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AIIMS PHYSICS – 2004

In case of questions of Assertion-Reason, follow the following instructions:

- If both Assertion and Reason are true and the reason is the correct explanation of the assertion, then mark a.

- If both Assertion and Reason are true but the reason is not the correct explanation of the assertion, then mark b.

- If Assertion is true statement but Reason is false, then mark c.

- If both Assertion and Reason are false statements, then mark d.

PHYSICAL WORLD AND MEASUREMENT

1. Using mass (M), length (L), time (T) and current (A) as fundamental quantities, the dimensional formula of permittivity is: a) [ML-2T2A] b) [M-1L-3T4A2] c) [MLT-2A] d) [ML2T-1A2]

Ans. (b) Hint: Putting the dimensions for quantities in

the expression containing 0. From Coulomb’s law, two stationary

point charges q1 and q2 attract/repel each other with a force F which is directly proportional to the product of charges and inversely proportional to the square of distance r between them

That is, F = 2

21

04π

1

r

q q

0 = r F

qq2

21

4π

1

Dimensions of permittivity

0 = 2 of dimensions of dimensions

of dimensions

rF

q

2

[0] = ]ATL[M ][L ] [MLT

]T[A 2431

22

22 -

MOTION IN ONE DIMENSION

1. vrms, vav and vmp are root mean square,

average and most probable speeds of molecules of a gas obeying Maxwellian velocity distribution. Which of the following statements is correct?

a) vrms < vav < vmp b) vrms > vav > vmp c) vmp < vrms < vav d) vmp > vrms > vav

Ans. (b) Hint: Root mean square speed. The root

mean square speed is used to measure the velocity of particles in a gas. It is given by

vrms = M

RT

M

RT 1.732

3 …(1)

where M is molar mass and R is gas constant, T is temperature. Most probable speed vp, is the speed

most likely to be possessed by any molecule in the system.

vav = M

RT

M

RT 1.41

2 … (2)

whereas mean speed is

vmp = M

RT

M

RT 1.6

8

π … (3)

From Eqs. (1), (2) and (3), we conclude that vrms > vav > vmp 2. Which of the following velocity-time graphs

shows a realistic situation for a body in motion?

Ans. (b) Hint: Except graph (b) the other three graphs

shows the motion of the body with more than one velocity which is not possible for realistic situation.

3. Assertion: The drive in a vehicle moving with a constant speed on a straight road is in a non-inertial frame of reference. Reason: A reference frame in which Newton’s laws of motion are applicable is non-inertial.

Ans. (c) Hint: If we take a body resting with respect

to two frames with origin O and O’ then the body will be at rest. But the frame (vehicle) is moving with constant speed, so, this frame can not be an inertial frame

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of reference or it is a non inertial frame of reference. But the frame in which Newton’s law of motion are applicable is an inertial frame.

WORK, ENERGY AND POWER

1. A bomb of mass 3.0 kg explodes in air into two pieces of masses 2.0 kg and 1.0 kg. The smaller mass goes at a speed of 80 m/s. The total energy imparted to the two fragments is: a) 1.07 kJ b) 2.14 kJ c) 2.4 kJ d) 4.8 kJ

Ans. (d) Hint: From law of conservation of momentum,

when no external force acts upon a system of two (or more) bodies then the total momentum of the system remains constant.

Momentum before explosion = momentum after explosion Since bomb v at rest, its velocity is zero, hence mv = m 1 v1 + m2v2

3 0 = 2v1 + 1 80

v1 = - s / m 40 - 2

80

Total energy imparted is

KE = 222

211

2

1

2

1vm vm

= 22 (80) 1 2

1 40) (- 2

2

1

= 1600 + 3200 = 4800 J = 4.8 kJ Note: Since velocity of second piece is

negative, it indicates that the piece is moving in opposite direction to the direction of the other piece.

ROTATIONAL MOTION

1. The direction of the angular velocity vector is along: a) The tangent to the circular path b) The inward radius

c) The outward radius d) The axis of rotation

Ans. (d) Hint: Angular velocity is the vector quantity

which represents the process of rotation (change of orientation) that occurs at as instant of time. For a rigid body it supplements translational velocity of the centre of mass to describe the full motion. The line of direction of the angular velocity is given by the axis of rotation and the right hand rule indicates the direction.

Note: If you curl the fingers of your right hand to follow the direction of the rotation, then the direction of the angular velocity vector is indicated by your curled fingers.

2. In an orbital motion, the angular momentum vector is: a) Along the radius vector b) Parallel to the linear momentum c) In the orbital plane d) Perpendicular to the orbital plane

Ans. (d) Hint: If a body is rotating about an axis, then

the sum of the moments of the linear moment of all the particles about the given axis is called the angular momentum of the body about that axis.

J = I = mrv

Since direction of velocity is perpendicular to

orbital plane and J v, therefore in an orbital motion the angular momentum vector is perpendicular to the orbital plane.

3. Assertion: There are very small sporadic changes in the speed of rotation of the earth. Reason: Shifting of large air masses in the earth’s atmosphere produce a change in the moment of inertia of the earth causing its speed of rotation to change.

Ans. (a) Hint: Earth and moon being considered one

system effect of moon on tides takes place, hence sporadic changes of speed of rotation of earth are observed.

Also from law of conservation of angular momentum we have when external torque is acting upon a body rotating about an axis, then the angular momentum of the body remains constant.

J = I = constant

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When large air masses in the earth’s atmosphere shift they cause a change in moment of inertia and since angular momentum is to be maintained constant, the angular velocity or speed of rotation changes.

PROPERTIES OF BULK MATTER

1. In old age arteries carrying blood in the human body become narrow resulting in an increase in the blood pressure. This follows from: a) Pascal’s law b) Stokes’ law c) Bernoulli’s principle d) Archimedes’ principle

Ans. (c) Hint: From Bernoulli’s principle in the horizontal

stream line motion of a liquid the sum of pressure (P) kinetic energy per unit volume of the liquid at any point is constant, hence

+ constant 2

12

ρv

It is clear from this equation that is a flowing liquid where the velocity of flow is less the pressure is large and vice-versa. Hence, arteries carrying blood become narrow resulting in increase in the blood pressure.

2. A sphere of mass M and radius R is falling in a viscous fluid. The terminal velocity attained by the falling object will be proportional to: a) R 2 b) R c) 1/R d) 1/R 2

Ans. (a) Hint: Since sphere is moving with constant

velocity, there is no acceleration in it. When the sphere of radius R is falling in

a liquid of density and coefficient of

viscosity it attains a terminal velocity v, under two forces

(i) Effective force acting downward

= V ( - ) g = 3

4 R 3 ( - ) g

Where is density of sphere.

(ii) Viscous forces acting upwards = 6 Rv

Since the sphere is moving with a constant velocity v, there is no acceleration in it, the net force acting on it must be zero. That is

6Rv = 3

4 R 3 ( - ) g

v = η

g σ) - (ρ 2

R

9

2

v R 2

Thus, terminal velocity is proportional to the square of its radius.

3. Assertion: Smaller drops of liquid resist deforming forces better than the larger drops. Reason: Excess pressure inside a drop is directly proportional to its surface area.

Ans. (c) Hint: Also excess pressure inside a bubble is

Where R is radius of bubble. Since excess pressure is inversely

proportional to radius, hence excess pressure inside a smaller drop is large as compared to larger drop due to which smaller drop resists deforming forces better than a larger drop

Also P = A R

W

R

T

Δ

44

A

W T Δ

P AΔ

1

Hence, excess pressure is inversely proportional to surface area.

HEAT AND THERMODYNAMICS

1. N moles of a monoatomic gas is carried round the reversible rectangular cycle ABCDA as shown in the diagram. The temperature at A is T0. The thermodynamic efficiency of the cycle

is:

a) 15% b) 50% c) 20% d) 25% Ans. (b) Hint: Maximum efficiency of reversible process is given by

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= 1 - ' T

T

At AP 0V 0 = nRT0 At B 2 P0V0 = nRT’ T ’ = 2T0

= 1 - 50% 2

1

0

0

T

T

2

2. Suppose the sun expands so that its radius becomes 100 times its present radius and its surface temperature becomes half of its present value. The total energy emitted by it then will increase by a factor of: a) 104 b) 625 c) 256 d) 16

Ans. (b) Hint: From Stefan’s law, if the emissive

power of a body at absolute temperature T be e, then the energy emitted by its unit

area per second is T 4 e, also if A is the surface area of the body, then

E = T 4eA

When R = 100 R and T = 2

T then

energy emitted is

E’ 4 (100R)2

4

2

T

424π TR

4

1002

E’ = E

2

4

100

625 E

E'

3. Assertion: The melting point of ice decreases with increase of pressure. Reason: Ice contracts on melting.

Ans. (b) Hint: On increasing the pressure, melting point

of ice decreases. Certain substances do not expand but contract when heated within a certain range of temperature. Water is one

such example. If ice is taken at 0C and start heating it we find that in the temperature

range 0C to 4C it contracts instead of expanding.

4. Assertion: Thermodynamic process in nature are irreversible. Reason: Dissipative effects cannot be eliminated.

Ans. (a) Hint: The thermodynamics process is

irreversible, as there always occurs a loss of energy due to energy spent in working against the dissipative force which is not recovered back. Other irreversible process also occurs in nature

such as friction where extra work is required to cancel out the effect of friction.

5. Assertion: In a pressure cooker the water is brought to boil. The cooker is then removed from the stove. Now on removing the lid of the pressure cooker, the water starts boiling again. Reason: The impurities in water bring down its boiling point.

Ans. (c) Hint: We known that the boiling point of

liquid increases with increase of vapour pressure. In a cooker water with the food to be cooked, is heated, such that confined water vapours raise the superincumbent pressure. As a result, water boils at temperature higher than

boiling point 100C. Now when the cooker removed from the stove and its lid is also removed, pressure decreases due to which its boiling point comes down

but it is still at 100C so, water again begins to boil.

When we add small impurities in water the boiling point is not affected and remains same.

OSCILLATIONS

1. Two springs are connected to a block of mass M placed on a frictionless surface as

shown below. If both the springs have a spring constant k, the frequency of oscillation of block

is:

Ans. (b) Hint: Let when the oscillating mass is at a

distance x towards right from its equilibrium position, the instantaneous extensions in the springs of force constants k, is

x = x1 + x2

Since, the springs are in series the restoring force exerted by each spring on mass m is same. Then

F = - kx 1 = - kx 2

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x 1 = - k

F -x

k

F 2 ,

and x = x 1 + x 2 = - F k

F

k

k

2 -11

F = - x k

2

Effective force constant is . k

2

Hence, frequency of oscillation is

n = M

k

π 22

1

WAVES

1. An organ pipe closed at one end has fundamental frequency of 1500 Hz. The maximum number of overtones generated by this pipe which a normal person can hear is? a) 4 b) 13 c) 6 d) 9

Ans. (c) Hint: A closed pipe produces only odd

harmonics. The frequency of note emitted from the

pipe for v being velocity of sound in air, is

f’ = n

l

v

4

and l is length of pipe.

f ’ = n fundamental frequency

We known that human ear can hear

frequencies upto 20, 000 Hz, hence

20, 000 = n 1500

n = 131500

000 20,

Maximum possible harmonics obtained are

1, 3, 5, 7, 9, 11, 13

Hence, man can hear upto 13th harmonic

= 7 - 1 = 6

So, number of overtones heard = 6

Note: First overtone of the closed pipe is called the third harmonic.

2. The waves produced by a motor boat sailing in water are: a) Transverse b) Longitudinal c) Longitudinal and transverse d) Stationary

Ans. (c) Hint: When the motor boat sails in water, the

motor boat disturbs the surface of water forming bow waves on the surface of water. Actually these are transverse waves which are produced on the surface of water. But inside the water the longitudinal waves produced due to vibration of the rudder.

ELECTROSTATICS

1. In the basic CsCl crystal structure, Cs+ and Cl ions are arranged in a bcc configuration as

shown in the figure. The net electrostatic force

exerted by the eight Cs+ ions on the Cl ion is:

a) 2

2

0 3

4

4π

1

a

e

b)

2

2

0 3

16

4π

1

a

e

c) 2

2

0 3

32

4π

1

a

e

d) Zero

Ans. (d) Hint: The given crystal structure is a body centered cubic structure. The electrostatic force.

F = 2

21

04π

1

r

q q

Due to one Cs+ ion is balanced by diagonally opposite Cs+ ion. Hence, net

electrostatic force on Cl ion due to eight Cs+ ions is zero.

2. A 40 F capacitor in a defibrillator is charged to 3000 V. The energy stored in the capacitor is sent through the patient during a pulse of duration 2 ms. The power delivered to the patient is: a) 45 kW b) 90 kW c) 180 kW d) 360 kW

Ans. (b) Hint: A capacitor is a device that stores

energy in the electric field created between a pair of conductors on which

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equal but opposite electric charges have been placed.

The energy stored in a capacitor = 2

2

1CV

Given, C = 40 F = 40 10 6 F, V = 3000 V

E = 2

6 - (3000) 10 40 2

1

= 180 J

Also 1 W = 1 J/s

2 ms = 2 10 3 s Hence, power

= kW 90 W 10 90 s10 2

180 3

3 -

J

3. Equipotential surface associated with an electric field which is increasing in magnitude along the x-direction are: a) Planes parallel to yz-plane b) Planes parallel to xy-plane c) Planes parallel to xz-plane d) Coaxial cylinders of increasing radii

around the x-axis Ans. (a) Hint: Equipotential surface are surfaces of

constant scalar potential. They are used to visualize an n-dimensional scalar potential function in (n – 1) dimensional space. The gradient of the potential, denoting the direction of greatest increase is perpendicular to the surfaces. Hence, equipotential surfaces associated with an electric field which is increasing in magnitude along the x-direction are planes parallel to yz-plane.

4. The electric field due to a uniformly charged non-conducting sphere of radius R as a function of the distance from its centre if

represented graphically by:

Ans. (b)

Hint: Outside the spherical charge, the intensity of electric field at a point P situated at a distance r from the centre of the charge is

E = 2

04

1

r

q

πε (if r > R) phph

On the surface of spherical charge the electric field is given by

E = 2

04

1

r

q

πε (if r = R)

and inside the spherical charge the electric field is

E = 3

04

1

R

qr

πε (if r < R)

Hence, the variation of E- r is as follows:

Note: At the centre of sphere, r = 0, so E = 0.

CURRENT ELECTRICITY

1. Eels are able to generate current with biological cells called electroplaques. The electroplaques in an eel are arranged in 100 rows, each row stretching horizontally along the body of the fish containing 5000 electroplaques. The arrangement is suggestively shown below. Each electroplaque has an emf of 0.15 V and

internal resistance of 0.25. The water surrounding the eel completes a circuit between the head and its tail. If the water

surrounding it has a resistance of 500 , the current an eel can produce in water is about:

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a) 1.5 A b) 3.0 A c) 15 A d) 300 A Ans. (a) Hint: Find the equivalent circuit of consisting

of net emf, net resistance. In each row all the 5000 electro-plaques

are connected in series and all the 100 rows are connected in parallel.

Hence, equivalent emf is 5000 0.15 = 750 V

Equivalent resistance = 0.25 5000

= 1250 Now, the circuit of eel is redrawn as,

Voltage across resistors connected in

parallel is same, also equivalent resistance is

Ω 12.5 100

1250

From Ohm’s law, V = IR

I = A46 1. 12.5 500

750

R

V

I 1. 5 A

2. The temperature (T) dependence of resistivity

() of a semiconductor is represented by:

Ans. (b) Hint: The variation of specific resistance with

temperature is as follows

= ref + [1 + (T – Tref)]

Where = specific resistance at temperature T

ref = specific resistance of reference

temperature

= temperature of coefficient of resistance. Rearranging the terms we have

= 0 + 0 (T – T0)

On comparing the equation with equation of straight line y = mx + c we have, the slope negative.

Note. The temperature coefficient of

resistance of semiconductors is negative that is their electrical resistance decreases (or conductivity increases) with rise in temperature. At absolute zero a semi-conductor behaves as an ideal insulator.

3. Assertion: A large dry cell has higher emf. Reason: The emf of a dry cell is proportional to its size.

Ans.(d) Hint: In order to maintain a continuous flow

of charge in an electrical circuit, work has to be done. This work is done by the cell. The energy liberated in the chemical reactions taking place in the cells maintains the flow of charge in the circuit. Thus, the cell converts the chemical energy of its electrodes and the electrolyte into the electrical energy. The energy given by the cell in the flow of unit charge in the whole circuit is called the emf of the cell. The emf is a characteristic of the cell which depends upon the nature of the plates and the electrolyte used in cell.

Hence, the emf does not depend upon the size of the cell.

MAGNETIC EFFECTS OF CURRENT

1. The cyclotron frequency of an electron gyrating in a magnetic field of 1 T is approximately: a) 28 MHz b) 280 MHz c) 2.8 GHz d) 28 GHz

Ans. (d) Hint: In a cyclotron, the centripetal force is

provided by the transverse magnetic field B, and the force on a particle travelling in a magnetic field (which cause it to curve) is equal to Bqv.

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In cyclotron, centripetal force = magnetic force

i.e., Bqv r

mv

2

m

Bq

r

v

where r

v = , therefore

= m

Bq

and frequency f = mπ

Bq 22π

ω

Putting the numerical values from the question, we have

B = 1 T, m = 9.1 10 – 31 kg, q = 1.6 10 -19 C

f = 31

19

-

-

10 9.1 3.14 2

1 10 1.6

= 28 109 Hz

f = 28 GHz 2. The magnetic moment of current (I) carrying

circular coil of radius (r) and number of turns (n) varies as : a) 1 / r2 b) 1 / r c) r d) r2

Ans. (d) Hint: The magnetic moment is a measure of

the strength of a magnetic source. The equation for magnetic moment in the current carrying loop, carrying current I and of area vector A for which the magnitude is given by

M = IA

Where I is current A, is area (= r 2)

M = I r 2

M r 2 Note: Magnetic moment and magnetic dipole moment is the same thing. 3. A circular coil of radius R carries an electric

current. The magnetic field due to the coil at a point on the axis of the coil located at a distance r from the centre of the coil, such that r >> R, varies as: a) 1/r b) 1/r 3/2 c) 1/r2 d) 1/ r 3

Ans. (d) Hint: For a circular coil, the component of the

field B perpendicular to the axis at P cancel each other while along the axis add up.

The resultant magnetic field at point P

will be due to the components along the axis. Hence,

B = dB sin = β sinθ

μπ

μ2

0 r

sin dl i

and as here angle between the element

dI and

r is 2

π every where and r is

same for all elements while sin = ,r

R so

Hence, we have

B = 3

20 2

μπ

μ

x

iRπ

Where x = (R 2 + r

2) 1 / 2

B = 3/22 )rR

iRπ 2

(

2

4π

μ 20

Given, r >> R then we have, neglecting R,

B = 3

20 2

4π

μ

r

iRπ

Also area = R 2

B = 3

0

2π

μ

r

Ai

B 3

1

r

4. The magnetic field due to a straight conductor of uniform cross-section of radius a and carrying a steady current is represented by:

Ans. (a) Hint: The magnetic field at a point outside

the straight conductor is given by

B = r

i

2π

μ 0

It means B r

1 (if r > a)

The magnetic field at a point inside the conductor is

B = 2

0

2π

μ

a

ri

or B r (if r < a)

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5. Two parallel beams of positrons moving in

the same direction will: a) Repel each other b) Will not interact with each other c) Attract each other d) Be deflected normal to the plane

containing the two beams Ans. (c) Hint: Direction of current is opposite the

direction of motion of electrons. Positron is the antiparticle or the

antimatter counterpart of the electron. The positron has an electron charge +e. Since direction of current is in the direction of positron, it is experimentally observed that two current carrying conductors attract each other.

6. A proton and an -particle, moving with the

same velocity, enter a uniform magnetic field perpendicular to the path of their motion. The ratio of the radii of the circular paths described

by the proton and -particle is: a) 1 : 2 b) 1 : 4 c) 1 : 16 d) 4 : 1

Ans. (a) Hint: The centripetal force to the two types of

particles is provided by magnetic force. Alpha particles are highly ionizing form of

particle radiation which have low penetration. They consist of two protons and two neutrons bound together identical to helium nucleus. Also proton is 1 H1. Hence,

m = 4mP and q = 2qp

When proton and -particle describe circular paths then centripetal and centrifugal forces balance each other

qvB r

mv

2

vm

eB

eB

vm

r

r

p

p

α

p

4

2

2

1

p

p

m

m

4

2

rp : r = 1 : 2

MAGNETISM

1. Liquid oxygen remains suspended between two pole faces of a magnet because it is: a) Diamagnetic b) Paramagnetic c) Ferromagnetic d) Antiferromagnetic

Ans. (b) Hint: Liquid oxygen has extra electrons,

which do not form complete pairs of opposite spin.

Liquid oxygen has a light blue colour and is highly paramagnetic. The atomic magnetic dipoles align with an external magnetic field. The effect occurs due to quantum mechanical spin as well as electron orbital angular momentum. When suspended between two pole faces of a magnet it acquires feeble magnetism hence remains suspended. Note: Paramagnetism requires that the atoms individually have permanent dipole moment even without an applied field, which typically implies a partially filled electron shell.

2. The magnetic Resonance Imaging (MRI) is based on the phenomenon of: a) Nuclear magnetic resonance b) Electron spin resonance c) Electron paramagnetic resonance d) Diamagnetism of human tissues

Ans. (a) Hint: Magnetic Resonance Imaging (MRI) which

is a spin technology has been applied to medical diagnostics with great success. The protons of various tissues of the human body are situated in many different local magnetic environments. When a part of body is immersed in a strong external magnetic field, these environmental differences can be detected by spin-flip techniques and translated by computer processing into an image. This process is generally based on the method nuclear magnetic resonance.

3. Assertion: The true geographic north direction is found by using a compass needle. Reason: The magnetic meridian of the earth is along the axis of rotation of the earth.

Ans. (d) Hint: A compass is simply a needle shaped magnet

that is mounted so it can rotate freely about a vertical axis. When it is held in a horizontal

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plane, the north pole end of the needle point, generally, towards the geomagnetic north pole (really a south magnetic pole remember). Thus, true geographic north direction can not be found by using a compass needle. Now vertical plane passing through the magnetic axis of earth’s magnet is called magnetic meridian.

4. Assertion: A disc-shaped magnet is leviated above a superconducting material that has been cooled by liquid nitrogen. Reason: Superconductors repel a magnet.

Ans. (a) Hint: Superconductor in a magnetic field will

completely repel all field lines. This is called the Meissner effect and is an example of perfect diamagnetism. The perfect exclusion of a magnetic field can be explained by using Faraday’s law. The integral over a closed loop of E with

ds

t - ds E Δ

Δ is equal to rate of

change of magnetic flux. Since no potential difference can exist is a superconductor (there is no resistance to make one possible), then the magnetic field inside a superconductor cannot change. When a magnet is placed above a superconductor and cooled by using nitrogen, then the magnetic field inside superconductor is exactly equal and opposite in direction to applied magnetic field. Hence, they are cancelled within the superconductor so these poles repel each other. Therefore, magnet is repelled due to this force of repulsion.

RAY OPTICS AND OPTICAL INSTRUMENTS

1. When a compact disc is illuminated by a source of white light, coloured lines are observed. This is due to: a) Dispersion b) Diffraction c) Interference d) Refraction

Ans. (b) Hint: Diffraction is the process of bending

and spreading of waves when they meet an obstruction. Also diffraction grating is a reflecting or transparent substance

whose surface contains fine parallel grooves or rulings that are equally spaced. When light is incident on grating coloured pattern are formed due to diffraction. When a compact disc is illuminated, then coloured lines are observed because as one surface of the CD has many small pits in the plastic, arranged with concentric rings, that surface has a thin layer of metal applied to make the pits more visible. These pits function as diffraction gratings.

2. An object is immersed in a fluid. In order that the object becomes invisible, it should: a) Behave as a perfect reflector b) Absorb all light falling on it c) Have refractive index one d) Have refractive index exactly matching

with that of the surrounding fluid Ans. (d) Hint: From the formula for refraction through

a lens, focal length (f) is

21

111) - (

1

R -

Rn

f

Where n is refractive index of the material of the lens and R1, R2 are the radii of curvature of its surfaces. If the lens is immersed in a liquid whose refractive index is equal to the refractive index of the material of the lens ( anl = an g ), then

l n g = 1 la

ga

n

n

1

- 1

1

2

0 1) - (1

RRf 1

That is focal length of lens becomes infinite.

Now the lens will behave just like a plane transparent plate and will become invisible.

Alternative: When the refractive index of two medium are exactly same, then there will not occur refraction and reflection at the common surface of separation due to which the visibility is observed.

3. A monochromatic beam of light is used for the formation of fringes on the screen by illuminating the two slits in the Young’s double slit interference experiment. When a thin film of mica is interposed in the path of one of the interfering beams then: a) The fringe width increases b) The fringe width decreases c) The fringe width remains the same but

the pattern shifts d) The fringe pattern disappears

Ans. (c)

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Hint: When a thin transparent plate of mica is introduced in the path of one of the two

interfering light beams (as shown) then the entire fringe pattern is displaced towards the beam in the path of which the plate is introduced, but the fringe width is not changed.

4. Sodium lamps are used in foggy conditions because: a) Yellow light is scattered less by the fog

particles b) Yellow light is scattered more by the fog

particles c) Yellow light is unaffected during its

passage through the fog d) Wavelength of yellow light is the mean of

the visible part of the spectrum Ans. (a)

Hint: Sodium light emits monochromatic light which has only one colour (deep yellow). This colour is scattered less by the fog particles. While the other light sources comprise lights of many different colours (of different wavelengths) so they are scattered most due to they provide some degree of colour rendering.

5. An endoscope is employed by a physician to view the internal parts of a body organ. It is based on the principle of: a) Refraction b) Reflection c) Total internal reflection d) Dispersion

Ans. (c) Hint: Endoscopy means looking inside and

refers to looking inside the human body for medical reasons. In an endoscope thousands of optical fibres are bundled together which is inserted into human body. Light can be directed down the fibres even if they are bent, this illuminates the area under observation. The surgeon can view this from a television camera linked to a monitor by other fibre.

Hence, it is based on the principle of total internal reflection.

6. Assertion: A red object appears dark in the yellow light. Reason: The red colour is scattered less.

Ans. (b) Hint: Red colour has the largest wavelength.

A red object looks red because it reflects only red colour and absorbs all other colours present in the white light. Hence, when red object is seen through yellow light then it absorbs yellow colour falling on it and appears dark. This assertion has got nothing to do with scattering. Also from Rayleigh’s criteria of scattering.

Scattering = 4h)(wavelengt

1

Since red colour has a larger wavelength it is scattered least.

7. Assertion: At the first glance, the top surface of the Morpho butterfly’s wing appears a beautiful blue-green. If the wind moves the colour changes. Reason: Different pigments in the wing reflect light at different angles.

Ans. (c) Hint: The morpho butterflies are among the

largest in the world, with a wingspan of 7.5 to 20 cm. The males have the beautiful metallic blue upper wings. Their iridescence is due to the microscopic structure of their wings. In these butterflies iridescence is caused by multiple slit interference. Sunlight contains a full range of light wavelengths. Interference occurs when light hitting the wing combines with light reflected off the wing. Light is a wave. If the crests and troughs of waves are aligned or “in phase” they will cause constrictive interference and iridescence.

8. Assertion: Crystalline solids can cause X-

rays to diffract. Reason: Interatomic distance in crystalline solids is of the order of 0.1 nm. Ans. (c) Hint: X-ray crystallography is technique in

which the pattern produced by the diffraction of X-rays through the closed spaced lattice of atoms in a crystal is recorded and then analyzed to reveal the nature of that lattice. The spacing in the crystal lattice can be determined using Bragg’s law. The regular spacing of the atoms in crystal is of the order of wavelength of X-rays (0.1 nm) because of which X-rays are diffracted from the crystals.

9. Assertion: A famous painting was painting by not using brush strokes in the usual manner, but rather a myriad of small colour dots. In this painting the colour you see at any given place on the painting changes as you move away. Reason: The angular separation of adjacent dots changes with the distance from the painting.

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Ans. (a)

Hint: Resolving power = 1.22d

λ , where is

Wavelength, d is distance. For the human eye the resolving power or angular separation of objects changes as distance from the object changes. Thus, when we are close to the painting our eyes can pick out the separate colours of dots. But when we move away from the painting, our eyes blend the dots and we see different colours.

ELECTRONS AND PHOTONS

1. A photon of energy 4 eV is incident on a metal surface whose work function is 2 eV. The minimum reverse potential to be applied for stopping the emission of electrons is: a) 2 V b) 4 V c) 6 V d) 8 V

Ans. (a) Hint: Kinetic energy of photoelectron is eV0

where V0 is stopping potential. From Einsteins photoelectric equation

Ek = W -hv mv 2max

2

1

Where Ek is maximum kinetic energy of electron, v is frequency and W is work function.

2eV 2eV -4eV 2

1 2max mv

but 02max

2

1eV mv

where V0 is stopping potential. Thus, eV0 = 2eV

V0 = 2 V 2. We wish to see inside an atom. Assuming

the atom to have a diameter of 100 pm, this means that one must be able to resolve a width of say 10 pm. If an electron microscope is used, the minimum electron energy required is about: a) 1.5 keV b) 15 keV c) 150 keV d) 1.5 MeV

Ans. (b) Hint: From de-Broglie equation, we have

mv

h

p

h λ

Where is wavelength, h is Planck’s constant, m is mass, p is momentum and v is velocity.

Given, = 10 pm = 10-11 m, m = 9.1 10 – 31

kg, h = 6.6 10 – 34 Js

v = mλ

h

= 1131

34

- -

-

10 10 9.1

10 6.6

= 7.25 10 7 m/s Energy of electron

= 19

2731

-

-

10 1.6

)10 (7.25 10 9.1

2

1

2

1

2mv

= 15 keV 3. Assertion: Photoelectric effect demonstrates

the wave nature of light. Reason: The number of photoelectrons is proportional to the frequency of light.

Ans. (d) Hint: Einstein explained the phenomenon of

photoelectric effect on the basis of Planck’s quantum theory, according to which light travels in the form of small bundles or packets of energy called photons. Hence, photoelectric effect explains the particle nature of light.

Also the photoelectric current or the rate of emission of photoelectrons is directly proportional to the intensity of the incident light.

ATOMIC PHYSICS

1. Assertion: In He-Ne laser, population inversion takes place between energy levels of neon atoms. Reason: Helium atoms have a meta-stable energy level.

Ans. (b) Hint: The transition in He-Ne laser is shown

below.

An electric discharge in the gas pumps the helium atom to higher energy states (which

is metastable state 10 -8 s). These He atoms collide with the ground state neon atoms and excite to higher state (levels) and produce in inverted population. Inversion causing a unidirectional photon beam (laser light).

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NUCLEAR PHYSICS

1. Carbon dating is best suited for determining the age of fossils if their age in years is of the order of: a) 103 b) 104 c) 105 d) 106

Ans. (b) Hint: Radiocarbon dating is a radio metric

dating method that uses the naturally occur isotope carbon-14 to determine the age of carbonaceous materials up to 60, 000 years that is age in the years is of the order of 104.

2. A nucleus of mass number A, originally at

rest, emits an -particle with speed v. The daughter nucleus recoils with a speed:

a) 4

2

A

v

b)

4

4

A

v

c) 4

4

A

v

d)

4

2

A

v

Ans. (c)

Hint: -Particle is equivalent to helium nucleus.

The emission of an -particle from the atom of an element reduces its atomic number by 2, and mass number by 4.

Hence, the radioactive emission is as follows:

- Particle

Z X A

Z 2 YA 4 + Z He4 (-Particle)

Also from law of conservation of momentum,

m 0 = my vy + m v

= (A – 4) v y + 4v

v y = - 4

4

A

v

Note: Negative sign of velocity indicates that it recoils in the direction opposite to

that of - particle. 3. When an electron-positron pair annihilates,

the energy released is about:

a) 0.8 10 13 J b) 1.6 10 13 J

c) 3.2 10 13 J d) 4.8 10 13 J Ans. (b) Hint: Positron is antiparticle of electron. Electron and positron are identical in all

respect except that charges on them are opposite when an electron and a positron meet together, they completely annihilate each other and two photons are produced.

e + + e 2 The rest mass energy of each particle,

electron and positron is 0.511 MeV.

Hence, 2 0.511 MeV energy of photon is converted to rest mass energy of two particles.

Total energy is

E ’ = 2 0.511 10 6 1.6 10 – 19

E’ = 1.6 10 –13 J

4. Assertion: 90Sr from the radioactive fall out from a nuclear bomb ends up in the bones of human beings through the milk consumed by them. It causes impairment of the production of red blood cells.

Reason: The energetic -particles emitted in the decay of 90Sr damage the bone marrow.

Ans. (a)

Hint: 38Sr90 decays to 39Y90 when -rays

emission is occurred. Sr gets absorbed in bones along with calcium which causes impairment of the production of red blood cells

Now Sr 90 Y 90

Sr decay to Yitrium Sr 90 emits -rays of very high energy. Bone marrow is damaged by

these high energetic -particles. 5. Assertion: Energy is released in nuclear fission.

Reason: Total binding energy of the fission fragments is larger than the total binding energy of the parent nucleus.

Ans. (a) Hint: When atoms of an element are bombarded

by neutrons, the atomic nuclei are disintegrated and emit lighter particles. This process is called nuclear fission. The mass of the nuclei obtained by fission is less than the mass of the disintegrated nucleus and lost mass reappears in the form of heat according to Einsteins mass energy relation. The total binding energy of the fission fragments is larger than the total binding energy of the parent nucleus, because fission occurs when total mass energy decreases

that is Ebn will increase. Note: This phenomenon is the basis of nuclear bombs and nuclear reactors. 6. Assertion: Heavy water is preferred over

ordinary water as a moderator in reactors. Reason: Heavy water, used for slowing down the neutrons, has lesser absorption probability of neutrons than ordinary water.

Ans. (a) Hint: Although the hydrogen atom are more

effective as a moderator but then more absorption of neutrons occurs. Even

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heavy water is most suitable moderator in which less absorption occurs. Hence, it is used as moderator compared to ordinary water.

SOLIDS AND SEMICONDUCTOR DEVICES

1. A Ge specimen is doped with Al. The

concentration of acceptor atoms is 1021 atoms/m3. Given that the intrinsic

concentration of electron-hole pairs is 1019/ m3, the concentration of electrons in the specimen is: a) 1017/m3 b) 1015 / m3 c) 104/m3 d) 102/ m3

Ans. (a) Hint: From law of mass- action

n hei n n 2

where ni is concentration of electron-hole pair and nh is concentration of acceptor or holes.

Given, ni = 1019 per m3, nh = 1021 per m3

(1019)2 = ne 1021

ne = 317

21

38

m per 10 10

10

2. The dependence of binding energy per nucleon, BN, on the mass number A, is represented by:

Ans. (a) Hint: Iron (Fe) is the most stable nucleus. From the concept of binding energy,

the binding energy per nucleon increases on an average and reaches a maximum of about 8.7 MeV per nucleon for A = 50 – 80 and generally a peak is found for 26Fe56, which is an iron isotope. This is also the most stable nucleus from all of them because it has even number of protons (26) and even number of neutrons (30) i.e., even- even pair.

Note: Stability of iron can also be judged from the fact that, maximum energy is needed to pull a nucleus away from it. Also

it has even number of protons and neutrons, hence it is most stable.

3. Which logic gate is represented by the following combination of logic gates?

a) OR b) NAND c) AND d) NOR Ans. (c) Hint: First two similar gates are NOT gates

and the third one is NOR gate.

The output of gate-1 = A

The output of gate- 2 = B The output of gate- 3

y = BA

Use Demorgan’s theorem

BA

= B A

Hence, y = B A

y = A B

This is the Boolean expression of AND gate. Note: We can define NOR gate as : The OR

gate followed by NOT gate as shown.

4. Assertion: In a transistor the base is made thin. Reason: A thin base makes the transistor stable.

Ans. (c) Hint: In a transistor the base is lightly doped

and very thin. This feature is main concept of transistor action on account of which only few holes (less than 5%) combine with the electrons in base region. Thickness of base region is so small that most of electrons diffusing through base region cross over through collector region.

By reducing the base width the power of transistor increase but it does not mean that transistor becomes stable.

5. Assertion: A transistor amplifier in common emitter configuration has a low input impedance. Reason: The base to emitter region is forward biased.

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Ans. (b) Hint: In transistor the emitter-base junction is

forward bias while the base-collector junction is given a large reverse bias. Under forward bias the charge carriers move towards base causing the flow of current.

The following shows the circuit of a common-emitter amplifier circuit using an n-p-n transistor. The input (base-emitter) circuit is forward biased by a low voltage battery VBE, so

that the resistance of the input circuit is

small. The output (collector-emitter circuit is reverse biased by means of a high voltage battery so that resistance of output is high. The weak input AC signal is applied across the base-emitter circuit and the amplified output signal is obtained across the collector-emitter circuit.

Input impedance of common-emitter configuration

constant

V

I

VCE

B

BE

Δ

Here, VBE = voltage across base and

emitter, IB = base current of the order of few micro ampere.