Physics 2004 Set 1 Sol

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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)

Why is shortwave band used for long distance radio broadcast?

Solution:

The shortwave band radio-waves are used for long distance broadcast because they are easily reflected

back to earth by the ionosphere.


Two metals A and B have work functions 2 eV and 5 eV respectively. Which metal has lower threshold

wavelength?

Solution:

i.e.,

Therefore, metal B with higher work function has lower threshold wavelength.


Draw the voltage-current characteristic of a zener diode.

Solution:

V-I characteristic for a zener diode is given below:

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A solenoid with an iron core and a bulb is connected to a d.c. source. How does the brightness of the

bulb change when the iron core is removed from the solenoid?

Solution:

The brightness of the bulb remains unchanged because inductive reactance in a d.c. circuit is zero.


Peak value of e.m.f of an a.c. source is E0. What is its r.m.s value?

Solution:


Solution:

OUT OF CURRET SYLLABUS


Solution:



Solution:


Section B


An electric dipole of length 4 cm, when placed with its axis making an angle of 60° with a uniform electric

field, experiences a torque of Nm. Calculate the (i) magnitude of the electric field (ii) potential energy

of the dipole, if the dipole has charges of ± 8 nC.

Solution:

(i) Here, length of the dipole, 2a = 4 cm = 4 × 10−2 m

Torque,τ = Nm

Charge, q = 8 × 10−9 C

∴τ = pE sinθ = (q × 2a) × E × sinθ

Or, = 8 × 10−9 × 4 × 10−2 × E × sin 60°



Or,

Or, E = 2.5 × 1010 NC−1

(ii) Potential energy,

Or, U = − 8 × 10−9 × 2 × 10−5 × 2.5 × 1010 cos 60°

Or, U = − 2 J


Explain how the resistivity of a conductor depends upon (i) number density ‘n’ of free electrons, and (ii)

relaxation time ‘λ’.

Solution:

Resistivity of a conductor is given by the formula,

Where, m = Mass of electron

n = Number density of free electrons

e = Charge on an electron

λ = Relaxation time

(i) It is evident from the formula given by equation (i) that the resistivity of a conductor is inversely

proportional to the number density ‘n’ of free electrons.

(ii) The resistivity of a conductor is inversely proportional to the relaxation time λ.


Two long parallel straight wires X and Y separated by a distance of 5 cm in air carry currents of 10 A and

5 A respectively in opposite directions. Calculate the magnitude and direction of the force on a 20 cm

length of the wire Y.

Or

A circular coil of 100 turns, radius 10 cm carries a current of 5 A. It is suspended vertically in a uniform

horizontal magnetic field of 0.5 T, the field lines making an angle of 60° with the plane of the coil. Calculate

the magnitude of the torque that must be applied on it to prevent it from turning.

Solution:

Here, r = 5 cm = 5 × 10−2 m

I1 = 10A, I2 = 5 A



I1 = 10A, I2 = 5 A

l = 20 cm × 10−2 m

Where is permeability of free space

The direction of the force is perpendicular to the length of wire Y and acts away from X (repulsion).

Or

Here,

Number of turns in the coil, = 100

Radius of the coil, r = 10 cm

Current, I = 5 A

Magnetic field, B = 0.5 T

θ = 90° − 60° = 30°

τ = IBA sinθ

τ = 100 ×5 × 0.5 × (3.14 × .10 × .10) × sin 30°

τ = 3.925 Nm


A bar magnet M is dropped so that it falls vertically through the coil C. The graph obtained for voltage

produced across the coil vs. time is shown in figure (b).

(i) Explain the shape of the graph.

(ii) Why is the negative peak longer than the positive peak?

Solution:

As the magnet M approaches coil C, magnetic flux linked with the coil increases and emf is induced in the

coil, which opposes the increase in flux. When magnet is inside the coil, magnetic flux linked with the coil

decreases and emf is induced in the coil, which opposes the decrease in flux. As velocity of magnet has

increased, induced emf is more. Therefore, negative peak is longer than the positive peak. When the

magnet has fallen through large distance, changing magnetic flux due to its movement vanishes. Induced

emf reduces to zero.



emf reduces to zero.


T.V. tower has a height of 400 m at a given place. Calculate as coverage range, if the radius of the earth is

6400 km.

Solution:

Coverage range of a T.V. tower is given by formula:

Where,

h = Height of tower = 400 m

R = Radius of earth = 6400 km = 6400 × 103 m


Draw a ray diagram of an astronomical telescope in the normal adjustment position. Write down the

expression for its magnifying power.

Solution:

Magnifying power,


Give the logic symbol for an OR gate. Draw the output wave form for input wave forms A and B for this

gate.

Solution:

Logic symbol of OR gate is given below:




State Gauss’ theorem in electrostatics. Using this theorem, derive an expression for the electric field

intensity due to an infinite plane sheet of charge density σ C/m2.

Solution:

Gauss’s Theorem:

It states that the total flux through a closed surface is times the total charge enclosed by the closed

surface.

Derivation:

Let us consider a non-conducting sheet of charge with surface charge density σ. Consider a cylinder of

length 2r and cross-sectional area A as Gaussian surface. From summitry, electric field points at right

angle to the end caps and away from the sheet. There is no contribution from the curved surface because

angle between and is 90°.

At the end faces, angle between and is zero.

From Gauss’s law,


A 10 µF capacitor is charged by a 30 V d.c. supply and then connected across an uncharged 50 µF

capacitor. Calculate (i) the final potential difference across the combination, and (ii) the initial and final

energies. How will you account for the difference in energy?

Solution:

Here,

Capacitance, C1 = 10 µF = 10 × 10−6 F



Voltage, V = 30 V

C2 = 50 µF = 50 × 10−6 F

(i) Charge on µF capacitor = CV = 10 × 10−6 × 30

= 3 × 10−4 C

∴Final potential difference across the combination

(ii) Initial electric energy of 10µF capacitor:

Final electrostatic energy of the combination

Electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation.


Solution:


Section C


The circuit diagram shows the use of a potentiometer to measure a small emf produced by a thermocouple

connected between X and Y. The cell C, of emf 2 V, has negligible internal resistance. The potentiometer

wire PQ is 1.00 m long and has a resistance of 5 Ω. The balance point S is found to be 400 mm from P.

Calculate the value of emf V, generated by the thermocouple.

Solution:

Here, E = 2 V

Length of the potentiometer wire = 1.00 m

∴Current flowing through wire

Potential drop across wire PQ = IR = 2 × 10−3 × 5

= 0.01 V



Potential gradient along wire PQ,

∴Potential drop across wire PS


Using Biot-Savart law, deduce an expression for the magnetic field on the axis of a circular current loop.

Draw the magnetic field lines due to a circular current carrying loop.

Or

A hydrogen ion of mass ‘m’ and charge ‘q’ travels with a speed ‘v’ in a circle of radius ‘r’ in a magnetic

field of intensity ‘B’. Write the equation in 4 terms of these quantities only, relating the force on the ion to

the required centripetal force. Hence derive an expression for its time.

Solution:

Consider a circular loop of radius ‘a’ carrying current I, held perpendicular to the plane of the paper. We

want to find the magnetic field at a point P, which is at a distance x from the centre of the coil. Let us take

a small element AB of length dl. From Biot-Savart’s law, the field at point P due to a small element is

given by:

The direction of the field is in a plane perpendicular to the plane containing dl and is at right angle to line

CP. Resolve this into two components along the x-axis and at right angles to the line

OP. It is clear from the figure that the resultant field along y-axis (i.e., components cancel each

other) is zero. Field due to is as shown. The resultant field at P will be only in x-direction.

The total field at P due to all such elements is given by:

Or

Substituting the values, we obtain



Substituting the values, we obtain

The magnetic field lines due to a circular current carrying loop are shown below:

Or

Magnetic force on the hydrogen ion = Centripetal force

Or

Where B → Applied magnetic field

q → Charge of hydrogen ion

m → Mass of ion

v → Speed

r → Radius of circle

Or,

Time period,


A uniform magnetic field gets modified as shown below, when two specimens X and Y are placed in it.

(i) Identify the two specimens X and Y.

(ii) State the reason for the behaviour of the field lines in X and Y.

Solution:

(i) The specimen X must be diamagnetic and the specimen Y must be paramagnetic or ferromagnetic.



(ii) This is because magnetic lines of force prefer not to pass through diamagnetic material, its permeability

being less than one.

On the other hand, through a paramagnetic or ferromagnetic material, magnetic lines of force prefer to

pass, their permeability being more than one.


Two narrow slits are illuminated by a single monochromatic source. Name the pattern obtained on the

screen. One of the slits is now completely covered. What is the name of the pattern now obtained on the

screen? Draw intensity pattern obtained in the two cases. Also write two differences between the patterns

obtained in the above two cases.

Solution:

With the narrow slits, an interference pattern is obtained.

When one slit is completely covered, diffraction patterns are obtained.

In case of interference, the following intensity distribution curve is obtained.

The intensity distribution curve for diffraction at a single slit is shown below:

Interference Diffraction

1.All bright fringes are of same

intensity.1.

Intensity of bright fringes decrease with the increase in

distance from the central bright fringe.

2.Widths of interferences may or

may not be equal.2. Widths of diffraction fringes are never equal.


Red light, however bright it is, cannot produce the emission of electrons from a clean zinc surface. But

even weak ultraviolet radiation can do so. Why?

X-rays of wavelength ‘λ’ fall on photosensitive surface, emitting electrons. Assuming that the work

function of the surface can be neglected, prove that the de-Broglie wavelength of electrons emitted will be

.

Solution:

The frequency of red light is less than the threshold frequency of zinc surface. Hence, it cannot cause

photoelectric emission from zinc surface, whatever may be its intensity. The frequency of ultraviolet



photoelectric emission from zinc surface, whatever may be its intensity. The frequency of ultraviolet

radiation is greater than the threshold frequency of zinc surface. Hence, even weak ultraviolet radiation can

cause photoelectric emission.

Given Φ0 = 0

Using Einstein’s photoelectric equation, we obtain

∴de Broglie wavelength of emitted photoelectron


Define the terms: ‘half-life period’ and ‘decay constant’ of a radioactive sample. Derive the relation

between these terms.

Solution:

Half life period:

Half life period is the time during which number of atoms left un-decayed in the sample is half the total

number of atoms present initially in the sample.

Decay constant:

Decay constant of a radioactive element is the reciprocal of the time during which the number of atoms left

in the sample reduces to times the original number of atoms in the sample.

Relation:

At

As = 0e−λt,

or

Or,

Taking natural logarithm,

Or,

Or,


When a deuteron of mass 2.0141 u and negligible kinetic energy is absorbed by a lithium nucleus of

mass 6.0155 u, the compound nucleus disintegrates spontaneously into two alpha particles, each of mass



mass 6.0155 u, the compound nucleus disintegrates spontaneously into two alpha particles, each of mass

4.0026 u. Calculate the energy in joules carried by each alpha particle. (1u = 1.66 × 10−27 kg)

Solution:

Total initial mass = 8.0296 u

Total final mass = 2 m = 2 × 4.0026

= 8 .0052 u

Mass defect, ∆m = 8.0296 − 8.0052

= 0.0244 u

= 0.0244 × 1.66 × 10−27 kg

Energy released

Q = ∆m × C2 = 0.0244 × 1.66 ×10−27 × (3 × 108)2

= 3.645 ×10−12 J

Energy of each α-particle = 1.8225 × 10−12 J


What is meant by ‘remote sensing’? Briefly explain how it is carried out. Mention any two applications of

remote sensing.

Solution:

Remote sensing is a technique of obtaining information about an object/area from a distance without being

in physical contact with it.

A satellite equipped with appropriate sensors to acquire data is placed in an orbit around the earth at any

height having a period of revolution. It takes photographs or collects any other information desired and

transmits it back to an earth station. This is known as remote sensing.

Applications of remote sensing are:

(i) In forestry

(ii) In ground water surveys


What is an optical detector? State its three essential characteristics. Name the factor which decides how

good a detector is.

Solution:

Optical detector is used to detect the optical signals at the receiving end and converts light into electrical

signal so that the transmitted information may be decoded.

Essential characteristics of an optical detector are:

(i) Size compatible with the fibre

(ii) High intensity at the desired optical wavelength



(ii) High intensity at the desired optical wavelength

(iii) High response time for fast speed data transmission and reception

The efficiency of generating electron-hole pairs in a photodiode decides how good that detector is.

Section D


With the help of a labelled circuit diagram, explain how an n-p-n transistor can be used as an amplifier in

common emitter configuration. Explain how the input and output voltages are out of phase by 180° for a

common-emitter transistor amplifier.

Or

For an n-p-n transistor in the common-emitter configuration, draw a labelled circuit diagram of an

arrangement for measuring the collector current as a function of collector-emitter voltage for at least two

different values of base current. Draw the shape of the curves obtained. Define the terms:

(i) output resistance and (ii) ‘current amplification factor’.

Solution:

The circuit details for using an n-p-n 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.

The potential difference VC when no a.c. signal is applied between the collector and the emitter is given by:

When an a.c. signal is applied 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 equation (i).

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



The output characteristics are drawn by plotting collection current IC versus collector emitter voltages

VCE, keeping the base current Ib constant.

Output resistance:

It is defined as the ratio of the collector emitter voltage (∆VCE) to the corresponding change in collector

current (∆IC) at constant base current Ib.

Current amplification factor:

It is defined as the ratio of the change in collector current to the change in base current.


What is induced emf? Write Faraday’s law of electromagnetic induction. Express it mathematically.

A conducting rod of length ‘l’ with one end pivoted, is rotated with a uniform angular speed ‘ω’ in a

vertical plane, normal to a uniform magnetic field ‘B’. Deduce an expression for the emf induced in this

rod.

In India, domestic power supply is at 220 V, 50Hz, while in USA it is 110 V, 50 Hz. Give one advantage

and one disadvantage of 220 V supply over 110 V supply.

Solution:

When the magnetic flux linked with a closed circuit changes, an emf is set up across it, which lasts only as

long as the change in flux is taking place. This emf is called induced emf.

Faraday’s Laws

First Law:

Whenever the amount of magnetic flux linked with a circuit changes, an emf is induced in the circuit. The

induced emf lasts so long as the change in magnetic flux continues.

Second Law:

The magnitude of emf induced in a circuit is directly proportional to the rate of change of magnetic flux

linked with the circuit.

Suppose the conducting rod completes one revolution in time T. Then,

Change in flux = B × Area swept

= B × πl2



Induced emf

However,

Advantage:

The power loss at 220 volt supply is less than at 110 V.

Disadvantage:

It is difficult to work with 220 V supply because its peak value (311 V) is much higher than the peak value

(155.5 V) of 110 V supply.


Complete the path of incident ray of light, showing the formation of a real image.

Hence derive the relation connecting object distance ‘u’, image distance ‘v’, radius of curvature R, and the

refractive indices n1 and n2 of the two media.

Briefly explain, how the focal length of a convex lens changes, with increase in wavelength of incident light.

Solution:

Sign conventions:

(i) All distances are measured from the pole of the spherical surface.

(ii) Distances measured in the direction of incident light are taken positive.

(iii) Distances measured in the opposite direction of incident light are negative.

Derivation:

Let a spherical refracting surface XY separate a rarer medium of refractive index n1 from a denser medium

of refractive index n2. Let P be the pole, C be the centre and R = PC be the radius of curvature of this

surface.

Consider a point object O lying on the principal axis of the surface.

From A, draw AM ⊥ OI

Let ∠AOM = α, ∠AIM = β, ∠ACM = γ

As external angle of a triangle is equal to sum of internal opposite angles, therefore, in



As external angle of a triangle is equal to sum of internal opposite angles, therefore, in

∆IAC,

r + β = γ

r = γ − β

Similarly, in ∆OBC, i = α + γ (i)

According to Snell’s law,

( angles are small)

∴ n1i = n2r

Using (i), we obtain

n1 (α + γ) = n2 (γ − β)

As angles α, β, and γ are small, using , we obtain

As aperture of the spherical surface is small, M is close to P. Therefore, MO ≈ PO, MI ≈ PI, MC≈ PC

From (3),

Using new Cartesian sign conventions, we put

This is the required relation.

As wavelength of incident light increases, µ decreases. Hence, focal length f increases.



Physics 2004 Set 1 Sol

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