(a unit of med-xel tutorials) aiims physics 2005drarvindsbiology.com/content/papers/2c_p13.pdf ·...

1 DR. ARVIND’S BIOLOGY CLASSES

(A Unit of Med-Xel Tutorials)

AIIMS PHYSICS – 2005

PHYSICAL WORLD AND MEASUREMENT

1. “Parsec” is the unit of: a) Time b) Distance c) Frequency d) Angular acceleration

Ans. (b) Hint: Parsec is unit of distance used in

astronomy. A star that is one parsec away from the earth has a parallax (apparent shift), due to the earth’s movement around the sun, of second of arc.

One parsec is approximately 3.08 1016 m. 2. Dimensions of electrical resistance are:

a) [ML2 T 3 A 1] b) [ML2 T 3 A 2]

c) [ML3 T 3 A 2] d) [ML 1 L 3 T 3 A2] Ans.(b) Hint: From ohm’s law V = IR Where V is voltage, R is resistance and I is current.

R = I

V

Dimensions of R = I

V

of Dimensions

ofDimensions

=

A

ATML132

= [ML2 T 3A 2]

Alternative: From definition of time constant t = RC, where R is resistance C

is capacitance.

R = ]ATL [M

[T]242--1

C

t

R = [ML2T

3 A

2]

MOTION IN TWO AND THREE DIMENSIONS

1. When a ball is thrown up vertically with velocity v0, it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocity:

a) 03 v b) 3 v0

c) 9 v0 d) 3/2 v0 Ans. (a) Hint: At maximum height velocity is zero. From equation of motion we have v2 = u2 – 2gh Where v is final velocity, u is initial velocity. Since ball reaches maximum height,

velocity at the highest point is zero. Therefore, we have

v = 0, u = v0

0 = v 20 - 2gh

v0 = gh2

when h = 3h then

v '0 = v 0'3233 2 ghhg

Note: If ball has to be thrown to a greater height its initial velocity should be more than the original one.

LAWS OF MOTION

1. In the figure given the position-time graph of a particle of mass 0.1 kg is

shown. The impulse at t = 2 s is:

a) 0.2 kg m s 1 b) – 0.2 kg m s1

c) 0.1 kg m s 1 d) – 0.4 kg m s1 Ans. (a) Hint: If a constant force F is applied on a

body for a short interval of time ∆t,

then impulse of this force is F ∆ t. From Newton’s second law

F = ma = m t

v

Δ

Δ

F ∆t = m ∆v = ∆ p

Impulse, I = ∆p = m t

x

Δ

Δ

Given, m = 0.1 kg, t

x

Δ

Δ = s /m

2

4

I = 0.1 2

4 = 0.2 kg ms 1

2. A person is standing in an elevator. In which situation he finds his weight less? a) When the elevator moves upward with

constant acceleration b) When the elevator moves downward

with constant acceleration c) When the elevator moves upward with

uniform velocity d) When the elevator moves downward

with uniform velocity



Ans. (b) Hint: Acceleration due to gravity always acts

in the downward. When elevator is moving downward with

constant acceleration i.e., when resultant acceleration is downward then

W = m(g – a) Is the resultant weight, which is less than

the original weight mg. Therefore, person finds his weightless.

3. Assertion: Frictional forces are conservative forces. Reason: Potential energy can be associated with frictional forces.

Ans. (d) Hint: A field of force such that the work done on

or by a body so that it is displaced in the field is independent of the path. Hence, if a body is moved in a closed path the net work done is zero. Gravity and an electro static field in vacuum are conservative. But any form of friction prevents the field from being conservative. Also potential energy cannot be associated with frictional forces.

4. A particle having charge q and mass m is

projected with velocity v = 2

i - 3

j in uniform

electric field E = E0

j . Change in momentum

∆P during any time interval t is given by:

a) M 13 b) q E0 t

c) m

tE q 0 d) Zero

Ans. (b) Hint: The product of force and time interval

is called impulse. Also from Newton’s second law, we have

F = m a = m

tΔ

Δv

F ∆t = m∆ v = ∆ P

The force on charge q due to electric field is qE0 .

Impulse = qE0 t .

WORK, ENERGY AND POWER

1. A block of mass 10 kg is moving in x-direction

with a constant speed of 10 m/s. It is subjected to a retarding force F = - 0.1 x J/m during its travel from x = 20 m to x = 30 m. Its final kinetic

energy will be: a) 475 J b) 450 J c) 275 J d) 250 J

Ans. (a) Hint: Apply work-energy theorem. When a force acts upon a moving body,

then the kinetic energy of the body increases and the increase is equal to the work done. This is work energy theorem.

Work done = if2

K -K mumv 2

1 -

2

1 2

Another definition of work done is

force displacement.

Fdx = Kf - 2imv

2

1

Where the subscripts f and i stand for final and initial.

F dx = Kf - 2

1 10 (10)2

F dx = Kf - 500

30

20

500 0.1) (-

x

x

-K dx x f

Using the formula have we ,1

1

n

xdx x

nn

- 0.1

30

20 2

2

x

x

x = Kf - 500

- 0.1 500 -2

(20) -

2

(30) 22

fK

K f – 500 = - 25

Kf = 500 – 25 = 475 J

2. Energy required to break one bond in DNA

is approximately:

a) 1 eV b) 0.1 eV

c) 0.01 eV d) 2.1 eV Ans. (a)

Hint: Energy required to break one bond

in DNA is a 1 eV.

ROTATIONAL MOTION

1. A solid sphere is rolling on a frictionless

surface, shown in figure with a translational

velocity v m/s. If it is to climb the inclined

surface then v should be:



a) gh 7

10 b) gh 2

c) 2gh d) 7

10gh

Ans. (a) Hint: Kinetic energy is converted to potential

energy. From law of conservation of energy,

energy can neither be created nor destroyed but it remains conserved. In the given case the sum of kinetic energy of rotation and translation is converted to potential energy.

Also moment of inertia of disc is

I = 2MR

5

2

energy) energy) energy) kinetic(Potential l(Rotationa onal(Translati

ω2 mghImv 2

1

2

1 2

mghR

vMRmv

5

2

2

222

2

1

2

1

where v = R, = angular velocity

mgh mv 2

10

7

v = gh 7

10

Hence, to climb the inclined surface

velocity should be greater than gh 7

10

2. A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time a viscous fluid of mass m is dropped at the centre and is allowed to spread out and finally fall. The angular velocity during this period: a) Decreases continuously b) Decrease initially and increases again c) Remains unaltered d) Increases continuously

Ans. (b)

Hint: From law of conservation of angular momentum if no external torque is acting upon a body rotating about an axis, then the angular momentum of the body remains constant that is

J = I = constant

Hence, if I decreases, increases and vice-versa. When liquid is dropped, mass increases hence I increases (I = mr2). So,

decreases, but as soon as the liquid

starts falling increases again. 3. A ladder is leaned against a smooth wall

and it is allowed to slip on a frictionless floor. Which figure represents the track of

its centre of mass?

Ans. (a) 4. Assertion: For system of particles under

central force field, the total angular momentum is conserved. Reason: The torque acting on such a system is zero.

Ans. (a) Hint: Torque is rate of charge of angular

momentum of a body. When a body rotates about an axis

under the action of an external torque , then the rate of change of angular momentum of the body is equal to the torque, that is

τ dt

dJ

If external torque is zero ( = 0), then

0 dt

dJ

dJ = 0

J = constant Thus, angular momentum remains conserved.

GRAVITATION

1. The condition for a uniform spherical mass m of radius r to be black hole is: [G = gravitational constant and g = accele-ration

due to gravity]



a) cr

Gm

2 /1

2 b) c

r

m

2 /1

2g

c) cr

Gm

2 /1

2 d) c

r

m

2 /1

g

Ans. (c) Hint: A black hole is an object so massive

that even light cannot escape from it. This requires the idea of a gravitational mass for a photon, which then allows the calculation of an escape energy for an object of that mass.

When gravitational potential energy of the photon is exactly equal to the photon energy, then

hv0 = 0v 2

rc

GMh … (1)

where G is gravitational constant, M is mass, r is radius, c is speed of light, h is Planck’s constant.

Then from Eq. (1) we have

r = 2

c

GM

Note that this condition is independent of

frequency v. Schwarzchild’s calculated gravitational radius differs from this result by a factor of 2 and is coincidently equal to the non-relativistic escape velocity expression.

v escape = c 2

r

GM

PROPERTIES OF BULK MATTER

1. For a constant hydraulic stress on an object,

the fractional change in the object’s volume (V

/ V) and its bulk modulus (B) are related as:

a) B V

V

Δ b)

B

V

V 1Δ

c) 2ΔB

V

V d) 2 - Δ

B V

V

Ans. (b) Hint: When strain is small the ratio of the

normal stress to the volume strain is called the bulk modulus of the material of the body.

B = strain volume

stress normal

B = V / V

P Δ

Δ

B

V

VΔ

1

BV

V 1

Δ

2. A candle of diameter d is floating on a liquid in a cylindrical container of

diameter D (D >> d) as shown in figure. If it is burning at the rate of 2 cm/h then

the top of the candle will:

a) Remain at the same height b) Fall at the rate of 1 cm / h c) Fall at the rate of 2 cm / h d) Go up at the rate of 1 cm/ h

Ans. (b) Hint: Weight of candle is equal to weight of

liquid displaced. From Archemedes’ principle when a

body is immersed in a liquid completely or partly then there is an apparent loss in its weight. This apparent loss in weight is equal to the weight of the liquid displaced by the body.

Also, volume of candle = Area length

= L 2

2

2

d

weight of candle = weight of liquid displaced

Vg = V g

ρ' L 4

π ρ 2 4

π22

dL

d

2

1

ρ'

ρ

Since, candle is burning at the rate of 2 cm/h, then after an hour, candle length is 2L – 2

(2L – 2) = (L – x )

1) - (2

ρ'

ρ

L

x -L

1)(2 -L

x -L

2

1



x = 1 cm

Hence, in one hour it melts 1 cm and so it falls at the rate of 1 cm / h.

3. A given shaped glass tube having uniform cross-section is filled with water and is mounted on a rotatable shaft as shown in figure. If the tube is rotated with a constant

angular velocity then:

a) Water levels in both sections A and B go up b) Water level in section A goes up and that

in B comes down c) Water level in section A comes down

and that in B it goes up d) Water levels remain same in both sections

Ans. (a) Hint: According to Bernoulli’s principle

P + 2ρ v

2

1 = constant

At the sides the velocity is higher, so the pressure is lower. But the pressure at a given horizontal level must be equal therefore the liquid rises at the sides to some height to compensate for this drop in pressure.

4. Assertion: Specific gravity of a fluid is a dimensionless quantity. Reason: It is the ratio of density of fluid to the density of water.

Ans. (a) Hint: The specific gravity or relative density

is defined as the ratio of the density of the substance to the density of water at

4C. Since specific gravity is a pure ratio, it has no unit and hence, no dimensions.

5. Assertion: For Reynold’s number Re > 2000, the flow of fluid is turbulent. Reason: Inertial forces are dominant compared to the viscous forces at such high Reynold’s numbers.

Ans.(a) Hint: Reynold’s number is the ratio of inertial

forces (vs ) to viscous force ( / L) and is used for determining whether a flow will be laminar or turbulent. It is the most important dimension less number in fluid dynamics and provides a criterion for determine dynamic similitude.

Re = μ

ρ L v s

Where vs = mean fluid velocity. L = characteristic length

= absolute dynamic fluid viscosity and

= fluid density. It is established from experiments that flow

is steady of laminar when Reynold’s number is less than about 2000, where in the region between 2000 to 3000 the flow is unstable, i.e., may change from laminar to turbulent or vice-versa. This case arises only when internal forces are dominant compared to viscous force.

HEAT AND THERMODYNAMICS

1. Assertion: Reversible systems are difficult to find in real world. Reason: Most processes are dissipative in nature.

Ans. (a) Hint: Heat and work can cause temperature

rise or change of state of a body. This heat given to body is used in increasing the temperature of body and rest against frictional forces. It is impossible by any means to recover the energy lost in doing work against dissipative forces. Usually all processes are dissipative in nature.

2. Assertion: Air quickly leaking out of a balloon becomes cooler. Reason: The leaking air undergoes adiabatic expansion.

Ans. (a) Hint: When a system undergoes a change

under the condition that no exchange of heat takes place between the system and surroundings, then such a process is called adiabatic process. The leaking air of balloon undergoes adiabatic expansion. In this expansion, due to work done against external pressure, the internal energy of air reduces. Thus, it becomes cooler.

3. Assertion: A body that is a good radiator is also a good absorber of radiation at a given wavelength. Reason: According to Kirchhoff’s law the absorptivity of a body is equal to its emissivity at a given wavelength.

Ans. (a) Hint: According to Kirchhoff’s law, the ratio of

the emissive power to the absorptive power for radiation of a given wavelength is the same for all bodies at the same temperature and is equal to the emissive power of a perfectly black body at that temperature.



A conclusion from Kirchhoff’s law is that, if surface is a good absorber of a particular wavelength of radiation it is also a good emitter of that wavelength of radiation.

4. Assertion: In pressure-temperature (P-T) phase diagram of water, the slope of the melting curve is found to be negative. Reason: Ice contracts on melting to water.

Ans. (a) Hint: The phase diagram that is pressure

temperature diagram of water is as shown. In phase diagram of water, the solid-liquid phase boundary has a negative slope. This reflects the fact that ice has a lower density than water which is an usual property i.e., ice contracts on melting to water.

5. Assertion: For higher temperature the peak emission wavelength of a black body shifts to lower wavelengths. Reason: Peak emission wavelengths of a black body is proportional to the fourth-power of temperature.

Ans. (c) Hint: Figure shows that how the energy of a

black body radiation varies with temperature and wavelength. As the temperature of the black body increases, two distinct behaviours are observed.

The first effect is that the peak of the

distribution shifts to shorter wavelengths. This shift is found to obey

the following relationship called Wien’s displacement law.

mT = constant The second effect is that the amount of energy, the black body emits per unit area per unit time increases with fourth power of absolute temperature T.

OSCILLATIONS

1. Which of the following functions represents a simple harmonic oscillation?

a) sin t – cos t b) sin2 t

c) sin t + sin 2 t d) sin t – sin 2 t Ans. (a) Hint: One of the conditions for SHM is that

restoring force (F) and hence acceleration (a) should be proportional to displacement (y).

Let, y = sin t – cos t

tt ω sin ω cos ω dt

dy

tt

y ω cos ω ωt sinω -

d

d 2 2

2

2

or a = - 2 (sin t - cos t)

a = - 2 y

a - y This satisfies the condition of SHM.

Other equations do not satisfy this condition.

ELECTROSTATICS

1. Two infinitely long parallel conducting plates

having surface charge densities + and - respectively, are separated by a small distance. The medium between the plates is

vacuum. It 0 is the dielectric permittivity of vacuum, then the electric field in the region between the plates is:

a) Zero b) / 2 0 V/m

c) / 0 V/m d) 2 /0 V/m Ans. (c) Hint: Given that conducting plates have surface

charge densities + and respectively. Since the sheet is large, the electric field E at energy point near the sheet will be perpendicular to the sheet.



The resultant electric field is given by E ’ = E + E = 2 E

If is surface charge density then, electric field

E ’ = 02ε

σ'

2 E = m V / ε

σ

ε 2

σ 2

00

2. Two concentric conducting thin spherical shells A and B having radii rA and rB (rB > rA)

are charged to QA and – QB (QB > QA ).

The electrical field along a line. (passing through the centre) is:

Ans. (c) Hint: As charge on shell B is negative so,

potential inside B is constant and negative. Also due to positive charge, potential inside A is positive and constant and greater than that of B. Therefore, as r increases potential decreases and as rA is crossed, potential becomes negative. At the surface of shell B, it is more negative. This variation is represented in graph (c). Note: This problem is considered as wrong, because question asked is for field and graphs are given by potential.

THERMAL AND CHEMICAL EFFECTS OF CURRENT

1. For ensuring dissipation of same energy in all three resistors (R1, R2, R3) connected as shown in figure, their values must be related as:

a) R1 = R2 = R3 b) R2 = R3 and R1 = 4R 2 c) R 2 = R3 and R1 = (1/4) R2 d) R1 = R2 + R3

Ans. (c) Hint: When resistors are connected in

parallel potential difference across them is same.

In the given circuit the resistor’s R2 and R3 are connected in parallel hence potential difference (V) across them is same. In order that they undergo same

energy loss, H = t, R

V2

R2 must be equal

to R3. i.e., R2 = R3

Now resistor R1 is in series with R2, hence energy through them is

H = i 2R1t = i 21 R 2t

Where i 1 is current across R2.

Since R2 = R3, therefore current through

them is 2

i = i1

i 2R1t = tR i

2

2

4

R1 = 4

2R

MAGNETISM

1. The magnetic moment () of a revolving electron around the nucleus varies with principal quantum number n as:

a) n b) 1 / n

c) n2 d) 1 / n2 Ans. (a) Hint: From Bohr’s model of atom, electrons

can revolve only in those orbits in which their angular momentum is an

integral multiple of '2π

h where h is

Planck’s universal constant. From Bohr’s Postulate

= 2π

nh

n

where n is integer (= 1, 2, 3, ………) and is called principal quantum number. Note: The discrete orbits in which electrons can revolve are known as stable orbits.



ELECTROMAGNETIC INDUCTION

1. A magnet is made to oscillate with a particular frequency, passing through a coil as shown in

figure. The time variation of the magnitude

of emf generated across the coil during one cycle is:

Ans. (a) Hint:When North Pole of magnet approaches

the one face of coil, then the force of the coil becomes a north pole to oppose this motion and current flows anticlockwise. Thus, in this case emf is developed in the coil and when it completes one half motion it is momentarily at rest and no emf is present. Now South Pole approaches the other face of coil making this face a south pole. The current now flows in clockwise direction and again an emf is developed in the coil. This variation is shown in figure (a).

2. A conducting ring of radius 1 m is placed in an uniform magnetic field B of 0.01 T

oscillating with frequency 100 Hz with its

plane at right angle to B. What will be the

induced electric field?

a) V / m b) 2 V/ m c) 10 V/ m d) 62 V/ m

Ans. (b) Hint: From Faraday’s law of electromagnetic

induction the induced emf is equal to negative rate of change of magnetic flux.

That is e = - Δt

Δ

Flux induced = 2 B A cos

Where B is magnetic field, A is area.

Given, = 0, ∆t = s 100

1

∆ = 2 0.01 (1)2 200 cos 0

e = 100

200 (1) 0.01 2 2 π

= - 4 Volt

Circumference of a circle of radius r is

2r.

Induced electric field E is

E = r

e

2π

= m V / 2

1

2

2

4

πr

π

ALTERNATING CURRENT AND ELECTROMAGNETIC WAVES

1. A 50 Hz AC source of 20 V is connected across R and C as shown in figure. The voltage across R is 12 V. The voltage across C is:

a) 8 V b) 16 V c) 10 V d) Not possible to determine unless values of R and C are given Ans. (b) Hint: Voltage across capacitor lags behind

current by 90 . The given circuit is a C-R series circuit,

VR is in phase with i, while VC lags

behind i by 90. Hence, resultant potential is

E = E0 Sin t



V = 2C

2R VV

Given, VR = 12 V, V = 20 V

(20)2 = (12)2 + V 2C

V 2C = (20)2 – (12)2

= 400 – 144 = 256

VC = V 16 256

2. The pressure exerted by an electromagnetic wave of intensity I (W/m2) on a non-

reflecting surface is: [c is the velocity of light] a) Ic b) Ic 2 c) I / c d) I/ c 2

Ans. (c) Hint: Force per unit area is the radiation

pressure. When a surface intercepts electro-

magnetic radiation, a force and a pressure are exerted on the surface. As the surface is non-reflecting, so, it is completely absorbed and in such case the force is,

F = c

IA

Where I is intensity of light. A is area of surface and c is velocity of electro-magnetic wave.

The radiation pressure is the force per unit area

i.e., P = c

I

A

F

3. In case of linearly polarised light, the magnitude of the electric field vector: a) Does not change with time b) Varies periodically with time c) Increases and decrease linearly with time d) Is parallel to the direction of propagation

Ans. (b) Hint: Light can be represented as a transverse

electromagnetic wave made up of mutually perpendicular fluctuating electric and magnetic fields. The diagram shows the electric field in the xy-plane, the magnetic field in xz-plane and the propagation of the wave in the x-direction.

Fig. (2) shows a line tracing out the electric

field vector as it propagates traditionally, only the electric field vector is dealt with because the magnetic field component is essentially the same.

4. The circuit shown below acts as:

a) Tuned filter b) Low pass filter c) High pass filter d) Rectifier Ans. (a) Hint: The circuit is a tuned filter circuit. It

utilizes one or more tuned circuits (resonant or anti-resonant circuits) to separate signals in relatively narrow bands of frequencies from signals wider frequency spectrums. Most transmitters and receivers incorporate many tuned filters. The given LCR tuned filter (also known as RLC tuned filter) is the most flexible able of all frequency or frequency non-linear circuits. It is preferred for its frequency selection property.

5. For sky wave propagation of 10 MHz signal, what should be the minimum electron density in ionosphere?

a) 1.2 1012m 3 b) 10 6 m 3

c) 1014 m 3 d) 1022 m 3 Ans. (a) Hint: The critical frequency of sky wave

undergoing reflection from a layer of atmosphere is

fc = 9 maxN

where N is electron density per m3.

N max = 81

2cf

= 81

)10 (10 26

= 1.2 1012 m – 3 6. Solid targets of different elements are

bombarded by highly energetic electron beams. The frequency (f) of the

characteristic X-rays emitted from different targets varies with atomic number Z as:

a) f Z b) f Z 2

c) f Z d) f Z 3 / 2



Ans. (b) Hint: Moseley’s law is an empirical law

concerning the characteristic electro-magnetic spectrum that is absorbed or emitted by atoms.

Using X-ray diffraction techniques Moseley found that most intense short wavelength line in the X-ray spectrum of a particular element is related to the element’s atomic number Z as follows

f = (2.48 1015) (Z – 1)2 Hz

f Z 2

7. Assertion: Television signals are received through sky-wave propagation. Reason: The ionosphere reflects electro-magnetic waves of frequencies greater than a certain critical frequency.

Ans. (d) Hint: In sky wave propagation the radio waves

which have frequency between 2 MHz to 30 MHz, are reflected back to the ground by the ionosphere. But radio waves having frequency greater than 30 MHz cannot be reflected by the ionosphere because at this frequency they penetrates the ionosphere. It makes the sky wave propagation less reliable for propagation of TV signal having frequency greater than 30 MHz. Critical frequency is defined as the highest frequency that is returned to the earth by the ionosphere. Thus, above this frequency a wave whether it is electromagnetic will penetrate the ionosphere and is not reflected by it.

RAY OPTICS AND OPTICAL INSTRUMENTS 1. What should be the maximum acceptance

angle at the air-core interface of an optical fibre if n1 and n2 are the refractive indices of

the core and the cladding, respectively?

a) sin 1 (n2 / n1) b) sin 1 22

21 n - n

c)

1

tann

n 21 d)

2

1

n

n 1tan

Ans. (b) Hint: When light is launched at one end of a

step index optical fibre then, the illustration of the path of light ray incident on the end of an optical fibre at an angle to the fibre axis is as shown

Acceptance angle is the maximum angle that a light ray can have relative to the axis of the fibre and propagate down the fibre. It is given by

= sin -1 22

21 n -n

2. A telescope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm. If this telescope is used to see a 50 m tall building at a distance of 2 km, what is the height of the image of the building formed by the objective lens? a) 5 cm b) 10 cm c) 1 cm d) 2 cm

Ans. (a) Hint: A telescope is an optical instrument

used to see distant objects.

Since, convex lens is used, from lens formula we have

000 uvf

1

1

1-

where vo and u0 are image and object distance respectively.

000 ufv

1

1

1

Given, f0 = 200 cm

u0 = - 2 km = - 2 105 cm

O = 50 m = 5 103 cm

3

0 10 200-

1

200

1

1

v

= 3

3

10 200

1 - 10

v0 = cm 999

10 200 3



Also magnification m = O

I

u

v

0

0

10 5

10 200 999

10 20033

3

I

I = cm 5 999

10 5 3

3. According to Hubble’s law, the red-shift (Z) of a receding galaxy and its distance r from earth are related as:

a) Z r b) Z 1 / r

c) Z 1/r2 d) Z r 3 / 2 Ans. (a) Hint: Hubble’s law is the statement in

physical cosmology that the red shift (Z) in light coming from different galaxies is proportional to their distance (r).

That is, Z r It is considered the first

observational basis for the expanding space and today serves as one of the most often cited pieces of evidence in support of the Big Bang.

4. When exposed to sunlight, thin films of oil on water often exhibit brilliant colours due to the phenomenon of: a) Interference b) Diffraction c) Dispersion d) Polarisation

Ans. (a) Hint: When a thin film of oil spreads over the

surface of water and it is seen in broad day light, brilliant colours are seen. These colours arise due to interference of sun-light reflected from the upper and lower surfaces of the film.

While bending of light rays at curvatures of obstacles is called diffraction. Dispersion is splitting of white light and polarisation is the restriction of the vibrations in a transverse wave.

5. The apparent depth of water in cylindrical water tank of diameter 2R cm is reducing at the rate of x cm/min when water is being

drained out at a constant rate. The amount of

water drained in cc per minute is : (n1 = refractive index of air, n2 = refractive index of water).

a) 2

12

n

nRπ x b)

1

22

n

nRπ x

c) 2

12

n

n Rπ d) R 2

x

Ans. (b) Hint: Let actual height of water of the tank be

h, then

1n 2 = depthapparent

depth actual

Also 1n 2 = 1

2

n

n

1

2

n

n =

x

h

where x is a apparent depth.

dt

dx

dt

dh

n

n

1

2

dt

dx

n

n

dt

dh

1

2

or change in actual rate of flow = 1

2

n

n

change in apparent rate of flow

or min / cmx n

n

dt

dh

1

2

Multiplying both sides by R2, we have

22

1

2 R x n

n R

dt

dh

Amount of water drained = xR2 1

2

n

n

6. Assertion: By roughening the surface of a glass sheet its transparency can be reduced. Reason: Glass sheet with rough surface absorbs more light.

Ans. (c) Hint: When glass surface is made rough then

the light falling on it is scattered in different



directions reducing its transparency. Also since rough surfaces scatter light more, it is absorbed less.

7. Assertion: Diamond glitters brilliantly. Reason: Diamond does not absorb sunlight.

Ans. (b) Hint: The faces of diamond are so cut that

the light ray entered inside diamond fall on the diamond air interface at an angle greater than critical angle and so suffer repeated total internal reflections and comes out of its faces only in some specific directions, because of which it shines.

All the light-entering in it comes out of diamond after number of reflections and no light is absorbed by it.

8. Assertion: The clouds in sky generally appear to be whitish. Reason: Diffraction due to clouds is efficient in equal measure at all wavelengths.

Ans. (c) Hint: A cloud is a large collection of very tiny

droplets of water or ice crystals. The droplets are so small and light that they can float in air. Clouds appear white because they reflect the light of the sun. Light is made up of colours of rainbow and when they are added up together, white colour is obtained clouds reflect all the colours by exactly the same amount so they generally look white.

ELECTRONS AND PHOTONS

1. Assertion: The energy (E) and momentum (p) of a photon are related by p = E / c. Reason: The photon behaves like a particle.

Ans. (a) Hint: When an electron/ proton behaves like a

particle, then its wavelength is given by

= p

h … (1)

where h is Planck’s constant, p is momentum. This wavelength is known as de-Broglie wavelength of the electron. Also

E = hv = λ

hc … (2)

where v is frequency, c is speed of light. From Eqs. (1) and (2), we get

p = c

E

ATOMIC PHYSICS

1. The ground state energy of hydrogen atom is – 13.6 eV. What is the potential energy of the electron in this state? a) Zero b) – 27.2 eV c) 1 eV d) 2 eV

Ans. (b) Hint: From the first postulate of Bohr’s atom

model.

2

22

r

KZe

r

mv

r

KZemv

2

2

1

2

1 2

i.e., K. E. of electron = r

KZemv

2

2

2

1 2 ..

(1) Potential due to the nucleus, in the orbit

in which electron is revolving = r

KZe

Potential energy of electron

= potential charge

= r

KZe (-e) = -

r

KZe2

… (2)

Total energy of electron in the orbit E = KE + PE

= r

KZe -

r

KZe22

2

1

= - r

KZe

2

2

…. (3)

This implies that PE = 2 E Given, that in ground state of hydrogen,

total energy E = - 13.6 eV

PE = - 2 13.6 = - 27.2 eV Note: Total energy of electron in outer

orbits is more than that in inner orbits. 2. Assertion: It is not possible to use 35Cl as

the fuel for fusion energy. Reason: The binding energy of 35Cl is too small.

Ans. (c) Hint: 35Cl has a large binding energy

therefore, it cannot be used as a fuel for fusion energy. This can be explained as we know that the binding energy for very light (and for very heavy) nuclides is small as compared to intermediate nuclides. It means light nuclides (and heavy) nuclides are less stable than that of intermediate nuclides.



NUCLEAR PHYSICS

1. A radioactive material has half-life of 10 days. What fraction of the material would remain after 30 days? a) 0.5 b) 0.25 c) 0.125 d) 0.33

Ans. (c) Hint: If N be the number of atoms of a

radioactive substance left at some instant of time, than

N = N0

n

2

1

Where N0 is the initial number of atoms and n is number of half lives.

n = 3 10

30

t

2 /T

3

2

1

0N

N

= 0.125 2

13

8

1

SOLIDS AND SEMICONDUCTOR DEVICES

1. The voltage gain of the following amplifier is: :

a) 10 b) 100 c) 1000 d) 9.9

Ans. (b) Hint: Voltage gain of the given operational

amplifier is defined as ratio of output voltage to input voltage.

AV = i

0

V

V

From ohm’s law V = iR, where i is current, R is resistance.

AV = 100 Ω 1

Ω 100

k

k

i

f

R

R

2. Which of the following is an amorphous solid?

a) Glass b) Diamond c) Salt d) Sugar

Ans. (a) Hint: An amorphous solid is a solid in which

there is no long range order of the positions of the atoms. Most classes of solid materials can be found or prepared in an amorphous form. For instance, common window glass is an amorphous ceramic. These materials are often prepared by rapidly cooling molten

material. The cooling reduces the mobility of materials molecules before they can pack into a more thermodynamically favourable crystalline state.

3. Which of the following logic gates is an universal gate? a) OR b) NOT c) AND d) NAND

Ans. (d) Hint: The NAND gate and the NOR gate can be said to

be universal gates since combinations of them can be used to accomplish any of the basic operations and can thus, produce and inverter, an OR gate or an AND gate. The non-inverting gates do not have this versatility since they can’t produce an invert.

Note: The basic symbol and truth table of NAND gate are shown.

4. Consider an n-p-n transistor amplifier in common-emitter configuration. The current gain of the transistor is 100. If the collector current changes by 1 mA, what will be the change in emitter current? a) 1.1 mA b) 1.01 mA c) 0.01 mA d) 10 mA

Ans. (b) Hint: Current gain is defined as ratio of

collector current to base current.

Current gain = current base in change

current collector in change

= B

C

i

i

Δ

Δ

Also i E = i B + i C ∆ i E = ∆ i B + ∆ i C

= CE

C

i -i

i

ΔΔ

Δ

Given, = 100, ∆i C = 1 mA

100 = 1 - Δ

1

Ei

∆i E – 1 = 0.01 100

1

∆i E = 1 + 0.01 = 1.01 mA.

Note: In common-emitter amplifier, the

output voltage signal is 180 out of phase with the input voltage signal.

5. In a semiconducting material the mobilities of

electrons and holes are e and h respectively. Which of the following is true?



a) e > h b) e < h

c) e = h d) e < 0; h > 0 Ans. (a) Hint: Electrons are lighter in weight compared

to protons. In a semiconductor, hole is equivalent to

a positively charged particle which moves in a direction opposite to that of an electron. Since electrons are lighter in mass compared to positively charged particles they move easily in the semiconducting material. Hence, the mobility of holes in p-type semiconductor is less than mobility of electrons in n- type semiconductor. i.e., ne > nh

Note: n-type semiconductor because of presence of excess electrons is more preferred compared to p-type semiconductor which has excess of holes.

6. Assertion: Diode lasers are used as optical sources in optical communication. Reason: Diode lasers consume less energy.

Ans. (b) Hint: A laser diode is a laser where the active medium

is a semiconductor similar to that found in a light emitting diode. Laser diode is formed by doping a very thin layer on the surface of a crystal water. They find wide use in optical communication as easily coupled light sources. Diode lasers are efficient sources of high power radiation and with suitable beam shaping technology the output can be easily launched into the cladding of a suitably designed fibre. Hence, preferred in optical communication. They also consume less power.

7. Assertion: The logic gate NOT can be built using diode. Reason: The output voltage and the input

voltage of the diode have 180 phase difference.

Ans. (d) Hint: NOT gate inverts the input signal. The

NOT gate has only one input and one output. It combines the input A with the output Y, from Boolean expression

A = Y We cannot realize a NOT gate by using

diodes.

Instead a transistor has to be used. An

electric circuit and truth table for a NOT gate using a n-p-n transistor is shown

8. Assertion: The resolving power of a

telescope is more if the diameter of the objective lens is more. Reason: Objective lens of large diameter collects more light.

Ans. (a) Hint: Telescope is an optical instrument used to

see distant objects. The image of the distance object formed by the telescope subtends a large visual angle at the eye, so that the object appears large to the eye. The objective consists of an achromatic convex lens of large focal length and large aperture. Aperture is taken large so that it may collect sufficient light and form bright image of very distant objects stars etc.

Note: Eyepiece of small aperture is taken so that the whole light may enter the eye.

9. Assertion: The number of electrons in a p-type silicon semiconductor is less than the number of electrons in a pure silicon semiconductor at room temperature. Reason: It is due to law of mass action.

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

n 2i = nenh

For p-type semiconductor, number of electrons is less than the number of electrons in a pure silicon semiconductor that is for p-type semiconductor, ne < nh.

Note: n-type semiconductor has excess of electrons.

10. Assertion: In a common-emitter transistor amplifier the input current is much less than the output current. Reason: The common-emitter transistor amplifier has very high input impedance.

Ans. (c) Hint: The common-emitter transistor amplifier

has input resistance equal to 1 k (approx.)

and output resistance equal to 10 k (approx.). The output current in CE amplifier is much larger than the input current.

(a unit of med-xel tutorials) aiims physics 2005drarvindsbiology.com/content/papers/2c_p13.pdf ·...

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