=
Chapter: Electrostatic Potential and capacitance
[Topic: Electric Flux & Gauss's Law]
⇒ Q49. The r.m.s. value of potential difference V shown in
√ the figure is
Chapter: Atoms
[Topic: Bohr Model & The Spectra of the Hydrogen
Atom]
Q46. Two bodies of same mass are projected with the
same velocity at an angle 30° and 60° respectively. The
ratio of their horizontal ranges will be (a) V0 (b)
√

76
(c) V0/2 (d) Q53. A bullet of mass 2 g is having a charge of 2µC.
√
Through what potential difference must it be accelerated,
Ans: (b)
starting from rest, to acquire a speed of 10 m/s?
. / (a) 50 V (b) 5 kV
Solution: Vrms = √ = .
√ (c) 50 kV (d) 5 V
Chapter: Alternating Current Ans: (c)
[Topic: Alternating Current, Voltage & Power]
Solution:
Q50. The mass number of He is 4 and that for sulphur is
32. The radius of sulphur nuclei is larger than that of ⇒
helium by ∴ V = 50 kV
(a) √ (b) 4 Chapter: Electrostatic Potential and capacitance
(c) 2 (d) 8 [Topic: Capacitors, Capacitance, Grouping of
Ans: (c) Capacitors & Energy Stored in a Capacitor.]
Q54. The time constant of C–R circuit is
Solution: . / . / (a) 1/CR
Chapter: Nuclei (b) C/R
[Topic: Composition and Size of the Nucleus] (c) CR
(d) R/C
PART 10. PHYSICS Ans: (c)
Solution: The time constant for resonance circuit,
QUESTION BANK = CR
Growth of charge in a circuit containing capacitance and
resistance is given by the formula, . /
Q51. Sand is being dropped on a conveyor belt at the CR is known as time constant in this formula.
rate of M kg/s. The force necessary to keep the belt Chapter: Alternating Current
moving with a constant velocity of v m/s will be: [Topic: Transformers & LC Oscillations]
(a) Mv newton (b) 2 Mv newton
Q55. The mass of proton is 1.0073 u and that of neutron
(c) newton (d) zero is 1.0087 u (u = atomic mass unit). The binding energy of
Ans: (a) is
( ) (a) 0.061 u (b) 0.0305 j
Solution: =
(c) 0.0305 erg (d) 28.4 MeV
∴ v is constant, Ans: (d)
∴ But =Mkg/s Solution: ∆m = (2 × 1.0074 + 2 × 1.0087 – 4.0015)
∴ F = vM newton. = 0.0307
Chapter: Dynamics Laws of Motion E = (∆m) × 931 MeV = 0.0307 × 931 = 28.5 MeV
[Topic: Ist, IInd & IIIrd Laws of Motion] Chapter: Nuclei
Q52. The total radiant energy per unit area, normal to the [Topic: Mass-Energy & Nuclear Reactions]
direction of incidence, received at a distance R from the Q56. A monkey of mass 20 kg is holding a vertical rope.
centre of a star of radius r, whose outer surface radiates The rope will not break when a mass of 25 kg is
as a black body at a temperature T K is given by: suspended from it but will break if the mass exceeds 25
kg. What is the maximum acceleration with which the
(a)
monkey can climb up along the rope ? (g = 10 m/s2)
(b) (a) 2.5 m/s2 (b) 5 m/s2
2
(c) 10 m/s (d) 25 m/s2
(c) Ans: (a)
(d) Solution: T = Tension caused in string by monkey
= m (g + a)
(where is Stefan's constant) 20 (10+a) ≤ 250
Ans: (a) or,
Solution: = Chapter: Dynamics Laws of Motion
[Topic: Motion of Connected Bodies, Pulleys]
= Q57. A system is taken from state a to state c by two
Chapter: Thermal Properties paths adc and abc as shown in the figure. The internal
[Topic: Calorimetry & Heat Transfer] energy at a is Ua = 10 J. Along the path adc the amount of
heat absorbed δQ1 = 50 J and the work done δW1 = 20 J

77
whereas along the path abc the heat absorbed δQ2 = 36 J. Therefore refractive index of the medium ( )
The amount of work done along the path abc is ( )
[NEET Kar. 2013] ( )

√
= =√
√
Chapter - Electromagnetic Waves
[Topic: Electromagnetic Waves, Conduction &
Displacement Current]
Q60. Two radioactive nuclei P and Q, in a given sample
decay into a stable nucleolus R. At time t = 0, number of
(a) 6 J (b) 10 J P species are 4 N0and that of Q are N0. Half-life of P (for
(c) 12 J (d) 36 J conversion to R) is 1 minute where as that of Q is 2
Ans: (a) minutes. Initially there are no nuclei of R present in the
Solution: From first law of thermodynamics sample. When number of nuclei of P and Q are equal, the
number of nuclei of R present in the sample would be
Qadc=∆Uadc + Wadc
50 J=∆Uadc + 20 J (a) 3N0 (b)
∆Uadc=30 J (c) (d) 2N0
Again,Qabc= ∆Uabc + Wabc
Ans: (b)
Wabc=Qabc – ∆Uabc
Solution: Initially P → 4N0
=Qabc – ∆Uadc
Q → N0
=36 J – 30 J
Half life TP = 1 min.
=6 J
TQ = 2 min.
Chapter: Thermal Properties
Let after time t number of nuclei of P and Q are equal,
[Topic: First Law of Thermodynamics]
that is
Q58. A 4µF conductor is charged to 400 volts and then
its plates are joined through a resistance of 1 kΩ. The
heat produced in the resistance is
(a) 0.16 J (b) 1.28 J ⇒
(c) 0.64 J (d) 0.32 J
Ans: (d) ⇒ 2t/1 = 4.2t/2
Solution: The energy stored in the capacitor 22.2t/2 = 2(2+t/2)
= ( ) ⇒
= 0.325 ⇒ t = 4 min
This energy will be converted into heat in the resistor ( )
Chapter: Electrostatic Potential and capacitance
[Topic: Electric Current, Drift of Electrons, Ohm's
Law, Resistance & Resistivity] at t = 4 min.
Q59. If ε0 and µ0 are the electric permittivity and
magnetic permeability in vacuum, ε and µ are
corresponding quantities in medium, then refractive index or population of R
of the medium is ( ) ( )
(a) √ =
Chapter: Nuclei
(b) √
[Topic: Radioactivity]
Q61. A body of mass 0.4 kg is whirled in a vertical
(c) √ circle making 2 rev/sec. If the radius of the circle is 1.2
m, then tension in the string when the body is at the top
(d) √ of the circle, is
Ans: (d) (a) 41.56 N (b) 89.86 N
Solution: We know that velocity of electromagnetic (c) 109.86 N (d) 115.86 N
Ans: (a)
wave in vacuum ( ) and velocity of Solution: Given : Mass (m) = 0.4 kg
√
electromagnetic wave in medium is ( ) . Its frequency (n) = 2 rev/sec
√

78
Radius (r) =1.2 m. We know that linear velocity of the >
body (v) = ωr = (2πn)r √ 1.414
= 2 × 3.14 × 1.2 × 2 = 15.08 m/s.
Therefore, tension in the string when the body is at the ⇒ µ = 1.50
top of the circle (T) Chapter - Ray Optics and Optical
( ) [Topic: Refraction of Light at Plane Surface & Total
= ( ) Internal Reflection]
=45.78-3.92=41.56N Q65. The mass of α-particle is
Chapter: Dynamics Laws of Motion (a) less than the sum of masses of two protons and two
[Topic: Circular Motion, Banking of Road] neutrons
Q62. A sample of gas expands from volume V1 to V2. (b) equal to mass of four protons
The amount of work done by the gas is greatest, when the (c) equal to mass of four neutrons
expansion is (d) equal to sum of masses of two protons and two
(a) adiabatic neutron
(b) isobaric Ans: (a)
(c) isothermal Solution: α-particle = It contains 2 p and 2 n. As
(d) equal in all cases some mass is converted into B.E., therefore, mass of α
Ans: (b) particle is slightly less than the sum of the masses of 2 p
Solution: In thermodynamics for same change in volume, and 2 n.
the work done is maximum in isobaric process because in Chapter: Nuclei
P – V graph, area enclosed by curve and volume axis is [Topic: Radioactivity]
maximum in isobaric process. Q66. When a long spring is stretched by 2 cm, its
So, the choice (b) is correct. potential energy is U. If the spring is stretched by 10 cm,
Chapter: Heat & Thermodynamics the potential energy stored in it will be
[Topic: Specific Heat Capacity & Thermodynamic (a) 25 U
Processes] (b) U/5
Q63. The current (I) in the given circuit is (c) 5 U (d) 10 U
Ans: (a)
Solution: If k be the spring constant, then
( )

( )
(a) 1.6 A (b) 2.0 A
(c) 0.32 A (d) 3.2 A ⇒
Ans: (b)
⇒
Solution: In circuit, RB and RC are in series, so, Rs = 6 + 6
Chapter: Work, Energy and Power
= 12 Ω. This 12 Ω resistance is in parallel with RA = 3 Ω,
[Topic: Energy]
So, equivalent resistance of circuit
Q67. When volume of system is increased twice and
temperature is decreased half of its initial temperature,
then pressure becomes
∴ Current in circuit, (a) 2 times (b) 4 times
= (c) times (d) times
. /
Chapter: Current Electricity Ans: (d)
[Topic: Combination of Resistances] Solution: Chapter: Kinetic Theory
[Topic: Kinetic Theory of an Ideal Gas & Gas Laws]
Q64. A ray of light travelling in a transparent medium of
refractive index μ, falls on a surface separating the Q68. Kirchhoff‟s first and second laws for electrical
medium from air at an angle of incidence of 45°. For circuits are consequences of
which of the following value of µ the ray can undergo (a) conservation of electric charge and energy
total internal reflection? respectively
[2010] (b) conservation of electric charge
(a) μ = 1.33 (b) μ = 1.40 (c) conservation of energy and electric charge
(c) μ = 1.50 (d) μ = 1.25 respectively
(d) conservation of energy
Ans: (c)
Ans: (a)
Solution: For total internal reflection,

79
Solution: Kirchhoff's first law deals with conservation of =√ = 20
electrical charge & the second law deals with m3v3 = 20 (momentum of third part)
conservation of electrical energy. or, m3 = = 5 kg
Chapter: Current Electricity
[Topic: Kirchhoff's Laws, Cells, Thermo emf & Chapter: Work, Energy and Power
Electrolysis] [Topic: Collisions]
Q69. A convex lens and a concave lens, each having Q72. The number of translational degrees of freedom for
same focal length of 25 cm, are put in contact to form a a diatomic gas is
combination of lenses. The power in diopters of the (a) 2 (b) 3
combination is (c) 5 (d) 6
(a) 50 (b) infinite Ans: (b)
(c) zero (d) 25 Solution: Number of translational degrees of freedom are
Ans: (c) same for all types of gases.
Solution: From the formula, Chapter: Kinetic Theory
[Topic: Degree of Freedom, Specific Heat Capacity &
Mean Free Path]
Q73. Fuse wire is a wire of
Power of combination (a) low resistance and high melting point
Chapter - Ray Optics and Optical (b) high resistance and high melting point
[Topic: Refraction at Curved Surface, Lenses & Power (c) high resistance and low melting point
of Lens] (d) low resistance and low melting point
Q70. A p-n photodiode is fabricated from a Ans: (c)
semiconductor with a band gap of 2.5 eV. It can detect a Solution: Fuse wire : It is used in a circuit to control the
signal of wavelength maximum current flowing in circuit. It is a thin wire
(a) 4000 nm (b) 6000 nm having high resistance and is made up of a material with
(c) 4000 Å (d) 6000 Å low melting point.
Ans: (c) Chapter: Current Electricity
Solution: [Topic: Heating Effects of Current]
Q74. Angle of deviation (δ) by a prism (refractive index
= µ and supposing the angle of prism A to be small) can
be given by
The wavelength detected by photodiode should be less
(a) ( )
than Hence it can detect a signal of wavelength
(b) ( )
4000Å.
Chapter: Semiconductor Electronics Materials, Devices (c)
[Topic: Solids, Semiconductors and P-N Junction
Diode] (d)
Q71. An explosion breaks a rock into three parts in a Ans: (a)
horizontal plane. Two of them go off at right angles to Solution: When the angle of prism is small, δ = (µ – 1) A
each other. The first part of mass 1 kg moves with a
Chapter - Ray Optics and Optical
speed of 12 ms–1 and the second part of mass 2 kg moves
[Topic: Optical Instruments]
with speed 8 ms–1. If the third part flies off with speed 4
ms–1 then its mass is Q75. To obtain a p-type germanium semiconductor, it
(a) 5 kg (b) 7 kg must be doped with
(c) 17 kg (d) 3 kg (a) arsenic
Ans: (a) (b) antimony
(c) indium
(d) phosphorus
Ans: (c)
Solution: p-type germanium semiconductor is formed
when it is doped with a trivalent impurity atom.
Chapter: Semiconductor Electronics Materials, Devices
[Topic: Solids, Semiconductors and P-N Junction
Diode]
Q76. A solid sphere of radius R is placed on a smooth
horizontal surface. A horizontal force F is applied at
Solution: height h from the lowest point. For the maximum
Presultant = √ acceleration of the centre of mass,

80
(a) h = R Solution: For dark fringe
(b) h = 2R
(c) h = 0 (d) The acceleration will be ( )
same whatever h may be
Ans: (d) ( ) ( )
Solution: As friction is absent at the point of contact, m = 6 × 10–5cm
Acceleration Chapter - Wave Optics
It is independent of h [Topic: Young's Double Slit Experiment]
Chapter: System of Particles and Rotational Motion Q80. The voltage gain of an amplifier with 9% negative
[Topic: Centre of Mass, Centre of Gravity & Principle feedback is 10. The voltage gain without feedback will be
of Moments] (a) 90 (b) 10
Q77. The composition of two simple harmonic motions (c) 1.25 (d) 100
of equal periods at right angle to each other and with a Ans: (d)
phase difference of π results in the displacement of the Solution: Negative feedback is applied to reduce the
particle along output voltage of an amplifier. If there is no negative
(a) circle feedback, the value of output voltage could be very high.
(b) figures of eight In the options given, the maximum value of voltage gain
(c) straight line is 100. Hence it is the correct option.
(d) ellipse Chapter: Semiconductor Electronics Materials, Devices
Ans: (c) [Topic: Junction Transistor]
Solution: x = a sin ωt Q81. If force (F) , velocity (V) and time (T) are taken as
and y = b sin (ωt + π) = – b sin ωt fundamental units, then the dimensions of mass are :
or, or (a) [F V T–1]
(b) [F V T–2]
It is an equation of a straight line. (c) [F V–1 T–1]
Chapter: Oscillation (d) [F V–1 T]
[Topic: Energy in Simple Harmonic Motion] Ans: (d)
Q78. In a Wheatstone‟s bridge all the four arms have Solution: Force = mass × acceleration
equal resistance R. If the resistance of the galvanometer ⇒[Mass]
arm is also R, the equivalent resistance of the
combination as seen by the battery is =0 1
(a) 2R
=6 7 = [F V–1 T]
(b)
Chapter: Units and Measurement
(c)
[Topic: Dimensions of Physical Quantities]
(d) R
Q82. If ⃗ is the force acting on a particle having position
Ans: (d)
vector and ⃗ be the torque of this force about the origin,
Solution: Since, Wheatstone's bridge is balanced, then
then:
resistance of galvanometer will be uneffective.
(a) ⃗ and⃗ ⃗ (b) ⃗ and⃗ ⃗
(c) ⃗ and⃗ ⃗ (d) ⃗ and⃗ ⃗
Ans: (b)
Solution: ⃗ ⃗ ⇒ ⃗ ⃗ ⃗
Since, ⃗ is perpendicular to the plane of and⃗ , hence the
dot product of ⃗ with and ⃗ is zero.
Chapter: Current Electricity Chapter: System of Particles and Rotational Motion
[Topic: Wheatstone Bridge & Different Measuring [Topic: Torque, Couple and Angular Momentum]
Instruments] Q83. The amplitude of a pendulum executing simple
Q79. In Young‟s experiment, two coherent sources are harmonic motion falls to 1/3 the original value after 100
placed 0.90 mm apart and fringe are observed one metre oscillations. The amplitude falls to S times the original
away. If it produces second dark fringe at a distance of 1 value after 200 oscillations, where S is
mm from central fringe, the wavelength of (a) 1/9 (b) 1/2
monochromatic light used would be (c) 2/3 (d) 1/6
(a) 60 × 10–4cm (b) 10 × 10–4cm Ans: (a)
–5
(c) 10 × 10 cm (d) 6 × 10–5cm Solution: In harmonic oscillator, amplitude falls
Ans: (d) exponentially.

81
After 100 oscillations amplitude falls to times.
∴ After next 100 oscillations i.e., after 200 oscillations
amplitude falls to . / times.
Chapter: Oscillation
[Topic: Time Period, Frequency, Simple Pendulum & Chapter: Semiconductor Electronics Materials, Devices
Spring Pendulum] [Topic: Digital Electronics and Logic Gates]
Q84. A 10 eV electron is circulating in a plane at right Q87. Dimensional formula of self inductance is
angles to a uniform field at magnetic induction 10–4 (a) [MLT–2A–2]
Wb/m2 (= 1.0 gauss). The orbital radius of the electron is (b) [ML2T–1A–2]
(a) 12 cm (b) 16 cm (c) [ML2T–2A–2]
(c) 11 cm (d) 18 cm (d) [ML2T–2A–1]
Ans: (c) Ans: (c)
Solution: K.E. of electron = 10 eV
Solution: . / . /
⇒
, -, -
⇒ ( ) or, , - , -, -
Chapter: Units and Measurement
⇒ [Topic: Dimensions of Physical Quantities]
⇒ v2 = 3.52 × 1012 ⇒ v = 1.88 × 106 m Q88. Four identical thin rods each of mass M and length
Also, we know that for circular motion l, form a square frame. Moment of inertia of this frame
⇒ cm about an axis through the centre of the square and
perpendicular to its plane is:
Chapter: Moving Charges and Magnetic Field
[Topic: Motion of Charged Particle in Magnetic Field (a)
& Moment] (b)
Q85. In the phenomenon of electric discharge through
gases at low pressure, the coloured glow in the tube (c)
appears as a result of: (d)
(a) excitation of electrons in the atoms Ans: (d)
(b) collision between the atoms of the gas Solution: Moment of inertia of a thin rod of length l
(c) collisions between the charged particles emitted from about an axis passing through centre and perpendicular to
the cathode and the atoms of the gas
(d) collision between different electrons of the atoms of the rod =
the gas Thus moment of inertia of the frame.
Ans: (a)
Solution: The coloured glow in the tube appears as a
result of excitations of electrons in the atoms. Total M.I. = 4 ×
Chapter - Dual Nature of Radiation and Matter
[Topic: Matter Waves, Cathode & Positive Rays] Chapter: System of Particles and Rotational Motion
Q86. In the following circuit, the output Y for all [Topic: Moment of Inertia, Rotational K.E. and Power]
possible inputs A and B is expressed by the truth table Q89. The time of reverberation of a room A is one
second. What will be the time (in seconds) of
reverberation of a room, having all the dimensions double
of those of room A ?
(a) 4 (b)
(a) ABY011011101110 (b) ABY001010100110
(c) ABY000011101111 (d) ABY000010100111 (c) 1 (d) 2
Ans: (c) Ans: (d)
Solution: Reverberation time is defined as the time
during which the intensity of sound in the auditorium
becomes one millionth of the initial intensity.
Solution: Sabine has shown that standard reverberation time for an
auditorium is given by the formula

Therefore truth table : Here, V is volume of the auditorium, S is the surface

area. So, (given)

82
 ∴ . / . /
(Assuming auditorium to be cubic in shape) S2 = 4S1
Chapter: Kinematics Motion in a Straight Line
[Topic: Non-uniform motion]
So, TR α l Q93. If a sphere is rolling, the ratio of the translational
If dimension is doubled, reverberation time t will be energy to total kinetic energy is given by
doubled. So, (a) 7: 10 (b) 2 : 5
New TR = 2 sec. (c) 10 : 7 (d) 5 : 7
Chapter: Waves Ans: (d)
[Topic: Basic of Waves] Solution:
Q90. A straight wire of diameter 0.5 mm carrying a
current of 1 A is replaced by another wire of 1 mm ( )
diameter carrying same current. The strength of magnetic
field far away is
(a) twice the earlier value
(b) same as the earlier value
(c) one-half of the earlier value
(d) one-quarter of the earlier value
Ans: (b) Chapter: System of Particles and Rotational Motion
Solution: and so it is independent of thickness. [Topic: Kepler's Laws of Planetary Motion]
Q94. When sound waves travel from air to water, which
The current is same in both the wires, hence magnetic
of the following remains constant?
field induced will be same.
(a) Velocity
Chapter: Moving Charges and Magnetic Field
(b) Wavelength
[Topic: Magnetic Field, Biot-Savart's Law & Ampere's
(c) Frequency
Circuital Law] (d) All of the above
Q91. The potential difference that must be applied to Ans: (c)
stop the fastest photoelectrons emitted by a nickel Solution: Chapter: Waves
surface, having work function 5.01 eV, when ultraviolet [Topic: Vibration of String & Organ Pipe]
light of 200 nm falls on it, must be:
Q95. A milli voltmeter of 25 milli volt range is to be
(a) 2.4 V (b) – 1.2 V
converted into an ammeter of 25 ampere range. The value
(c) – 2.4 V (d) 1.2 V
(in ohm) of necessary shunt will be :
Ans: (d)
[2012]
Solution: Kmax = (a) 0.001 (b) 0.01
= (c) 1 (d) 0.05
( ) Ans: (a)
–5.01 = 6.1875 – 5.01 = 1.17775 Solution: Galvanometer is converted into ammeter, by
= 1.2 V connected a shunt, in parallel with it.
Chapter - Dual Nature of Radiation and Matter
[Topic: Electron Emission, Photon Photoelectric Effect
& X-ray]
Q92. A particle starts its motion from rest under the
action of a constant force. If the distance covered in first
10 seconds is S1 and that covered in the first 20 seconds is
S2, then:
[2009]
(a) S2 = 3S1 (b) S2 = 4S1
(c) S2 = S1 (d) S2 = 2S1
Ans: (b)
Solution: u = 0, t1=10s, t2 = 20s Here S << G so
S = 0.001 Ω
Using the relation, S = ut + at2 Chapter: Moving Charges and Magnetic Field
Acceleration being the same in two cases, [Topic: Galvanometer and Its Conversion into Ammeter
& Voltmeter]
Q96. If the threshold wavelength for a certain metal is
2000 Å, then the work-function of the metal is

83
(a) 6.2 J (b) 6.2 eV Gravitational potential V =
(c) 6.2 MeV (d) 6.2 keV
Ans: (b) V0 = – – – –
Solution: Threshold wavelength (λ) = 2000 Å – 2G 0 1
= 2000 × 10–10 m. Work function
( ) ( ) = – 2G × = – 2G × = – 4 G.
( )
Chapter: Gravitation
[Topic: Gravitational Field, Potential and Energy]
Q99. Two sound sources emitting sound each of
Chapter - Dual Nature of Radiation and Matter wavelength λ are fixed at a given distance apart. A
[Topic: Electron Emission, Photon Photoelectric Effect listener moves with a velocity u along the line joining the
& X-ray] two sources. The number of beats heard by him per
Q97. If a ball is thrown vertically upwards with speed u, second is
the distance covered during the last t seconds of its ascent (a)
is
(a) (u + gt) t (b)
(b) ut (c)
(c)
(d)
(d) Ans: (b)
Ans: (c) Solution: Frequency received by listener from the rear
Solution: Let body takes T sec to reach maximum height. source,
Then v = u – gT
v = 0, at highest point. Frequency received by listener from the front source,
…(1)

No. of beats = n'' – n'

= =
Chapter: Waves
[Topic: Beats, Interference & Superposition of Waves]
Q100. The magnetic moment of a diamagnetic atom is
(a) equal to zero (b) much greater than one
(c) 1 (d) between zero and one
Ans: (a)
Velocity attained by body Solution: The magnetic moment of a diamagnetic atom is
in (T – t) sec v = u – g (T – t) equal to zero.
= u – gT + gt = u – Chapter: Magnetism and Matter
or v = gt…(2) [Topic: The Earth's Magnetism, Magnetic Materials
and their Properties]
∴Distance travelled in last t sec of its ascent
( ) PART 11. PHYSICS
Chapter: Kinematics Motion in a Straight Line
[Topic: Motion Under Gravity] QUESTION BANK
Q98. Infinite number of bodies, each of mass 2 kg are
situated on x-axis at distances 1m, 2m, 4m, 8m, .....
respectively, from the origin. The resulting gravitational Q1. Consider 3rd orbit of He+ (Helium), using non-
potential due to this system at the origin will be relativistic approach, the speed of electron in this orbit
will be [given K = 9 × 109 constant, Z = 2 and h (Plank's
(a) G
Constant) = 6.6 × 10–34 J s]
(b) G [2015]
(c) – 4 G (d) – G (a) 1.46 × 106 m/s (b) 0.73 × 106 m/s
8
Ans: (c) (c) 3.0 × 10 m/s (d) 2.92 × 106 m/s
Ans: (a)
Solution: Speed of electron in nth orbit
Solution:
Vn =

84
V = (2.19 × 106 m/s) = √
6
V = (2.19 × 10 ) (Z = 2 & n = 3) ∴ √
V = 1.46 × 106 m/s
Chapter: Electrostatic Potential and capacitance
Chapter: Atoms
[Topic: Electric Field, Electric Field Lines & Dipole]
[Topic: Bohr Model & The Spectra of the Hydrogen
Atom]
Q5. The magnetic flux through a circuit of resistance R
changes by an amount ∆θ in a time ∆t. Then the total
Q2. Find the torque of a force ⃗ ̂ ̂ acting quantity of electric charge Q that passes any point in the
at the point ̂ ̂ ̂. circuit during the time ∆t is represented by
(a) ̂ ̂ (b) ̂ ̂
(a)
(c) ̂ ̂ (d) ̂ ̂
Ans: (d) (b)
Solution: ⃗ ̂ ̂; ̂ ̂ ̂ (c)
Torque (⃗ ) = ⃗
(d)
=( ̂ ̂ ) ( ̂ ̂)
Ans: (c)
= ̂ ( )̂ ( ) ̂ ̂ ( )
Solution: ⇒ ( ) = QR
= ̂ ̂
Chapter: Kinematics Motion in a Plane
[Topic: Vectors]
Chapter: Electromagnetic
Q3. The escape velocity from the surface of the earth is
[Topic: Magnetic Flux, Faraday's & Lenz's Law]
ve.The escape velocity from the surface of a planet whose
mass and radius are three times those of the earth, will be Q6. When hydrogen atom is in its first excited level, its
(a) ve (b) 3ve radius is
(c) 9ve (d) 1/3ve (a) four times its ground state radius
Ans: (a) (b) twice
Solution: Escape velocity on surface of earth ( ) (c) same
(d) half
√ √ . Ans: (a)
Solution:
√ √ Radius in ground state =
Radius in first excited state = (∵ n =2)
√ √ Hence, radius of first excited state is four times the radius
in ground state.
or, vP= ve. Chapter: Atoms
Chapter: Gravitation [Topic: Bohr Model & The Spectra of the Hydrogen
[Topic: Motion of Satellites, Escape Speed and Orbital Atom]
Velocity] Q7. A particle moves so that its position vector is given
Q4. Two positive ions, each carrying a charge q, are by ̂ ̂ . Where ω is a constant. Which
separated by a distance d. If F is the force of repulsion of the following is true?
between the ions, the number of electrons missing from (a) Velocity and acceleration both are perpendicular to
each ion will be (e being the charge of an electron) (b) Velocity and acceleration both are parallel to
(c) Velocity is perpendicular to and acceleration is
(a)
directed towards the origin
(b) √ (d) Velocity is perpendicular to and acceleration is
directed away from the origin
Ans: (c)
(c) √
Solution: Given: Position vector
(d) = cos ωt ̂ + sin ωt ̂
∴Velocity, ⃗ = – ωsin ωt ̂ + ωcos ωt ̂
Ans: (c) and acceleration,
Solution: Let n be the number of electrons missing. ⃗ = –ω2 cos ωt ̂ – ω2sin ωt ̂ = – ω2
. ⃗ = 0 hence ⊥ ⃗ and
⃗ is directed towards the origin.
Chapter: Kinematics Motion in a Plane

85
 [Topic: Relative Velocity in2D & Circular Motion] (c)
Q8. A certain number of spherical drops of a liquid of (d) E0I0
radius „r‟ coalesce to form a single drop of radius „R‟ and Ans: (c)
volume „V‟. If „T‟ is the surface tension of the liquid, Solution: The average power in the circuit over one cycle
then : of a.c. is given by
(a) energy = . / is released Pav = erms ×irms × cosθ
(b) energy = . / is absorbed = cosf cosf
√ √
Chapter: Alternating Current
(c) energy = . / is released
[Topic: Alternating Current, Voltage & Power]
(d) energy is neither released nor absorbed Q11. The nucleus which has radius one-third of the
Ans: (c) radius of Cs189 is
Solution: As surface area decreases so energy is released. (a)
Energy released = 4πR2T[n1/3 – 1]
(b)
where R = n1/3r
(c)
= 0 1= 0 1 (d)
Chapter: Mechanical Properties of Fluids Ans: (b)
[Topic: Surface Tension, Surface Energy & Capillarity] Solution: Chapter: Nuclei
Q9. A hollow cylinder has a charge q coulomb within it. [Topic: Composition and Size of the Nucleus]
If  is the electric flux in units of voltmeter associated Q12. The two ends of a rod of length L and a uniform
with the curved surface B, the flux linked with the plane cross-sectional area A are kept at two temperatures T1
surface A in units of voltmeter will be and T2 (T1 > T2). The rate of heat transfer, through the
rod in a steady state is given by:
( )
(a)
(b) ( )
(a) ( )
(c)
(b) (d)
( )

(c) Ans: (c)

( )
(d) . / Solution:
Ans: (d) [(T1–T2) is the temperature difference]
Solution: Since , Chapter: Thermal Properties
[Topic: Calorimetry & Heat Transfer]
where q is the total charge. Q13. Each corner of a cube of side l has a negative
As shown in the figure, flux associated with the curved charge, –q. The electrostatic potential energy of a charge
surface B is θ = θB q at the centre of the cube is
Let us assume flux linked with the plane surfaces A and C
be (a)
√
θA = θC = θ' √
(b)
Therefore,
(c)
√

(d)
⇒ ( ) √
Ans: (d)
Chapter: Electrostatic Potential and capacitance Solution: Length of body diagonal = √
[Topic: Electric Flux & Gauss's Law] ∴ Distance of centre of cube from each corner
Q10. In an a.c circuit the e.m.f. (e) and the current (i) at
√
any instant are given respectively by
e = E0 sin ωt
P.E. at centre
i = I0 sin (ωt – θ)
= 8 × Potential Energy due to A
The average power in the circuit over one cycle of a.c. is
=8× ( )
(a)
= ( )
√ √
(b)
Chapter: Electrostatic Potential and capacitance

86
 [Topic: Capacitors, Capacitance, Grouping of Solution: Initial and final condition is same for all
Capacitors & Energy Stored in a Capacitor.] process
Q14. A transformer having efficiency of 90% is working ∆U1 = ∆U2 = ∆U3
on 200V and 3kW power supply. If the current in the from first law of thermodynamics
secondary coil is 6A, the voltage across the secondary ∆Q = ∆U + ∆W
coil and the current in the primary coil respectively are : Work done
(a) 300 V, 15A (b) 450 V, 15A ∆W1 > ∆W2 > ∆W3 (Area of P.V. graph)
(c) 450V, 13.5A (d) 600V, 15A So ∆Q1 > ∆Q2 > ∆Q3
Ans: (b) Chapter: Heat & Thermodynamics
( ) [Topic: First Law of Thermodynamics]
Solution: Efficiency η =
Q18. The resistance of a wire is 'R' ohm. If it is melted
⇒ Vs = 450 V and stretched to 'n' times its original length, its new
As VpIp = 3000 so resistance will be :-
Ip = (a)
Chapter: Alternating Current (b) n2R
[Topic: Transformers & LC Oscillations] (c)
Q15. For a nuclear fusion process, the suitable nuclei are (d) nR
(a) any nuclei Ans: (b)
(b) heavy nuclei
(c) light nuclei Solution: We know that, R =
(d) nuclei lying in the middle of the periodic table orR =
Ans: (c)
Solution: For nuclear fusion process the nuclei with low According to question l 2 = nl 1
mass are suitable. =
Chapter: Nuclei
[Topic: Mass-Energy & Nuclear Reactions] or,
Q16. A lift weighing 1000 kg is moving upwards with ⇒R2 = n2R1
an accelertion of 1 m/s2. The tension in the supporting Chapter: Current Electricity
cable is [Topic: Electric Current, Drift of Electrons, Ohm's
(a) 980 N (b) 10800 N Law, Resistance & Resistivity]
(c) 9800 N (d) 8800 N Q19. The oscillating electric and magnetic field vectors
Ans: (b) of electromagnetic wave are oriented along
Solution: T – (1000 × 9.8)= 1000 × 1 (a) the same direction and in phase
T = 10800 N (b) the same direction but have a phase difference of 90º
Chapter: Dynamics Laws of Motion (c) mutually perpendicular directions and are in phase
[Topic: Motion of Connected Bodies, Pulleys] (d) mutually perpendicular directions but has a phase
Q17. An ideal gas goes from state A to state B via three difference of 90º
different processes as indicated in the P-V diagram : Ans: (c)
Solution: The direction of oscillations of E and B fields
are perpendicular to each other as well as to the direction
of propagation. So, electro-magnetic waves are transverse
in nature.
The electric and magnetic fields oscillate in same phase.
Chapter - Electromagnetic Waves
[Topic: Electromagnetic Waves, Conduction &
Displacement Current]
If Q1, Q2, Q3 indicate the heat a absorbed by the gas along
Q20. The activity of a radioactive sample is measured as
the three processes and ∆U1, ∆U2, ∆U3 indicate the
N0 counts per minute at t = 0 and N0/e counts per minute
change in internal energy along the three processes
at t = 5 minutes. The time (in minutes) at which the
respectively, then
activity reduces to half its value is
(a) Q1> Q2> Q3 and ∆U1= ∆U2= ∆U3
(b) Q3> Q2> Q1 and ∆U1= ∆U2= ∆U3 (a) (b)
(c) Q1= Q2= Q3 and ∆U1> ∆U2> ∆U (c) 5 log 102 (d) 5 loge 2
(d) Q3> Q2> Q1 and ∆U1> ∆U2> ∆U3 Ans: (d)
Ans: (a) Solution: N =
Here, t = 5 minutes

87
 Chapter: Current Electricity
[Topic: Combination of Resistances]
, Q24. 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
Now, T1/2 = =5 (a) 1.50 (b) 3.00
(c) 1.33 (d) 1.40
Chapter: Nuclei
Ans: (b)
[Topic: Radioactivity]
Solution: By using
Q21. What will be the maximum speed of a car on a
Here,
road turn of radius 30 m if the coefficient of friction
between the tyres and the road is 0.4 (Take g = 9.8 m/s2)
(a) 10.84 m/s (b) 9.84 m/s m/s
(c) 8.84 m/s (d) 6.84 m/s Refractive index of the material,
Ans: (a)
Solution: r = 30 m and µ = 0.4.
Chapter - Ray Optics and Optical
√ √ =10.84m/s
[Topic: Refraction of Light at Plane Surface & Total
Chapter: Dynamics Laws of Motion
Internal Reflection]
[Topic: Circular Motion, Banking of Road]
Q25. The half life of radium is 1600 years. The fraction
Q22. An ideal gas undergoing adiabatic change has the of a sample of radium that would remain after 6400 years
following pressure-temperature relationship (a) 1/4 (b) 1/2
(a) =constant [1996] (c) 1/8 (d) 1/16
(b) constant Ans: (d)
(c) constant
(d) constant Solution: . / . /
Ans: (d) Chapter: Nuclei
Solution: We know that in adiabatic process, [Topic: Radioactivity]
PVγ = constant....(1)
Q26. If the kinetic energy of a particle is increased by
From ideal gas equation, we know that
300%, the momentum of the particle will increase by
PV = nRT
(a) 20% (b) 200%
V= ....(2) (c) 100% (d) 50%
Puttingt the value from equation (2) in equation (1) , Ans: (c)
. / = constant Solution: New K.E., E' = 4E
√ and √
P(1–γ) T γ = constant
Chapter: Heat & Thermodynamics
[Topic: Specific Heat Capacity & Thermodynamic √
Processes]
Q23. The current in the following circuit is [on substrating 1 in both sides.]
=(2-1)×100=100%
Chapter: Work, Energy and Power
[Topic: Energy]
(a) 1A Q27. The equation of state for 5 g of oxygen at a
(b) pressure P and temperature T, when occupying a volume
V, will be
(c) (a) PV = (5/16) RT
(d) (b) PV = (5/32) RT
(c) PV = 5 RT
Ans: (a) (d) PV = (5/2) RT
Solution: Resistance of ACB, R' = 3Ω + 3Ω = 6Ω. where R is the gas constant.
For net resistance between A and B; R' = 6Ω and 3Ω are Ans: (b)
in parallel.
Solution: PV= RT [ ∵ PV = nRT]
Chapter: Kinetic Theory
Current in circuit ( ) [Topic: Kinetic Theory of an Ideal Gas & Gas Laws]

88
Q28. Two cells, having the same e.m.f., are connected in [Topic: Collisions]
series through an external resistance R. Cells have Q31. If for a gas, , the gas is made up of
internal resistances r1 and r2 (r1 > r2) respectively. When
the circuit is closed, the potential difference across the molecules which are
first cell is zero. The value of R is (a) diatomic
(b) mixture of diatomic and polyatomic molecules
(a) (c) monoatomic
(b) (d) polyatomic
(c) Ans: (c)
(d) Solution: Since ⇒ hence gas is
Ans: (d) monoatomic.
Solution: Current in the circuit Chapter: Kinetic Theory
[Topic: Degree of Freedom, Specific Heat Capacity &
Mean Free Path]
P.D. across first cell = E – ir1 Q32. If 25W, 220 V and 100 W, 220 V bulbs are
connected in series across a 440 V line, then
( ) [2001]
Now, (a) only 25W bulb will fuse
( )
(b) only 100W bulb will fuse
⇒ ⇒ (c) both bulbs will fuse
R = r1 – r2 (d) none of these
Chapter: Current Electricity Ans: (a)
[Topic: Kirchhoff's Laws, Cells, Thermo emf & Solution: As for an electric appliance . / , so for
Electrolysis] same specified voltage Vs
Q29. Sodium has body centred packing. Distance
between two nearest atoms is 3.7 Å. The lattice parameter
is i.e, with
(a) 4.3Å (b) 3.0 Å Now in series potential divides in proportion to
(c) 8.6 Å (d) 6.8 Å resistance.
Ans: (a)
√ So, ( )
Solution: d = a
√
i.e.,
3.7 = a
and ( )
a=
√
i.e.,
Chapter: Semiconductor Electronics Materials, Devices
[Topic: Solids, Semiconductors and P-N Junction From this, it is clear that voltage across 100 W bulb (= 88
Diode] V) is lesser than specified(220 V) while across 25 W bulb
Q30. A solid cylinder of mass 3 kg is rolling on a (= 352 V) is greater than specified (220 V), so, 25 W
bulb will fuse.
horizontal surface with velocity 4 ms–1. It collides with a
horizontal spring of force constant 200 Nm–1. The Chapter: Current Electricity
maximum compression produced in the spring will be : [Topic: Heating Effects of Current]
(a) 0.5 m (b) 0.6 m Q33. A astronomical telescope has objective and
(c) 0.7 m (d) 0.2 m eyepiece of focal lengths 40 cm and 4 cm respectively.
Ans: (b) To view an object 200 cm away from the objective, the
Solution: At maximum compression the solid cylinder lenses must be separated by a distance :
will stop so loss in K.E. of cylinder = gain in P.E. of (a) 37.3 cm (b) 46.0 cm
spring (c) 50.0 cm (d) 54.0 cm
Ans: (d)
⇒
Solution: Given: Focal length of objective, f0 = 40cm
⇒ . / Focal length of eye – piece fe = 4 cm
image distance, v0 = 200 cm
⇒ Using lens formula for objective lens
⇒ ( ) ⇒
⇒ ⇒ x = 0.6 m ⇒
Chapter: Work, Energy and Power

89
⇒v0 = 50 cm volt/m
Tube length = |v0| + fe = 50 + 4 = 54 cm. Chapter: Current Electricity
Chapter - Ray Optics and Optical [Topic: Wheatstone Bridge & Different Measuring
[Topic: Optical Instruments] Instruments]
Q34. When arsenic is added as an impurity to silicon, the Q38. The Young‟s double slit experiment is performed
resulting material is with blue and with green light of wavelengths 4360Å and
(a) n-type semiconductor 5460Å respectively. If x is the distance of 4th maxima
(b) p-type semiconductor from the central one, then
(c) n-type conductor (a) x (blue) = x (green)
(d) insulator (b) x (blue) > x (green)
Ans: (a) (c) x (blue) < x (green)
Solution: Arsenic contains 5 electrons in its outermost ( )
(d) ( )
shell. When Arsenic is mixed with silicon there is one
electron extra in silicon crystal. Hence, such type of semi Ans: (c)
conductor is n-type semi conductor. Solution: Distance of nth maxima,
Chapter: Semiconductor Electronics Materials, Devices As
[Topic: Solids, Semiconductors and P-N Junction
Diode]
Chapter - Wave Optics
Q35. The centre of mass of a system of particles does [Topic: Young's Double Slit Experiment]
not depend upon
(a) masses of the particles
Q39. A transistor is operated in common emitter
(b) forces acting on the particles configuration at constant collector voltage Vc = 1.5V such
(c) position of the particles that a change in the base current from 100 µA to 150 µA
(d) relative distances between the particles produces a change in the collector current from 5 mA to
10 mA. The current gain (β) is
Ans: (b)
(a) 75 (b) 100
Solution: Centre of mass of system depends upon
(c) 50 (d) 67
position and masses of particle. Also, it depends upon
relative distance between particles. Ans: (b)
Chapter: System of Particles and Rotational Motion Solution:
[Topic: Centre of Mass, Centre of Gravity & Principle
of Moments]
Q36. A particle moving along the X-axis, executes Chapter: Semiconductor Electronics Materials, Devices
simple harmonic motion then the force acting on it is [Topic: Junction Transistor]
given by Q40. The pair of quantities having same dimensions is
(a) – A kx (a) Young‟s modulus and energy
(b) A cos (kx) (b) impulse and surface tension
(c) A exp (– kx) (c) angular momentum and work
(d) Akx (d) work and torque
where, A and k are positive constants. Ans: (d)
Ans: (a) Solution: Work = Force × displacement
Solution: For simple harmonic motion, F = – Kx. Torque = Force × force arm
Here, K = Ak. = mass × acceleration × length
Chapter: Oscillation = [M] × [LT–2] × [L] = [M L2T–2]
[Topic: Energy in Simple Harmonic Motion] Chapter: Units and Measurement
Q37. If specific resistance of a potentiometer wire is 10–7 [Topic: Dimensions of Physical Quantities]
Ωm, the current flow through it is 0.1 A and the cross- Q41. A uniform rod AB of length l, and mass m is free
sectional area of wire is 10–6 m2 then potential gradient to rotate about point A. The rod is released from rest in
will be the horizontal position. Given that the moment of inertia
(a) 10–2 volt/m (b) 10–4 volt/m of the rod about A is , the initial angular acceleration
–6
(c) 10 volt/m (d) 10–8 volt/m of the rod will be
Ans: (a) [2007]
Solution: Potential gradient = Potential fall per unit
length. In this case resistance of unit length.

Potential fall across R is

volt/m

90
 Ans: (c)
Solution: The electron moves with constant velocity
without deflection. Hence, force due to magnetic field is
equal and opposite to force due to electric field.
qvB = qE ⇒ v = 40m/s
Chapter: Moving Charges and Magnetic Field
[Topic: Motion of Charged Particle in Magnetic Field
(a) & Moment]
(b) Q44. A particle of mass 1 mg has the same wavelength
as an electron moving with a velocity of 3×106 ms–1. The
(c) velocity of the particle is:
(d) (a) 2.7× 10–18 ms–1 (b) 9 × 10–2ms–1
(c) 3 × 10–31 ms–1 (d) 2.7×10–21 ms–1
Ans: (c) (mass of electron = 9.1×10 kg)–31

Ans: (d)
Solution: Wavelength of particle (λ1) = ( )
where v is the velocity of the particle.
Wave length of electron
( )
Solution: ( ) ( )
Weight of the rod will produce torque, But λ1= λ2

( ) ( ) ( )
Also, ⇒v= = 2.73×10 ms –21 –1

where, I is the moment of inertia = Chapter - Dual Nature of Radiation and Matter
and α is the angular acceleration [Topic: Matter Waves, Cathode & Positive Rays]
Q45. The following figure shows a logic gate circuit
⇒ with two inputs A and B and the output C. The voltage
Chapter: System of Particles and Rotational Motion waveforms of A, B and C are as shown below
[Topic: Torque, Couple and Angular Momentum]
Q42. A simple pendulum has a metal bob, which is
negatively charged. If it is allowed to oscillate above a
positively charged metallic plate, then its time period will
(a) increase
(b) decrease
(c) become zero
(d) remain the same
Ans: (b)
Solution: We know that the time period (T)
= √ g' = geffective
√
Since the negatively charged bob is attracted by the The logic circuit gate is
positively charged plate, therefore acceleration due to (a) NAND gate
gravity will increase and time period will decrease. (b) NOR gate
Chapter: Oscillation (c) OR gate
[Topic: Time Period, Frequency, Simple Pendulum & (d) AND gate
Spring Pendulum] Ans: (d)
Q43. A beam of electrons is moving with constant Solution: On the basis of given graph, following table is
velocity in a region having simultaneous perpendicular possible.
electric and magnetic fields of strength 20 Vm–1 and 0.5 T
respectively at right angles to the direction of motion of
the electrons. Then the velocity of electrons must be
(a) 8 m/s (b) 20 m/s
(c) 40 m/s (d) m/s

91
 (a) 3 : 2 (b) 2 : 3
(c) 9 : 4 (d) 4 : 9
Ans: (c)
Solution: Intensity = Energy/sec/unit area
Area r2 ⇒ I 1/r2
⇒
It is the truth table of AND gate. Chapter: Waves
Chapter: Semiconductor Electronics Materials, Devices [Topic: Basic of Waves]
[Topic: Digital Electronics and Logic Gates] Q49. At what distance from a long straight wire carrying
Q46. Of the following quantities, which one has a current of 12 A will the magnetic field be equal to
dimension different from the remaining three? Wb/ ?
(a) Energy per unit volume (a) (b)
(b) Force per unit area (c) (d)
(c) Product of voltage and charge per unit volume
Ans: (a)
(d) Angular momentum.
Solution: Current (I) = 12 A and magnetic field (B) = 3 ×
Ans: (d)
Solution: For angular momentum, the dimensional 10–5 Wb/m2. Consider magnetic field ⃗ at distance r.
formula is ML2T–1. For other three, it is ML–1T–2. Magnetic field ,
Chapter: Units and Measurement ( )
⇒
[Topic: Dimensions of Physical Quantities] ( )
Q47. A thin rod of length L and mass M is bent at its Chapter: Moving Charges and Magnetic Field
midpoint into two halves so that the angle between them [Topic: Magnetic Field, Biot-Savart's Law & Ampere's
is 90°. The moment of inertia of the bent rod about an Circuital Law]
axis passing through the bending point and perpendicular Q50. The number of photo electrons emitted for light of
to the plane defined by the two halves of the rod is: a frequency ν (higher than the threshold frequency ν 0) is
(a) proportional to:
(a) Threshold frequency (ν 0)
(b) (b) Intensity of light
(c) Frequency of light (ν )
(c) (d) ν – ν 0
√ Ans: (b)
(d)
Solution: The number of photoelectrons emitted is
Ans: (b)
proportional to the intensity of incident light. Saturation
Solution: Mass of each part = M/2
current intensity.
Length of each part = L/2
Chapter - Dual Nature of Radiation and Matter
[Topic: Electron Emission, Photon Photoelectric Effect
& X-ray]

PART 12. PHYSICS

QUESTION BANK
Q51. The distance travelled by a particle starting from
rest and moving with an acceleration , in the third
Total M.I. = Sum of M.I.s of both parts
second is:
= . /. / . /. / (a) 6 m (b) 4 m
(c)
= .
Chapter: System of Particles and Rotational Motion (d)
[Topic: Moment of Inertia, Rotational K.E. and Power] Ans: (c)
Q48. A point source emits sound equally in all directions Solution: Distance travelled in the nth second is given by
in a non–absorbing medium. Two points P and Q are at ( )
distances of 2 m and 3 m respectively from the source.
The ratio of the intensities of the waves at P and Q is put u = 0, ,n=3

92
∴d=0+ (2 × 3 –1) = Ans: (c)
Solution: To keep the main current in the circuit
Chapter: Kinematics Motion in a Straight Line
unchanged, the resistance of the galvanometer should be
[Topic: Non-uniform motion]
equal to the net resistance.
Q52. Kepler's third law states that square of period of
revolution (T) of a planet around the sun, is proportional ( )
to third power of average distance r between sun and
planet i.e. T2 = Kr3 here K is constant. If the masses of
sun and planet are M and m respectively then as per
Newton's law of gravitation force of attraction between
them is F = here G is gravitational constant. The
relation between G and K is described as
(a) GMK = 4π2 (b) K = G
(c) K = (d) GK = 4π2
Ans: (a) Chapter: Moving Charges and Magnetic Field
[Topic: Galvanometer and Its Conversion into Ammeter
Solution: As we know, orbital speed, √ & Voltmeter]
Time period T = √ Q55. Kinetic energy of an electron, which is accelerated
√ in a potential difference of 100 V is
Squarring both sides, (a) (b) J
√ (c) (d)
T2 = . /
√
Ans: (a)
⇒ Solution: Potential difference (V) = 100 V. The kinetic
⇒GMK = 4π2. energy of an electron = 1 eV = 1 × (1.6 × 10–19)
Chapter: Gravitation = 1.6 × 10–19J. Therefore kinetic energy in 100 volts =
[Topic: Kepler's Laws of Planetary Motion] (1.6 × 10–19) × 100 = 1.6 × 10–17J.
Q53. Equation of a progressive wave is given by [Alt : K.E. = qV
= 1.6 × 10–19 × 100 J = 1.6 × 10–17]
[ ( )] Chapter - Dual Nature of Radiation and Matter
Then which of the following is correct? [Topic: Electron Emission, Photon Photoelectric Effect
(a) v = 5 cm & X-ray]
(b) λ = 18 cm Q56. A man throws balls with the same speed vertically
(c) a = 0.04 cm upwards one after the other at an interval of 2 seconds.
(d) f = 50 Hz What should be the speed of the throw so that more than
Ans: (b) two balls are in the sky at any time? [Given g = 9.8 m/s2]
Solution: The standard equation of a progressive wave is (a) Only with speed 19.6 m/s
(b) More than 19.6 m/s
[ ( ) ] (c) At least 9.8 m/s
The given equation can be written as (d) Any speed less than 19.6 m/s
Ans: (b)
[ ( ) ]
Solution: Let the required speed of throw be u ms–1. Then
∴ a = 4 cm, T = 10 s, λ = 18 cm and θ = π/6 time taken to reach maximum height,
Hence, (b) is correct. t=
Chapter: Waves
For two balls to remain in air at any time, t must be
[Topic: Vibration of String & Organ Pipe]
greater than 2.
Q54. A galvanometer of resistance, G is shunted by a
∴ m/s
resistance S ohm. To keep the main current in the circuit
unchanged, the resistance to be put in series with the Chapter: Kinematics Motion in a Straight Line
galvanometer is [Topic: Motion Under Gravity]
(a) ( Q57. Two waves of lengths 50 cm and 51 cm produce 12
) beats per sec. The velocity of sound is
(b) ( )
(a) 306 m/s (b) 331 m/s
(c) 340 m/s (d) 360 m/s
(c) Ans: (a)
(d) ( Solution: Given : Wavelength of first wave (λ1)
)
= 50 cm = 0.5 m

93
Wavelength of second wave (λ2) Q61. A satellite in force free space sweeps stationary
= 51 cm = 0.51m interplanetary dust at a rate dM/dt = αv where M is the
frequency of beats per sec (n) = 12. mass and v is the velocity of the satellite and α is a
We know that the frequency of beats, constant. What is the deceleration of the satellite?
n = 12 = (a)
(b)
⌈ ⌉ ⇒ [ ]
, - (c)
or, =306m/s (d)
[where, v = velocity of sound] Ans: (c)
Chapter: Waves Solution: . / . /
[Topic: Beats, Interference & Superposition of Waves]
Q58. If a diamagnetic substance is brought near the ∴ Retardation =
north or the south pole of a bar magnet, it is: Chapter: Gravitation
(a) repelled by the north pole and attracted by the south [Topic: Motion of Satellites, Escape Speed and Orbital
pole Velocity]
(b) attracted by the north pole and repelled by the south Q62. An electron is moving round the nucleus of a
pole hydrogen atom in a circular orbit of radius r. The
(c) attracted by both the poles
Coulomb force ⃗ between the two is
(d) repelled by both the poles
Ans: (d) (a)
Solution: Diamagnetic substances do not have any
(b) ̂
unpaired electron. And they are magnetised in direction
opposite to that of magnetic field. Hence, when they are (c) ̂
brought to north or south pole of a bar magnet, they are
repelled by poles. (d) . /
Chapter: Magnetism and Matter Ans: (d)
[Topic: The Earth's Magnetism, Magnetic Materials Solution: Charges (–e) on electron and (e) on proton
and their Properties] exert a force of attraction given by
Q59. Two particles of masses m1, m2 move with initial Force =
( )(( )( ))
̂ .̂
⃗
/
velocities u1 and u2. On collision, one of the particles get | |
excited to higher level, after absorbing energy ε. If final Note : Magnitude of Coulomb force is given by
velocities of particles be v1 and v2 then we must have ,
(a) but in vector form ⃗
(b) Chapter: Electrostatic Potential and capacitance
(c) [Topic: Electric Field, Electric Field Lines & Dipole]
(d) Q63. A magnetic field of 2 × 10–2 T acts at right angles
Ans: (b) to a coil of area 100 cm2, with 50 turns. The average
Solution: By law of conservation of energy, e.m.f. induced in the coil is 0.1 V, when it is removed
K.Ef = K.Ei – excitation energy (ε) from the field in t sec. The value of t is
(a) 10 s (b) 0.1 s
or (c) 0.01 s (d) 1 s
Chapter: Atoms Ans: (b)
[Topic: Bohr Model & The Spectra of the Hydrogen ( ) ( )
Solution:
Atom]
Q60. Which of the following is not a vector quantity?
[1996]
(a) displacement Chapter: Electromagnetic
(b) electric field [Topic: Magnetic Flux, Faraday's & Lenz's Law]
(c) work Q64. The spectrum obtained from a sodium vapour lamp
(d) acceleration is an example of
Ans: (c) (a) band spectrum
Solution: Chapter: Kinematics Motion in a Plane (b) continuous spectrum
[Topic: Vectors] (c) emission spectrum
(d) absorption spectrum

94
 Ans: (c)
Solution: A spectrum is observed, when light coming
directly from a source is examined with a spectroscope.
Therefore spectrum obtained from a sodium vapour lamp
is emission spectrum. (a) EL2/2 (b) zero
Chapter: Atoms (c) EL2 (d) EL2/ (2ε0)
[Topic: Bohr Model & The Spectra of the Hydrogen Ans: (b)
Atom] Solution: Flux ⃗ ⃗
Q65. A ship A is moving Westwards with a speed of 10 ⃗ is electric field vector & ⃗ is area vector.
km h–1 and a ship B 100 km South of A, is moving
Northwards with a speed of 10 km h–1. The time after Here, angle between is 90º.
which the distance between them becomes shortest, is : So, ⃗ ⃗ ; Flux = 0
(a) 5 h (b) √ Chapter: Electrostatic Potential and capacitance
(c) √ (d) 0 h [Topic: Electric Flux & Gauss's Law]
Ans: (a) Q68. In an A.C. circuit with voltage V and current I the
power dissipated is
Solution: ⃗ ( ̂)
(a) dependent on the phase between V and I
⃗ ( )̂
(b)
⃗ ̂ √ km/h √

Distance OB = 100 cos 45° = √ km (c)

√
(d) VI
Ans: (a)
Solution: Power dissipated = Erms. Irms = (Erms) (Irms) cos θ
Hence, power dissipated depends upon phase difference.
Chapter: Alternating Current
[Topic: Alternating Current, Voltage & Power]
Q69. The mass density of a nucleus varies with mass
number A as
(a) A2 (b) A
(c) constant (d) 1/A
Ans: (c)
Time taken to reach the shortest distance between Solution: The nuclear radius r varies with mass number
√ A according to the relation
A and ⃗⃗⃗⃗⃗⃗⃗⃗ √ or
Chapter: Kinematics Motion in a Plane
Now, density
[Topic: Relative Velocity in2D & Circular Motion]
Q66. The wetability of a surface by a liquid depends Further, mass A and volume
primarily on constant
(a) surface tension Chapter: Nuclei
(b) density [Topic: Composition and Size of the Nucleus]
(c) angle of contact between the surface and the liquid Q70. If a cricketer catches a ball of mass 150 gm moving
(d) viscosity with a velocity of 20 m/s, then he experiences a force of
Ans: (c) (Time taken to complete the catch is 0.1 sec.)
Solution: Wetability of a surface by a liquid primarily (a) 300 N (b) 30 N
depends on angle of contact between the surface and (c) 3 N (d) 0.3 N
liquid. Ans: (b)
If angle of contact is acute liquids wet the solid and vice-
versa. Solution: Net force experienced =
Chapter: Mechanical Properties of Fluids =
[Topic: Surface Tension, Surface Energy & Capillarity] Chapter: Dynamics Laws of Motion
Q67. A square surface of side L metres is in the plane of [Topic: Ist, IInd & IIIrd Laws of Motion]
the paper. A uniform electric field ⃗ (volt/m), also in the Q71. A black body at 227°C radiates heat at the rate of 7
plane of the paper, is limited only to the lower half of the cals/cm2s. At a temperature of 727°C, the rate of heat
square surface (see figure). The electric flux in SI units radiated in the same units will be:
associated with the surface is [2009]
(a) 50 (b) 112

95
(c) 80 (d) 60
Ans: (b)
Solution: According to Stefan‟s law
T1 = 500 K
T2 = 1000 K

( ) ( )
(a) g / 3 (b) g / 2
∴ E2 = 16 × 7 = 112 cal / cm2s (c) g
Chapter: Thermal Properties (d) g / 5
[Topic: Calorimetry & Heat Transfer] (where g is acceleration due to gravity)
Q72. A particle of mass m and charge q is placed at rest Ans: (a)
in a uniform electric field E and then released. The
kinetic energy attained by the particle after moving a
distance y is
(a) qEy2 (b) qE2y
(c) qEy (d) q2Ey
Ans: (c)
Solution: K.E. = Force × distance = qE.y
Chapter: Electrostatic Potential and capacitance Solution:
[Topic: Capacitors, Capacitance, Grouping of Let T be the tension in the string.
Capacitors & Energy Stored in a Capacitor.] ∴ 10g – T = 10a....(i)
Q73. The primary of a transformer when connected to a T – 5g = 5a....(ii)
dc battery of 10 volt draws a current of 1 mA. The Adding (i) and (ii),
number of turns of the primary and secondary windings 5g = 15a
are 50 and 100 respectively. The voltage in the secondary Chapter: Dynamics Laws of Motion
and the current drawn by the circuit in the secondary are [Topic: Motion of Connected Bodies, Pulleys]
respectively Q76. The internal energy change in a system that has
(a) 20 V and 0.5 mA (b) 20 V and 2.0 mA absorbed 2 kcals of heat and done 500 J of work is:
(c) 10 V and 0.5 mA (d) Zero and therefore no (a) 6400 J (b) 5400 J
current (c) 7900 J (d) 8900 J
Ans: (d) Ans: (c)
Solution: A transformer is essentially an AC device. DC Solution: According to first law of thermodynamics
source so no mutual induction between coils Q = ∆U + W
⇒ E2 = 0 and I2 = 0 ∆U = Q – W
Chapter: Alternating Current = 2 × 4.2 × 1000 – 500 = 8400 –500
[Topic: Transformers & LC Oscillations] = 7900 J
Q74. It is possible to understand nuclear fission on the Chapter: Heat & Thermodynamics
basis of the [Topic: First Law of Thermodynamics]
(a) liquid drop model of the nucleus Q77. Across a metallic conductor of non-uniform cross
(b) meson theory of the nuclear forces section a constant potential difference is applied. The
(c) proton-proton cycle quantity which remains constant along the conductor is :
(d) independent particle model of the nucleus (a) current
Ans: (a) (b) drift velocity
Solution: According to liquid drop model of nucleus, an (c) electric field
excited nucleus breaks into lighter nuclei just like an (d) current density
excited drop breaks into tiny drops. Ans: (a)
Chapter: Nuclei Solution: Here, metallic conductor can be considered as
[Topic: Mass-Energy & Nuclear Reactions] the combination of various conductors connected in
Q75. Two blocks m1 = 5 gm and m2 = 10 gm are hung series. And in series combination current remains same.
vertically over a light frictionless pulley as shown here.
What is the acceleration of the masses when they are left
free?

Chapter: Current Electricity

96
 [Topic: Electric Current, Drift of Electrons, Ohm's Ans: (b)
Law, Resistance & Resistivity] Solution: Initial temperature (T1) = 18°C = 291 K
Q78. The frequency of electromagnetic wave, which is Let Initial volume (V1) = V
best suited to observe a particle of radius 3 × 10–4 cm is of Final volume (V2) =
the order of According to adiabatic process,
(a) 1015 (b) 1014 TVγ–1 = constant
13 12
(c) 10 (d) 10
Ans: (b) According to question,
Solution: Size of particle ⇒ ( )

( ) = 293 × 2.297 = 668.4K

ν = 1014 Hz
[ ]
However, when frequency is higher than this, wavelength
is still smaller. Resolution becomes better. Chapter: Heat & Thermodynamics
Chapter - Electromagnetic Waves [Topic: Specific Heat Capacity & Thermodynamic
[Topic: Electromagnetic Spectrum] Processes]
Q79. The decay constant of a radio isotope is λ. If A1 and Q82. What will be the equivalent resistance of circuit
A2 are its activities at times t1 and t2 respectively, the shown in figure between two points A and D
number of nuclei which have decayed during the time (t1 [1996]
– t2 ) :
(a) ( )
(b)
(c)
(d)
Ans: (d)
Solution: Activity is given by
(a) 10Ω (b) 20Ω
(c) 30Ω (d) 40Ω
Activity at time t1 is Ans: (c)
Solution: Equivalent Circuit
and activity at time t2 is A2 = – l N2
As t1 > t2, therefore, number of atoms remained after time
t1 is less than that remained after time t2. That is, N1 < N2.
∴ number of nuclei decayed in (t1 – t2)
( )
=
Chapter: Nuclei
[Topic: Radioactivity] Equivalent Resistance of circuit
Q80. A particle of mass M is moving in a horizontal =
circle of radius R with uniform speed V. When it moves Chapter: Current Electricity
from one point to a diametrically opposite point, its [Topic: Combination of Resistances]
(a) kinetic energy changes by MV2/4
Q83. A small coin is resting on the bottom of a beaker
(b) momentum does not change
filled with liquid. A ray of light from the coin travels
(c) momentum changes by 2 MV
upto the surface of the liquid and moves along its surface.
(d) kinetic energy changes by MV2
How fast is the light travelling in the liquid?
Ans: (c)
Solution: On the diametrically opposite points, the
velocities have same magnitude but opposite directions.
Therefore, change in momentum is MV – (– MV) = 2MV
Chapter: Dynamics Laws of Motion
[Topic: Circular Motion, Banking of Road]
Q81. A diatomic gas initially at 18ºC is compressed
adiabatically to one eighth of its original volume. The
temperature after compression will be (a) 2.4 × 108 m/s (b) 3.0 × 108 m/s
(a) 18ºC (b) 668.4ºK (c) 1.2 × 108 m/s (d) 1.8 × 108 m/s
(c) 395.4ºC (d) 144ºC

97
 Ans: (d) ⇒ √ √ ( )
Chapter: Work, Energy and Power
[Topic: Energy]
Q86. A gas at 27ºC temperature and 30 atmospheric
pressure is allowed to expand to the atmospheric
pressure. If the volume becomes 10 times its initial
volume, then the final temperature becomes
Solution: (a) 100ºC (b) 173ºC
Hypotenuse comes out to be 5 cm. (c) 273ºC (d) –173ºC
Since, Ans: (d)
Solution: Given : Initial temperature of gas
(T1) = 27°C = 300 K
Initial pressure (P1) = 30 atm
Speed, m/s Initial volume (V1) = V
Final pressure (P2) = 1 atm
Chapter - Ray Optics and Optical Final volume (V2) = 10 V.
[Topic: Refraction of Light at Plane Surface & Total We know from the general gas equation that
Internal Reflection]
Q84. The nucleus absorbs an energetic neutron and or,
emits a beta particle (β) . The resulting nucleus is or, T2 = 100 K = –173°C.
(a) 7N14 (b) 7N13 Chapter: Kinetic Theory
13
(c) 5B (d) 6C13 [Topic: Speeds of Gas, Pressure & Kinetic Energy]
Ans: (b) Q87. Two batteries, one of emf 18 volt and internal
Solution: resistance 2Ω and the other of emf 12 volt and internal
Energy resistance 1Ω, are connected as shown. The voltmeter V
Chapter: Nuclei will record a reading of
[Topic: Radioactivity]
Q85. In a simple pendulum of length l the bob is pulled
aside from its equilibrium position through an angle θ
and then released. The bob passes through the
equilibrium position with speed (a) 30 volt (b) 18 volt
(a) √ ( ) (c) 15 volt (d) 14 volt
(b) √ Ans: (d)
(c) √ Solution: V = =
(d) √ ( )
(Since the cells are in parallel).
Ans: (d)
Chapter: Current Electricity
Solution: If l is length of pendulum and θ be angular
amplitude then height [Topic: Kirchhoff's Laws, Cells, Thermo emf &
Electrolysis]
Q88. A convex lens is dipped in a liquid whose
refractive index is equal to the refractive index of the
lens. Then its focal length will
(a) remain unchanged
(b) become zero
(c) become infinite
(d) become small, but non-zero
Ans: (c)
h=AB-AC = l – l cos θ = l(1 – cos θ) Solution: ( ). /
At extreme position, potential energy is maximum and where, is given.
kinetic energy is zero; At mean (equilibrium) position
potential energy is zero and kinetic energy is maximum, ⇒ ( ). / ⇒
so from principle of conservation of energy. Chapter - Ray Optics and Optical
(KE+PE)atP=(KE+PE)atB [Topic: Refraction at Curved Surface, Lenses & Power
of Lens]

98
Q89. A p-n photodiode is made of a material with a band Q92. A battery of 10 V and internal resistance 0.5Ω is
gap of 2.0 eV. The minimum frequency of the radiation connected across a variable resistance R. The value of R
that can be absorbed by the material is nearly for which the power delivered is maximum is equal to
(a) 10 × 1014 Hz (b) 5 ×1014Hz (a) 0.25Ω (b) 0.5Ω
(c) 1 × 1014 Hz (d) 20 × 1014 Hz (c) 1.0Ω (d) 2.0Ω
Ans: (b) Ans: (b)
–19
Solution: Eg = 2.0 eV = 2 × 1.6 × 10 J Solution: Power is maximum when r = R, R = r = 0.5Ω.
Eg = hv Chapter: Current Electricity
∴ = [Topic: Heating Effects of Current]
Q93. In an astronomical telescope in normal adjustment
= 0. 4833 × 1015 s–1 = 4.833 × 1014 Hz a straight black line of lenght L is drawn on inside part of
Chapter: Semiconductor Electronics Materials, Devices objective lens. The eye-piece forms a real image of this
[Topic: Solids, Semiconductors and P-N Junction line. The length of this image is l. The magnification of
Diode] the telescope is :
Q90. Two spheres A and B of masses m1 and m2
(a) (b)
respectively collide. A is at rest initially and B is moving
with velocity v along x-axis. After collision B has a (c) (d)
velocity in a direction perpendicular to the original Ans: (c)
direction. The mass A moves after collision in the Solution: Magnification by eye piece
direction. m=
(a) Same as that of B
(b) Opposite to that of B , ( )-
or,
(c) θ = tan–1 (1/2) to the x-axis
Magnification, M =
(d) θ = tan–1(–1/2) to the x-axis
Ans: (c) Chapter - Ray Optics and Optical
Solution: m2m1 [Topic: Optical Instruments]
Q94. Which of the following, when added as an
impurity, into the silicon, produces n-type semi-
conductor?
u=0 (a) Phosphorous
conservation of linear momentum along x-direction (b) Aluminium
m2v = m1vx ⇒ (c) Magnesium
along y-direction (d) Both b and c
Ans: (a)
⇒
Solution: Phosphorous (P) is pentavalent and silicon is
Note: Let A moves in the direction, which makes an tetravalent. Therefore, when silicon is doped with
angle θ with initial direction i.e. pentavalent impurity, it forms a n-type semiconductor.
Chapter: Semiconductor Electronics Materials, Devices
tan θ = [Topic: Solids, Semiconductors and P-N Junction
Diode]
tan θ =
Q95. In carbon monoxide molecule, the carbon and the
⇒θ = tan–1. / to the x-axis. oxygen atoms are separated by a distance 1.12 × 10–10m.
Chapter: Work, Energy and Power The distance of the centre of mass, from the carbon atom
[Topic: Collisions] is
Q91. For hydrogen gas, Cp – Cv = a and for oxygen gas, (a) 0.64 × 10–10m (b) 0.56 × 10–10m
–10
Cp– Cv = b, so the relation between a and b is given by (c) 0.51 × 10 m (d) 0.48 × 10–10m
(a) a = 16 b (b) 16 b = a Ans: (a)
(c) a = 4 b
(d) a = b
Solution:
Ans: (d) From definition of centre of mass.
Solution: Both are diatomic gases and Cp – Cv = R for all
gases.
Chapter: Kinetic Theory
[Topic: Displacement, Phase, Velocity & Acceleration = 0.64 × 10–10 m.
of SHM] Chapter: System of Particles and Rotational Motion

99
 [Topic: Angular Displacement, Velocity and (c) collector is positive and emitter is negative with
Acceleration] respect to the base
Q96. The particle executing simple harmonic motion has (d) collector is positive and emitter is at same potential as
a kinetic energy . The maximum values of the the base
potential energy and the total energy are respectively Ans: (c)
(a) K0/2 and K0 (b) K0 and 2K0 Solution: When the collector is positive and emitter is
(c) K0 and K0 (d) 0 and 2K0. negative w.r.t. base, it causes the forward biasing for each
Ans: (c) junction, which causes conduction of current.
Solution: We have, U + K = E Chapter: Semiconductor Electronics Materials, Devices
where, U = potential energy, K = Kinetic energy, E = [Topic: Junction Transistor]
Total energy. Q100. The dimensions of ( ) are
Also, we know that, in S.H.M., when potential energy is (a) [L1/2 T–1/2]
maximum, K.E. is zero and vice-versa. (b) [L–1 T]
⇒ (c) [L T–1]
Further, (d) [L–1/2 T1/2]
Ans: (c)
Solution: (µ0ε0)–1/2 = : speed of light
But by question, √
where ε0 = permittivity of free space
µ0 = permeability of free space
Hence, total energy, So dimensions are [LT–1]
Chapter: Units and Measurement
& [Topic: Dimensions of Physical Quantities]
Chapter: Oscillation
[Topic: Energy in Simple Harmonic Motion]
Q97. Potentiometer measures potential more accurately
PART 13. PHYSICS
because
(a) it measures potential in the open circuit
QUESTION BANK
(b) it uses sensitive galvanometer for null deflection
(c) it uses high resistance potentiometer wire
(d) it measures potential in the closed circuit Q1. A particle of mass m moves in the XY plane with a
Ans: (a) velocity v along the straight line AB. If the angular
Solution: Potentiometer measures potential current more momentum of the particle with respect to origin O is LA
accurately because it measure potential in open circuit when it is at A and LB when it is at B, then
and hence error in potential due to internal resistance is
removed.
Chapter: Current Electricity
[Topic: Motion of Charged Particle in Magnetic Field
& Moment]
Q98. In Young‟s double slit experiment, the fringe width
is found to be 0.4 mm. If the whole apparatus is
immersed in water of refrative index , without (a) LA = LB
disturbing the geometrical arrangement, the new fringe (b) the relationship between LA and LB depends upon the
width will be slope of the line AB
(a) 0.30 mm (b) 0.40 mm (c) LA < LB
(c) 0.53 mm (d) 450 microns (d) LA > LB
Ans: (a) Ans: (a)
Solution: mm Solution: Angular momentum = Linear momentum ×
distance of line of action of linear momentum about the
Chapter - Wave Optics origin.
[Topic: Diffraction, Polarization of Light & Resolving
Power]
Q99. An n-p-n transistor conducts when
(a) both collector and emitter are negative with respect to
the base
(b) both collector and emitter are positive with respect to
the base

100
 electrons and produce positively charged ions. They
become part of positive rays.
Chapter - Dual Nature of Radiation and Matter
[Topic: Matter Waves, Cathode & Positive Rays]
Q5. The output of OR gate is 1
(a) if either input is zero
(b) if both inputs are zero
LA = pA × d, LB = pB × d (c) if either or both inputs are
As linear momenta are pA and pB equal, (d) only if both inputs are 1
therefore, LA = LB. Ans: (c)
Chapter: System of Particles and Rotational Motion Solution: Output will be one if A or B or both are one. Y=
[Topic: Torque, Couple and Angular Momentum] A+B
Chapter: Semiconductor Electronics Materials, Devices
Q2. Masses MA and MB hanging from the ends of strings
[Topic: Digital Electronics and Logic Gates]
of lengths LA and LB are executing simple harmonic
motions. If their frequencies are fA = 2fB, then Q6. The dimensional formula for angular momentum is
(a) LA = 2LB and MA = MB/2 (a) [M0L2T–2]
(b) LA = 4LB regardless of masses (b) [ML2T–1]
(c) LA = LB/4 regardless of masses (c) [MLT–1]
(d) LA = 2LB and MA = 2MB (d) [ML2T–2]
Ans: (c) Ans: (b)
Solution: [Angular momentum ]
Solution: √ = [Momentum of inertia] × [Angular velocity]
= ML2 × T–1
and √ = ML2T–1
Chapter: Units and Measurement
∴ √ √ [Topic: Dimensions of Physical Quantities]
Q7. The moment of inertia of a uniform circular disc of
radius R and mass M about an axis touching the disc at
⇒ √ = , its diameter and normal to the disc is
regardless of mass. (a) MR2 (b) MR2
Chapter: Oscillation
(c) MR2 (d) MR2
[Topic: Time Period, Frequency, Simple Pendulum &
Spring Pendulum] Ans: (c)
Q3. An electron enters a region where magnetic field (B) Solution: M.I. of a uniform circular disc of radius „R‟
and electric field (E) are mutually perpendicular, then and mass „M‟ about an axis passing though C.M. and
(a) it will always move in the direction of B normal to the disc is
(b) it will always move in the direction of E
(c) it always possesses circular motion
(d) it can go undeflected also
Ans: (d)
Solution: When the deflection produced by electric field
is equal to the deflection produced by magnetic field,
then the electron can go undeflected.
Chapter: Moving Charges and Magnetic Field
[Topic: Motion of Charged Particle in Magnetic Field
& Moment]
Q4. In a discharge tube ionization of enclosed gas is From parallel axis theorem,
produced due to collisions between
(a) negative electrons and neutral atoms /molecules
(b) photons and neutral atoms /molecules Chapter: System of Particles and Rotational Motion
(c) neutral gas atoms/molecules [Topic: Moment of Inertia, Rotational K.E. and Power]
(d) positive ions and neutral atoms/molecules Q8. The phase difference between two waves,
Ans: (a) represented by
Solution: When electrons emitted from cathode collide y1 = 10–6 sin{100 t + (x/50) + 0.5} m
with gas molecules or atoms, they knock out outer y2 = 10–6 cos{100 t + (x/50)} m

101
where x is expressed in metres and t is expressed in
seconds, is approximately
(a) 1.5 radians (b) 1.07 radians
(c) 2.07 radians (d) 0.5 radians
Ans: (b)
Solution: y1 = 10–6 sin (100 t + x/50 + 0.5)m
–6
= 10 cos (100 t + x/50 – π/2 + 0.5)m
y2 = 10–6 cos (100 t + x/50)m
∴ θ = π/2 – 0.5 = 1.07 rad A particle shows distance - time curve as given in this
Chapter: Waves figure. The maximum instantaneous velocity of the
[Topic: Basic of Waves] particle is around the point:
Q9. The magnetic field at a distance r from a long wire (a) B
carrying current i is 0.4 tesla. The magnetic field at a (b) C
distance 2r is (c) D
(a) 0.2 tesla (b) 0.8 tesla (d) A
(c) 0.1 tesla (d) 1.6 tesla Ans: (b)
Ans: (a) Solution: The slope of the graph is maximum at C and
Solution: or hence the instantaneous velocity is maximum at C.
When r is doubled, the magnetic field becomes half, i.e., Chapter: Kinematics Motion in a Straight Line
now the magnetic field will be 0.2 T. [Topic: Non-uniform motion]
Chapter: Moving Charges and Magnetic Field Q12. The figure shows elliptical orbit of a planet m
[Topic: Magnetic Field, Biot-Savart's Law & Ampere's about the sun S. The shaded area SCD is twice the shaded
Circuital Law] area SAB. If t1 is the time for the planet to move from C
Q10. The figure shows a plot of photo current versus to D and t2 is the time to move from A to B then :
anode potential for a photo sensitive surface for three
different radiations. Which one of the following is a
correct statement?

(a) t1 = 4t2 (b) t1 = 2t2

(c) t1 = t2 (d) t1 > t2
Ans: (b)
Solution: According to Kepler‟s law, the areal velocity of
a planet around the sun always remains constant.
SCD : A1– t1 (areal velocity constant)
SAB : A2 – t2
(a) Curves (1) and (2) represent incident radiations of
same frequency but of different intensities.
(b) Curves (2) and (3) represent incident radiations of t 1 = t2 . (given A1 = 2A2)
different frequencies and different intensities.
(c) Curves (2) and (3) represent incident radiations of = t2 .
same frequency having same intensity. ∴t1 = 2t2
(d) Curves (1) and (2) represent incident radiations of
Chapter: Gravitation
different frequencies and different intensities.
[Topic: Kepler's Laws of Planetary Motion]
Ans: (a)
Solution: Retarding potential depends on the frequency Q13. The velocity of sound in any gas depends upon
of incident radiation but is independent of intensity. (a) wavelength of sound only
(b) density and elasticity of gas
Chapter - Dual Nature of Radiation and Matter
(c) intensity of sound waves only
[Topic: Electron Emission, Photon Photoelectric Effect
(d) amplitude and frequency of sound
& X-ray]
Ans: (b)
Q11. Solution: Velocity of sound in any gas depends upon
density and elasticity of gas.
Chapter: Waves
[Topic: Vibration of String & Organ Pipe]

102
Q14. A galvanometer has a coil of resistance 100 ohm =
and gives a full-scale deflection for 30 mA current. It is to
work as a voltmeter of 30 volt range, the resistance
required to be added will be
(a) (b) 1800
(c) (d) 1000Ω
Ans: (a)
Solution: Let the resistance to be added be R, then
30 = Ig (r + R)
∴ Chapter: Gravitation
= 1000 – 100 = 900 [Topic: Gravitational Field, Potential and Energy]
Chapter: Moving Charges and Magnetic Field Q18. Two waves of the same frequency and intensity
[Topic: Galvanometer and Its Conversion into Ammeter superimpose each other in opposite phases. After the
& Voltmeter] superposition, the intensity and frequency of waves will
Q15. In photoelectric effect the work function of a metal (a) increase
is 3.5 eV. The emitted electrons can be stopped by (b) decrease
applying a potential of –1.2 V. Then (c) remain constant
[1994] (d) become zero
(a) the energy of the incident photon is 4.7 eV Ans: (c)
(b) the energy of the incident photon is 2.3 eV Solution: We know that interference is said to be
(c) if higher frequency photon be used, the photoelectric constructive at point where resultant intensity is
current will rise maximum (are in phase) and destructive at points where
(d) when the energy of photon is 3.5 eV, the photoelectric resultant intensity is minimum or 0 (are in opposite
current will be maximum phase). Therefore, after the superposition, frequency and
Ans: (a) intensity of waves will remain constant.
Solution: hν = W0 + Ek = 3.5 + 1.2 = 4.7 eV Chapter: Waves
Chapter - Dual Nature of Radiation and Matter [Topic: Beats, Interference & Superposition of Waves]
[Topic: Electron Emission, Photon Photoelectric Effect Q19. Curie temperature is the temperature above which
& X-ray] (a) ferromagnetic material becomes paramagnetic
Q16. If a ball is thrown vertically upwards with a material
velocity of 40 m/s, then velocity of the ball after two (b) paramagnetic material becomes diamagnetic material
seconds will be (g = 10 m/s2) (c) paramagnetic material becomes ferromagnetic
(a) 15 m/s (b) 20 m/s material
(c) 25 m/s (d) 28 m/s (d) ferromagnetic material becomes diamagnetic
Ans: (b) material.
Solution: Initial velocity (u) = 40 m/s Ans: (a)
Acceleration a = –g m/s2 = –10 m/s2 Solution: Curie temperature is the temperature above
Time = 2 seconds which ferromagnetic material becomes paramagnetic
By Ist equation of motion, material.
v = u + at Chapter: Magnetism and Matter
v = 40 – 10 (2) = 20 m/s [Topic: The Earth's Magnetism, Magnetic Materials
Chapter: Kinematics Motion in a Straight Line and their Properties]
[Topic: Motion Under Gravity] Q20. In the spectrum of hydrogen, the ratio of the
Q17. A particle of mass M is situated at the centre of longest wavelength in the Lyman series to the longest
spherical shell of mass M and radius a. The magnitude of wavelength in the Balmer series is
the gravitational potential at a point situated at a/2 [2015 RS, 2013]
distance from the centre, will be (a)
(a) (b)
(b) (c)
(c) (d)
(d) Ans: (d)
Ans: (b) Solution: For Lyman series (2 → 1)
Solution: =R0 1=

103
For Balmer series (3 → 2) (c) contact is broken
=R0 1= (d) won't become bright at all
Ans: (c)
⇒ = = . /= Solution: When a circuit is broken, the induced e.m.f. is
largest. So the answer is (c) .
Chapter: Atoms Chapter: Electromagnetic
[Topic: Bohr Model & The Spectra of the Hydrogen [Topic: Motional and Static EMI & Applications]
Atom] Q25. When a hydrogen atom is raised from the ground
Q21. Which of the following is not a vector quantity? state to an excited state,
(a) speed (a) P.E decreases and K.E. increases
(b) velocity (b) P.E. increases and K.E decreases
(c) torque (c) both K.E. and P.E. decrease
(d) displacement (d) absorption spectrum
Ans: (a) Ans: (b)
Solution: A vector quantity has both magnitude and Solution: and ,where, r is the
direction. In the given options, speed has only magnitude,
therefore, it is non- vector or scalar quantity. radius of orbit which increases as we move from ground
to an excited state. Therefore, when a hydrogen atom is
Chapter: Kinematics Motion in a Plane
raised from the ground state, it increases the value of r.
[Topic: Vectors]
As a result of this, P.E. increases (decreases in negative)
Q22. The escape velocity from earth is 11.2 km/s. If a and K.E. decreases.
body is to be projected in a direction making an angle 45° Chapter: Atoms
to the vertical, then the escape velocity is [Topic: Bohr Model & The Spectra of the Hydrogen
(a) 11.2 × 2 km/s (b) 11.2 km/s
Atom]
(c) 11.2 / √ km/s (d) 11. √ km/s
Q26. The position vector of a particle ⃗ as a function of
Ans: (b)
time is given by:
Solution: Escape velocity does not depend on the angle
⃗ ( ) ( ) ̂
of projection.
Chapter: Gravitation Where R is in meter, t in seconds and ̂ and ̂ denote unit
[Topic: Motion of Satellites, Escape Speed and Orbital vectors along x-and y-directions, respectively.
Velocity] Which one of the following statements is wrong for the
motion of particle?
Q23. When air is replaced by a dielectric medium of
force constant K, the maximum force of attraction (a) Magnitude of acceleration vector is , where v is the
between two charges, separated by a distance velocity of particle
(a) decreases K-times (b) Magnitude of the velocity of particle is 8
(b) increases K-times meter/second
(c) remains unchanged (c) path of the particle is a circle of radius 4 meter.
(d) becomes times (d) Acceleration vector is along ⃗
Ans: (a) Ans: (b)
Solution: In air, Fair = Solution: Here,x = 4sin(2πt)...(i)
y = 4cos(2πt)...(ii)
In medium, Fm = Squaring and adding equation (i) and (ii)
x2 + y2 = 42 ⇒ R = 4
(decreases K-times) Motion of the particle is circular motion,
Chapter: Electrostatic Potential and capacitance accelerationvector is along –⃗ and its magnitude =
[Topic: Electric Field, Electric Field Lines & Dipole]
Velocity of particle, V = ωR = (2π) (4) = 8π
Q24. In the circuit of Fig, the bulb will become suddenly Chapter: Kinematics Motion in a Plane
bright if
[Topic: Relative Velocity in2D & Circular Motion]
Q27. The angle of contact between pure water and pure
glass, is
(a) 0º (b) 45º
(c) 90º (d) 135º
Ans: (a)
Solution: We know that angle of contact is the angle
(a) contact is made or broken between the tangent to liquid surface at the point of
(b) contact is made contact and solid surface inside the liquid. In case of pure
water and pure glass, the angle of contact is zero.

104
 Chapter: Mechanical Properties of Fluids Solution: Change in momentum along the wall
[Topic: Thermometry, Thermocouple& Thermal = mv cos60º – mv cos 60º = 0
Expansion] Change in momentum perpendicular to the wall
Q28. A charge q is located at the centre of a cube. The = mv sin60º – (– mv sin60º)= 2mv sin60º
electric flux through any face is ∴ Applied force =
(a) ( ) =
(b) √
( ) = √
(c) ( ) = √ newton
(d) Chapter: Dynamics Laws of Motion
( )
[Topic: Ist, IInd & IIIrd Laws of Motion]
Ans: (c)
Solution: Cube has 6 faces. Flux through any face is Q32. An electric kettle takes 4A current at 220 V. How
given by much time will it take to boil 1 kg of water from
temperature 20° C? The temperature of boiling water is
θ ( ) 100° C.
Chapter: Electrostatic Potential and capacitance (a) 6.3 min (b) 8.4 min
[Topic: Electrostatic Potential & Equipotential (c) 12.6 min (d) 4.2 min
Surfaces] Ans: (a)
Q29. In an a.c. circuit, the r.m.s. value of current, irms is Solution: Heat required to raise the temperature of 1kg
related to the peak current, i0 by the relation water from 20°C to 100°C is given by Q = ms∆θ = (1×
[1994] 4200 × 80) J
(a) √ Power of kettle (P) = VI = (220 × 4)W
(b) ∴ Time taken =
(c) = 381.81 sec = 6.36 min
Chapter: Thermal Properties
(d) [Topic: Calorimetry & Heat Transfer]
√
Ans: (d) Q33. A capacitor is charged by a battery. The battery is
Solution: removed and another identical uncharged capacitor is
√
connected in parallel. The total electrostatic energy of
Chapter: Alternating Current
resulting system :
[Topic: Alternating Current, Voltage & Power]
(a) decreases by a factor of 2 (b) remains the same
Q30. The constituents of atomic nuclei are believed to be (c) increases by a factor of 2 (d) increases by a factor of
(a) neutrons and protons 4
(b) protons only Ans: (a)
(c) electrons and protons Solution: When battery is replaced by another uncharged
(d) electrons, protons and neutrons. capacitor
Ans: (a) As uncharged capacitor is connected parallel
Solution: Nucleus contains only neutrons and protons. So, C' = 2C
Chapter: Nuclei
and Vc =
[Topic: Composition and Size of the Nucleus]
Q31. A 3 kg ball strikes a heavy rigid wall with a speed Vc =
of 10 m/s at an angle of 60º. It gets reflected with the
same speed and angle as shown here. If the ball is in ⇒Vc =
contact with the wall for 0.20s, what is the average force Initial Energy of system, Ui = … (i)
exerted on the ball by the wall?
Final energy of system, Uf = ( ). /
= . / …(ii)
From equation (i) and (ii)
Uf=
i.e., Total electrostatic energy of resulting system
decreases by a factor of 2
(a) 150 N
Chapter: Electrostatic Potential and capacitance
(b) zero
[Topic: Capacitors, Capacitance, Grouping of
(c) √ (d) 300 N Capacitors & Energy Stored in a Capacitor.]
Ans: (c)

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Q34. A 220 volts input is supplied to a transformer. The ∵l′=2l
output circuit draws a current of 2.0 ampere at 440 volts. ∴ A′ =
If the efficiency of the transformer is 80%, the current
drawn by the primary windings of the transformer is ∴ R′ = 4R = 4 × 4 Ω = 16 Ω
(a) 3.6 ampere (b) 2.8 ampere
Therefore the resistance of new wire becomes 16 Ω
(c) 2.5 ampere (d) 5.0 ampere
Chapter: Current Electricity
Ans: (d)
[Topic: Electric Current, Drift of Electrons, Ohm's
Solution: Law, Resistance & Resistivity]
Q39. The energy of the em waves is of the oder of 15
⇒ keV. To which part of the spectrum does it belong?
V1 = 220 V, I2 = 2.0 A, V2 = 440 V (a) Infra-red rays
(b) Ultraviolet rays
= =5A
(c) γ-rays
Chapter: Alternating Current (d) X-rays
[Topic: Transformers & LC Oscillations] Ans: (d)
Q35. In nuclear reactions, we have the conservation of Solution: Energy of x-ray is (100 ev to 100 kev)
(a) mass only Hence energy of the order of 15 kev belongs to x-rays.
(b) energy only Chapter - Electromagnetic Waves
(c) momentum only [Topic: Electromagnetic Spectrum]
(d) mass, energy and momentum Q40. The number of beta particles emitted by a
Ans: (d) radioactive substance is twice the number of alpha
Solution: Chapter: Nuclei particles emitted by it. The resulting daughter is an
[Topic: Mass-Energy & Nuclear Reactions] (a) isomer of parent
Q36. A mass of 1 kg is suspended by a thread. It is (b) isotone of parent
(i) lifted up with an acceleration 4.9 m/s2, (c) isotope of parent
(ii)lowered with an acceleration 4.9 m/s2. (d) isobar of parent
The ratio of the tensions is Ans: (c)
(a) 3 : 1 (b) 1 : 2 Solution: Isotopes of an element have the same atomic
(c) 1 : 3 (d) 2 : 1 number but different mass number. A radioactive
Ans: (a) substance when emits one alpha particles ( ) , its mass
Solution: In case (i) we have number reduces by 4 and charge no. reduces by 2 and
( ) = 1 × 4.9 after emission of two β-particles its charge no. increase
T1 = 9.8 + 4.9 = 14.7 N by 2 thus the charge no. i.e. atomic number remains the
In case (ii), l × g – T2 = 1 × 4.9 same.
T2 = 9.8 – 4.9 = 4.9 N Chapter: Nuclei
∴ [Topic: Radioactivity]
Q41. When milk is churned, cream gets separated due to
Chapter: Dynamics Laws of Motion
(a) centripetal force
[Topic: Motion of Connected Bodies, Pulleys] (b) centrifugal force
Q37. 110 joules of heat is added to a gaseous system (c) frictional force
whose internal energy is 40 J. Then the amount of (d) gravitational force
external work done is Ans: (b)
(a) 150 J (b) 70 J Solution: Cream gets separated from a churned milk due
(c) 110 J (d) 40 J to centrifugal force.
Ans: (b) Chapter: Dynamics Laws of Motion
Solution: ∆Q = ∆U + ∆W [Topic: Work]
⇒ ∆W = ∆Q – ∆U = 110 – 40 = 70 J Q42. An ideal gas A and a real gas B have their volumes
Chapter: Heat & Thermodynamics increased from V to 2V under isothermal conditions. The
[Topic: Specific Heat Capacity & Thermodynamic increase in internal energy
Processes] (a) will be same in both A and B
Q38. A wire of resistance 4 Ω is stretched to twice its (b) will be zero in both the gases
original length. The resistance of stretched wire would be (c) of B will be more than that of A
(a) 4 Ω (b) 8 Ω (d) of A will be more than that of B
(c) 16 Ω (d) 2 Ω Ans: (b)
Ans: (c) Solution: Under isothermal conditions, there is no
Solution: Resistance R = change in internal energy.

106
 Chapter: Heat & Thermodynamics
[Topic: Specific Heat Capacity & Thermodynamic ( )
Processes] Chapter: Nuclei
Q43. Two wires of the same metal have same length, but [Topic: Radioactivity]
their cross-sections are in the ratio 3:1. They are joined in Q46. Two bodies with kinetic energies in the ratio 4 : 1
series. The resistance of thicker wire is 10Ω. The total are moving with equal linear momentum. The ratio of
resistance of the combination will be their masses is
(a) 10 Ω (b) 20 Ω (a) 1 : 2 (b) 1 : 1
(c) 40 Ω (d) 100 Ω (c) 4 : 1 (d) 1 : 4
Ans: (c) Ans: (d)
Solution: Length of each wire = l ; Area of thick wire ( )
(A1) = 3A; Area of thin wire (A2) = A and resistance of Solution: ( )
⇒
thick wire (R1) = 10 Ω. Resistance ( ) (if l is ( )
⇒( ⇒
constant) )

⇒
[Given: p1=p2]
or, R2 = 3R1 = 3 × 10 = 30 Ω
Chapter: Work, Energy and Power
The equivalent resistance of these two resistors in series
[Topic: Energy]
= R1 + R2= 30 + 10 = 40Ω.
Chapter: Current Electricity
Q47. The equation of state, corresponding to 8g of O2 is
(a) PV = 8 RT
[Topic: Combination of Resistances]
(b) PV = RT /4
Q44. A beam of light composed of red and green rays is (c) PV = RT
incident obliquely at a point on the face of rectangular (d) PV = RT/2
glass slab. When coming out on the opposite parallel Ans: (b)
face, the red and green rays emerge from Solution: 8g of oxygen is equivalent to (1/4) mole.
(a) one point propagating in the same direction
(b) two points propagating in two different non-parallel So, PV =
directions Chapter: Kinetic Theory
(c) two points propagating in two different parallel [Topic: Speeds of Gas, Pressure & Kinetic Energy]
directions Q48. A battery is charged at a potential of 15V for 8
(d) one point propagating in two different directions hours when the current flowing is 10A. The battery on
Ans: (c) discharge supplies a current of 5A for 15 hours. The
Solution: Since refractive index for both the light are mean terminal voltage during discharge is 14V. The
different, so they emerge out moving in two different “watt-hour” efficiency of the battery is
parallel directions. (a) 87.5% (b) 82.5%
(c) 80% (d) 90%
Ans: (a)
Solution: Efficiency is given by
or 87.5 %
Chapter: Current Electricity
Chapter - Ray Optics and Optical [Topic: Kirchhoff's Laws, Cells, Thermo emf &
[Topic: Refraction of Light at Plane Surface & Total Electrolysis]
Internal Reflection]
Q49. A body is located on a wall. Its image of equal size
Q45. A radioactive element has half life period 800 is to be obtained on a parallel wall with the help of a
years. After 6400 years what amount will remain? convex lens. The lens is placed at a distance 'd' ahead of
[1989] second wall, then the required focal length will be
(a) (a) only
(b) (b) only
(c)
(c) more than but less than
(d)
(d) less than
Ans: (d)
Ans: (b)
Solution: No. of half lives,
Solution: Using the lens formula

107
Given v = d, for equal size image | | | |
By sign convention u = –d
or
Chapter - Ray Optics and Optical
[Topic: Refraction at Curved Surface, Lenses & Power
of Lens]
Q50. For a cubic crystal structure which one of the
following relations indicating the cell characteristics is
correct?
(a) a ≠ b ≠ c and α = β = γ = 90°
(b) a = b = c and α ≠ β ≠ γ = 90°
(c) a = b = c and α = β = γ = 90°
(d) a ≠ b ≠ c and α ≠ β and γ ≠ 90°
Ans: (c)
Solution: For a cubic crystal,
a = b = c and α = β = γ = 90°
Chapter: Semiconductor Electronics Materials, Devices
[Topic: Solids, Semiconductors and P-N Junction
Diode]

108