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SIROHI CLASSES PH-9810272244/9810252244/34

: /Series SKS 1 Code No. 55/1/1

Candidates must write the Code on

the title page of the answer-book. R.No.

Please check that this question paper contains 6 printed pages.

Code number given on the right hand side of the question paper should be written on the

title page of the answer-book by the candidate.

Please check that this question paper contains 26 questions.

Please write down the Serial Number of the question before attempting it.

15 minutes time has been allotted to read this question paper. The question paper will be

distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the question paper only

and will not write any answer on the answer-book during this period.

PHYSICS (Theory)

Time allowed : 3 hours Maximum Marks : 70

General Instructions

1. All questions are compulsory. There are 26 questions in all.

2. This question paper has five sections: Section A, Section B, Section C, Section D and

Section E.

3. Section A contains five questions of one mark each, Section B contains five questions of two

marks each, Section C contains twelve questions of three marks each, Section D contains one

value based question of four marks and Section E contains three questions of five marks

each.

4. There is no overall choice. However, an internal choice has been provided in one question of

two marks, one question of three marks and all the three questions of five marks weightage.

You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary.

c = 3 × 108 m/s

h = 6.63 × 10–34

Js

e = 1.6 × 10–19

C

µo = 4 ×10–7

T m A–1

0 = 8.854 × 10–12

C2 N

–1 m

–2

0

1

4 9 10

9 N m

2 C

–2

me = 9.1 10–31

kg

mass of neutron = 1.675 × 10–27

kg

mass of proton = 1.673 × 10–27

kg

Avogadro Numbers = 6.023 ×1023

per gram mole

Boltzmann constant = 1.38 × 10–23

JK–1


Section A ( 5×1= 5 Marks)

1) Name the physical quantity having unit V-m. Is it scalar or vector? (1)

2) The compass needle , pivoted about the horizontal axis and free to move in the magnetic

meridian ,is observed to point along the

a) Vertical direction at a place A

b) Horizontal direction at a place B

give the value of angle of dip at these two places . (1)

3) Why long distance radio broadcasts use short wave band ? (1)

4) A wavefront AB passing through a system C emerges as DE. Name the system C. (1)

5) A point charge Q is placed at point O as shown in the figure. Is the potential difference VA – VB

positive, negative or zero, if Q is (a) positive (b) negative? (1)

Section B (5×2= 10 Marks)

6) (a)The given network has two cells of emf's E1 and E2 and two resistances R1 and R2 . The

ammeter A reads zero. What would the voltmeter V read ?

(b) Name the conservation law related to Kirchhoff’s junction rule .

The adjoining figure shows part of electrical circuit. Compute i . (2)


7) In a Young's double slit experiment, the intensity at a point where the path difference is

6

( , being the wave-length of the light used) is I If I0 denotes the maximum intensity, Find

0

I

I.

(2)

8) A nucleus of mass M. initially at rest splits into two fragments of masses '

3

Mand

2 '( ')

3

MM M . Find the ratio of de-Broglie wavelengths of the two fragments. (2)

OR

Calculate the ratio of de-Broglie wavelengths associated with a deutron moving with velocity 2v and

an alpha particle moving with velocity v.

9) At a point due to a point-charge, the electric field intensity and potential are 32 N C –1

and

16 JC–1

respectively. Calculate : (i) the distance of the charge from the point of observation, and

(ii) magnitude of the charge. (2)

10) In the potentiometer circuit shown in Fig. , the balance (null) point is at X. State with reason,

where the balance point will be shifted when

i) resistance R is increased, keeping all parameters unchanged.

ii) resistance S is increased, keeping R constant. (2)

Section C (12×3 = 36 Marks)

11) State the theorem which relates the enclosed charge, inside a closed surface. with the electric flux

through it. Use this theorem to obtain the electric field due to a uniformly charged thin spherical

shell at an (i) outside point (ii) inside point. (3)

12) Two monochromatic beams A and B of equal intensity I, hit a screen . The number of photons

hitting the screen by beam A is twice that by beam B Then , when inference can you make about

their frequencies. (3)

13) The output of an unregulated dc power supply needs to the regulated. Name the device that can be

used for this purpose and draw the relevant circuit diagram. (3)

14) A cell of emf (E) and internal resistance (r) is connected across a variable external resistance (R).

Plot graphs to show variation of (i) E with R, (ii) Terminal p.d. of the cell (V) with R. (3)


15) Name the regions of E.M. Spectrum in which Lyman and Paschen series lies. If 1 and 2 are

the wavelengths of the first members of the Lyman and Paschen series respectively, then find

1 : 2. (3)

16) What is Brewster’s Law ? The polarising angle of a transparent medium is 60°. Determine (i) the

refractive index of the medium, (ii) the refracting angle. (3)

17) State the underline principle of a cyclotron. Give its limitations and two practical uses. (3)

18) Define the terms (i) disintegration constant and (ii) half-life for a radioactive nucleus. obtain the

relation between the two. (3)

19) A beam of light converges to a point P. A lens is placed in the path of the convergent beam 12 cm

from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20

cm, and (b) a concave lens of focal length 16 cm ? (3)

OR

A converging lens which has a focal length of 20 cm is placed 60 cm to the left of a concave

mirror of focal length 30 cm. An object is placed 40 cm to the left of the lens. Find the position,

nature and magnification of the final image.

20) What is meant by detection of a signal in a communication system? With the help of a block

diagram explain the detection of A.M. signal. (3)

21) A parallel plate capacitor, each with plate area A and separation d, is charged to a potential

difference V. The battery used to charge it is then disconnected. A dielectric slab of thickness d

and dielectric constant K is now placed between the plates. What change, if any will take place in

a) charge on the plate

b) electric field intensity between the plates

c) capacitance of the capacitor?

Justify your answer in each case. (3)

22) (a) What are electromagnetic waves ? A plane electromagnetic wave travels in vacuum along

z-direction. What can you say about the directions of its electric and magnetic field vectors? If the

frequency of the wave is 30 MHz, what is its wavelength?

(b) What is displacement current ? A variable frequency AC source is connected to a capacitor .

How will the displacement current change with decrease in frequency ? (3)

Section D ( 1×4 = 4 Marks)

23) Ritu saw her aunt suffering from severe joint pain. Her aunt could not take any pain killer as she

was allergic to them. Ritu in her quest to help her aunt found the use of magnets. She read Dr.

Philpott’s work on magnetic therapy, that most people are negative magnetic field deficient due to

electromagnetic pollution. Supplementing the body with negative field energy has shown to


restore balance and encourage healing. Ritu takes her aunt to the doctor daily without fail for the

treatment. Her aunt is improving at a phenomenal speed.

i) What values does Ritu have?

ii) A (hypothetical) bar magnet (AB) is cut into two equal parts. One part is now kept over

the other, so that pole C2 is above C1. If M is the magnetic moment of the original

magnet, what would be the magnetic moment of the combination so formed ? (4)

Section E (3×5 = 15 Marks)

24) State the principle of a step-up transformer. Explain, with the help of a labelled diagram, its

working.

Describe briefly any two energy losses, giving the reasons for their occurrence in actual

transformers. How these losses can be minimised ?

How is the transformer used in large scale transmission and distribution of electrical energy over

long distance? (5)

OR

(a) In a series LCR circuit, the voltages across an inductor, a capacitor and a resistor are 30 V, 30

V and 60 V respectively. What is the phase difference between the applied voltage and the current

in the circuit?

(b) In the following circuit, calculate

(i) the capacitance 'C' of the capacitor, if the power factor of the circuit is unity, and

(ii) also calculate the Q-factor of the circuit.

25) Draw a circuit diagram to study the input and output characteristics of a n-p-n transistor in its

common -emitter configuration . Draw the typical input characteristics and explain how these

graphs and used to calculate

a) Input resistance

b) Output resistance and


c) Current amplification factor of the transistor. (5)

OR

(a) If the resistance Ri is increased (see figure.), how will the readings of the ammeter and

voltmeter change?

(b) (i) Two car garages have a common gate which needs to open automatically when a car

enters either of the garages or cars enter in both. Devise a circuit that resembles this situation.

Give truth table also.

(ii)The truth table of logic gate has the form given here. Name this gate and draw its symbol.

26) Draw the labelled ray diagram for the formation of image by an astronomical telescope. Derive

the expression for its magnifying power in normal adjustment. Write two basic features which can

distinguish between a telescope and a compound microscope. (5)

OR

(a) Draw a ray diagram for the formation of image by a compound microscope.

Define its magnifying power. Deduce the expression for the magnifying power of the microscope.

(b) Explain

( i) Why must both the object and the eyepiece of a compound microscope have short focal

lengths?

( i i) While viewing through a compound microscope, why should our eyes be positioned not

on the eyepiece but a short distance away from it for best viewing?


SOLUTIONS

1) Name the physical quantity having unit V-m. Is it scalar or vector? (1)

SOL: Electric Flux . It is scalar.

2) The compass needle , pivoted about the horizontal axis and free to move in the magnetic

meridian ,is observed to point along the

a) Vertical direction at a place A

b) Horizontal direction at a place B

give the value of angle of dip at these two places . (1)

SOL: (a) Angle of dip at A is 900 . (b) Angle of dip at B is 0

0 .

3) Why long distance radio broadcasts use short wave bands ? (1)

SOL: This is because ionosphere reflects waves in these band.

4) A wavefront AB passing through a system C emerges as DE. Name the system C. (1)

SOL: A prism.

5) A point charge Q is placed at point O as shown in the figure. Is the potential difference VA –

VB positive, negative or zero, if Q is (a) positive (b) negative? (1)

SOL: Let the distance of points A and B from charge Q be rA and rB respectively.

Potential difference between points A and B

VA - VB depends on the nature of charge Q.

(a)VA – VB is positive when Q > 0.

(b)VA – VB is negative when Q < 0.

6) a)The given network has two cells of emf's E1 and E2 and two resistances R1 and R2 . The

ammeter A reads zero. What would the voltmeter V read ? (1)


SOL: No current is being drawn from the cell E2 . The voltmeter will read the emf E2 of this cell.

b) Name the conservation law related to Kirchhoff’s junction rule .

The adjoining figure shows part of electrical circuit. Compute i . (1)

SOL: Conservation of charge . 1 A.

7) In a Young's double slit experiment, the intensity at a point where the path difference is

6

( , being the wave-length of the light used) is I If I0 denotes the maximum intensity, Find

0

I

I. (2)

SOL:

8) A nucleus of mass M initially at rest splits into two fragments of masses '

3

Mand

2 '( ')

3

MM M . Find the ratio of de-Broglie wavelengths of the two fragments. (2)

SOL: h

p

From law of conservation of linear momentum


Ratio 1 : 1

OR

Calculate the ratio of de-Broglie wavelengths associated with a deutron moving with velocity 2v

and an alpha particle moving with velocity v.

SOL:

1

2

1 2

1 2

2 2 4

4 4

v v

v v

p m m

p m m

p p

Ratio 1 : 1

9) At a point due to a point-charge, the electric field intensity and potential are 32 N C –1

and

16 JC–1

respectively. Calculate : (i) the distance of the charge from the point of observation,

and (ii) magnitude of the charge. (2)

SOL:

= 0.88×10

–9 C

10) In the potentiometer circuit shown in Fig. , the balance (null) point is at X. State with

reason, where the balance point will be shifted when

i) resistance R is increased, keeping all parameters unchanged.

ii) resistance S is increased, keeping R constant. (2)

SOL: (i) When R is increased, the current from the driver cell is decreased. This decreases the

potential gradient. So, the balance point will shift to the right.

(ii) Increasing resistance S shall have no effect on balance point. This is because there will be no

flow of current in this branch when the potentiometer is balanced.

11) State the theorem which relates the enclosed charge, inside a closed surface. with the electric

flux through it. Use this theorem to obtain the electric field due to a uniformly charged thin

spherical shell at an (i) outside point (ii) inside point. (3)

SOL: Gauss's theorem


Gauss theorem states that the total flux through a closed surface is 1/ 0 times the net charge

enclosed by the closed surface.

Mathematically, it can be expressed as

0

0

0

1

1

1

q

Ed S q

E dS q

0

.

cos

,cos cos0 10

for =0

E d S

EdS

Where q q = ( total charge enclosed within the Gaussian surface)

Electric field due to a uniformly charged thin spherical shell (hollow sphere):

Consider a thin spherical shell of charge of radius R with uniform surface charge density . From

symmetry, we see that the electric field r at any point is radial and has same magnitude at points

equidistant from the centre. To determine electric field at any point P at a distance r from O, we

choose a concentric sphere of radius r as the Gaussian surface.

When point P lies outside the spherical shell (r>R): The total charge q inside the Gaussian surface

is the charge on the shell of radius R and area 4 R2.

q = 4 R2

According to Gauss’s theorem flux through

Gaussian surface

0

1q

0

2

0

2

0

1

1.4

1

4

E dS q

E r q

qE

r

This field is the same as that produced by a

charge q placed at the centre O. Hence for points

outside the shell, the field due to a uniformly

charged shell is as if the entire charge of the shell

is concentrated at its centre

Substituting 24q R

2

2

0

1 4

4

RE

r

. 2

2

0

RE

r


Radially away from the shell

where surface charge density.

24

q

R

When point P lies inside the spherical shell(r<R):

As is clear from Fig. , the charge enclosed by the

Gaussian surface is zero, i.e., q =0.

Flux through the Gaussian surface,

Hence electric field due to a uniformly charged spherical shell is zero at all points inside the shell.

12) Two monochromatic beams A and B of equal intensity I, hit a screen . The number of

photons hitting the screen by beam A is twice that by beam B Then , when inference can you

make about their frequencies. (3)

SOL: IntensityA = IntensityB

The number of photons of beam A = nA

The number of photons of beam B = nB

According to question, nA = 2nB

Let VA be the frequency of beam A and vB be the frequency of beam B.

13) The output of an unregulated dc power supply needs to the regulated. Name the device that

can be used for this purpose and draw the relevant circuit diagram. (3)

SOL: Zener diode .


14) A cell of emf (E) and internal resistance (r) is connected across a variable external resistance

(R). Plot graphs to show variation of (i) E with R, (ii) Terminal p.d. of the cell (V) with R.(3)

SOL:

.

15) Name the regions of E.M. Spectrum in which Lyman and Paschen series lies. If 1 and 2

are the wavelengths of the first members of the Lyman and Paschen series respectively, then

find 1 : 2. (3)

SOL: Lyman series lies in the ultra violet region.

Paschen series lies in the infrared region .

1 : 2 = 7: 108


16) What is Brewster’s Law ? The polarising angle of a transparent medium is 60°. Determine (i)

the refractive index of the medium, (ii) the refracting angle. (3)

SOL: BREWSTER'S LAW

It states that when light is incident at polarising angle at the interface of a refracting medium, the

refractive index of the medium is equal to the tangent of the polarising angle.

If ip is polarising angle and is the refractive index of the refracting medium, then

according to Brewster's law,

tan i p ………………..(1)

When light is incident at polarising angle, the reflected light is found to be completely plane

polarised.

sin i

sin

p

r

………………….(2)

When light is incident at polarising angle i p on a refracting medium of refractive index , let r be

the angle of refraction. Then, according to Snell's law,

From the equations (1) and (2), we have

0 0

0

sin i sin i[ 180 (90 ) (90 )]

sin cos i

cos i sin sin sin(90 i ) 90 i

i 90

p p

p p

p

p p p

p

r i ir

r or r or r

r

Now, angle of reflection i.e. NBC = i p, the angle of incidence. Since r + i p

= 90°, it follows that

CBD = 90° i.e. reflected ray BC is perpendicular to refracted ray BD.

Hence, when a ray of light is incident at polarising angle, the reflected ray is at right angle to the

refracted ray.

(i) From Brewster’s law; the refractive index of the medium is given by

ptan i

Where p is the polarizing angle ( or Brewster angle).

otan60 3 1.732

(ii) Since the reflected light is plane-polarised; the incident angle is polarizing angle. We know that if

the polarizing angle is p and refracting angle r , then

p

p

i 90

90 i 90 60 30

o

o o o o

r

r

17) State the underline principle of a cyclotron. Give its limitations and two practical uses. (3)


SOL: CYCLOTRON:

Principle. A charged particle can be accelerated to very high energies by making it pass through a

moderate electric field a number of times. This can be done with the help of a perpendicular magnetic

field which throws the charged particle into a circular motion, the frequency of which does not depend

on the speed of the particle and the radius of the circular

orbit.

Construction. As shown in Fig. , a cyclotron consists of

two small, hollow, metallic half-cylinders D1 and D2, called

dees. An alternating voltage of few hundred kilovolts is

applied across the gap between the two dees. The dees are

placed between the poles of a strong electromagnet. A

source of charged particles or ions is placed near the centre

of the dees. These ions move in a circular fashion in the

dees, D1 and D2, on account of the uniform perpendicular

magnetic field B. The whole assembly is evacuated to

Minimise collisions between the ions and the air molecules.

Theory. As a particle of charge q and mass m follows a

circular path under the effect of perpendicular magnetic

field B, so

Magnetic force on charge q = Centripetal force on charge q

Clearly, this frequency is independent of both the velocity of the particle and the radius of the

orbit and is called cyclotron frequency or magnetic resonance frequency This is the key fact which

is made use of in the operation of a cyclotron.

Limitations of a Cyclotron

i) A cyclotron cannot be used to accelerate electrons , because its mass is extremely small .

when we increase the velocity beyond a certain limit it becomes more and more out of

step.

According to Einstein's special theory of relativity, the mass of a particle increases with its

velocity.

0

2

21

v

mm

c

At high velocities, the cyclotron frequency will decrease due to increase in mass. This will

throw the particle out of resonance with the oscillating field.

ii) Neutrons, being electrically neutral, cannot be accelerated in a cyclotron.

Uses: It is a device used to accelerate charged particles like protons, deutrons, -particles,

etc…., to very high energies.

18) Define the terms (i) disintegration constant and (ii) half-life for a radioactive nucleus. obtain

the relation between the two. (3)


SOL: RADIOACTIVE DECAY CONSTANT The decay constant of a radioactive sample is a characteristic of the radioactive substance. The

radioactive substances having large value of the decay constant , decay rapidly in comparison to those

having a small value of decay constant.

According to radioactive decay law,

Hence, radioactive decay constant of a substance (radioactive) may be defined as the ratio of

its instantaneous rate of disintegration to the number of atoms present at that time.

Hence, radioactive decay constant of a substance may also be defined as the reciprocal of the time,

after which the number of atoms of a radioactive substance decreases to 0.368 (or 36.8%) of their

number present initially.

HALF-LIFE

The half-life of a radioactive substance is defined as the time during which the nuclei of half of the

atoms of the radioactive substance will disintegrate. It is denoted by T.

Consider that a radioactive sample contains No atoms at time t = 0. Then, the number of atoms

left behind after time t is given by

t

oN N e ……..(1)

From the definition of half-life, it follows that 2

oNN

When t = T (half-life of the sample),

Setting the above condition in equation (1), we have

Thus, half-life of a radioactive substance is inversely proportional to its decay constant and is a

characteristic property of its nucleus. It cannot be altered by any known method.


19) A beam of light converges to a point P. A lens is placed in the path of the convergent beam

12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal

length 20 cm, and (b) a concave lens of focal length 16 cm ? (3)

SOL

In this problem, the object is virtual.

The image is formed on the right of the lens (7.5 cm from the lens) and is real.

(b) u = + 12 cm, f = – 16 cm, v = ?

The image is formed on the right of the lens (48 cm from the lens) and is real.

OR

A converging lens which has a focal length of 20 cm is placed 60 cm to the left of a concave

mirror of focal length 30 cm. An object is placed 40 cm to the left of the lens. Find the

position, nature and magnification of the final image.

SOL:


20) What is meant by detection of a signal in a communication system? With the help of a block

diagram explain the detection of A.M. signal. (3)

SOL: In a communication system detection is the process of recovering the modulating signal from

the modulated carrier wave. Following are the essential steps for detection.

1. Selection of desired signal which may be mixed with unwanted noise using tuned circuit

comparising of an inductor L and a variable capacitor.

2. Rectification of the signal.

3. Envelope detection [using a RC circuit in which capacitor works as a bypass for carriers waves].

21) A parallel plate capacitor, each with plate area A and separation d, is charged to a potential

difference V. The battery used to charge it is then disconnected. A dielectric slab of

thickness d and dielectric constant K is now placed between the plates. What change, if any

will take place in

a) charge on the plate

b) electric field intensity between the plates

c) capacitance of the capacitor?

Justify your answer in each case. (3)

SOL: Before introduction of slab

After introduction of dielectric slab between two plates of parallel plate capacitor.

(a) Since, capacitor is disconnected by battery, therefore, charge of capacitor remain conserved i.e.,

q2 = q1

i.e., No change in charge.

(External electric field reduces due to electric field due to induced charge)

i.e., Electric field will reduce to 1

K times of original value.

(c) m

a

C

C = K

C2 = KC1

Capacitance increase to K times of original value.


22) (a) What are electromagnetic waves ? A plane electromagnetic wave travels in vacuum

along z-direction. What can you say about the directions of its electric and magnetic field

vectors? If the frequency of the wave is 30 MHz, what is its wavelength?

(b) What is displacement current ? A variable frequency AC source is connected to a

capacitor . How will the displacement current change with decrease in frequency ? (3)

SOL: (a) ELECTROMAGNETIC WAVES: The electric field (varying along Y-axis) and the

magnetic field (varying along Z-axis) radiated from an oscillating electric dipole combine together to

form an electromagnetic wave which propagates along X-axis.

Thus, an electric dipole is a basic source of e. m. waves.

As we know that the direction of electromagnetic wave is perpendicular to both electric and magnetic

fields. Here, electromagnetic wave is travelling in Z-direction, then electric and magnetic fields are in

X-Y direction and are perpendicular to each other.

If Frequency of waves f = 30 MHz = 30 × 106 Hz

Speed c =3 × 108 m/s

Using the formula, c = f

Wavelength of electromagnetic waves

Thus, the wavelength of electromagnetic waves is 10 m.

(b) Displacement current is that current which comes into existence, in addition to the conduction

current, whenever the electric field and hence the electric flux changes with time.

A displacement current, 0E

d

di

dt

was added to the current term of Ampere circuital law to make

logically consistent so that the modified form of Ampere's circuital law is

0. ( )dB dl i i

23) Ritu saw her aunt suffering from severe joint pain. Her aunt could not take any pain killer

as she was allergic to them. Ritu in her quest to help her aunt found the use of magnets. She

read Dr. Philpott’s work on magnetic therapy, that most people are negative magnetic field

deficient due to electromagnetic pollution. Supplementing the body with negative field

energy has shown to restore balance and encourage healing. Ritu takes her aunt to the

doctor daily without fail for the treatment. Her aunt is improving at a phenomenal speed.

i) What values does Ritu have?


ii) A (hypothetical) bar magnet (AB) is cut into two equal parts. One part is now kept

over the other, so that pole C2 is above C1. If M is the magnetic moment of the

original magnet, what would be the magnetic moment of the combination so

formed? (4)

SOL: (i) Love, sympathy, punctuality and regularity, diligent, maturity and responsibility.

(ii) Zero

24) State the principle of a step-up transformer. Explain, with the help of a labelled diagram, its

working.

Describe briefly any two energy losses, giving the reasons for their occurrence in actual

transformers. How these losses can be minimised ?

How is the transformer used in large scale transmission and distribution of electrical energy

over long distance? (5)

SOL: Transformers

It is a device for converting low ac voltage ( at high current ) into high voltage (at low current ) and

vice-versa .

Principle: It works on the principle of mutual induction, i.e., when a changing current is passed

through one of the two inductively coupled coils, an induced emf is set up in the other coil.

Construction. A transformer essentially consists of two coils of insulated copper wire having

different number of turns and wound on the same soft iron core. The coil P to which electric energy is.


supplied is called the primary and the coil from which energy is drawn or output is obtained is called

the secondary. To prevent energy losses due to eddy currents, a laminated core is used.

Working. As the alternating current flows through the primary, it generates an alternating magnetic

flux in the core which also passes through the secondary. This changing flux sets up an induced emf

in the secondary, also a self-induced emf in the primary. If there is no leakage of magnetic flux, then

flux linked with each turn of the primary will be equal to that linked with each turn of the secondary.

Theory. Consider the situation when no load is connected to the secondary, i.e., its terminals are

open. Let NP and NS be the number of turns in the primary and secondary respectively

Then Induced emf in the primary coil, P P

de N

dt

Induced emf in the primary coil, S S

de N

dt

where is the magnetic flux linked with each turn of the primary or secondary at any instant. Thus

s s

P P

e N

e N

Let e be the emf applied to the primary. By Lenz's law, self-induced emf Pe opposes e, in the

primary coil.

Resultant emf in the primary = e– Pe

This emf sends current i1, through the primary coil of resistance R .

e – e P =R iP

But R is very small, so the term R iP can be neglected.

Then e = Pe

Thus Pe may be regarded as input emf and

Se as the output emf.

power in the secondary = power in the primary

s s p p

p s s

s p p

V i V i

i V N

i V N

Output emf

Input emf

s s s

P P P

e V Nr

e V N

The ratio s

P

N

N, of the number of turns in the secondary to that in the primary, is called the turns ratio

of the transformer. It is also called transformation ratio.

In a step up transformer, s PN N , i.e., the turns ratio is greater than 1 and therefore

s Pe e .

The output voltage is greater than the input voltage.

In a step down transformer, s PN N ,i.e., the turns ratio is less than 1 and therefore

s Pe e .

The output voltage is less than the input voltage.

Two energy losses in the actual transformers

(I) Hysterisis loss Loss of energy in magnetizing and demagnetizing the core of transformer in every

cycle.

It can be minimized by taking case of soft iron having low coercivity and retentivity and hysterisis

loop.


(II) Eddy current Energy loss in the form of heat due to eddy current. It can be minimized by taking

laminated core consisting of insulated rectangular sheets piled up one over other.

Step-up transformer are used at generating stations so as to transmit the power at high voltage to

minimise the loss in the form of heat whereas series of step-up transformer are used at receiving ends

as shown in schematic diagram below.

Utility of Transformers in Long-distance Power Transmission : If electric power generated at the

power station, say 22,000 W, be transmitted over long distances at the same voltage as required by

consumers, say 220 V, a number of disadvantages will arise :

(i ) The current i flowing through the line wires will be very high (= 22,000/220 = 100 A). Hence a

large amount of energy (i 2Rt ) will be wasted as heat during transmission, R being the resistance of

the line wires.

(i i ) The voltage drop along the line wire (iR) will be considerable. Hence the voltage at the

receiving station will be considerably lower than the voltage at the generating station.

(i i i) The line wires, which are to carry the high current, will have to be made thick. Such wires will

be expensive and require stronger poles to support them.

If, however, the power is transmitted at a high voltage, (say 11.000 V), all the above disadvantages

almost disappear. The current flowing through the line wires will then be only 22,000/11,000 = 2 A.

This will cause much less heating, much less voltage-drop along the line wire and will require much

thinner line wires.

Hence, the electric power generated at the power station is stepped-up to a very high voltage by

means of a step-up transformer and transmitted to distant places. At the place where the supply is

required, it is again stepped-down by a step-down transformer.

Only alternating-current is suitable for this transmission because only alternating-voltage can be

stepped-up or stepped-down by means of transformers.

OR

(a) In a series LCR circuit, the voltages across an inductor, a capacitor and a resistor are 30 V,

30 V and 60 V respectively. What is the phase difference between the applied voltage and the

current in the circuit?

(b) In the following circuit, calculate

(i) the capacitance 'C' of the capacitor, if the power factor of the circuit is unity, and

(ii) also calculate the Q-factor of the circuit.


SOL: (a)

So, there is no phase difference between the applied voltage and the current.

(b)

25) Draw a circuit diagram to study the input and output characteristics of a n-p-n transistor in

its common -emitter configuration . Draw the typical input characteristics and explain how

these graphs and used to calculate

a) Input resistance

b) Output resistance and

c) Current amplification factor of the transistor. (5)

SOL: Common-Emitter Transistor Characteristics

The common-emitter characteristics of a transistor are the graphs representing the variation of current

with voltage when the emitter is kept as a common terminal which is earthed. The base is made as the

input terminal and the collector as the output terminal.

The important transistor characteristics of the common-emitter configuration are the input

characteristics and the output characteristics.

(i) Input Characteristics : A graph between the base-emitter voltage VBE and the base-current iB at a

constant collector-emitter voltage VCE is called the 'input characteristic' of the transistor.

(ii) Output Characteristics : A graph between the collector-emitter voltage VCE and the collector

current is at a constant base-current iB is called the 'output characteristic' of the transistor.

The circuit for obtaining the common-emitter characteristics of an n-p-n transistor is shown in

Fig. The base is connected, through a rheostat, to the positive terminal of a battery VBB whose


negative terminal is connected to the emitter. The base-to-emitter voltage VBE is read by a voltmeter,

and the small base current by a microammeter µ A. The collector is connected, through another

rheostat, to the positive terminal

of another battery Vcc whose negative terminal is connected to the emitter. The collector-to-emitter

voltage VCE is read by another voltmeter, and the collector current iC by a milliammeter mA.

There are four variables : base-to-emitter voltage VBE and base-current iB for the input circuit; and

collector-to-emitter voltage VCE and collector current is for the output circuit. The base-emitter circuit

is a low-current circuit microamperes), and the collector-emitter circuit is a relatively high-current

circuit. To draw the input (or base) characteristics, the collector-to- emitter voltage VCE is adjusted

to 1 volt (say) by the potential-divider in the

collector-emitter circuit. The base-emitter

voltage VBE is

now increased from zero onwards by means of

the potential-divider in the base-emitter

circuit, and the corresponding base current iB is

noted. A graph between VBE and iB is then

plotted

(Fig.). Another graph is plotted for VCE = 10

volt (say).

These characteristics lead to the following

conclusions.

(i) The base current remains almost zero so long the base-emitter voltage VBE is less than the barrier

voltage (= 0.3 V or so). As soon as VBE exceeds the barrier voltage, the current increases first slowly

and then rapidly. Now the curve resembles that of a forward-biased diode. Most of the emitter

electrons reach the collector to form the collector current, the base current being very small.

(ii) The input characteristics are only slightly dependent upon collector-to-emitter voltage VCE.

(iii) The input resistance' Rin is defined as the ratio of change in base-emitter voltage to the change in


base-current, that is

As the input characteristic is non-linear, Rin varies. At any point on the characteristic curve, Rin equals

the slope of the curve at that point and is of the order of kilo-ohms.

The input resistance is higher compared to that in the common-base configuration.

To obtain the output (or collector) characteristics,

the base voltage supply VBB in Fig. is adjusted to

set the base current iB at a fixed value. Then, the

collector-emitter voltage VCE is increased

gradually from zero and the corresponding

collector current ic is noted. A graph between VCE

and ic is then plotted

(Fig.). In this manner, a family of curves is

obtained for various constant base currents.

These characteristics lead to the following

conclusions :

(i) The collector current ic varies rapidly with collector-emitter voltage VCE for VCE between 0 and

about 1 volt only. The value of VCE upto which ic

changes is called the 'knee voltage'.

(ii) Above knee voltage, iC is almost constant, varying very slightly and linearly with VCE . The linear

part of the characteristic is used in audio-frequency amplifier circuits to obtain undistorted output.

(iii) For a given value of VCE , the collector current ic increases with increasing iB .

(iv) There is a small collector current even when the base current iB is zero. This is due to the intrinsic

conduction inherent in semiconductors and is highly temperature-dependent.

The a.c. output resistance Rout of the transistor in common-emitter configuration is defined, in

the linear part of the characteristics, as the ratio of change in collector-emitter voltage (VCE) to the

change in collector-current ( iC) at constant base-current. That is

The output resistance is lower compared to that in the common-base configuration.

Determination of Current-gain : It is evident from these characteristics that there is a current-

amplification in the common-emitter configuration. The input current iB is measured in microamperes,

and the output current ic is measured in milliamperes. For the common-emitter configuration, the

current-amplification is defined as the ratio of change in the collector-current to the change in the


base-current for a constant collector to emitter voltage and is denoted by . Thus

The value of β can be determined from the output characteristic curves Fig. For example, at

VCE 10 volt, we have

For a typical transistor, is about 50.

OR

(a) If the resistance Ri is increased (see figure.), how will the readings of the ammeter and

voltmeter change?

(b) (i) Two car garages have a common gate which needs to open automatically when a

car enters either of the garages or cars enter in both. Devise a circuit that resembles this

situation. Give truth table also.

(ii)The truth table of logic gate has the form given here. Name this gate and draw its

symbol.

SOL: (a) As we know the formula for base current, BB BEB

i

V VI

R

As Ri is increased, IB is decreased.

Now, the current in ammeter is collector current IC.

C BI I as IB decreased IC also decreased and the reading of voltmeter and ammeter also decreased.

(b) (i) As car enters in the gate, any one or both are opened.

The device is shown.


(ii) Logic gate is NAND gate

26) Draw the labelled ray diagram for the formation of image by an astronomical telescope.

Derive the expression for its magnifying power in normal adjustment. Write two basic

features which can distinguish between a telescope and a compound microscope. (5)

SOL: ASTRONOMICAL TELESCOPE

An astronomical telescope is an optical instrument which is used for observing distinct images of

heavenly bodies like stars, planets etc.

It consists of two lenses (or lens systems), the objective lens, which is of large focal length and large

aperture and the eye lens, which has a small focal length and small aperture. The two lenses are

mounted co-axially at the free ends of the two tubes. The distance between these lenses can be

adjusted using a rack and pinion arrangement.

In normal adjustment of telescope, the final image is formed at infinity.

The course of rays in normal adjustment of telescope is shown in Fig. A parallel beam of light from

an astronomical object (at infinity) is made to fall on the objective lens of the telescope. It form a real,


inverted and diminished image A'B' of the object. The eye piece is so adjusted that A'B' lies just at

the focus of the eye piece. Therefore, a final highly magnified image is formed at infinity. The final

image is erect w.r.t., A'B' and is inverted w.r.t. the object.

However, in astronomical telescope, final image being inverted w.r.t. the object does not matter, as

the astronomical objects are usually spherical.

Magnifying Power of an astronomical telescope in normal adjustment is defined as the ratio of 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 lie at infinite distance from the eye.

(b) As the object lies at very huge distance, therefore, angle subtended by the object at C2 (where eye

is held) is almost the same as the angle subtended by the object at C1 (because C1 is close to C2). Let it

be α, i.e. A' C1 B' = α. Rays coming from the final image at infinity make A'C2 B = on the

eye. Therefore, by definition,

Magnifying power, m =

……..(1)

As angles α and are small,

≈ tan and ≈ tan

From (1) tan

tanm

………..(2)

In A’B’C2 ,

2

' 'tan

'

A B

C B

In A’ 'B C1 ,

1

' 'tan

'

A B

C B

Putting in (2), we get 1 1

2 2

' ''

' ' ' '

C B C BA Bm

C B A B C B

o

e

fm

f

where C1B' = fo = focal length of objective lens,

C2B' = – fe = focal length of eye lens.

Negative sign of m indicates that final image is inverted.

Thus, to increase magnifying power of an astronomical telescope in normal adjustment, focal length

of objective lens should be large and focal length of eye lens should be small.

In normal adjustment of telescope, distance between the objective lens and eye lens

C1C2 = L = vo + ue = ( fo + fe )

(c) (i) Telescope has objective of large aperture and large focal length whereas microscope have

objective of small aperture and focal length.


(ii) The relative distance between objective and eye lens may change in telescope whereas the

separation between objective and eye lens in compound microscope remain fixed.

OR

(a) Draw a ray diagram for the formation of image by a compound microscope.

Define its magnifying power. Deduce the expression for the magnifying power of the

microscope.

(b) Explain

( i ) Why must both the object and the eyepiece of a compound microscope have short focal

lengths?

( i i ) While viewing through a compound microscope, why should our eyes be positioned not

on the eyepiece but a short distance away from it for best viewing?

SOL: COMPOUND MICROSCOPE

A compound microscope is an optical instrument used for observing highly magnified images of tiny

objects.

Construction: A compound microscope consists of two converging lenses (or lens systems); an

objective lens O of very small focal length and short aperture and an eye piece E of moderate focal

length and large aperture. The two lenses are held co-axially at the free ends of a tube, at a suitable

fixed distance from each other. The distance of the objective lens from the object can be adjusted by

rack and pinion arrangement.

The course of rays through a compound microscope is shown AB is a tiny object held perpendicular to

the common principal axis, in front of the objective lens beyond its principal focus Fo. A real, inverted

and enlarged image A'B' of this object is formed by the objective lens.

Now A'B' acts as an object for the eye lens, whose position is so adjusted that A'B' lies between

optical centre C2 of eye lens and its principal focus Fe . A virtual and magnified image A" B" is

formed by the eye lens. This image is erect w.r.t. A'B' but inverted w.r.t. AB. The final image A" B"


is seen by the eye held close to eye lens. The adjustments are so made that A" B" is at the least

distance of distinct vision (d) from the eye i.e. C2 B" = d.

Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the

eye by the final image to the angle subtended at the eye by the object, when both the final image

and the object are situated at the least distance of distinct vision from the eye.

In Fig., C2 B" = d. Imagine the object AB to be shifted to Al B" so that it is at a distance d from the

eye. If A" C2 B" = and A1 C2 B" = , then by definition,

Magnifying power, m

…..(1)

For small angles expressed in radians, tan ≈

≈ tan and ≈ tan

From (1) tan

tanm

………..(2)

In A’’B’’C2 ,

2

'' ''tan

''

A B

C B

In A1 ''B C2 , 1

2 2

''tan

'' ''

A B AB

C B C B

Putting in (2), we get 2

2

'''' " " " " " ' '

" ' '

C BA B A B A B A Bm

C B AB AB A B AB

e om m m ……….(3)

Where " "

' 'e

A Bm

A B , magnification produced by eye lens,

And ' '

o

A Bm

AB , magnification produced by objective lens.

Now 1e

e

dm

f

where d is C2 B" = least distance of distinct vision, fe is focal length of eye lens. And

1

1

' ' '' ' 1

1

distance of image fromC

distance of object from C

oo

o

vA B C BA Bm

AB AB C B u

Putting these values in (3), we get

………(4)

As the object AB lies very close to Fo, the focus of objective

1 1o oo

o e o e

v vd dm

u f u f


lens, therefore,

ou = C1B ≈ C1Fo = fo = focal length of objective lens.

As A'B' is formed very close to eye lens whose focal length is also short, therefore,

v0 = C1B' ≈ C1 C2 = L = length of microscope tube.

Putting in (4), we get

….. (5)

(i) fo and fe of compound microscope must be small so as to

have large magnifying power as

(ii) When eyes are positioned at short distance away from eyepiece, then the image formed at infinity

can be seen which is more suitable and comfortable for viewing by the relaxed eye.

1 1o

o e o e

L d L dm

f f f f

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