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# CBSE Physics 2016 Exam Questions

21. Draw a schematic ray diagram of a reflecting telescope showing how rays coming from a distant object are received at the eyepiece. Write its two important advantages over a refracting telescope.

A reflecting telescope is as shown below:

In a reflecting telescope, an image is formed by reflection from a curved mirror. The image formed is then magnified by a secondary mirror.

Advantages of reflecting telescope over a refracting telescope are:

1. There is no chromatic aberration for reflecting telescopes as the objective is a mirror.

2. Spherical aberration is reduced in the case of reflecting telescopes by using mirror objective in the form of a paraboloid.

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22. (i) Write the function of a transformer. State its principle of working with the help of a diagram. Mention various energy losses in this device.

(ii) The primary coil of an ideal step-up transformer has 100 turns and the transformation ratio is also 100. The input voltage and power are 220 V and 1100 W, respectively. Calculate the:

a) number of turns in secondary

b) current in primary

c) voltage across secondary

d) current in secondary

e) power in secondary

i) A transformer is a device that changes a low alternating voltage into a high alternating voltage or vice versa.

Transformer works on the principle of mutual induction.

A changing alternate current in primary coil produces a changing magnetic field, which induces a changing alternating current in secondary coil.

Energy losses in transformer:

Flux leakage due to poor structure of the core and air gaps in the core.

Loss of energy due to heat produced by the resistance of the windings.

Eddy currents due to alternating magnetic flux in the iron core, which leads to loss due to heating.

Hysteresis, frequent and periodic magnetisation and demagnetization of the core, leading to loss of energy due to heat.

ii)

a)  Number of turns in secondary coil is given by,

b) Current in primary is given by,

c) Voltage across secondary  is given by,

d) Current in secondary is given by,

e) In an ideal transformer,

Power in secondary = Power in primary = 1,100 W
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23. (i) In Young's double-slit experiment, deduce the condition for (a) constructive and (b) destructive interference at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position 'x' on the screen.

(b) Compare the interference pattern observed in Young's double-slit experiment with single-slit diffraction pattern, pointing out three distinguishing features.

Expression for fringe width in Young's Double Slit Experiment

Let S1 and S2 be two slits separated by a distance d.

GG' is the screen at a distance D from the slits S1 and S2.

Point C is equidistant from both of the slits.

The intensity of light will be maximum at this point because the path difference of the waves reaching this point will be zero.

At point P, the path difference between the rays coming from the slits is given by,

S1 = S2 P - S1

Now, S1 S2 = d, EF = d, and S2 F = D

In
S2P =

For constructive interference, the path difference is an integral multiple of wavelengths, that is, path difference is n.

Graph of intensity distribution in young's double slit experiment is,

ii)

Three distinguishing features observed in Young's Double Slit experiment as compared to single slit diffraction pattern is,

1. In the interference pattern, all the bright fringes have the same intensity. The bright fringes are not of the same intensity in a diffraction pattern.

2. In interference pattern, the dark fringes have zero or small intensity so that the bright and dark fringes can be easily distinguished. While in diffraction pattern, all the dark fringes are not of zero intensity.

3. In interference pattern, the width of all fringes are almost the same, whereas in diffraction pattern, the fringes are of different widths.

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24. Meeta's father was driving her to school. At the traffic signal, she noticed that each traffic light was made of many tiny lights instead of a single bulb. When Meeta asked this question to her father, he explained the reason for this.

Answer the following questions based on above information:

(i) What were the values displayed by Meeta and her father?

(ii) What answer did Meeta's father give?

(iii) What are the tiny lights in traffic signals called and how do these operate?

i)

The values displayed by Meeta and her father are:

Meeta: Curious mind
Meeta's Father: knowledge and patience

ii)

The answer that Meeta's father had given would be the advantages of using LED lights over a single bulb.

a) LED's consume very less power as compared to an incandescent bulb.

b) The cost of tiny LED is much less than of a bulb. This reduces the maintenance cost of LED bulbs.

c) The working of the traffic will remain unhindered even if one of the bulbs is not working.

iii)

The tiny lights are called LED (Light Emitting Diode) and they work on the principle of de-excitation of electrons in a forward biased semiconductor upon passing electricity through them.

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25. (i) Define the term drift velocity.

(ii) On the basis of electron drift, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend?

(iii) Why alloys like constantan and manganin are used for making standard resistors?

i)

Drift velocity is the average velocity of the free electrons in the conductor with which they get drifted towards the positive end of the conductor under the influence of an external electric field.

ii)

Free electrons are in continuous random motion. They undergo changethey undergo change in direction at each collision and the thermal velocities are randomly distributef in all directions.

Electric field, E = -eE

Acceleration of each electron,          ... (2)
Here,

m = mass of an electron
e = charge on an electron

Drift velocity is given by,

Electrons are accelerated because of the external electric field.

They move from one place to another and current is produced.

For small interval dt, we have

I dt = -q ; where q is the total charge flowing

Let, n be the free electrons per unit area. Then, total charge crossing area A in time dt is given by,

Idt = neAvd dt

Substituting the value of vd, we get

I.dt = neA
Current density, J  =
From ohm's law, we have

J =
Here,  is the conductivity of the material through whic the current is flowing,

Thus,

iii)

Alloys like constantan and manganin are used for making standard resistors because:

a) they have high value of resistivity

b) temperature coefficient of resistance is less.

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26. (i) State the principle of working of a potentiometer.

(ii) In the following potentiometer circuit, AB is a uniform wire of length 1 m and resistance 10 Ω. Calculate the potential gradient along the wire and balance length AO (= l)

i)

Principale of potentiometer: The potential difference across any two points of current carrying wire, having uniform cross-sectional area and material, of the potentiometer is directly proportional to the length between the two points.

That is,                                            V

Proof:

V = IR = I

i.e., V =

For unifrom current and cross- sectional area, we have

ii)

Given,

E = 2 V; R = 15  ; RAB = 10

Potential difference across the wire, =
Therefore, potential gradient = 0.8/1 = 0.8 V/ m

Potential difference across AO =

Therefore,

Length, AO =
= ; which is the required balance length of the wire.

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27. (i) State Bohr's quantization condition for defining stationary orbits. How does the de Broglie hypothesis explain the stationary orbits?

(ii) Find the relation between three wavelengths λ1, λ2 and λ3  from the energy-level diagram shown below.

Bohr's Quantisation Rule:

According to Bohr, an electron can revolve only in certain discrete, non-radiating orbits for which the total angular momentum of the revolving electron is an integral multiple of  ; where h is the Planck's constant.

That is,
b)

Using Rydberg's formula for spectra of hydrogen atom, we have

Hence, the relation between 3 wavelengths from the energy-level diagram is obtained.

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28. (i) An a.c. source of voltage V = Vo sin ωt is connected to a series combination of L, C and R. Use the phasor diagram to obtain an expression for impedance of a circuit and the phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called?

ii) In a series LR circuit, XL = R and the power factor of the circuit is P1. When capacitor with capacitance C, such that XL = XC is put in series, the power factor becomes P2. Calculate P1 / P2.

Voltage of the source is given by,

V = Vo sin ωt

Let current of the source be I = Io sin ωt

Maximum voltage across R is VR = Vo R, represented along OX

Maximum voltage across L = VL  = IO XL, represented along OY and is 90o ahead of Io.

Maximum voltage across C = VC = Io XC, represented along OC and is lagging behind Io by 90o

Hence, reactive voltage is VL - VC, represented by OB'

the vector sum of VR, VL and VC is resultant of OA and OB', represented along OK.

OK = Vo
i.e., Vo

Impedance, Z =

When, X= XC ,  the voltage and current are in the same phase.

In such a situation, the circuit is known as non-inductive circuit.

ii)

Given,

Power factor, P1 = R/Z

Thus,

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29. (i) Plot a graph to show variation of the angle of deviation as a function of angle of incidence for light passing through a prism. Derive an expression for refractive index of the prism in terms of angle of minimum deviation and angle of prism.

(ii) What is dispersion of light? What is its cause?

(iii) A ray of light incident normally on one face of a right isosceles prism is totally reflected, as shown in fig. What must be the minimum value of refractive index of glass? Give relevant calculations.

i)

If the angle of incidence is increased gradually, then the angle of deviation first decreases, attains a minimum value () and then again starts increasing.

When angle of deviation is minimum, the prism is at a minimum deviation position.

ii)

The splitting of light into its component colors is known as dispersion of light.

When a narrow beam of light is incident on a glass prism, the emergent light splits into seven colours called VIBGYOR.

Reason for dispersion: The colors in the spectrum have different wavelengths. The wavelength of violet light is smaller than that of red light. the refractive index of a material in terms of the wavelength of the light is given by the Cauchy's expression.

; where a, b and c are constants.

Thus the refractive index is different for different colours and hence, dispersion occurs.

iii)

The light which is incident normally on one face of a right-angled isosceles triangle prism is totally reflected.

Therefore, minimum value of the refractive index of the glass is 1.41
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