Subject

Physics

Class

CBSE Class 12

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsLong Answer Type

31.

State the two Kirchhoff's laws. Explain briefly how these rules are justified.


Kirchhoff’s First Law or Junction Rule states that “The sum of the currents flowing towards a junction is equal to the sum of currents leaving the junction.”

This is in accordance with the conservation of charge which is the basis of Kirchhoff’s current rule. 


Here, I1I2 I3, and I4 are the currents flowing through the respective wires.

Convention: The current flowing towards the junction is taken as positive and the current flowing away from the junction is taken as negative. 

I3 + (− I1) + (− I2) + (− I4) = 0 

Kirchhoff’s Second Law or Loop Rule states that In a closed loop, the algebraic sum of the emf is equal to the algebraic sum of the products of the resistances and the currents flowing through them. 

 OR

“The algebraic sum of all the potential drops and EMFs along any closed path in a network is zero.”



For the closed loop BACB:

E1 − E2 = I1R1 + I2R2 − I3R3

For the closed loop CADC:

E2 = I3R3 + I4R4 + I5R5

This law is based on the law of conservation of energy.

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32. Distinguish between unpolarized light and linearly polarized light. How does one get the linearly polarised light with the help of a polar?

In a beam of unpolarized light, the vibrations of light vectors are in all directions in a plane perpendicular to the direction of propagation. 

In the linearly polarised light, the vibrations of light take place in a particular direction, perpendicular to the direction of wave motion.

Polarized light can be distinguished, from unpolarized light, when it is allowed to pass through a Polaroid. Polarized light does can show the change in its intensity, on passing through a Polaroid; intensity remains same in case of unpolarized light.

When the unpolarised light wave is incident on a polaroid, then the electric vectors along the direction of its aligned molecules, get absorbed; the electric vector, oscillating along a direction perpendicular to the aligned molecules, pass through. This light is called linearly polarized light.

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33.

(a) Draw a labelled diagram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of number of turn and currents in the two coils.

(b) A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 300 turns. Find the number of turns in the secondary to get the power output at 220 V.


 Labelled diagram of a step-down transformer:


 
Principle:
The 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
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34.

The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r as shown in the figure. Obtain the expression for (i) the current draw from the cell and (ii) the power consumed in the network.


From the above figure, we can use the horizontal symmetry to deduce the current.

As both resistance in the circuit, as well as the internal resistance in the circuit, have the same resistance 'r'

The resistance of the loop I will be
1 over straight R subscript straight I space equals space 1 over straight r space plus fraction numerator 1 over denominator 2 straight r end fraction
1 over straight R subscript straight i space equals space fraction numerator 3 straight r over denominator 2 straight r squared end fraction
straight R subscript straight I space equals space fraction numerator 2 straight r over denominator 3 end fraction

Because the circuit I and II are same we can say
RII  = 2r/3
Combined resistance will be
R = RI + RII
1 over straight R space equals space 1 over straight R subscript 1 space plus 1 over straight R subscript II
1 over straight R space equals space fraction numerator 1 over denominator 2 straight r divided by 3 end fraction space plus fraction numerator 1 over denominator 2 straight r divided by 3 end fraction
1 over straight R space equals space 2 space straight x space open parentheses fraction numerator 1 over denominator 2 straight r divided by 3 end fraction close parentheses
straight R space equals space straight r over 3

Now this circuit is series with internal resistance 'r'

Resultant resistance will be
The current drawn from the cell will be
I=3E/4r

And power consumed by the network
straight P space equals space straight I squared straight R
space equals space fraction numerator 9 straight E squared over denominator 16 space straight r squared end fraction space straight x fraction numerator 4 straight r over denominator 3 end fraction
space straight P space equals space fraction numerator 3 straight E squared over denominator 4 straight r end fraction

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35.

Derive an expression for drift velocity of electrons in a conductor. Hence deduce Ohm's law.


Let an electric field E be applied the conductor. Acceleration of each electron is
straight a space equals negative eE over straight m
Velocity gained by the electron

straight v space equals space minus eE over straight m straight t
Let the conductor contain n electrons per unit volume. The average value of time't', between their successive collisions, is the relaxation time, 't'.

Hence average drift velocity straight v subscript straight d space equals space fraction numerator negative eE over denominator straight m end fraction straight tau

The amount of charge, crossing area A, in time Δt is
=neAvdΔt = IΔt

Substituting the value of vd, we get
straight I space increment straight t equals space neA space open parentheses eEτ over straight m close parentheses increment straight t
therefore space straight I space equals space open parentheses fraction numerator straight e squared straight A space τn over denominator straight m end fraction close parentheses space straight E space equals space σE
open parentheses straight sigma space equals space fraction numerator straight e squared τn over denominator straight m end fraction close parentheses
But I = JA, where J is the current density
rightwards double arrow straight J space equals open parentheses fraction numerator straight e squared τn over denominator straight m end fraction close parentheses straight E
rightwards double arrow space straight J space equals σE

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36.

(a) Define the term 'self-inductance' and write its S.I. unit.

(b) Obtain the expression for the mutual inductance of two long co-axial solenoids S1 and S2 wound one over the other, each of length L and radii r1 and r2 and n1 and n2 number of turns per unit length when a current I is set up in the outer solenoid S2.


a) Mutual inductance of two coils is equal to the e.m.f induced in one coil when the rate of change of current through the other coil is unity.

SI unit of mutual inductance is Henry.

b) Consider two long solenoids S1 and S2 of same length ‘l’ such that S2  surrounds S1 completely.



 

Let,

n1  = Number of turns per unit length of S1

n2 = Number of turns per unit length of S2

I1 = Current passing through solenoid S1

straight empty set subscript 21= Flux linked with S2  due to current flowing in S1

straight ϕ subscript 21 straight space proportional to straight space straight I subscript 1 

rightwards double arrow space space straight ϕ subscript 21 space equals space straight M subscript 21 straight I subscript 1 space semicolon space straight M subscript 21 is the coefficient of the mutual inductance of two solenoids.

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

Magnetic field inside solenoid S1 is given by, straight B subscript 1 space equals space straight mu subscript straight o straight n subscript 1 straight I subscript 1

Magnetic flux linked with each turn of the solenoid  =  straight S subscript 2 space equals space straight B subscript 1 straight A

Total magnetic flux linked with S2 is given by,

straight ϕ subscript 21 space equals space straight B subscript 1 straight A space cross times space straight n subscript 2 straight I space equals space straight mu subscript straight o straight n subscript 1 straight I subscript 1 cross times space straight A cross times space straight n subscript 2 straight I

rightwards double arrow space             straight space straight ϕ subscript 21 space equals space space straight mu subscript straight o straight n subscript 1 space end subscript straight n subscript 2 AI subscript 1

therefore space space space space space space space space space space space space straight M subscript 21 space equals space straight mu subscript straight o straight n subscript 1 space end subscript straight n subscript 2 AI 

Similarly, mutual inductance between two solenoids, when current is passed through S2 and emf induced in solenoid S1 is given by, 

straight M subscript 12 equals straight mu subscript straight o straight n subscript 1 straight n subscript 2 AI

therefore space space space straight M subscript 12 space equals space straight M subscript 21 equals straight M 

Hence, coefficient of mutual induction between the two long solenoids is given by, 

straight M equals straight mu subscript straight o straight n subscript 1 straight n subscript 2 AI

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37.

(i) A ray of light incident on face AB of an equilateral glass prism shows the minimum deviation of 30°. Calculate the speed of light through the prism.


(ii) Find the angle of incidence at face AB so that the emergent ray grazes along the face AC.




At the minimum deviation, the refracted ray inside the prism becomes parallel to its base.
Angle of minimum deviation is given as Dm = 30°
Since, the prism is equilateral, So, A = 60° 
Refractive index of the prism
straight mu space equals space fraction numerator Sin space open parentheses begin display style fraction numerator straight A space plus space straight D subscript straight m over denominator 2 end fraction end style close parentheses over denominator Sin space begin display style straight A over 2 end style end fraction
space equals space fraction numerator Sin space begin display style 90 over 2 end style over denominator Sin space begin display style 60 over 2 end style end fraction
space equals space fraction numerator begin display style fraction numerator 1 over denominator square root of 2 end fraction end style over denominator 1 divided by 2 end fraction space equals space square root of 2

We know that μ = v1/v2
Hence the speed of light in prism would be 1/√2 times the speed of light in air i.e = 3 x108 /1.414 = 2.121 x108 m/s

(ii) 

From Snell's law, we know that fraction numerator sin space straight i over denominator sin space straight r end fraction space equals space straight mu subscript 12

For the emergent ray to graze at the face AC, the angle of refraction should be 90
So, applying snell's law at face AC, we get
fraction numerator sin space straight i subscript AC over denominator sin space straight r subscript AC end fraction space equals space straight mu subscript 21 space rightwards double arrow space fraction numerator sin begin display style space end style begin display style begin display style straight i end style subscript AC end style over denominator sin begin display style space end style begin display style begin display style 90 end style to the power of straight o end style end fraction space equals space fraction numerator 1 over denominator square root of 2 end fraction space or space straight i subscript AC space equals space 45 to the power of straight o

From figure, we can see that angle of refraction at face AB is 15
So applying Snell's law we get
Sin iAB = sin rAB x μ12
or iAB = sin-1 (√sin 15°)

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38. Draw a labelled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil.

Ans (a) Principle − Based on the phenomenon of electromagnetic induction

Construction

Main parts of an ac generator:

  1. Armature − Rectangular coil ABCD

  2. Filed Magnets − Two pole pieces of a strong electromagnet

  3. Slip Rings − The ends of coil ABCD are connected to two hollow metallic rings R1 and R2.

  4. Brushes − B1 and B2 are two flexible metal plates or carbon rods. They are fixed and are kept in tight contact with R1 and R2 respectively.

Theory and Working − As the armature coil is rotated in the magnetic field, angle θ between the field and normal to the coil changes continuously. Therefore, magnetic flux linked with the coil changes. An emf is induced in the coil. According to Fleming’s right-hand rule, current induced in AB is from A to B and it is from C to D in CD. In the external circuit, current flows from B2 to B1.

To calculate the magnitude of emf induced:

Suppose

A → Area of each turn of the coil

N → Number of turns in the coil

straight B with rightwards arrow on top→ Strength of magnetic field

θ → Angle which normal to the coil makes with at any instant t

∴ Magnetic flux linked with the coil in this position:

straight capital phi space equals space straight N space left parenthesis straight B with rightwards arrow on top. straight A with rightwards arrow on top right parenthesisNBA cosθNBA cosωt …(i)

Where, ‘ω’ is angular velocity of the coil

As the coil rotates, angle θchanges. Therefore, magnetic flux Φ linked with the coil changes and hence, an emf is induced in the coil. At this instant t, if e is the emf induced in the coil, then

straight e space equals space minus space dθ over dt space equals negative straight d over dt space left parenthesis NAB space cos space ωt right parenthesis
space equals space minus NAB straight d over dt left parenthesis cos space ωt right parenthesis
equals negative NAB space left parenthesis negative sin space ωt right parenthesis space straight omega

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39. Describe the working principle of a moving coil galvanometer. Why is it necessary to use (i) a radial magnetic field and (ii) a cylindrical soft iron core in a galvanometer? Write the expression for the current sensitivity of the galvanometer.

Can a galvanometer as such be used for measuring the current? Explain.


Moving coil galvanometer is an instrument used for the detection and measurement of small currents.



Principle: The working of moving coil galvanometer is that when a current carrying coil is placed in a varying magnetic field, it experiences torque.

Consider a rectangular coil for which no. of turns = N

Are of cross-section is A = lb

Intensity of the uniform magnetic field = B

Current through the coil = I

Therefore,

Deflecting torque is given by,

BIL x b = BIA

For N number of turns,

straight tau= NBIA

Restoring torque in the spring = kstraight theta

Therefore, 
NBIA space equals space kθ
This space implies
straight I space equals space open parentheses straight k over NBA close parentheses straight theta
That space is space
straight I proportional to straight theta
 i) The soft iron coil in a galvanometer will make the field radial. Also, it increases the strength of the magnetic field.

ii) Current sensitivity in the galvanometer is given by, 
straight theta over straight I space equals space fraction numerator N B A over denominator k end fraction
Voltage sensitivity in the galvanometer is given by, 

straight theta over straight V space equals space fraction numerator theta over denominator I R end fraction space equals space open parentheses fraction numerator n B A over denominator k end fraction close parentheses. begin inline style 1 over R end style
Yes a galvanometer can be used for measuring the current.

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