Spectrum Physics - March 2016 (gnv64).pdf

FROM THE EDITOR'S DESK

EXAM TACTICS:

Tackling JEE

Made Easy

The present structure of JEE includes a two tier examination

system i.e. JEE Mains and JEE Advanced. Let us explore them

one by one

JEE MAINS PAPER

In Physics

, most of the problems are numerical based along

with some graphs and other figure based questions.

Although, the pattern or frame of the paper is not fixed, but

the analysis of past three examinations suggest following

weightage (approximately) of topics in the

Mechanics 6-12 questions

Fluid Mechanics 1-2 questions

Heat and Thermodynamics 3-5 questions Waves and Oscillations 2-4 questions Electrostat and Electric Current 2-11 questions Magnetism and Magnetic Effects of Current 2-5 questions

Optics 3-6 questions

EMI and AC 2-4 questions

Modern Physics 2-5 questions

Miscellaneous 2-5 questions

The class coverage wise analysis of 2015 JEE Mains suggests

44% questions from class 11th syllabus and 56% questions

from class 12th syllabus.

The difficulty level wise analysis of 2015 JEE Mains paper

suggests the presence of 26% easy questions, 16% tough

questions and 58% medium level questions.

MANTRAS TO CRACK JEE MAINS

The two tier examination pattern was introduced in 2013.

The weightage of board percentage in final merit was one of

the features of this system. The JEE Mains paper of this

examination system is comparable with the IIT JEE

screening examination of late nineties and early twenties.

For successfully facing this kind of examination try to focus on

following points:

•

One must prepare for final exam (JEE Advanced) only,

i.e. do not target Mains examinations first. In my

opinion, “Chote exams target karne se kabhi bhi

bade results nahin milte”

, i.e. targeting smaller

exams never gives big results. If you consider JEE

Advanced as your target then Mains will be cleared

automatically.

In my opinion, Mains is only a screening to differentiate between

serious and non-serious aspirants. Thus, if you are among the

serious lot than Mains will be of no problem to you.

•

The analysis of 2013, 2014 and 2015 JEE Mains paper clearly

suggests the importance of class 11th syllabus for this paper, i.e.

46% in Physics, 33% in Chemistry and 57% in Mathematics. So, a

real grip over class 11th syllabus is required to succeed in JEE

Mains paper.

•

Nobody can deny the importance of NCERT Textbooks for this

examination. Thus, a good command over NCERT textbooks is

required for a great performance in this examination.

•

For capturing NCERT line by line very recently, I saw Master the

NCERT. A book, in which objective type questions are framed on

each and every line of NCERT. This book can be very helpful to get

success in this examination.

•

Choose offline or online paper in accordance of your comfort zone

and fine preparation. Both the examinations have the own

advantages or disadvantages. The difficulty levels are also more or

less same in both the formats of JEE Mains but on analysis it is

quite visible that the online paper is framed on a slightly tougher

side.

•

Although, the questions paper contains all the three subjects and

merit formation includes total marks obtained in the paper but

remember subject wise cut off also exists in JEE. So, a balance in

preparation is required subject wise to save you from such a

subject wise cut off.

As all the questions are single answer type MCQs hence, use

elimination technique in those questions where you don’t have

clue about the correct option, but you have clue about incorrect

options. Eliminate 3 incorrect options to get the most probable

correct option.

•

As negative marking is there, hence be cautious in answering the

questions. Do not take so much of risks.

JEE ADVANCED PAPER

The JEE Advanced paper is based upon old IIT JEE paper till 2013. It has

2 papers in total, i.e. first in the morning session (9 am to 12 pm) and

second in the afternoon session (2 pm-5pm). Both the papers contain 20

questions per subject, i.e. total 60 questions per paper. The question

paper pattern is not at all fixed.

In 2015 Advanced paper following pattern was followed:

PAPER 1

•

Section 1 with 8 Integer Type Questions

•

Section 2 with 10 MCQs having One or More than One

Correct Option

•

Section 3 with 2 Matrix Match Questions.

PAPER 2

•

Section 1 with 8 Integer Type Questions

•

Section 2 with 8 MCQs having One or More than One

Correct Option

•

Section 3 with 2 Paragraphs having 2 questions each.

In both the papers, section 2 and 3 had a negative marking of

50%.

However, in 2014 JEE Advanced questions paper following

pattern was followed :

PAPER 1

•

10 Multiple Choice Questions which are one or more than

one answer correct.

•

10 Integer Type Questions. All the 20 questions with 3

marks each with no negative marking.

PAPER 2

•

10 Multiple Choice Questions with only one answer

correct.

•

3 Paragraphs with 2 question each having only One

Answer Correct.

•

A Matching Type MCQs with one answer correct.

•

All the questions with 3 marks each and there is the

negative marking of 1/3, i.e. +3 for correct answer and –1

for wrong answer.

Thus, the uniqueness of JEE Advanced paper lies in 2 facts i.e.

Its surprising element

Every year JEE Advanced paper

incorporates one or more surprise elements e.g. in 2014,

negative marking in paper 2 was the surprised element.

Similarly, the frame of question paper was changed in 2013 as

compared to the one seen in 2012 IIT-JEE paper.

Its variety of questions

In both 2015 and 2014 JEE Advanced

paper following type of questions were asked

•

Single answer correct (MCQs)

•

More than one answer correct (MCQs)

•

Integer type question

•

Linked comprehension type

•

Matching type

The matrix match type which was the usual component of IIT

JEE papers before 2013, become reintroduced in 2015. Likewise

comprehension linked questions were 2 per passage in 2015

and 2014, while it was 3 per passage in IIT JEE before 2013.

So, we can say the nature of JEE Advanced paper is very dynamic.

No pattern is fixed for the paper. The syllabus wise distribution of

marks is more or less same as that seen in JEE Mains paper.

MANTRAS TO CRACK JEE ADVANCED

I personally feel that JEE Advanced is one of the toughest exam,

among those of same nature we have worldwide. I also believe in the

fact that more organised examinations are always easier to crack. So,

although it is toughest pattern or frame wise, but if we planned

scientifically than we can hit the bull’s eye.

For scientific planning following points must be kept in mind :

•

Dedicate your 2 years study to this particular exam only with

keeping in mind that the board weightage is one of the

considerations in this examination for final merit hence a dual

target planning has to be executed as given in chapter 5.

•

Phase wise learning and practice is a must to crack these kinds

of examinations. After having the analysis of previous year

exams some conclusions can be drawn easily like, those about

(i) the weightage of class 11th syllabus.

(ii) the difficulty level distribution among the questions of

previous year papers.

(iii) the topic wise, question wise distribution of questions

(iv) the topic wise, marks wise distribution

Such conclusions will help in strategy development according to

your strength and weaknesses.

•

If you will be approaching phase wise towards this

examination, then it is very clear that till Mains you would be

having very less opportunities to practice different question

types which are the features of JEE Advanced only. Utilise the

period between JEE Mains and JEE Advanced for this purpose.

A mass of students waste their some of this precious time in the

wait of Mains result. Don’t ever commit this kind of mistake.

Be optimistic, if you have JEE advanced as the target from day 1

in your mind than you would clear JEE Mains with surety.

•

The surprise attack is one of the most lethal weapon of this

examination, hence, those aspirants who don’t have any kind

of premature frame about the pattern of question paper in

their mind, has better chances of hitting the target. So, never

create the premature frame of question paper in your mind.

•

The large time span (9am to 5pm) is a very crucial factor in this

examination. Such a large time span creates tiredness of all

kind (i.e. physical as well as mental) in any person. So during

the gap of 2 hrs between Paper 1 and 2 try to relax as much as

you can. Remember! The more you relax, better you perform.

Last but not the least, a balance between the approaches towards

Subjective and Objective formats is required to succeed in IIT as a

whole as in present format the final merit also includes the

weightage of your Board examinations.

Always keep in mind the words of Sir Winston Churchill

“The pessimist sees difficulty in every opportunity.

The optimist sees opportunity in every difficulty.”

For mulae at a Glance

Gravitation

● Newton’s law of gravitation, F G m m

r

= 1 2

2 , where G is the universal gravitational constant, m1 and m2 are point mass bodies.

● _{Acceleration due to gravity, g} GM

R

= 2, where M and R are the mass and radius of the earth.

● _{Effect of altitude,} ′ = + g gR R h 2 2 ( ) and ′ = − ⎡ ⎣⎢ ⎤ ⎦⎥ g g h R 1 2 , where h is height of an object. ● _{Effect of depth,} ′ = ⎡ ₋ ⎣⎢ ⎤ ⎦⎥ g g d R

1 , where d is depth of the earth.

● _{Effect of rotation of earth, g}′ = −_{g Rω}2_cos2_λ

● _{Intensity of gravitational field, I} GM

r

= 2 , where distance r from the centre of the body of mass M.

● _{Gravitational potential, V} W M GM r = Work done ( )= − Test mass ( 0)

● _{Gravitational potential energy, U}= Gravitational potential × Mass of

the body = −GM×

r m

● _{Orbital speed of a satellite, v R} g

R h

O = + where, h is height of a satellite.

● _{Time period of a satellite, T}

R R h

g

=2π ( + )3

● _{Height of satellite above the earth’s surface, h}=⎡T R g _R

⎣ ⎢ ⎤ ⎦ ⎥ − 2 2 2 1 3 4_π /

● _{Total energy of satellite, E}=_PE+_KE = −

+

GMm R h

2( )

where, M is mass of the earth and m is the mass of the satellite.

● _{Binding energy of satellite, − =}

+ E GMm R h ( ) ● _{Escape speed, v} GM R gR e = = 2 2

Mechanical Properties of Solids

● _{Young’s modulus, Y} F A Mg r = Normal stress = = Longitudinal strain / / Δl l Δ l π2 _l ● _{Bulk modulus, B} p V V pV V = Normal stress = = − Volumetric strain _{− Δ} / _Δ ● _{Compressibility =} 1 _{= −} Bulk modulus ΔV pV ● _{Modulus of rigidity, G} F A L L FL A L =Tangential stress= = = Shearing strain / / Δ Δ F Aθ

● _{If a spring or a wire follows Hooke’s law, then spring constant or}

force constant is given by K= F =YA

Δl l, where l is the length of wire

and A is the area of cross-section of the wire.

● _{The number of atoms having interatomic distance r}

0, in length l of a wire is N= l /r0

● _{Poisson’s ratio} σ = Lateral strain = − = −

Longitudinal strain Δ Δ Δ R R/ R / l l l l R_Δ

● _{Elastic potential energy stored per unit volume of a strained body,}

u_{= ×}1

2 Stress × Strain = = × (Stress)2 (Strain)2

2Y 2

Y

where, Y is the Young’s modulus of elasticity of a solid body.

● _{Work done in a stretched wire,}

W_{= ×}1 _{× ×}

2 Stress Strain Volume

=1 × = × × 2F ΔL

1

2 Load Extension

● _{Relation between Y B G, , and σ}

(i) y=3B(1 2− σ) (ii) y=2G(1+σ) (iii) σ = − + 3 2 2 6 B G G B (iv) 9 1 3 y=B+G

● _{Thermal stress} = ×_{y Thermal strain} = ∝_y Δθ

where, α is the coefficient of linear expansion of the solid rod.

Mechanical Properties of Fluids

● _{1 bar} = 106

dyne/cm2 _●

1 torr = 1 mm of mercury

● _{1 pascal} = N/m2

● _Thrust = _{Pressure Area h g}× = ρ ×_A ● _{Relative density of the substance} =

° Density of substance Density of water at 4 C

● _{Apparent weight of the body of density ρ when immersed in a liquid}

of density _{σ = Actual weight − Upward thrust}

● _{Bernoulli’s equation is given by p}+1 _v + _gh=

2 2 ρ ρ constant where, p_{= pressure,} 1 2 2

ρv = kinetic energy per unit volume and ρgh = potential energy per unit volume.

● _{The velocity of efflux is given by v}= 2_gh

where, h= depth of the hole below the free surface.

● _{Viscous force is given by F A}dv

dy

= − η

where, _{η = coefficient of viscosity, A = area of layer of fluid in contact} and dv

dy= velocity gradient between the layers.

● _{Poiseuille’s formula V} pR L = π η 4 8

where, V_{= volume of liquid coming out of tube per second} L_{= length of the capillary tube}

R= radius of the capillary tube p= pressure difference across the tube

● _{Terminal velocity, v} r g T = − 2 9 2_{(ρ σ)} η

where, r= radius of the body, ρ = density of material, σ = density of fluid, _{η = coefficient of viscosity.}

● _{Surface tension, T= / lF}

where, F_{= force acting on one side of imaginary line of length l.}

● _{Ascent formula, h} S

r g

=2 cosθ ρ

where, S= surface tension of liquid, θ = angle of contact, r = radius of capillary tube, ρ = density of liquid, g = acceleration due to gravity.

● _{Excess pressure in a liquid drop of radius r, p} T

r

=2 where, T_{= surface tension of the liquid drop.}

● _{Excess pressure in a soap bubble of radius r, p} T

r

=4

● _{Excess pressure in a soap bubble of radius r, when it is inside a}

liquid, p T

r

=2

● _{If two droplets of radii r}

1, and r2 in vacuum coalesce under isothermal conditions, then the radius of the big drop is given by r= r1 +r

3 2

3 3

● _{The viscous force ( )F acting on a spherical body of radius r moving}

with velocity v through a medium of coefficient of viscosity η is given by Stoke’s law as F= 6 π ηr v

● _{Critical velocity of a liquid (v )}

c is given by v R D c e = η ρ where, _{η = coefficient of viscosity of liquid, ρ = density of liquid}

and D_{= diameter of tube and R}e = Reynold’s number.

1

General Instructions

1. This test consists of 30 questions.

2. Each question is allotted 4 marks for correct response.

3. 1/4 marks will be deducted for indicating incorrect response of each question. No deduction from the total score will

be made if no response is indicated in each question.

4. There is only one correct response for each question. Filling up more than one response in any question will be treated

as wrong response and marks for wrong response will be deducted according as per instructions.

1.

The speed of ripples ( )v on the surface of water depends

on the surface tension ( ),σ density ( )ρ and wavelength ( )λ . The square of speed ( )v is proportional to

(a) σ

ρλ (b) ρσλ (c) λσρ (d) ρ λσ

2.

Two bodies of masses m1 and m2 are initially at rest at

infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approaching at a separation distance r is (a) 2 1 2 1 2 G m m r ( − ) / ⎡ ⎣⎢ ⎤ ⎦⎥ (b) 2 1 2 1 2 G r (m m ) / + ⎡ ⎣⎢ ⎤ ⎦⎥ (c) r G m m 2 _{1 2} 1 2 ( ) / ⎡ ⎣ ⎢ ⎤_⎦⎥ (d) 2 1 2 1 2 G r m m ⎡ ⎣⎢ ⎤ ⎦⎥ /

3.

A long horizontal rod has a bead which can slide along its length and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about

A with constant angular acceleration α. If the

coefficient of friction between the rod and the bead is μ, and gravity is neglected, then the time after which the beads starts slipping is

(a) μ

α (b) μα (c) 1

μα (d) μα

4.

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 acceleration due to gravity) will be

(a) mg cosθ (b) mg sinθ (c) mg (d) mg

cosθ

5.

A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses (m₁=0 32. kg) and (m₂ =0 72. kg). Taking

g= 10 m/s2, find the work done (in joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest

(a) 6 J (b) 5 J (c) 8 J (d) 4 J

6.

A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to

(a) t1 2/ (b) t3 4/ (c) t3 2/ (d) t2

7.

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by U x a

x b x

( )= ₁₂ − ₆, where a and b are constants and x is the distance between the atoms. If the dissociation energy of the molecule is

D=[ (U x= ∞ −) U_{at equilibrium}], where D is (a) b a 2 6 (b) b a 2 2 (c) b a 2 12 (d) b a 2 4

30

m1 m2 T T a a

8.

Three rods of identical area of cross-section and made from the same metal from the sides of an isosceles triangle ABC, right-angled at B. The points A and B are maintained at temperatures T and 2 T, respectively. In the steady state, the temperature of the point C is TC.

Assuming that only heat conduction takes place, T

T C _is equal to (a) 1 2 1 ( ₊ ) (b) 3 2 1 ( ₊ ) (c) 1 2( 2₋1) (d) 1 3( 2₊1)

9.

The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process from an initial state (p V T₁, ₁, ) to the final state (p₂,V T₂, ) is equal to (a) zero (b) RlnT (c) R V V ln 1 2 (d) R V V ln 2 1

10.

Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V . The mass of the gas in A is mA and that in B is mB. The gas in each

cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found to be Δp and 1.5 Δp, respectively. Then,

(a) 4m_A=9m_B (b) 2m_A=3m_B (c) 3mA=2mB (d) 9mA=3mB

11.

For the cir cuit shown be low E1 = . V, R4 0 1= Ω,2

E₂ = . V, R6 0 2 = Ω and R4 3 = Ω. The current I2 1 is

(a) 1.6 A (b) 1.8 A (c) 1.25 A (d) 1.0 A

12.

A wire of resistance 10 Ω is bent to form a circle. P and Q are points on the circumference of the circle dividing it into a quadrant and are connected to a battery of 3 V and internal

resistance 1Ω as shown in the figure. The currents in the two parts of the circle are

(a) 6 23 A and 18 23 A (b) 5 26 A and 15 26 A (c) 4 25 A and 12 25 A (d) 3 25 A and 9 25 A

13.

Two equal negative charge −q are fixed at the fixed points ( , )0 a and ( ,0 − a on the Y-axis. A positive charge q) is released from the point (2a, )0 on the X-axis. The charge q will

(a) execute simple harmonic motion about the origin (b) move to the origin and remains at rest

(c) move to infinity

(d) execute oscillatory but not of simple harmonic motion

14.

Let there be a spherically symmetric charge distribution with charge density varying as ρ ρ( )r 0 r R 5 4− ⎛

⎝⎜ ⎞⎠⎟ upto r= , and ρ ( )R r = 0 for r R> , where r is the distance from the origin. The electric field at a distance r (r< from the origin is given byR)

(a) ρ ε0 0 4 5 4 r r R − ⎛ ⎝⎜ ⎞⎠⎟ (b) 4 3 5 3 0 π ρ ε0 r r R − ⎛ ⎝⎜ ⎞⎠⎟ (c) ρ ε0 0 4 5 3 r r R − ⎛ ⎝⎜ ⎞⎠⎟ (d) 4 3 5 4 0 ρ ε0 r r R − ⎛ ⎝⎜ ⎞⎠⎟

15.

A particle of mass M and charge q is at the mid-point between two other fixed similar charges each of magnitude Q placed at a distance 2d apart. The system is collinear as shown in the figure. The particle is now displaced by a small amount x x( < < along the lined) joining the two charges and is left to itself. It will now oscillate about the mean position with a time period (ε0 = permittivity of free space)

(a) 2 3 0 π M dε Qq (b) 2 2 0 3 π M dε Qq (c) 2 3 0 3 π M dε Qq (d) 2 3 0 3 π Mε Qq d

16.

The field normal to the plane of a wire of n turns and radius r which carries a current i is measured on the axis of the coil at a small distance h from the centre of the coil. This is smaller than the field at the centre by the fraction (a) 3 2 2 2 h r (b) 2 3 2 2 h r (c) 3 2 2 2 r h (d) 2 3 2 2 r h

17.

A solenoid has a core of material with relative permeability 500 and its windings carry a current of 1 A. The number of turns of the solenoid is 500 per metre. The magnetisation of the material is nearly

(a) 2 5. ×103Am−1 (b) 2 5. ×105Am−1

(c) 2 0. ×103Am−1 (d) 2 0. ×105Am−1

18.

A magnet is suspended in the magnetic meridian with an untwisted wire. The upper end of wire is rotated through 180° to deflect the magnet by 30° from magnetic meridian. When this magnet is replaced by another magnet, the upper end of wire is rotated through 270° to deflect the magnet 30° from magnetic meridian. The ratio of magnetic moments of magnets is

(a) 1 : 5 (b) 1 : 8 (c) 5 : 8 (d) 8 : 5

19.

Each atom of an iron bar ( 5 cm × 1 cm × 1 cm) has a magnetic moment 18. ×10−23 Am−2. Knowing that the density of iron is 7 78. ×103 kg- m−3, atomic weight is 56 and Avogadro’s number is 6 02. ×1023, the magnetic moment of bar in the state of magnetic saturation will be (a) 4.75 Am2 (b) 5.74 Am2 (c) 7.54 Am2 _{(d) 75.4 Am}2

31

R =2 1 Ω R =2 3 Ω R =4 2 Ω E =4 V1 E =6 V2 I1 I2 d d Q q Q P Q 3 V 1 Ω

20.

An inductor of inductance L= 400 mH and resistors of resistances R1= Ω and R2 2 = Ω are connected to a2

battery of emf 12V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t= 0. The potential drop across L as a function of time is (a) 6_e−5t _{V (b)} 12 3 t e t − _V (c) 6 1⎛ − 0 2 ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ − e t . _{V (d) 12e}−5t _V

21.

For an oscillation of L-C circuit, the maximum charge on the capacitor is q. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is (a) q 2 (b) q 2 (c) q 3 (d) q 3

22.

A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle 2θ. The earth’s magnetic field component in the direction perpendicular to swing is B. Maximum potential difference induced across the pendulum is

(a) 2 2 1 2 BLsin⎛θ (gL)/ ⎝⎜ ⎞⎠⎟ (b) BLsin 2θ (gL) ⎛ ⎝⎜ ⎞⎠⎟ (c) BLsin θ (gL)/ 2 3 2 ⎛ ⎝⎜ ⎞⎠⎟ (d) BLsin 2θ (gL) 2 ⎛ ⎝⎜ ⎞⎠⎟

23.

An α-particle of 5 MeV energy strikes with a nucleus of uranium at stationary at an scattering angle of 180°. The nearest distance upto which α-particle reaches the nucleus will be of the order of

(a) 1 Å (b) 10−10 _{cm (c) 10}−12 _{cm (d) 10}−15 _cm

24.

When a certain metal surface is illuminated with light of frequency ν, the stopping potential for photoelectric current is V0. When the same surface is illuminated by

light of frequency ν

2, the stopping potential is

V₀

4. The threshold frequency for photoelectric emission is (a) ν 6 (b) ν3 (c) 2 3 ν _(d) 4 3 ν

25.

Figure shows a circuit in which three identical diodes are used. Each diode has forward resistance of 20 Ω and infinite backward resistance. Resistors

R₁=R₂ =R₃ =50Ω. Battery voltage is 6 V. The current through R3is

(a) 50 mA (b) 100 mA (c) 60 mA (d) 25 mA

26.

A mixture consists of two radioactive materials A1 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

(a) 60 s (b) 80 s (c) 20 s (d) 40 s

27.

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength λ. If R is the Rydberg’s constant, the principal quantum number n of the excited state is

(a) λ λ R R− 1 (b) λ λR − 1 (c) λ λ R R 2 1 − (d) λ λ R − 1

28.

Diameter of a plano-convex lens is 6 cm and thickness at the centre is 3 mm. If the speed of light in the material of the lens is 2×108 m/s, find the focal length of the lens.

(a) 15 cm (b) 20 cm (c) 30 cm (d) 10 cm

29.

A car is fitted with a convex side-view mirror of focal length 20 cm. A second car 2.8 m behind the first car is overtaking the first car at a relative speed of 15 m/s. The speed of the image of the second car as seen in the mirror of the first one is

(a) 1

10 m/s (b)

1 15 m/s

(c) 10 m/s (d) 15 m/s

30.

In Young’s double slit experiment, the two slits acts as coherent sources of equal amplitude A and wavelength λ. In another experiment with the same set up; the two slits are of equal amplitude A but are incoherent. The ratio of the intensity of light at the mid-point of the screen in the first case to that in the second case is

(a) 1 : 2 (b) 2 : 1 (c) 4 : 1 (d) 1 : 1

32

h θ L R1 R2 R3 6 V D2 D3 D1 + – E R2 R1 S L

1. (a) As speed of ripples ( )v _{∝ σ ρ λ}a b c

Equating dimensions on both sides, we get [M L T0 1 −1]_∝[MT−2] [a ML−3] [ ]b L c

[M L0 1T−1]∝[ ]Ma+b[ ]L−3b+c[ ]T−2a

Equating the powers of M, L and T on both sides, we get

a+ b= 0,−3b+c= −1, 2a= −1 Solving above equations, we get

a₌1 2/ ,b_{= −}1 2/ ,c _{= −}1 2/ ∴ v∝σ1 2/ ρ−1 2/ λ−1 2/

v2∝ σ

ρλ

2. (b) Let velocities of their masses at distance r from each other

be v₁ and v₂, respectively.

By conservation of momentum, m v_{1 1}−m v_{2 2}=0

⇒ m v_{1 1=}m v_{2 2} ...(i) By conservation of energy, Change in potential energy = Change in kinetic energy

G m m r m v m v 1 2 1 1 2 2 2 2 1 2 1 2 = + ⇒ m v m v Gm m r 1 12 2 22 1 2 2 + = …(ii)

On solving Eqs. (i) and (ii), we get

v Gm r m m 1 2 2 1 2 2 = + ( ) and v Gm r m m 2 1 2 1 2 2 = + ( )

∴ Relative velocity of approach v v v G

r m m

app=| |1 +| |2 = ( 1+ 2)

2

3. (a) Let the bead starts slipping after time t .

For critical condition, Frictional force provides the centripetal force, i.e.

mω2L=μR=μm×a_t =μ αLm

⇒ m(αt)2L₌μ αmL

Time taken by bead to start slipping, ⇒ t = μ

α [As, ω α= t]

4. (d) When the whole system is accelerated towards left, then

pseudo force (ma) works on a block towards right.

For the condition of equilibrium, mgsinθ=macosθ⇒ a=gsin

cos θ θ ∴ Force exerted by the wedge on the block

R=mgcosθ+masinθ or R=mg + m⎛g ⎝⎜ ⎞⎠⎟ cos sin cos sin θ θ θ θ =mg (cos +sin ) cos 2_θ 2_θ θ R= ⎡mg ⎣⎢ ⎤ ⎦⎥

cos_θ [Q sin cos

2_θ+ 2_{θ= ]1}

5. (c) In the given condition, tension in the string

T m m m m g = + ⋅ 2 1 2 1 2 ( ) = × × _{× =} 2 0 36 072 108 10 4 8 . . . . N

and acceleration of each block

a m m m m g = − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × 2 1 1 2 072 0 36 072 0 36 . . . . 10 = 10 3 m/s 2

Let s is the distance covered by block of mass 0.36 kg in first second. s=ut + 1 at 2 2 _{⇒ 0} 1 2 10 3 1 10 6 2 + ⎛ ⎝⎜ ⎞⎠⎟× = m ∴ Work done by the string ( )W = Ts =4 8×10=

6 8

. J

6. (c) As, we know power

P Fv mav m dv dt = = = ⎛ ⎝⎜ ⎞⎠⎟ ⇒ P mdt =vdv ⇒ P m t v × = 2 2 ⇒ v P m t = ⎛ ⎝⎜ ⎞⎠⎟ 2 1 2 1 2 / / ( ) ⇒ Now, s v dt P m t dt = = ⎛ ⎝⎜ ⎞⎠⎟

∫

∫

2 1 2 1 2 / / ∴ s P m t = ⎛ ⎝⎜ ⎞⎠⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 2 3 1 2/ 3 2/ ⇒ s t∝ 3 2/

7. (d) As, potential energy function

U x a x b x ( )= − 12 6 F dU dx a x b x = − =12₁₃ −6₇ ⇒ x a b = ⎛ ⎝⎜ ⎞⎠⎟ 2 1 6/ U x( = ∞ = 0) U a a b b a b b a equilibrium = ⎛ ⎝⎜ ⎞⎠⎟ − ⎛ ⎝⎜ ⎞⎠⎟ = − 2 2 2 4 2 ∴ U x U b a b a ( = ∞ −) = − −⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = equilibrium 0 4 4 2 2 .

33

L R mg sin θ mg θ mg cos θ ma sin + θ ma ma cos θ θ θ a

8. (b) Q T_{B> ⇒ Heat will flow B to A via two paths (i) B to A (ii)}T_A

and along BCA as shown Rate of flow of heat in path BCA will be same, i.e.

Q t Q t BC CA ⎛ ⎝⎜ ⎞⎠⎟ = ⎛⎝⎜ ⎞⎠⎟ ⇒ K T T A a C ( 2 − ) =K T −T A a C ( ) 2 ⇒ T T C ₌ + 3 2 1 ( )

9. (d) The change in entropy of an

ideal gas ΔS ΔQ T

= …(i)

In isothermal process, temperature does not change, i.e. internal energy which is a function of temperature will remains same, i.e. ΔU = 0.

First law of thermodynamics gives

ΔU₌ΔQ_{− or 0 =}W _{ΔQ W or ΔQ W− =}

i.e., ΔQ = Work done by gas in isothermal process which went through from ( ,p V T1 1, ) to (p V T2, 2, )

⇒ ΔQ RT V V e = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ μ log 2 1 …(ii)

For 1 mole of an ideal gas, μ = 1, so from Eq. (i) and Eq. (ii), we get ΔS R V V R V V e = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ log 2 ln 1 2 1

10. (c) As, process is isothermal. Therefore, T is constant,

p V ∝ ⎛ ⎝⎜ ⎞⎠⎟ 1

volume is increasing, therefore, pressure will

decreases. In chamber A Δp p p RT V RT V RT V i f A A A = − =μ −μ =μ 2 2 ...(i) In chamber B 1.5 Δp p p RT V RT V RT V i f B B B = − =μ −μ =μ 2 2 ...(ii)

From Eqs. (i) and (ii), we get μ μ A B = 1 = 15 2 3 . ⇒ m M m M A B / / = 2 3 ⇒ 3mA=2mB

11. (b) Ap ply ing Kirchhoff’s law for the loops (1) and (2) as shown

in the figure.

For Loop (1), ₋2i₁₋2(i₁₋i₂)₊4₌0

⇒ 2i₁₋i₂₌2 …(i) For Loop (2), −2(i₁−i₂)+4i₂−6 =0

⇒ −i₁+ 3i₂=3 ...(ii) On solving Eq (i) and Eq (ii), we get i_{1= . A}18

12. (a) In the fol low ing figure,

Resistance of part PNQ, R₁ 10 4 2 5 = = . Ω and Resistance of part PMQ, R₂ 3 4 10 7 5 = × = . Ω Req= × + = × + = R R R R 1 2 1 2 7 5 2 5 7 5 15 8 2.5 . ( . . ) Ω Main cur rent, i =

+ ⎛ ⎝⎜ ⎞⎠⎟ = 3 15 8 1 24 23 A So, i i R R R 1 2 1 2 24 23 7 5 2 5 7 5 18 23 = × + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = . . . A and i₂ i i₁ 24 23 18 23 6 23 =( − )= − = A

13. (d) By symmetry of problem, the components of force on Q

due to charges at A and B along Y-axis will cancel each other along X-axis will add up and will be along CO.

Under the action of this force change q will move towards O. If at any time, charge q is at a distance x from O. Net force on charge q. F F qq a x net = = − + × 2 2 1 4 2 2 cos . ( ) θ πε0 x a x ( 2+ 2)1/2 i.e. F qqx a x net o = − ⋅ + 1 4 2 2 2 3 2 πε ( )/

Therefore, the restoring force F_net is not linear, motion will be oscillatory (with amplitude 2 a) but not Simple Harmonic Motion. 14. (c) Total charge, q = = ⎛ − ⎝⎜ ⎞⎠⎟

∫

ρdv

∫

ρ r π R r dr r r 0 0 2 0 5 4 4 = ⎛ − ⎝ ⎜ ⎞ ⎠ ⎟ = ⎡ − ⎣ ⎢ ⎤ ⎦ ⎥

∫

4 5 4 4 5 12 4 0 2 3 0 3 4 0 π ρ r r π ρ R dr r r R r E Kq r r r r R = = ⎡ − ⎣ ⎢ ⎤ ⎦ ⎥ 2 2 0 3 4 1 4 4 5 12 4 πε0 π ρ . ₌ ⎡ ₋ ⎣⎢ ⎤ ⎦⎥ ρ ε 0 0 4 5 3 r r R

34

R =2 1 Ω R =2 3 Ω R =4 2 Ω E =4 V1 E =6 V2 I1 I2 i1 i2 (i –i )1 2 2 1 P Q 3 V, 1 Ω N M i1 i2 i a a A B –q –q O θ q F C F x 2a (T) A √2 a√2 C (T )C a a TB B

15. (c) Restoring force on displacement of x F K Qq d x Qq d x KQq d x d x = − − + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = ⎡ ₋ − ₊ ⎣ ⎢ ⎤ ⎦ ( )2 ( )2 ( )2 ( )2 1 1 ⎥ = − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ K Qq dx d x 4 2 2 2 ( ) = ⎡ ⎣⎢ ⎤ ⎦⎥ K Qq dx d 4 4 if (d> > x) = ⎡ ⎣⎢ ⎤ ⎦⎥ K Qq x d 4 3 Acceleration, a = F = M KQq Md x 4 3 ⇒ ω 2 3 4 = KQq Md Time Period, T =2 =2 = 4 2 3 3 3 0 π ω π π ε Md KQq Md Qq

16. (a) Field at the centre B in

r 1 0 4 2 =μ × π π ₌μ0_⋅ 1 2 n r

Field at a distance h from the centre

B nir r h 2 0 2 2 2 3 2 4 2 = + μ π π . ( )/ = ⋅ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ μ0 2 3 2 2 3 2 2 1 nir r h r / = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − B h r 1 2 2 3 2 1 / = ⎛ − ⎝ ⎜ ⎞ ⎠ ⎟ B h r 1 2 2 1 3

2. [By Binomial theorem]

Hence, B₂ is less than B₁ by a fraction =3 2 2 2 h r . 17. (b) Here, n= 500 turns/m I= 1 A , μ_r = 500

As _{μr = +}1 _χ, where _{χ is the magnetic susceptibility of the} material or _χ=(_{μr −}1)

Magnetisation (M) =χ_H =(μ_r− ⋅ =1) H (500− ×1) 500 Am−1 =499×500 Am−1

=2 495. ×105 Am−1 ~ .−2 5×105 Am−1

18. (c) Let M1 and M2 be the mag netic mo ments of mag nets and

H is the hor i zon tal com po nent of earth’s field.

We have τ= MHsin . If φ is the twist of wire, then τθ = c , cφ being restoring couple per unit twist of wire.

⇒ cφ=MHsinθ Here, φ₁ 180 30 150 150 π 180 =( ° − ° =) ° = × rad φ₂ 270 30 240 240 π 180 =( ° − ° =) ° = × rad

So, cφ₁=M H₁ sin [For deflection θ θ =30 of I magnet]° cφ₂=M H₂ sin [For deflection θ θ =30 of II magnet]° Dividing φ φ 1 2 1 2 = M M M M 1 2 1 2 150 180 240 180 15 24 5 8 = = × × = = φ φ π π ⇒ M M1: 2=5 8:

19. (c) The number of atoms per unit volume in a specimen is

n N A A = ρ For iron ρ =7 8. ×103kgm−3 N_A=6 02. ×1026/kg mol A_{= 56} ⇒ n=7 8×10 ×6 02 ×10 = × − 56 8 38 10 3 26 28 3 . . . m

Total number of atoms in the bar is

N₀₌nV_{=8 38. ×10}28_×(5×10−2_{× 1×10}−2_{× ×1 10}−2₎

N0=419. ×1023

The saturated magnetic moment of bar =_419. ×₁₀23×_18. ×₁₀−23

= 7 54. Am2

20. (d) Electric field across BC, E

= LdI dt R I 1 2 2 + I I e t t 2= 0(1− −/0) I E R 0 2 12 2 6 = = = A τ = = = × − t L R 0 3 400 10 2 = 02. s ∴ I e t 2=6 1( − −/ .0 2)

Potential drop across L =E−R I2 2

=₁₂− ×_{2 6 1(} −_e−t/ .0 2₎

=_12e−t/ .0 2=_12e−5t_V

21. (b) Maximum energy stored in the capacitor

U q C

max= 2

2

The energy is stored equally in electric and magnetic fields. So, energy in electric field, E q

C = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 2 2 Now, q C q C ′ = 2 1 2 2 2

Net charge inside a capacitor,

⇒ q′ = q

2

22. (a) From the given figure,

⇒ h₌L(1₋cos )_θ …(i)

Maximum velocity at equilibrium is given by v2₌₂gh₌₂gL_{(1−cos )θ} = ⎛ ⎝⎜ ⎞⎠⎟ 2 2 2 2 gL sin θ ⇒ v= 2 gL 2 sin θ

Thus, maximum potential difference

V_max=B× =L B× 2 gLsin L 2 θ = ⎛ ⎝⎜ ⎞⎠⎟ 2 2 1 2 BLsin θ (gL)/

35

12 V R2 R1 S L I2 I1 D C A B h θ L

23. (c) At closest distance of approach

i.e. Kinetic energy = Potential energy

5 10 16 10 1 4 16 19 × × _. × − = ×( )( ) πε0 Ze Ze r For uranium, Z= 92, so r =5 3. ×10−12 cm

24. (b) According to Einstein’s photoelectric emission

K_max=hν φ− ₀

where, ν is the incident frequency and φ₀ is the work done of the metal.

As, Kmax =eV0, where V0 is the stopping potential.

Therefore, eV₀=hν φ− ₀ …(i)

and eV0 h

0

4 = 2 −

ν _{φ …(ii)}

From Eqs. (i) and (i), we get

h_{ν φ} h_ν 4 4 2 0 − = or _φ0 φ0 ν ν 4 2 4 = − =h −h ⇒ 3 4φ0 4 ν =h or _φ0 ν 3 =h Therefore, threshold frequency (ν₀)

ν0 φ0 ν ν 3 1 3 = = × = h h h .

25. (a) As diode is conducting in forward bias condition and now

conducting in reverse bias condition.

Diode D₁ is in forward bias, and diode D₂ is in forward bias but diode D₃ is reverse bias.

So, the figure can be drawn as Here, 20Ω, R₁Ω and R₃Ω are in series. Equivalent resistance =50+ 50+20=120Ω ∴ I V R = = 6 120 = 1 20 A ⇒ I_{= 50 mA}

26. (d) According to radioactive decay of half-life of an element,

we have N N N N t t 1 01₂₀ 2 02₁₀ 2 2 = = ( )/ , ( )/ , i.e. N1=N2 ⇒ 40 2 160 2 20 10 ( )t/ =( )t/ ⇒ 2 2 20 10 2 t t / ₌ − ⎛ ⎝ ⎜ ⎞_⎠⎟ ⇒ t t 20=10− ⇒ 2 t t 20−10= − or 2 t 20= ⇒ t = 40 s2

27. (a) According to Rydberg’s formula, 1 1₂ 1₂

λ = − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ R n_f n_i Here, n_f =1,n_i=n ∴ 1 1 1 1 2 2 λ= − ⎡ ⎣⎢ ⎤ ⎦⎥ R n ⇒ 1 1 1 2 λ= ⎡⎣⎢ − ⎤ ⎦⎥ R n …(i)

Multiplying Eq. (i) by λ on both sides, we get

1 1 1 2 = ⎡ − ⎣⎢ ⎤ ⎦⎥ λR n ⇒ 1 1 1 2 λR = − n ⇒ 1 1 1 2 n = −λR ⇒ 1₂ 1 n R R = λ − λ ⇒ n R R = − λ λ 1

28. (c) According to lens formula,

1 1 1 1 1 2 f = − R − R ⎡ ⎣⎢ ⎤ ⎦⎥ (μ )

The lens is plano-convex, i.e. R1= and RR 2= ∞

Hence, 1 1 f = R − (_μ ) ⇒ f = R − (μ 1)

Speed of light in medium of lens, v= ×2 108 m/s _{μ = =} × × = = c v 3 10 2 10 3 2 15 8 8 .

If r is the radius and Y is the thickness of lens (at the centre), the radius of curvature R of its curved surface in accordance with the figure is given by

R2=r2+(R−y) 2⇒ r2+ y2−2Ry=0 Neglecting y2_{, we get R} r y = = × = 2 2 2 6 2 2 0 3 15 ( / ) . cm

Hence, focal length of the lens,

f=

− = 15

15. 1 30 cm

29. (b) According to mirror formula,

1 1 280 1 20 v + − = 1 1 20 1 280 v = + ⇒ 1 14 1 280 v = + ⇒ v =280 15 v v u v I = − ⎛ ⎝⎜ ⎞⎠⎟ 2 0

∴ Speed of the image

vI = − × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − × 280 15 280 15 15 15 15 2 m/s _{= −} 1 15m/s

30. (b) Resultant intensity, I = +I₁ I₂+2 I I_{1 2}cos_ϕ At central position with coherent source (and I₁=I₂= )I₀

Icoh = 4I0 …(i)

In case of incoherent at a given point, φ varies randomly with time so (cos )φ_av = 0 ...(ii) ∴ I_{incoh = +}I₁ I₂₌2I₀ Hence, I I coh incoh =2 1

36

R r B A C O y (R–y) R1 R3 6 V 20 Ω + – 15 cm 2.8 m f = 20

arihantbooks.com

More than 10,000 Practice

Questions as a whole

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Questions to Measure Your Problem Solving Skills

General Instructions

1. This test consists of 30 questions.

2. Each question is allotted 4 marks for correct response.

3. 1/4 marks will be deducted for indicating incorrect response of each question. No deduction from the total score will

be made if no response is indicated in each question.

4. There is only one correct response for each question. Filling up more than one response in any question will be treated

as wrong response and marks for wrong response will be deducted according as per instructions.

1.

The SI unit of energy is J =kgm s2 -2, that of speed v is ms- 1 and of acceleration a is ms- 2. Which one of the formula for kinetic energy given below is correct on the basis of dimensional arguments?

[Given, m stands for mass of the body]

(a) K =m v2 2 (b) K=ma (c) K =1mv +ma 2 2 _{(d) K=}1_mv 2 2

2.

A body of mass 5 10´ -3 kg is launched upon a rough inclined plane making an angle of 30° with the horizontal. Find the coefficient of friction between the body and the plane if the time of ascent is half of the time of descent.

(a) 0.346 (b) 0.436 (c) 0.463 (d) 0.364

3.

A uniform chain of length L and mass M over hangs a horizontal table with its two-third part on the table. The friction coefficient between the table and the chain is m. The work done by the friction during the period, the chain slips off the table is

(a) _-1 4m MgL (b) -2 9m MgL (c) -4 9m MgL (d) -6 7m MgL

4.

A neutron travelling with velocity u and kinetic energy K collides head on elastically with the nucleus of an atom of mass number A at rest. The fraction of its kinetic energy retained by the neutron even after the collision is

(a) 1 1 2 -+ æ è ç ö ø ÷ A A (b) A A + -æ è ç ö ø ÷ 1 1 2 (c) A A -æ èç öø÷ 1 2 (d) A A + æ èç öø÷ 1 2

5.

One end of a uniform rod of mass m1, uniform area of

cross-section A is suspended from the roof and the mass m2 is suspended from the other end. What is the

stress at the mid-point of the rod?

(a) (m m) .g A 1+ 2 _(b) (m m) .g A 1- 2 (c) (m / ) m) . A g 1 2 + 2 é ëê ù ûú (d) m m A g 1+ 2 2 é ëê ù ûú ( / ) .

6.

Two moles of an ideal monoatomic gas at 27°C occupies a volume of V. If the gas is expanded adiabatically to the volume 2V, then the work done by the gas will be Yæ = R=

èç öø÷

5

3, 8 31. J mol-K/

(a) + 2767.23 J (b) 2627.23 J (c) 2500 J (d) - 2500 J

7.

Use Moseley’s law with b= 1 to find the frequency of the Ka X-rays of La (Z= 57 if the frequency of)

Ka X-rays of Cu(Z= 29 is known to be 188 10) . ´ 18Hz

(a) 7.52 10 Hz´ 18 (b) 3.25 10 Hz´ 16 (c) 8.51 10 Hz_´ 19 _{(d) 9.1 10 Hz´} 15

8.

A small particle of mass m is

projected at an angle q with the

X-axis with an initial velocity v₀ in the X-Y plane as shown in the figure. At a time t v

g

< 0sinq_,

the angular momentum of the particle is

θ

X Y

(a) 1 2 0 2 mg v t cosq i$ (b) _{- mg v t0} 2_{cosq j}$ (c) mg v t₀ cosq k$ (d) -1 2 0 2 mgv t cosq k$

9.

A ball falls on an inclined plane of inclination q from a height h above the point of impact and makes a perfectly elastic collision where will it hit the inclined plane again

(a) 8h

sin_q (b) 4h sinq (c) 8h sinq (d) 4h sin_q

10.

A thin uniform annular disc of mass M has outer radius 4R and inner radius 2R. The work required to take a unit mass from point P on its axis to infinity is (a) 2 7 4 2 5 GM R ( - ) (b) -2 -7 4 2 5 GM R ( ) (c) GM R 4 (d) 2 5 2 1 GM R ( - )

11.

Four rods of identical cross-sectional area and made from the same metal form the sides of square. The temperature of two diagonally opposite points are T and 2 T respectively, in the steady state. Assuming that only heat conduction takes place, what will be the temperature difference between other two points? (a) ( 2 1) 2 + T (b) 2 2 1 ( + ).T (c) 0 (d) None of these

12.

An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus on film?

(a) 7.2 (b) 2.4 (c) 3.2 (d) 5.6

13.

A sphere and a cube of same material and same volume. One heated upto the same temperature and allowed to cool in the same surroundings. The ratio of the amounts of radiation emitted will be

(a) 1 : 1 (b) 4 3 1 p_{: (c) p} 6 1 1 3 æ èç öø÷ / : (d) 1 2 4 3 1 2 3 p æ èç öø÷ / :

14.

In Young’s double slit experiment, when violet light of a wavelength 4358 Å is used, then 84 fringes are seen in the field of view, but when sodium light of certain wavelength is used, then 62 fringes are seen in the field of view, the wavelength of a sodium light is

(a) 6893 Å (b) 5904 Å (c) 5523 Å (d) 6429 Å

15.

A transparent solid cylinderical rod has a refractive index of 2/ 3. It

is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle q for which the light ray grazes along the wall of the rod is

(a) sin- æ èç öø÷ 1 1 2 (b) sin - æ è ç ö ø ÷ 1 3 2 (c) sin - æ èç öø÷ 1 2 3 (d) sin - æ èç öø÷ 1 1 3

16.

When two tunning forks (fork 1 and fork 2) are sounded simultaneously, 4 beat s- 1 are heard. Now, some tape is attached on the prong of the fork 2. When the tunning forks are sounded again, 6 beats s-1 are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?

(a) 200 Hz (b) 202 Hz (c) 196 Hz (d) 204 Hz

17.

A particle of charge q and mass m starts moving from the origin under the action of an electric field,

E=E₀ $i and B B= ₀ $i with a velocity, v v= ₀ $j. The

speed of the particle will becomes 5 2/ v after a0

time (a) mv qE 0 _(b) mv qE 0 2 (c) 3 2 0 mv qE (d) 5 2 0 mv qE

18.

The current in the primary circuit of a potentiometer is 0.2 A. The specific resistance and cross-section of the potentiometer wire are 4´10-7W m and 8´10-7m2, respectively. The potential gradient will be equal to

(a) 0.2 V/m (b) 1 V/m (c) 0.3 V/m (d) 0.1 V/m

19.

Two straight long conductors AOB and COD are perpendicular to each other and carry currents i1 and

i₂. The magnitude of the magnetic induction at a point P at a distance a from the point O in a direction perpendicular to the plane ABCD, is

(a) m p 0 1 2 2 a(i + i ) (b) mp 0 1 2 2 a(i -i ) (c) m p 0 12 22 1 2 2 a(i i ) / + (d) m p 0 1 2 1 2 2 a i i i i . ( + )

20.

Two conductors have the same resistance of 0°C, but their temperature coefficients of resistance are a1

and a2. The respective temperature, coefficients of

their series and parallel combinations are nearly

(a) (a1 a2) 2 + _{, (a a )} 1+ 2 (b) (a1+a2), (_{a1 a2}) 2 + (c) (a1+a2), a a a a 1 2 1 2 . ( + ) (d) (a₁ a₂) 2 + _, (a₁ a₂) 2 +

21.

A parallel plate capacitor C with plates of unit area and the separation d is filled with a liquid of dielectric constant

K= 2, the level of liquid is d/3, initially. Suppose, the

liquid level decreases at a constant speed v, the time constant as a function of time t is

(a) 6 5 3 0 e R d+ vt (b) (15 9 ) 2 3 9 0 2 2 2 d vt R d dvt v t + - -e (c) 6 5 3 0 e R d- vt (d) (15 9 ) 2 3 9 0 2 2 2 d vt R d dvt v t -+ -e

39

2R 4R 4R θ d d/3 C R

22.

A 10mF capacitor and a 20 mF capacitor are connected in series across 200 V supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. What is the potential difference across each capacitor?

(a) 800

9 V (b) 400 V (c) 800

3 V (d) 200 V

23.

A candidate connects a moving coil voltmeter

V and a moving coil

ammeter A and resistor

R as shown in the

figure. If the voltmeter reads 10 V and the ammeter reads 2 A, then R is

(a) equal to 5W (b) greater than 5_W (c) less than 5W

(d) greater or less than 5W depending upon its material

24.

A rectangular loop has a sliding connector PQ of

length l and a resistance RW and it is moving with the speed v as shown.

The set up is placed in the uniform magnetic field going into the plane of the paper. The three currents

I₁,I₂ and I are (a) l₁ l₂ l 6 = =B v R, l l =B v R 3 (b) l l l 1= - 2= B v R, l l =B v R (c) _{l1 l2} l 3 = = B v R, l l =2 3 B v R (d) l l l l 1= 2= = B v R

25.

A solid sphere of radius R has a charge q distributed in its volume with a charge density r = kra, where k and a are constants and r is the distance from its centre. If the electric field at r=R

2 is 1

8 times that are

r= , then the value of a isR

(a) 2 (b) 4 (c) 6 (d) 8

26.

Two positive charges of magnitude q are placed at the end of a side 1 of a square of side 2a. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge q moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is

(a) 1 4 2 1 1 5 0 pe qq a -æ èç öø÷ (b) zero (c) 1 4 2 1 1 5 0 pe qq a + æ èç öø÷ (d) 1 4 2 1 2 5 0 pe qq a -æ èç öø÷

27.

Two inductors L1 and L2 are connected in parallel and

a time varying current flows as shown in the figure. The ratio of current, i

i 1 2 at any time is (a) L L 2 1 (b) L L 1 2 (c) L L L 1 2 1 22 ( + ) (d) L L L 2 2 1 22 ( + )

28.

A p-n junction (D) shown in the figure can act as a rectifier. An alternating current source ( )V is

connected in the circuit. The current ( )I in the

resistor ( )R can be shown by

29.

The graph given below represents the I-V characteristics of a Zener diode. Which part of the characteristic curve is most relevant for its operation as a voltage regulator?

(a) ab (b) bc (c) cd (d) de

30.

A fully charged capacitor C with initial charge q₀ is connected to a coil of self-inductance L at t= 0. The time at which the equation is stored equally between the electric and magnetic field is

(a) p 4 LC (b) p LC (c) 2p LC (d) LC

40

i i L1 L2 i1 i2 A R D C V R D V

Answers

1. (d) 2. (a) 3. (b) 4. (a) 5. (c) 6. (a) 7. (a) 8. (a) 9. (c) 10. (a) 11. (c) 12. (d) 13. (c) 14. (b) 15. (d) 16. (c) 17. (b) 18. (d) 19. (c) 20. (d) 21. (a) 22. (a) 23. (c) 24. (c) 25. (a) 26. (a) 27. (a) 28. (b) 29. (d) 30. (a)

v R Ω R Ω R Ω I I I1 I2 P Q (a) I t (b) I t (c) I t (d) I t Forward bias Reverse bias V d c b a e Current VZ I (mA) I (μA)

Paper 1

One or More Than One Option Correct Type

1.

A solid sphere of radius R and density r is attached to one end of a massless spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density 3r. The complete arrangement is placed in liquid of density 2r and is allowed to reach equilibrium. Choose the correct statement (s).

(a) The net elongation of spring is 4

3

3

p rR g k (b) The net elongation of spring is 8

3

3

p rR g k (c) The light sphere is partially submerged

(d) The elongation of spring due to small sphere is

4 3

3

p rR g k

2.

In a modified Young’s double slit experiment, a monochromatic uniform and parallel beam of light of wavelength 6000 Å and intensity (10/p) W/m2 is incident normally on two circular apertures A and B of radii 0.001 m and 0.002 m, respectively. A perfectly transparent film of thickness 2000 Å and refractive index 1.5 for wavelength 6000 Å is placed in front of aperture A . Calculate the power (in watts) received at the focal spot F of the lens. The lens is symmetrically placed with respect to the

aperture. Assume that 10% of the power received by each aperture goes in the original direction and is brought to

the focal point. Also, find the phase difference.

(a) The power received at the focal spot F of the lens is 7_´10-6 _W

(b) The value of phase difference is _p/3

(c) The power received at the focal spot F of the lens is

9 6. ´10-7 W

(d) None of the above

3.

Calculate the angular frequency of the system shown in the figure. Friction is absent everywhere and the threads, spring and pulleys are massless. Given that

m_A =m_B=m

(a) w = 3k/4m (b) w = 4k/5m (c) w = 2k/3m (d) w = 6k/5m

4.

A wooden rod weighing 25 N is mounted on a hinge below the free surface of water as shown. The rod is 3 m long and uniform in cross-section and the support is 1.6 m below the free surface. The cross-section of rod is 9.5 10´ -4 m2 in area. The density of water is 1000 kg/m

3

. Assume buoyancy to act at centre of immersion

g =9.8 m/ s2. Also, find the reaction on the hinge in this position. (a) 1.5 N (b) 1.1 N (c) 2.6 N (d) 1.6 N

41

A B L F B A k 3 m 1.6 m α