physics 2008 set 1 sol

Physics 2008 Set 1 Close

Subjective Test

(i) All questions are compulsory.

(ii) There are 30 questions in total.

Questions 1 to 8 carry one mark each,

Questions 9 to 18 carry two marks each,

Question 19 to 27 carry three marks each and

Question 28 to 30 carry five marks each.

(iii) There is no overall choice. However, an internal choice has been provided.

(iv) Wherever necessary, the diagrams drawn should be neat and properly labelled.

(v) Use of calculators is not permitted.

Question 1 ( 1.0 marks)

State two characteristic properties of nuclear force.

Solution:

Properties of nuclear force:

(i) Nuclear force is the strongest force in nature.

(ii) Nuclear force is only effective on a very short range.


How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced with

red light?

Solution:

δviolet > δred

Thus, when incident violet light is replaced with red light, the angle of minimum deviation of a glass

decreases.


The instantaneous current and voltage of an a.c. circuit are given by

i = 10 sin 300 t A and

V = 200 sin 300 t V.

What is the power dissipation in the circuit?

Solution:

2/3/2011 Subjective Test Paper - Physics - Meritn…

meritnation.com/…/tx4ggrzWFFjnmFNN… 1/22

It is given that:

i = 10 sin 300 t A

V = 200 sin 300 t V

∴ i0 = 10 A and V0 = 200 V

∴ Average power dissipation = V0i0

= 200 × 10 = 2000 W


Why should the spring/suspension wire in a moving coil galvanometer have low torsional constant?

Solution:

The moving coil galvanometer has low torsional constant (restoring torque per unit twist) to make it very

sensitive.


Why does the bluish colour predominate in a clear sky?

Solution:

A blue colour has a shorter wavelength than red. Therefore, blue colour is scattered much more strongly.

Hence, the sky looks blue.


Which orientation of an electric dipole in a uniform electric field would correspond to stable equilibrium?

Solution:

When the electric dipole is in the direction of the electric field, it is said to be in stable equilibrium.


Two lines, A and B, in the plot given below show the variation of de Broglie wavelength, λ versus ,

where V is the accelerating potential difference, for two particles carrying the same charge.

Which one of two represents a particle of smaller mass?

Solution:



(Slope of λ and graph)

Slope of B > Slope of A

Therefore, line B represents a particle of smaller mass.


State the reason why GaAs is most commonly used in making of a solar cell.

Solution:

GaAs (gallium arsenide) is most commonly used in the making of a solar cell because it absorbs relatively

more energy from the incident solar radiations, being of relatively higher absorption coefficient.


Draw a labelled ray diagram of an astronomical telescope in the near point position. Write the expression

for its magnifying power.

Solution:

Astronomical telescope in the near point position:

Magnifying power,


The following figure shows the variation of intensity of magnetisation versus the applied magnetic field

intensity, Hj for two magnetic materials A and B:

(a) Identify the materials A and B.



(b) Why does the material B have a larger susceptibility than A for a given field at constant temperature?

Solution:

(a) Material A is diamagnetic.

Material B is paramagnetic.

(b) Material B has a larger susceptibility than A because paramagnetic substances have a tendency to pull

in magnetic field lines when placed in a magnetic field.


Two metallic wires of the same material have the same length but cross-sectional area is in the ratio 1:2.

They are connected (i) in series and (ii) in parallel. Compare the drift velocities of electrons in the two

wires in both the cases (i) and (ii).

Solution:

It is given that A1:A2 = 1: 2 ⇒

(i) When two wires are connected in series, the current in both wires A and B will be the same.

(ii) When two wires are connected in series, the potential difference across the wires A and B will be the

same.


Draw a block diagram of a simple amplitude modulation. Explain briefly how amplitude modulation is

achieved.

Solution:

A conceptually simple method to produce AM wave is shown in the following block diagram.

Here the modulating signal Am sin wmt is added to the carrier signal Ac sin wct to produce the signal x (t).

This signal x (t) = Am sin wmt + Ac sin wct is passed through a square law device, which produces an

output.

y (t) = B x (t) + Cx2 (t)

Where, B and C are constants



Where, B and C are constants

This signal is then passed through a band pass filter, which rejects dc. The output of the band pass filter is,

therefore, an AM wave.


Calculate the energy released in MeV in the following nuclear reaction:

[Mass of

Mass of

Mass of

1u = 931.5 MeV/c2]

Solution:

Given nuclear reaction is:

Mass defect = MU − MTH − MHe

= 238.05079 − 234.043630 − 4.002600

= 0.00456 u

Energy released = (0.00456) × (931.5 MeV/c2)

= 4.25 MeV


Using Ampere’s circuital law, obtain an expression for the magnetic field along the axes of a current

carrying solenoid of length l and having number of turns.

Solution:

Magnetic field due to solenoid: Consider a rectangular amperian loop PQRS near the middle of solenoid

as shown in figure where PQ = l

Let the magnetic field along the path PQ be B and zero along RS. As the paths QR and PS are

perpendicular to the axis of solenoid, the magnetic field component along these paths is zero. Therefore,

the path QR and PS will not contribute to the line integral of magnetic field B.

Total number of turns in length l = l

The line integral of magnetic field induction B over the closed path PQRS is



Here,

And,

Also, ( outside the solenoid, B = 0)

From Ampere’s circuital law,

Total current through the rectangle PQRS

= µ0 × Number of turns in rectangle × Current

BL = µ0nlI

Or

This relation gives the magnetic field induction at a point well inside the solenoid.


Derive an expression for the resistivity of a good conductor, in terms of the relaxation time of electrons.

Solution:

Relation between the resistivity and relaxation time: We know that drift velocity of electron is given by:

However,

∴ According to Ohm’s law,

But resistivity is given by:

Comparing (i) and (ii), we obtain

The required relationship between resistivity and relaxation time of electrons is




The circuit arrangement given below shows that when an a.c. passes through the coil A, the current starts

flowing in the coil B.

(i) State the underlying principle involved.

(ii) Mention two factors on which the current produced in the coil B depends.

Solution:

(i) The principal involved in the given circuit arrangement is mutual induction. Mutual induction is the

property of two coils by virtue of which each opposes any change in the strength of current flowing

through the other by developing an induced emf.

(ii) The current produced in the coil B depends on

(a) number of turns in the coil

(b) nature of material

(c) distance between two coils


State one feature by which the phenomenon of interference can be distinguished from that of diffraction.

A parallel beam of light of wavelength 600 nm is incident normally on a slit of width ‘a’. If the distance

between the slits and the screen is 0.8 m and the distance of 2nd order maximum from the centre of the

screen is 15 mm, calculate the width of the slit.

Solution:

Difference between interference and diffraction: Interference is due to superposition of two distinct waves

coming from two coherent sources. Diffraction is produced as a result of superposition of the secondary

wavelets coming from different parts of the same wavefront.

Numerical: Here, λ = 600 nm = 600 × 10−19 = 6 × 10−7 m

D = 0.8 m, x = 15 mm = 1.5 × 10−3 m,

n = 2, a = ?


Two point charges q1 = 10 × 10−8 C and q2 = −2 × 10

−8 C are separated by a distance of 60 cm in air.

(i) Find at what distance from the 1st charge, q1, would the electric potential be zero.



1

(ii) Also calculate the electrostatic potential energy of the system.

OR

Two point charges 4Q, Q are separated by 1 m in air. At what point on the line joining the charges is the

electric field intensity zero? Also calculate the electrostatic potential energy of the system of charges,

taking the value of charge, Q = 2 × 10−7 C.

Solution:

(i) Here, q1 = 10 × 10−8 C, q2 = −2 × 10

−8 C

And AB = 60 cm = 0.60 = 0.6 m

Let AP = x

Then, PB = 0.6 − x

Potential P due to charge

Potential P due to charge

Potential at

∴Distance from first charge = 0.5 = 50 cm

(ii) Electrostatic potential energy of the system

OR

Let the point be at a distance x from 4Q charge.

Electric field at P due to 4Q = Electric field at P due to Q



x = 2 − 2x or − x = 2 − 2x

x + 2x = 2 or − x + 2x = 2

3x = 2 or x = 2

or x = 2

x = 2 m is not possible

m

Electrostatic potential energy of the system is


Identify the following electromagnetic radiations as per the wavelengths given below. Write one application

of each.

(a) 10−3 nm

(b) 10−3 m

(c) 1 nm

Solution:

(a) Wavelength range: 10−3 nm = 10−3 × 10−9 = 10−12 m

Type of EMW: X-rays

Applications:

X-rays are used in surgery for the detection of fractures, foreign bodies such as bullets, diseased organs

and stones in human body.

(b) Wavelength range: 10−3 m

Type of EMW: Microwave

Application: Microwaves are used in Radar systems for aircraft navigation.

(c) Wavelength range: 1 nm

Type of EMW: Infrared



Application: Infrared waves are used for taking photographs during the conditions of fog, smoke, etc.


Explain why high frequency carrier waves are needed for effective transmission of signal. A message signal

of 12 KHz and peak voltage 20 V is used to modulate a carrier wave of frequency 12 MHz and peak

voltage 30 V. Calculate the (i) modulation index (ii) side-band frequencies.

Solution:

For the effective transmission of signals, the high frequency carrier waves are used because these high

frequency carrier waves travel through space or medium with the speed of light and they are not

obstructed by earth’s atmosphere.

Numerical: Here, Vm = 12 KHz, Em = 20 V

Vc = 12 MHz = 1200 KHz, Ec = 30 V

(i) Modulation index,

(ii) The side bands are:

USB = Vc + Vm = 12000 + 12 = 12012 KHz

LSB = Vc + Vm = 12000 − 12 = 11988 KHz


Distinguish between unpolarised and plane polarised light. An unpolarised light is incident on the boundary

between two transparent media. State the condition when the reflected wave is totally plane polarised.

Find out the expression for the angle of incidence in this case.

Solution:

The independent light waves whose planes of vibrations are randomly oriented about the direction of

propagation are said to be unpolarised.

When the light waves transmit only one component parallel to a special axis, the resulting light is called

plane polarised or linearly polarised light.

When the reflected wave is perpendicular to the refracted wave, the reflected wave is totally polarised

wave. The angle of incidence in this case is called Brewster’s angle and is denoted by iB.

The relation between incidence angle iB is given by:

µ = tan iB


Draw the labelled circuit diagram of a common-emitter transistor amplifier. Explain clearly how the input

and output signals are in opposite phase.

OR

State briefly the underlying principle of a transistor oscillator. Draw a circuit diagram showing how the

feedback is accomplished by inductive coupling. Explain the oscillator action.

Solution:

The circuit details for using an npn transistor as common emitter amplifier are shown in the figure:



The input (base-emitter) circuit is forward biased and the output (collector-emitter) circuit is reverse

biased.

When no ac signal is applied, the potential difference Vc between the collector and the emitter is given by:

Vc = Vce − IcRL ... (i)

Where Vce is the voltage of battery VCE

When an ac signal is fed to the input circuit, the forward bias increases during the positive half cycle of the

input. This results in an increase in Ic and a consequent decrease in Vc as is clear from (1). Thus, during

positive half cycle of the input, the collector becomes less positive.

During the negative half cycle of the input, the forward bias is decreased, resulting in a decrease in Ie and

hence Ic. Therefore, from equation (i), Vc would increase, making the collector more positive. Hence, in a

common-emitter amplifier, the output voltage is 180° out of phase with the input voltage.

OR

Transistor as an oscillator: In an oscillator, the output at a desired frequency is obtained without applying

any external input voltage.

The common emitter n-p-n transistor as an oscillator is shown in the following figure.

A variable capacitor C of suitable range is connected in parallel to coil L to give the variation in frequency.

Oscillator action:

As in an amplifier, the base-emitter junction is forward biased while the base collector junction is reverse

biased. When the switch S is put on, a surge of collector flows in the coil T2. The inductive coupling

between coil T2 and T1 cause a current to flow in the emitter circuit i.e., feedback from input to output. As

a result of positive feedback, the collector current reaches at maximum. When there is no further feedback

from T2 to T1, the emitter current begins to fall and collector current decreases. Therefore, the transistor

has reverted back to its original state. The whole process now repeats itself.

The resonance frequency (f) of the oscillator is given by:

The tank of tuned circuit is connected in the oscillator side. Hence, it is known as tuned collector



The tank of tuned circuit is connected in the oscillator side. Hence, it is known as tuned collector

oscillator.


The ground state energy of hydrogen atom is − 13.6 eV.

(i) What is the kinetic energy of an electron in the 2nd excited state?

(ii) If the electron jumps to the ground state from the 2nd excited state, calculate the wavelength of the

spectral line emitted.

Solution:

(i) The energy of the electron in the nth orbit is given by the relation:

For second excited state, we have n = 3

(ii) The energy radiated when an electron jumps from n = 3 to n = 1 is given by the relation:

∴Wavelength of the spectral line emitted is


The following graph shows the variation of stopping potential V0 with the frequency ν of the incident

radiation for two photosensitive metals X and Y:

(i) Which of the metals has larger threshold wavelength? Give reason.

(ii) Explain giving reason, which metal gives out electrons, having larger kinetic energy, for the same

wavelength of the incident radiation.

(iii) If the distance between the light source and metal X is halved, how will the kinetic energy of electrons

emitted from it change? Give reason

Solution:



Solution:

(i) Let = Frequency of incident radiations of metal Y

= Frequency of incident radiations of metal X

Therefore, metal X has larger threshold wavelength.

(ii) Since the kinetic energy of the emitted electrons is directly proportional to the frequency of incident

radiation, metal Y having larger incident frequency will have larger kinetic energy.

∴ Metal Y has larger kinetic energy.

(iii) Kinetic energy of the emitted photoelectrons is independent of the intensity of the incident light. Hence,

kinetic energy of the emitted photoelectrons remains unchanged if the distance between the light source

and metal X is halved.


A circular coil of 200 turns and radius 10 cm is placed in a uniform magnetic field of 0.5 T, normal to the

plane of the coil. If the current in the coil is 3.0 A, calculate the

(a) Total torque on the coil

(b) Total force on the coil

(c) Average force on each electron in the coil due to the magnetic field

Assume the area of cross-section of the wire to be 10−5 m2 and the free electron density is 1029/m3.

Solution:

Here n = 200, r = 10 cm = 0.1 m

B = 0.5 T, I = 3 A

(a) Torque is given by the relation:

(b) The total force on a current loop placed in a magnetic field is always zero.

(c) Given = 1029 /m3, A = 10−5 m2

Average force on an electron of charge (e), moving with drift velocity (Vd) in the magnetic field (B), is

given by:

F = BeVd




The inputs A and B are inverted by using two NOT gates and their outputs are fed to the NOR gate as

shown below.

Analyze the action of the gates (1) and (2) and identify the logic gate of the complete circuit so obtained.

Give its symbol and the truth table.

Solution:

The output of the gates is

(i) The gates (1) and (2) act as NOT gate.

(ii) The logic gate of the complete circuit is AND gate.

Symbol:

Truth Table

Input Output

Y = ABA B

0 0 0

0 1 0

1 0 0

1 1 1


(i) Calculate the equivalent resistance of the given electrical network between points A and B.

(ii) Also calculate the current through CD and ACB, if a 10 V d.c source is connected between A and B,

and the value of R is assumed as 2 Ω.

Solution:

(i) Equivalent circuit is



This is a balanced Wheatstone bridge. Therefore, the resistance of the arm CD is ineffective.

∴ Resistance of arm ACB, R1 = R + R = 2R and the resistance of arm ADB, R2 = R + R = 2R

∴ Effective resistance between A and B is

(ii) The points C and D are at the same potential. Therefore, no current flows through the arm CD.

Resistance of arm ACB = 2R = 2 × 2 = 4 Ω

∴Current through branch ACB

∴ I = 2.5 A


Derive an expression for the energy stored in a parallel plate capacitor.

On charging a parallel plate capacitor to a potential V, the spacing between the plates is halved, and a

dielectric medium of is introduced between the plates, without disconnecting the d.c source.

Explain, using suitable expressions, how the (i) capacitance, (ii) electric field and (iii) energy density of the

capacitor charge.

OR

(a) Define electric flux. Write its SI unit.

(b) The electric field components due to a charge inside the cube of side 0.1 m are as shown:

Where α = 500 N/C − m

Ey = 0, Ez = 0

Calculate (i) the flux through the cube and (ii) the charge inside the cube



Calculate (i) the flux through the cube and (ii) the charge inside the cube

Solution:

Energy stored in a parallel plate capacitor:

At any intermediate stage, suppose charge on conductor 1 is + q and charge on conductor 2 is − q.

∴Potential difference between conductors 1 and 2 is q/C, where C is the capacity of the capacitor.

Suppose the capacitor is charged gradually and at any stage, the charge on the capacitor is q.

∴ Potential of capacitor =

Small amount of work done giving an additional charge dq to the capacitor is

Total work done in giving a charge Q to the capacitor

As electrostatic force is conservative, this work is stored in the form of potential energy (U) of the

capacitor.

Put Q = CV

(i) Let be the charge in capacitance.



(ii) Charge of field:

E

Charge of energy density:

OR

(a) Electric flux:

It is the number of electric field lines passing through a surface normally.

S.I unit of flux = Nm2 C−1

(b) Here, Ex = αx, Ey = 0, Ez = 0

α = 500 N/C-m, Side of cube a = 0.1 m

Since the electric field has only x-component,

for each of four faces of cube ⊥ to y-axis and z-axis.

∴ Electric flux is only for left and right face along x-axis of the cube.

(i) Electric field at the left face, x = a, is

and electric field at the right face, x = a + a = 2a, is

∴ Net flux through the cube =

(ii) By Gauss’s law,


Derive the lens formula, for a concave lens, using the necessary ray diagram.



Two lenses of powers 10 D and − 5 D are placed in contact.

(i) Calculate the power of the new lens.

(ii) Where should an object be held from the lens, so as to obtain a virtual image of magnification 2?

OR

(a) What are coherent sources of light? Two slits in Young’s double slit experiment are illuminated by two

different sodium lamps emitting light of the same wavelength. Why is no interference pattern observed?

(b) Obtain the condition for getting dark and bright fringes in Young’s experiment. Hence write the

expression for the fringe width.

(c) If S is the size of the source and its distance from the plane of the two slits, what should be the criterion

for the interference fringes to be seen?

Solution:

Derivation of lens formula:

ABis an object held perpendicular to the principal axis of the lens. A virtual, erect, and smaller image

is formed due to refraction through concave lens as shown in figure.

As ∆s and ABC are similar,

Again as ∆s andCDF are similar,

However, CD = AB

From (i) and (ii),

Using new Cartesian sign conventions, let



Dividing both sides by uvf, we obtain

This is the required lens formula.

(i) Power of new lens, P = P1 + P2

∴P = 10 − 5 = + 5 D

(ii) Here, u =?

cm = 20 cm

m = 2 i.e., or v = 2u

Using lens formula,

∴ Object distance = 10 cm

OR

(a) Coherent sources of light:

The sources of light, which emit continuous light waves of the same wavelength, same frequency, and in

same phase, are called coherent sources of light.

If the two slits in Young’s double slit experiment are illuminated by two different sodium lamps emitting

light of the same wavelength, then no interference pattern will be observed because phase difference

between the light waves emitted from two different sodium lamps will change continuously.

(b) For bright fringes (maxima), path difference,

where n = 0, 1, 2, 3, …

For dark fringes (minima),

Path difference,

, where n = 0, 1, 2, 3, …

The separation between the centres of two consecutive bright fringes is the width of a dark fringe.

∴Fringe width, P = Xn − Xn−1



(c) Criterion is


An a.c source generating a voltage is connected to a capacitor of capacitance C. Find the

expression for the current i, flowing through it, plot a graph of v and i versus ωt to show that the current is

, ahead of the voltage.

A resistor of 200 Ω and a capacitor of 15 µF are connected in series to a 220 V, 50 Hz a.c source.

Calculate the current in the circuit and the rms voltage across the resistor and the capacitor. Is the

algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox.

OR

Explain briefly, with the help of a labelled diagram, the basic principle of the working of an a.c generator.

In an a.c generator, coil of turns and area A is rotated at V revolutions per second in a uniform magnetic

field B.

Write the expression for the emf produced. A 100-turn coil of area 0.1 m2 rotates at half a revolution per

second. It is placed in a magnetic field 0.01 T perpendicular to the axis of rotation of the coil. Calculate

the maximum voltage generated in the coil.

Solution:

A.C source containing capacitor:

Alternating emf supplied is:

Potential difference across the plates of capacitor

At every instant, the potential difference V must be equal to the emf applied i.e.,

Or, q = C

It I is instantaneous value of current in the circuit at instant t, then



Numerical:

Here, r = 200 Ω, C = 15 µF = 15 × 10−6 F

E = 220 V, f = 50 Hz, I =?

This is because these voltages are not in same phase and they cannot be added like ordinary numbers.

OR

A.C generator principle:

Whenever a closed coil is rotated in a uniform magnetic field about an axis perpendicular to the field, the

magnetic flux linked with coil changes and an induced emf is set up across its ends.

The construction of an ac generator is shown in the figure. Initially, the coil ABCD is horizontal. The coil

starts rotating clockwise and the arm AB moves up whileCD moves down. By Fleming’s right hand rule,

the induced current flows along ABCD.

In second half rotation, the arm CD moves up and AB moves down. The induced current flows in the

opposite direction that is along DCBA. Thus, an alternating current flows in the circuit.

The magnetic flux linked with the coil at any instant is



Φ = BA cos ωt

Induced emf will be

Or E = E0 sin ωt, where E0 = BA ω = Peak value of induced emf

Numerical:

= 100, A = 0.1 m2, B = 0.01T

∴ Maximum voltage,



physics 2008 set 1 sol

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