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Physics

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CBSE Class 12

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CBSE Physics 2016 Exam Questions

Long Answer Type

21.

(i) Use Gauss’s law to find the electric field due to a uniformly charged infinite plane sheet. What is the direction of field for positive and negative charge densities?

(ii) Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitances in the ratio 1 : 2 so that the energy stored in the two cases becomes the same.


i) Electric field due to a uniformly charged infinite plane sheet:


Consider an infinite thin plane sheet of positive charge with a uniform charge density straight sigma on both sides of the sheet. Let a point be at a distance a from the sheet at which the electric field is required.

The gaussian cylinder is of area of cross section A.

Electric flux crossing the gaussian surface,

straight ϕ space equals space straight E space straight x spaceArea of the cross section of the gaussian cylinder.
Here, electric lines of force are parallel to the curved surface area of the cylinder, the flux due to the electric field of the plane sheet of charge passes only through two circular sections of the cylinder.


straight ϕ space equals space straight E space straight x space 2 straight A space space bold space bold space bold. bold. bold. bold space bold left parenthesis bold i bold right parenthesis

According to Gauss's Theorem,


straight ϕ space equals space straight q over straight epsilon subscript straight o

Here, charge enclosed by the gaussian surface,

space space space space space space space straight q space equals space straight sigma space straight A
therefore space space space straight ϕ space equals space σA over straight epsilon subscript straight o space space space space space bold space bold. bold. bold. bold space bold left parenthesis bold ii bold right parenthesis

From equations (i) and (ii), we get

straight E space straight x space 2 straight A space equals space σA over straight epsilon subscript straight o
space space space space space space space space space straight E space equals space fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction

The direction of electric field for positive charge is in the outward direction and perpendicular to the plane of infinite sheet.

Direction of electric field for negative charge is in the inward direction and perpendicular to the sheet.

ii) Given: Two capacitors are in the ratio of 1:2.

That is, C2 = 2C1

When the capacitors are connected in parallel,

Total capacitance will be, CP = C1 + C2 = 3 C1

Energy stored in the capacitor,

straight E space equals space 1 half straight C subscript straight P space straight V subscript straight P squared
space space space space equals space fraction numerator 3 straight C subscript 1 space straight V subscript straight P squared over denominator 2 end fraction
When the capacitors are connected in series,

1 over straight C subscript straight S equals 1 over C subscript 1 plus 1 over C subscript 2

C subscript S space equals space fraction numerator 2 C subscript 1 over denominator 3 end fraction
Energy stored in the capacitor,

straight E space equals space 1 half straight C subscript straight S straight V subscript straight S squared

space space space equals space fraction numerator straight C subscript 1 straight V subscript straight S squared over denominator 3 end fraction

Given that, energy stored in both the cases is same.

That is,

fraction numerator 3 space straight C subscript 1 straight V subscript straight P squared over denominator 2 end fraction space equals space fraction numerator C subscript 1 V subscript S squared over denominator 3 end fraction
space space space space space space space space space V subscript P over V subscript s space equals space fraction numerator square root of 2 over denominator 3 end fraction

Hence, the result.
9681 Views

22.

(a) Explain the meaning of the term mutual inductance. Consider two concentric circular coils, one of radius r1 and the other of radius r2 (r1 < r2) placed coaxially with centres coinciding with each other. Obtain the expression for the mutual inductance of the arrangement.

(b) A rectangular coil of area A, having number of turns N is rotated at ‘ f ’ revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is 2straight pif NBA.


a) Mutual induction is the property of a pair of coils due to which an emf induced in one of the coils is due to change in the current in the other coil.

Mathematically,

straight M space equals space fraction numerator negative straight e subscript 2 over denominator begin display style bevelled di subscript 1 over dt end style end fraction
Consider two circular coils S1 and S2 of the same length l, such that both their centres coincide with each other.

Let,

n1 is the number of turns per unit length of S1
n2 is the number of tuns per unit length of S2
I1 is the current passed through the solenoid S1

straight ϕ subscript 21 is the flux linked with S2 due to current flowing through S1

We have,

space space space space space straight ϕ subscript 21 space proportional to space straight I subscript 1
rightwards double arrow space straight ϕ subscript 21 space equals space straight M subscript 21 space straight I subscript 1

where,

M21 is the coefficient of mutual induction of the two coils.

When current is passed through S1 an emf is induced in S2.

Magnetic field produced inside S1 on passing current through it is given by,

straight B subscript 1 space equals space straight mu subscript straight o space straight n subscript 1 space straight I subscript 1
Magnetic flux linked with each turn of S2 will be equal to B1 times the area of cross section of S1.

Magnetic flux linked with each turn of S2 = B1A

Therefore,

Total magnetic flux linked with each turn of the S2 is,

space space space space space straight phi subscript 21 space equals space straight B subscript 1 space xA space straight x space straight n subscript 2 straight l
space space space space space space space space space space space space equals straight mu subscript straight o space straight n subscript 1 space straight l subscript 1 space straight x space straight A space straight x space straight n subscript 2 straight l

space space space space space space straight phi subscript 21 space equals straight mu subscript straight o space straight n subscript 1 space straight n subscript 2 space straight A italic space l italic space straight I subscript 1

therefore space straight M subscript 21 space equals space straight mu subscript straight o space straight n subscript 1 space straight n subscript 2 space straight A italic space l italic space

Similarly, mutual induction between the two coils, when current is passed through coil S2 and induced emf is produced in coil S1 is given by,

space space space space straight M subscript 12 space equals space straight mu subscript straight o space straight n subscript 1 space straight n subscript 2 space straight A space l

italic therefore italic space M subscript italic 12 italic space italic equals italic space M subscript italic 21 italic space italic equals italic space M italic space italic left parenthesis s a y italic right parenthesis

Hence, coefficient of mutual induction between the two coil will be,

straight M space equals space straight mu subscript straight o space straight n subscript 1 space straight n subscript 2 space straight A italic space l
b) Flux is given by,

straight ϕ space equals space NBA space cos space straight theta
We know that,

straight e space equals space fraction numerator negative dϕ over denominator dt end fraction space
space space space equals space open parentheses negative NBA left parenthesis negative sinθ right parenthesis dθ over dt close parentheses
For space maximum space emf comma
sinθ space equals space 1
therefore space
straight e subscript max space equals space left parenthesis negative NBA space sin space straight theta space dθ over dt right parenthesis
space space space space space space space space space space equals space NBA space left parenthesis 2 πf space right parenthesis

Hence proved.
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23.

(i) Draw a labelled diagram of a step-down transformer. State the principle of its working.

(ii) Express the turn ratio in terms of voltages.

(iii) Find the ratio of primary and secondary currents in terms of turn ratio in an ideal transformer.

(iv) How much current is drawn by the primary of a transformer connected to 220 V supply when it delivers power to a 110 V- 550 W refrigerator?


i) Labelled diagram of a step-down transformer:                       soft iron core
 
Principle:
Transformer is based on the principle of electromagnetic mutual induction.

When the current flowing through the primary coil changes, an emf is induced in the secondary coil due to the change in magnetic flux linked with the primary coil.

ii) Turn ratio in terms of voltage is,

n = straight N subscript straight s over straight N subscript straight P space equals space V subscript s over V subscript P
iii) For an ideal transformer, according to the law of conservation of energy,

Input electrical power = Output electrical power.

IP VP = IS VS

i.e., straight V subscript straight S over straight V subscript straight P space equals space I subscript P over I subscript S
therefore space space space space space straight I subscript straight P over straight I subscript straight s space equals space straight V subscript straight s over straight V subscript straight P equals straight N subscript straight s over straight N subscript straight P
rightwards double arrow space space space space straight I subscript straight P over straight I subscript straight s space equals space straight n

iv) Given:

VS = 110 V ; Power,P = 550 W

Power, P = VP IP

rightwards double arrow space straight I subscript straight P space equals space straight P over straight V subscript straight P space equals space 550 over 220 space equals space 2.5 space straight A
3139 Views

Short Answer Type

24.

For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kstraight capital omega is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kV.


Given,

Output voltage, Vo = 2 V

Output resistance, Ro = 2 kstraight capital omega
Base resistance, Ri = 1 kstraight capital omega
Current amplification factor, straight beta space equals space 100
Then input signal voltage is,

space space space straight V subscript straight o over Vf space equals space R subscript o over R subscript f space cross times beta
space space 2 over V subscript f space equals space 2 over 1 x 100
rightwards double arrow space V subscript i space equals space 10 space m V
Now, collector current is,

ICstraight V subscript straight o over straight R subscript straight o space equals space 2 over 2 space equals space 1 space m A
Therefore,

Base current, IB straight I subscript straight C over straight beta space equals space fraction numerator 1 space m A over denominator 100 end fraction equals 10 mu A

2699 Views

25.

(i) Explain with the help of a diagram the formation of depletion region and barrier potential in a p-n junction.

(ii) Draw the circuit diagram of a half wave rectifier and explain its working.


i) 

In a p-n junction, a p-type and an n-type material are joined together. P-type has a higher concentration of holes and n-type has a higher concentration of electrons. Hence, there is a concentration gradient between the p-type and n-type materials. Therefore, holes drift from p-side to n-side by the process of diffusion due to concentration gradient. In a similar manner, electrons move from n-side to p-side.

As a result of the movement of holes from p-side, they leave ionised spaces (negative charge) on p-side near the junction. Similarly, a positive space charge region is formed on the n-side due to the diffusion of electrons from n-side. The two space charge regions on either side of the junction is called as depletion region.

A potential difference is developed across the junction of two regions, which is known as the barrier potential. The reverse polarity of this potential opposes further flow of carriers.

ii) Half wave rectifier:




Working: When an alternating voltage is applied across a diode in series with a load, a pulsating voltage will appear across the load only during that half cycle on the AC input during which the diode is forward biased.

In the positive half cycle of AC input, a current flows through load resistance RL and we get the output voltage.

In the negative half cycle, there is no output. Therefore, the output voltage is restricted to only one direction and is said to be rectified.
3995 Views

26.

(i) Which mode of propagation is used by shortwave broadcast services having frequency range from a few MHz upto 30 MHz ? Explain diagrammatically how long distance communication can be achieved by this mode.

(ii) Why is there an upper limit to frequency of waves used in this mode?


i) Sky-wave propagation is used for long distance communication. Long distance communication is achieved by reflection of radio waves by the ionosphere, back towards the earth. This ionosphere layer acts as a reflector for certain range of frequencies from few MHz to 30 MHz.


ii) There is an upper limit to the frequency of waves used in sky wave propagation because electromagnetic waves of frequencies higher than 30MHz, penetrate the ionosphere and escape whereas waves less than 30 MHz are reflected back to the earth by the ionosphere.
4152 Views

Long Answer Type

27.

(i) If two similar large plates, each of area A having surface charge densities 1s and 2s are separated by a distance d in air, find the expressions for

(a) field at points between the two plates and on outer side of the plates. Specify the direction of the field in each case.
(b) the potential difference between the plates.
(c) the capacitance of the capacitor so formed.

(ii) Two metallic spheres of radii R and 2R are charged so that both of these have same surface charge density s. If they are connected to each other with a conducting wire, in which direction will the charge flow and why ?


i)
  

Given, each plate has an area A and surface charge densities 1s and 2s respectively.

The plates 1 and 2 be separated by a small distance d.

For plate 1:

Surface charge density, straight sigma space equals space straight Q over straight A
For plate 2:

Surface charge density, straight sigma space equals space fraction numerator negative straight Q over denominator straight A end fraction equals negative straight sigma
Electric field in different regions:

Outside region 1,

E = fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction minus fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction equals 0

Outside region 2,

straight E equals space space fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction minus fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction equals 0

Inner region:

In the inner region between plates 1 and 2, electric fields due to the two charged plates add up.

That is,

straight E space equals space fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction plus space fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction
space space space space equals straight sigma over straight epsilon subscript straight o
space space space space equals space fraction numerator straight Q over denominator straight epsilon subscript straight o straight A end fraction

Direction of electric field is from positive to the negative plate.

b) Potential difference between the plates, V = Ed = 1 over straight epsilon subscript straight o fraction numerator Q d over denominator A end fraction
c) Capacitance of the capacitor so formed, C = straight Q over straight V equals fraction numerator epsilon subscript o A over denominator d end fraction
ii) Potential on and inside of charged sphere is given by,

straight V space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction. straight q over straight r

space space space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction. fraction numerator 4 πr squared space straight sigma over denominator straight r end fraction

therefore space straight V space proportional to space straight r

That is, the sphere having larger radius will be at a higher potential.

Therefore, charge will flow from bigger sphere to smaller sphere.
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Short Answer Type

28.

Define the term wave front. State Huygen’s principle.

Consider a plane wave front incident on a thin convex lens. Draw a proper diagram to show how the incident wave front traverses through the lens and after refraction focusses on the focal point of the lens, giving the shape of the emergent wave front.


                                                                     OR


Explain the following, giving reasons :

(i) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency.

ii) When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a reduction in the energy carried by the wave?

(iii) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity in the photon picture of light?


Wavefront is defined as the locus of all the points in space that reach a particular distance by a propagating wave at the same instant.

Huygen's principle:

i) Each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions. These travel with the same velocity as that of the original wavefront.

ii) The shape and position of the wave-front, after time 't', is given by the tangential envelope to the secondary wavelets.

A plane wavefront is incident on a thin convex lens:



                                                    OR



i) Reflection and refraction arise through interaction of incident light with atomic constituents of matter which vibrate with the same frequency as that of the incident light. Hence, frequency remains unchanged.

ii) No, energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation.

iii) For a given frequency, intensity of light in the photon picture is determined by the number of photon incident normally on crossing unit area per unit time.

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Long Answer Type

29.

(i) Derive the mathematical relation between refractive indices n1 and n2 of two radii and radius of curvature R for refraction at a convex spherical surface. Consider the object to be a point since lying on the principle axis in rarer medium of refractive index n1 and a real image formed in the denser medium of refractive index n2. Hence, derive lens maker’s formula.

(ii) Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of light source from the glass surface is 100 cm. At what position is the image formed ?


i)
The figure given above shows the geometry of formation of real image I of an object O and the principal axis of a spherical surface with centre of curvature c and radius of curvature R.

Assumptions:

1. The aperture of the surface is small compared to other distance involved.

2. NM is taken to be nearly equal to to the length of the perpendicular from the point N on the principal axis.

tan space angle N O M space equals space fraction numerator M N over denominator O M end fraction
tan space angle N C M space equals space fraction numerator M N over denominator M C end fraction
tan space angle N I M space equals space fraction numerator M N over denominator M I end fraction
F o r space increment N O C comma
i space i s space t h e space e x t e r i o r space a n g l e.

Assuming the incident ray is very close to the principal axis, all the angles are very small. Hence, for small angles,

tan x = x = sin x

Therefore,

straight i space equals space angle NOM plus angle NCM

straight i space equals space MN over OM plus MN over MC space bold space bold. bold. bold. bold space bold left parenthesis bold i bold right parenthesis

Similarly comma

space straight r space equals space angle NCM space minus space angle NIM

straight r equals space MN over MC space minus space MN over MI space bold space bold. bold. bold. bold space bold left parenthesis bold ii bold right parenthesis

Using space Snell apostrophe straight s space law comma

straight n subscript 1 space sin space straight i space equals space straight n subscript 2 space sin space straight r

For space small space angles comma

straight n subscript 1 straight i space equals space straight n subscript 2 space straight r

Putting the values of i and r from eqns. from (i) and (ii), we have

straight n subscript 1 open parentheses MN over OM plus MN over MC close parentheses space equals space straight n subscript 2 open parentheses MN over MC minus MN over MI close parentheses

straight n subscript 1 over OM plus straight n subscript 2 over MI equals fraction numerator space straight n subscript 2 minus straight n subscript 1 over denominator MC end fraction space bold. bold. bold. bold space bold left parenthesis bold iii bold right parenthesis
Applying new cartesian sign conventions,

OM = -u, MI = + v, MC = +R

Putting these values in equation (iii), we have

straight n subscript 2 over straight v minus n subscript 1 over u space equals space fraction numerator n subscript 2 minus n subscript 1 over denominator R end fraction

ii) According to the question,


space space space straight n subscript 2 over straight v minus n subscript 1 over u equals fraction numerator n subscript 2 minus n subscript 1 over denominator R end fraction

rightwards double arrow space fraction numerator 1.5 over denominator V end fraction minus fraction numerator 1 over denominator left parenthesis negative 100 right parenthesis end fraction equals fraction numerator 1.5 minus 1 over denominator 20 end fraction
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 space space space space space space space space equals space 1 over 40
rightwards double arrow space fraction numerator 1.5 over denominator V end fraction equals 1 over 40 minus 1 over 100
space space space space space space space space space space space space space space equals fraction numerator 5 minus 2 over denominator 200 end fraction
space space space space space space space space space space space space space space equals space 3 over 200
rightwards double arrow space v space equals space fraction numerator 1.5 space x 200 over denominator 3 end fraction
space space space space space space space space space equals 100 space c m
Therefore, the image is formed at a distance of 100 cm in the denser medium.
3713 Views

30.

(a) Draw a labelled ray diagram to obtain the real image formed by an astronomical telescope in normal adjustment position. Define its magnifying power.

(b) You are given three lenses of power 0.5 D, 4 D and 10 D to design a telescope.

(i) Which lenses should he used as objective and eyepiece ? Justify your answer.

ii) Why is the aperture of the objective preferred to be large?


a) The image formed by an astronomical telescope in normal adjustment position is given below,


In normal adjustment, the image is formed at infinity.

Magnification power is defined as the angle subtended at the eye, by the final image, to the angle subtended at the eye, by the object directly, when the final image and the object both are at infinity.

Mathematically, it is given by

Magnification, m = straight beta over straight alpha
Since straight alpha space and space straight beta are very small,
straight beta space equals space tan space straight beta
straight alpha space equals space tan space straight alpha
rightwards double arrow straight m equals space fraction numerator tan space straight beta over denominator tan space straight alpha end fraction

b) Given, three lenses of power 0.5 D, 0.4 D and 10 D.

i) The formula for magnification, M = straight f subscript straight o over straight f subscript straight e equals P subscript e over P subscript o
Therefore, an objective with power = 0.5 D and eyepiece of power 10 D should be used.

This choice would give a higher magnification.

ii) The aperture is preferred to be large so that the telescope can collect maximum amount of light coming from the distant object. 
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