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Physics

Class

CBSE Class 12

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CBSE Physics 2016 Exam Questions

Short Answer Type

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. 

4429 Views

Long Answer Type

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, 


space space space space space space straight N subscript straight s over straight N subscript straight P space equals space n
rightwards double arrow space space N subscript S over 100 space equals space 100 space
rightwards double arrow space space space space space N subscript S space equals space 10 comma 000 

b) Current in primary is given by, 

straight I subscript straight P space straight V subscript straight P space equals space straight P

rightwards double arrow space straight I subscript straight P space equals space 1100 over 220 space equals space 5 space straight A

c) Voltage across secondary  is given by, 

space space space space space space space fraction numerator straight V subscript straight S over denominator space straight V subscript straight P end fraction space equals space N subscript S over N subscript P space equals space n

rightwards double arrow space space space straight V subscript straight S space equals space 100 space straight x space 220 space equals space 22 comma 000 space straight V

d) Current in secondary is given by, 

space space space space space straight V subscript straight S space straight I subscript straight S space equals space straight P

rightwards double arrow space straight I subscript straight S space equals space straight P over straight V subscript straight S space equals space 1100 over 22000 space equals space 0.05 space straight A

e) In an ideal transformer,

Power in secondary = Power in primary = 1,100 W
2030 Views

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 increment straight S subscript 2 PF
S2P = S subscript 2 P space equals space square root of straight S subscript 2 straight F squared plus PF squared end root space

S subscript 2 P italic space italic equals italic space open square brackets D to the power of italic 2 italic space italic plus italic space open parentheses x italic plus begin display style d over italic 2 end style close parentheses to the power of italic 2 over D to the power of italic 2 close square brackets to the power of begin inline style bevelled italic 1 over italic 2 end style end exponent

italic space italic space italic space italic space italic space italic space italic space italic space italic space italic equals italic space D italic space open square brackets italic 1 italic plus open parentheses x italic plus begin display style d over italic 2 end style close parentheses to the power of italic 2 over D to the power of italic 2 close square brackets to the power of begin inline style bevelled italic 1 over italic 2 end style end exponent

S i m i l a r l y italic comma italic space i n italic space italic increment S subscript italic 1 P E italic comma

S subscript italic 1 P italic space italic equals italic space D italic space open square brackets italic 1 italic plus italic 1 over italic 2 open parentheses x italic plus d over italic 2 close parentheses to the power of italic 2 over D to the power of italic 2 close square brackets italic space italic minus italic space D open square brackets italic 1 italic plus italic 1 over italic 2 open parentheses x italic minus d over italic 2 close parentheses to the power of italic 2 over D to the power of italic 2 close square brackets

O n italic space e x p a n d i n g italic space b i n o m i a l l y italic comma italic space w e italic space g e t

S subscript italic 2 P italic space italic minus italic space S subscript italic 1 P italic space italic equals italic space fraction numerator italic 1 over denominator italic 2 D end fraction open square brackets italic 4 x d over italic 2 close square brackets italic space italic equals italic space fraction numerator x d over denominator D end fraction 

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

Therefore comma space

nλ space equals space xd over straight D

straight x space equals space nλD over straight d semicolon space where space straight n space equals space 0 comma 1 comma 2 comma 3 comma 4 comma....

Similarly space for space destructive space interference comma space

straight x subscript straight n space equals space left parenthesis 2 straight n minus 1 right parenthesis space straight lambda over 2 straight D over straight d

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. 

7698 Views

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. 

Advantages of LED: 

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. 

2941 Views

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. 

straight u space equals space fraction numerator straight u subscript 1 space space plus space straight u subscript 2 space plus space.... plus space straight u subscript straight n over denominator straight n space end fraction space equals space 0 space space space space space space space space space space... left parenthesis 1 right parenthesis



Electric field, E = -eE 

Acceleration of each electron, straight a with rightwards harpoon with barb upwards on top space equals space fraction numerator negative e E over denominator m end fraction         ... (2) 
Here, 

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

Drift velocity is given by, straight v subscript straight d space equals space fraction numerator straight v subscript 1 space plus space straight v subscript 2 space plus space.... plus space straight v subscript straight n over denominator straight n end fraction

straight v subscript straight d space equals space fraction numerator left parenthesis straight u subscript 1 space space plus aτ subscript 1 right parenthesis plus space left parenthesis straight u subscript 2 plus aτ subscript 2 right parenthesis plus....... plus thin space left parenthesis straight u subscript straight n space plus space aτ subscript straight n right parenthesis over denominator straight n end fraction space

straight v subscript straight d space equals space fraction numerator left parenthesis straight u subscript 1 space plus space straight u subscript 2 plus.... plus space straight u subscript straight n right parenthesis over denominator straight n end fraction plus fraction numerator straight a left parenthesis straight tau subscript 1 space plus space straight tau subscript 2 plus.... plus straight tau subscript straight n right parenthesis over denominator straight n end fraction

Since comma space straight v subscript straight d space equals space aτ space space space space space space space space space space space space space space space space space space... space left parenthesis 3 right parenthesis space

Here comma space

straight tau space equals space fraction numerator straight tau subscript 1 space plus space straight tau subscript 2 plus..... plus space straight tau subscript straight n over denominator straight n end fraction comma space is space the space average space time space elapsed
Substituting space for space straight a space from space equation space left parenthesis 2 right parenthesis comma space

straight v subscript straight d space equals space fraction numerator negative eE over denominator straight m end fraction space straight tau space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 4 right parenthesis 

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 open parentheses eE over straight m close parentheses d t 
Current density, J  = straight n space straight e squared over straight m open vertical bar straight E close vertical bar straight tau
From ohm's law, we have

J = straight sigma space straight E
Here, straight sigma is the conductivity of the material through whic the current is flowing,

Thus, 

straight sigma space equals space straight n space straight e squared over straight m straight tau
straight sigma space equals space 1 over straight rho space or space straight rho space equals space 1 over straight sigma

Substituting space the space value space of space conductivity comma space we space have

straight rho space equals space fraction numerator straight m over denominator ne squared straight tau end fraction space semicolon space straight tau space is space the space relaxation space time

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. 

2532 Views

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 Error converting from MathML to accessible text.

Proof:

V = IR = I Error converting from MathML to accessible text. 

i.e., V = Error converting from MathML to accessible text.

For unifrom current and cross- sectional area, we have

Error converting from MathML to accessible text.


ii) 

Given, 

E = 2 V; R = 15 straight capital omega ; RAB = 10 straight capital omega

Potential difference across the wire, = 2 over 25 space x space 10 space equals space 0.8 space V divided by m e t e r
Therefore, potential gradient = 0.8/1 = 0.8 V/ m

Potential difference across AO = fraction numerator 1.5 over denominator 1.5 end fraction space x space 0.3 space equals space 0.3 space V

Therefore, 

Length, AO = fraction numerator 0.3 over denominator 0.8 end fraction cross times 100 
               = 3 over 2 space x space 25 space equals space 37.5 space c m; which is the required balance length of the wire. 

3583 Views

Short Answer Type

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 fraction numerator straight h over denominator 2 straight pi end fraction ; where h is the Planck's constant. 

That is,    mvr space equals space fraction numerator nh over denominator 2 straight pi end fraction
b) 

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

1 over straight lambda subscript 1 space equals space R space open parentheses 1 over n subscript 2 squared space minus space 1 over n subscript 3 squared close parentheses space space space space space space space space space space space space... space left parenthesis 1 right parenthesis thin space

1 over straight lambda subscript 2 space equals space R space open parentheses 1 over n subscript 1 squared space minus space 1 over n subscript 2 squared close parentheses space space space space space space space space space space space space... space left parenthesis 2 right parenthesis thin space

1 over straight lambda subscript 3 equals space R space open parentheses 1 over n subscript 1 squared space minus space 1 over n subscript 3 squared close parentheses space space space space space space space space space space space space space... space left parenthesis 3 right parenthesis thin space

Now comma space adding space left parenthesis 1 right parenthesis thin space and space left parenthesis 2 right parenthesis comma space we space get

1 over straight lambda subscript 1 space plus space 1 over straight lambda subscript 2 space equals space space R space open parentheses 1 over n subscript 1 squared space minus space 1 over n subscript 3 squared close parentheses space equals space 1 over straight lambda subscript 3 space semicolon space

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

4463 Views

Long Answer Type

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 = Vosquare root of OA squared plus OB squared end root
i.e., Vosquare root of straight V subscript straight R squared plus left parenthesis straight V subscript straight L minus straight V subscript straight C right parenthesis squared end root space equals space square root of left parenthesis I subscript O R right parenthesis squared plus left parenthesis I subscript o X space minus space V subscript C right parenthesis squared end root

rightwards double arrow space straight V subscript straight o space equals space straight I subscript straight o space square root of straight R squared space plus space left parenthesis straight X subscript straight L minus straight X subscript straight c right parenthesis squared end root

Impedance, Z = straight V subscript straight o over straight I subscript straight o space equals space R square root of straight R squared space plus space left parenthesis straight X subscript straight L minus straight X subscript straight c right parenthesis squared end root

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

rightwards double arrow space straight P subscript 1 space equals space fraction numerator straight R over denominator square root of straight R squared space plus space straight X squared end root end fraction space equals space fraction numerator straight R over denominator square root of 2 straight R squared end root end fraction space equals space fraction numerator 1 over denominator square root of 2 end fraction

space space space space space straight P subscript 2 space equals space straight R over straight Z

rightwards double arrow space straight P subscript 2 space equals space fraction numerator straight R over denominator square root of straight R squared plus left parenthesis straight X subscript straight L minus straight X subscript straight C right parenthesis squared end root end fraction space equals space 1 space
Thus, 

straight P subscript 1 over straight P subscript 2 space equals space fraction numerator 1 over denominator square root of 2 end fraction





3757 Views

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 (straight delta subscript straight m) and then again starts increasing.

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

When space straight delta space equals space straight delta subscript straight m comma space

straight e space equals space straight i space and space straight r subscript 2 space equals space straight r subscript 1 space equals space straight r space space space space space space space space space space space... space left parenthesis straight i right parenthesis

because space straight r subscript 1 space plus space straight r subscript 2 space equals space straight A

From space left parenthesis straight i right parenthesis comma space we space get

straight r space plus space straight r space equals space straight A

straight r space space equals space straight A over 2
Also comma space we space have

straight A space plus space straight delta space equals space straight i space plus space straight e

Substituting space straight delta space equals space straight delta subscript straight m space and space straight e space equals space straight i comma

straight A space plus space straight delta subscript straight m space equals space straight i space plus space straight i space

straight i space equals space fraction numerator open parentheses straight A space plus space straight delta subscript straight m close parentheses over denominator 2 end fraction

because space straight mu space equals space fraction numerator sin space straight i over denominator sin space straight r end fraction

Therefore comma space

straight mu space equals space fraction numerator sin space open parentheses fraction numerator straight A space plus space straight delta subscript straight m over denominator 2 end fraction close parentheses over denominator sin space open parentheses begin display style straight A over 2 end style close parentheses end fraction
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. 

straight mu space equals space straight a space plus space straight b over straight lambda squared space plus space straight c over straight lambda to the power of 4 ; 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.

straight mu space equals space fraction numerator 1 over denominator sin space straight c end fraction space semicolon space where space straight c space is space the space critical space angle space equals space 45 to the power of straight o

Therefore comma space

straight mu space equals space fraction numerator 1 over denominator sin space 45 to the power of straight o end fraction

straight mu space equals space square root of 2 space equals space 1.41 space

Therefore, minimum value of the refractive index of the glass is 1.41
4023 Views

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