physics (cbse 2008)jayroy.in/cbse/2008 qp with solns.pdf · mass, and so the line b represents the...

1. State two characteristic properties of nuclear force. [1]

Solution (a) Nuclear force is a short range force. (b) The nuclear force is charge independent. 2. How does the angle of minimum deviation of a glass

prism vary, if the incident violet light is replaced with red light? [1]

Solution The angle minimum deviation increase with increase in

frequency Therefore:

δ δviolet red>

That is the angle of minimum deviation decreases. 3. The instantaneous current and voltage of an a.c. circuit

are given by i = 10 sin 300 t and V = 200 sin 300 t.

What is the power dissipation in the circuit? [1]

Solution The power dissipated in the circuit

P I V= rms rms

P = × =102

2002

1000 W

4. Why should the spring/suspension wire in a moving coil galvanometer have low torsional constant? [1]

Solution The moving coil galvanometer springs have low torsional

constants so that they respond to weak electric current, hence the instrument becomes sensitive.

5. Why does the bluish color predominate in a clear sky? [1]

Solution A blue color has a shorter wavelength than red. Therefore,

blue color is scattered much more strongly. Hence, the sky looks blue.

6. Which orientation of an electric dipole in a uniform electric field would correspond to stable equilibrium? [1]

Solution When the electric dipole is oriented in the direction of

applied electric field, then the configuration is said to be in stable equilibrium.

7. Two lines, A and B, in the plot given below show the

variation of de Broglie wavelength, l versus 1V

, where

V is the accelerating potential difference, for two particles carrying the same charge.

Which one of two represents a particle of smaller mass? [1]

B

λ A

1/√V

Solution The de Broglie wavelength

λ = h

2meV

or λ =

he m V2

1 1

Therefore, the wavelength decreases with increase in mass, and so the line B represents the particle with smaller mass.

8. State the reason why GaAs is most commonly used in making of a solar cell. [1]

Physics (CBSE 2008)Time: 3 hours Max. Marks: 70

General Instructions

1. All questions are compulsory.2. There are 30 questions in total. Questions 1 to 8 carry one mark each, questions 9 to 18 carry two marks each, questions 19 to 27

carry three marks each and questions 28 to 30 carry five marks each.3. 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 three questions of five marks each. You have to attempt only one of the given choices in such questions.4. Use of calculators is not permitted.

Physics Special market Book 2_2008 CBSE.indd 1 2011-12-22 1:15:23 PM

Physics (CBSE 2008)2

Solution The band gap of the GaAs is best suits for the absorption

of the visible light, and also they have a high absorption coefficient along with its good electrical conductivity makes it ideal for the solar cells.

9. Draw a labeled ray diagram of an astronomical telescope in the near point position. Write the expression for its magnifying power. [2]

Solution

Magnifying power

A

B

fe

fo

EO

Objective Eyepiece

a a

bh

Magnifying power

mff

fDe

= +

o e1

Where D is the distance between the eyepiece and the final image.

10. The following figure shows the variation of intensity of magnetization versus the applied magnetic field intensity, Hi for two magnetic materials A and B:

I

B

H

A

(a) Identify the materials A and B. (b) Why does the material B have a larger susceptibility

than A for a given field at constant temperature? [2]

Solution (a) The material A is diamagnetic whereas B is

paramagnetic in nature. (b) The susceptibility of the paramagnetic material

is more because the magnetic moment of paramagnetic atom is due to the electron spin, but in diamagnetic materials the magnetic properties arise due to the orbital motion of the electron. Since the spin magnetic moment is higher than the orbital magnetic moment paramagnetic materials show higher susceptibility than the diamagnetic materials.

11. Two metallic wires of the same material have the same length but cross-sectional area is in the ratio 1: 2. They are connected (i) in series and (ii) in parallel. Compare the drift velocities of electrons in the two wires in both the cases (i) and (ii). [2]

Solution While the wires are connected in series current passing

through them is same.

I neAvd=

Therefore ne A v ne A vd d1 1 2 2=

vv

AA

d

d

1

2

2

1

=

or v vd d1 2 2 1: :=

When the wires are connected in parallel the voltage across the wires are equal.

VvL

d=μ

Where m is the mobility of the electron and L is the length of the wire.

Since the applied voltage across both conductor are equal

vL

vL

d d1 2

μ μ=

or v vd d1 2 1 1: :=

12. Draw a block diagram of a simple amplitude modula-tion. Explain briefly how amplitude modulation is achieved. [2]

Solution

m(t) x(t) y(t)

Bx(t)+Cx(t)2c(t)

Ac sinwct(carrier)

Am sinwmt(Modulating

Signal)

Squarelaw device

Bandpassfilter centred

at wc

AM Wave+

The modulating signal Amsinwmt added to the carrier wave Acsinwct produce the signal x t A t A tm m c c( ) sin sin= +ω ω . This signal is passed through a square law device which produce an output in the form of y t Bx t Cx t( ) ( )= + 2

The signal is then pass through a band pass filter which reject d.c. So, the output will be AM modulated wave.

13. Calculate the energy released in MeV in the following nuclear reaction:

92238

90234

24U Th He Q→ + +

[Mass of 92238 238 05079U u= .

Mass of 90234 234 043630Th u= .

Mass of 24 4 002600He u= .

1u = 931.5 MeV/c2] [2]


Physics (CBSE 2008) 3

Solution Energy released

E mc= ∆ 2

∆ = - +m M M MU Th He( )

= - + =238 05079 234 04363 4 002600 0 00456. ( . . ) .u u u u

Therefore Q mc= ∆ = ×2 0 00456 931 5. . MeV

Q = 4 24764. MeV

14. Using Ampere’s circuital law, obtain an expression for the magnetic field along the axes of a current carrying solenoid of length l and having N number of turns. [2]

Solution

Magnetic field due to solenoid: Consider a rectangular Amperian loop PQRS near the middle of solenoid shown

in figure where PQ = l

a

P

d c

w

b

hQ

B

Let the magnetic field along the path ab be B and zero along cd. As the paths bc and da are perpendicular to the axis of solenoid, the magnetic field component along these paths is zero. Therefore, the path bc and da will not contribute to the line integral of magnetic field B

The total number of turns in the length l =Nl The line integral of magnetic field induction on B over the

closed path PQRS is

B d B d B d B d B d

a

b

d

a

c

d

b

c

i i i i il l l l l= + + +∫∫ ∫∫∫B nI= μ0

15. Derive an expression for the resistivity of a good conduc-tor, in terms of the relaxation time of electrons. [2]

Solution

The resistivity is defined as the resistance of uniform wire of unit area and unit length.

Generally the resistivity and the resistance are related by the relation

Resistivity ρ = RAl

Where R is the resistance of the material, A area of cross section and l is the length of the resistance wire.

The S.I unit of the resistivity is Ω-meter. Consider a conductor of length land area of cross

section A. Let a potential V be applied across it. An electric filed is developed across the conductor that derives the electrons in one direction and constitutes current I in it.

E

Vl

= (1)

But current

I ne A d= υ (2)

Where n is number density of free electrons Vd is the drift velocity Force acting on the electron is

F eE=

ma eE=

mVeEd

ττ= where is the relaxation time.

v

eEmd = τ

(3)

Therefore, (1) become

I neAeEm

neAeVml

=

=τ τ

From ohm‘s law

VI

Rml

ne A= = 2 τ

On comparing the equations

RA

mne

= =ρτ

l l2

or ρτ

= mne2

16. The circuit arrangement given below shows that when an a.c. passes through the coil A, the current starts flowing in the coil B.

Coil A Coil B

(a) State the underlying principle involved. (b) Mention two factors on which the current produced

in the coil B depends. [2]



Solution The current produced in the coil B due to the electro-

magnetic induction. The current produced in the coil depends on 1. Nature of the coil 2. Number of the turns of the coils 3. Distance between the coils.

17. State one feature by which the phenomenon of inter-ference can be distinguished from that of diffraction.

A parallel beam of light of wavelength 600 nm is incident normally on a slit of width ‘a’. If the distance between the slits and the screen is 0.8 m and the distance of 2nd order maximum from the centre of the screen is 15 mm, calculate the width of the slit. [2]

Solution

Interference is produced due to the superposition of two wavefronts whereas the diffraction is due to the superposition of secondary wavelets coming from the same wavefront.

dn D

x= λ

x n D= = = = × -15 2 0 8 600 10 9mm m m, . λ

d = × × ××

= ×-

--2 600 10 0 8

15 106 4 10

9

34.

. m

18. Two point charges q1 = 10 × 10−8 C and q2 = −2 × 10−8 C are separated by a distance of 60 cm in air.

(a) Find at what distance from the charge, q1, would the electric potential be zero.

(b) Also calculate the electrostatic potential energy of the system. [2]

OR Two point charges 4Q and Q are separated by 1 m in air. At

what point on the line joining the charges is the electric field intensity zero? Also calculate the electrostatic potential energy of the system of charges, taking the value of charge, Q = 2 × 10−7 C. [2]

Solution

(a)

L x Pq1 q2

Let the potential at P is zero

kqL x

kqx

1 2 0+

- =

10 2L x x+

=

or 8 24

x L xL= =or

We have L = 60 cm Therefore

x = =604

15 cm.

The point P is at a distance L + x distance from q1

Distance from q1 = L + x = 60 cm + 15 cm = 75 cm

Electrostatic energy

Ekq q

L= 1 2

= × × × × - ××

- -

-

9 10 10 10 2 1060 10

9 8 8

2

U = - × -3 10 5 J

OR

(b)

xP

L = 1mQ4Q

The electric field at x distance from the 4Q charge is zero.

E E1 2 0+ =

41

02 2

kQx

kQx

--

=( )

4 11

4 8 42 22 2

x xx x x=

-- + =

( )or

x = ± - × ×

×86

8 4 3 42 3

2

x x= =2

23

or

But beyond 1 m the field due to both charges are in the same direction. Therefore the Solution is x = 2/3. Electrostatic energy of the system

U

kQL

= 4 2

2

= × × × × -4 9 10 2 109 7 2( )

U = × -1 44 10 3. J

19. Identify the following electromagnetic radiations as per the wavelengths given below. Write one application of each.

(a) 10−3 nm (b) 10−3 m (c) 1 nm [3]

Solution (a) The 10−3 nm wavelength is X-rays. Application X-rays are used in surgery for detection of fractures. (b) 10−3 is Microwave radiation Application Microwaves are used in radar navigation. (c) 1nm radiation is in infrared region. Infrared photography is used in bad weather and

night vision.

20. Explain why high frequency carrier waves are needed for effective transmission of signal. A message signal of 12 kHz and peak voltage 20 V is used to modulate a carrier wave of frequency 12 MHz and peak voltage 30 V. Calculate the (i) modulation index (ii) side-band frequencies. [3]



Solution The three main reason why the high frequency carrier

waves is used to send signal is • Size of Antenna: For efficient transmission and

reception, the transmitting and receiving antenna must have a length equal to the one fourth of the wavelength. For a 20 kHz audio signal the length of antenna come upto 15 km, therefore the modulation is used.

• Effective power radiated by antenna: The power

radiated by the antenna of length l Pl∝

λ

2

. There-

fore the power radiated by shorter wavelength is high. • Mixing up of signal: Signal mixing up can be avoided

using carrier wave windows called bandwidths. The modulation index

μ= MA

Where M and A are the peak voltages of the signal and carrier respectively.

μ = =2030

0 67.

The Upper side band Frequency

U S B. . = + = + =ν νc m 12000 12 12012 kHz.

The lower side band

L S B. . = - = - =ν νc m 12000 12 11988 kHz.

21. Distinguish between unpolarised and plane polarised light. An unpolarised light is incident on the boundary between two transparent media. State the condition when the reflected wave is totally plane polarised. Find out the expression for the angle of incidence in this case. [3]

Solution If the plane of vibration of the electric field of a light beam is

randomly oriented , then the light is said to be unpolarised. In plane polarized light the plane of vibration of the

electric field is restricted to one plane. The reflected light is fully polarized when the light is

incident at Brewster angle. At Brewster angle the refractive index and the angle of

incident are related.

n i= tan B

22. Draw the labeled circuit diagram of a common-emitter transistor amplifier. Explain clearly how the input and output signals are in opposite phase. [3]

OR State briefly the underlying principle of a transistor

oscillator. Draw a circuit diagram showing how the feedback is accomplished by inductive coupling. Explain the oscillator action. [3]

Solution Common-emitter amplifier, the output voltage is 180° out

of phase with the Input voltage.

RB

IC

IE

IB

VBBvi

vi

RC

VCC

B C

E

ui = input signal, RB = Input resistance, VBB = Base biase, RC = load resistance. VCC = collector biase.

The change in IC due to a change in IB causes a change in VCE and the voltage drop across the resistor RL because VCC is fixed.

The change can be given as

V V I RC CC C L= - (1)

When an ac signal is fed to the input circuit, the forward bias increases during the positive half cycle of the input. This results in an increase in IC and a consequent decrease in VC. Thus, during positive half cycle of the input, the collector becomes less positive. During the negative half cycle of the input, the forward bias is decreased, resulting in a decrease in IE and hence IC. Therefore, from equation (i) VC would increase, making the collector more positive. Hence, in a common-emitter amplifier, the output voltage is 180° out of phase with the Input voltage.

OR

S1 (Switch)

Mutual inductance(Coupling throughmagnetic field)

T 2 OutputL4

3

2

n-p-n

1T1

T2C

An oscillator gives output without giving any inputs. A common emitter amplifier which working as oscillator is shown in the figure above. in which the feedback is accomplished by inductive coupling from one coil winding (T1) to another coil winding (T2). Note that the coils T2 and T1 are wound on the same core and hence are inductively coupled through their mutual inductance. As in an amplifier, the base-emitter junction is forward biased while the base-collector junction is reverse biased.

The oscillator frequency is same as that of the LC circuit

fLC

= 12π

23. The ground state energy of hydrogen atom is −13.6 eV. (a) What is the kinetic energy of an electron in the 2nd

excited state?


6 Physics (CBSE 2008)

(b) If the electron jumps to the ground state from the 2nd excited state, calculate the wavelength of the spectral line emitted. [3]

Solution The energy of the electron at nth orbit is

Enn = -13 6

2

. eV

Therefore, energy at 2nd excited state or n = 3

E3 2

13 63

1 51= - = -..

eVeV

The energy of photon emitted.

h E Eν = -1 2

= -

=13 6

11

19

12 08. .eV eV

Therefore, the wavelength

λ = = - =hchν

124012 08

102 6eV nm

eV.. nm

24. The following graph shows the variation of stopping potential V0 with the frequency hn of the incident radiation for two photosensitive metals X and Y:

0.5

X Y

Vo

1.0u (´1015s–1)

(a) Which of the metals has larger threshold wavelength? Give reason.

(b) Explain giving reason, which metal gives out electrons, having larger kinetic energy, for the same wavelength of the incident radiation.

(c) If the distance between the light source and metal X is halved, how will the kinetic energy of electrons emitted from it change? Give reason. [3]

Solution (a) From the graph the threshold frequency of the metal

X is less than the metal Y. Therefore the metal X has larger threshold wavelength.

(b) The work function of metal X is less than the work function of metal Y. Therefore metal Y gives more energetic electron.

(c) The kinetic energy of the electron is independent of the intensity of light, therefore the kinetic energy do not change.

25. A circular coil of 200 turns and radius 10 cm is placed in a uniform magnetic field of 0.5 T, normal to the plane of the coil. If the current in the coil is 3.0 A, calculate the

(a) Total torque on the coil (b) Total force on the coil (c) Average force on each electron in the coil due to the

magnetic field

Assume the area of cross-section of the wire to be 10−5 m2 and the free electron density is 1029/m3. [3]

Solution (a) Total torque τ = nIAB sinθ But θ = 90° therefore the net torque is

t = 200 ×3 × 2 × 3.14 × (0.1)2 × 0.5 =18.84 Nm.

(b) Force on the coil is given by

F = nIBl = 3 × 0.5 × 2 × 3.14 × 0.1 × 200 = 188.4 N

(c) The total number of electrons in the entire length of the wire can be calculated as

n = length of wire × area of wire × 1029/m3

= 200 × 2 ×3.14 × 0.1 ×10-5 × 1029

= 1.256 ×1026

Therefore average force is 188.41.256 10

=1.51 10 N.×

×2624-

26. The inputs A and B are inverted by using two NOT gates and their outputs are fed to the NOR gate as shown below.

(1)

(2)

A

Y

B

Analyze the action of the gates (1) and (2) and identify the logic gate of the complete circuit so obtained. Give its symbol and the truth table. [3]

Solution The truth table of gate 1

A Output0 1

1 0

Therefore both (1) and (2) act as NOT gates.

Y A B A B A B= +( ) = =. .

The logic gate of complete circuit is AND gate its symbol and truth table are given below.

B

A

O

A B O0 0 01 0 00 1 01 1 1



27. (a) Calculate the equivalent resistance of the given electrical network between points A and B.

(b) Also calculate the current through CD and ACB, if a 10 V d.c source is connected between A and B, and the value of R is assumed as 2 Ω.

C

RR

R

R

RD E

BA

Solution [3]

The given circuit is a balanced Whetstones bridge. Therefore we can omit the resistance between C and D. Equivalent resistance

R R R Req = =2 2||

The current passing through CD is zero. Since the circuit is symmetric half the current passing

through the circuit pass through the arm ACB and the rest is through ADB.

The net current through the circuit is

IVR

= = =102

5A

Therefore 2.5 A pass through the arm ACB.

28. Derive an expression for the energy stored in a parallel plate capacitor. On charging a parallel plate capacitor to a potential V, the spacing between the plates is halved, and a dielectric medium of ∈r = 10 is introduced between the plates, without disconnecting the d.c source. Explain, using suitable expressions, how the (a) capacitance, (b) electric field and (c) energy density of the capacitor charge. [5]

OR (a) Define electric flux. Write its SI unit. (b) The electric field components due to a charge inside

the cube of side 0.1 m are as shown:

X

a = 0.1m0.1 m

O

Y

Z

Ex = ax, where α = 500 N/C m

Ey = 0, Ez = 0.

Calculate (a) the flux through the cube and (b) the charge inside the cube. [5]

Solution

+

+

+

+

+

+

+

+

+

−

−

−

−

−

−

−

−

−

Q

1 2

E

−Q

Let q is the charge on the plate 1 and –q is the charge on the plate 2. And the potential difference between 1 and 2 is q/C. Let

d dWqC

= q

is the work done to add an extra charge dq on the plates. Therefore the total work done

WqC

qQC

Q= =∫0

212

d

But Q = CV Therefore,

W CV= 12

2

The capacitance of the parallel plate capacitor is given by

CK A

d= ε0

Initially the capacitance was CA

d= ε0 , and after the

changes

CA

d′ = =10

2

200εC

Therefore the capacitance increases by 20 times. Electric field

EVd

=

So reducing the plate separation increases the electric field by 2 times.

The energy density

U E= 12 0

2ε

So the new energy density

U E′ = × × =12

10 2 4002ε ( ) U

The energy density increases by 40 times.

OR The number of electric field lines passing through a

surface normally is called electric flux. Unit of electric flux is Nm2/C.



Since the electric field is directed along the x-axis only left and right faces of the cube contribute to the flux.

φL E da aAx= = = × × = -

. . . .500 0 01 0 1 0 5 2 1Nm C

Similarly

φR E da aAx= = = × × = -

. . . .500 0 01 0 1 0 5 2 1Nm C

Therefore the net flux is

φ = -1 2 1Nm C

From Gauss’s law

E AQ⋅ =ε0

Q C= ε0

29. Derive the lens formula, 1 1 1f v u

= - for a concave lens,

using the necessary ray diagram. Two lenses of powers 10 D and − 5 D are placed in contact. (a) Calculate the power of the new lens. (b) Where an object should be held from the lens, so as

to obtain a virtual image of magnification ? [5]

OR (a) What are coherent sources of light? Two slits in

Young’s double slit experiment are illuminated by two different sodium lamps emitting light of the same wavelength. Why is no interference pattern observed?

(b) Obtain the condition for getting dark and bright fringes in Young’s experiment. Hence write the expression for the fringe width.

(c) If S is the size of the source and its distance from the plane of the two slits, what should be the criterion for the interference fringes to be seen? [5]

Solution

O lR1

R2

n2n1

Consider a paraxial image at the point O on the axis of thin lens.

The lens placed in a medium of refractive index n1 and the refractive index of the lens is n2. Let R1 and R2 are the radii of curvature of surface one and two respectively. In order to get the Image consider we consider successive refraction at these two surfaces. The image formed at the first refracting surface is given by the equation

nv

nu

n nR

2

1

1 2 1

1

- = - (1)

U1 is negative for the objective point shown in figure. This images acts as the object of the second surfaces.

nv

nv

n nR

1 2

1

1 2

2

- = - (2)

Adding (1) and (2)

1 11

1 1

1 2v un

R R- = - -

( )

Where nnn

= 2

1

The focal length f of the lens is defined as

11

1 1

1 2fn

R R= - -

( )

Then

1 1 1v u f

- =

Power of lens in contact = sum of the power of lenses.

Ptotal = 10D – 5D = 5D

The effective focal length of the combination is

fP

= =1 15

m

The magnification

mf

f u= -

+

= -+

=

15

15

2u

or 1 51

23

10+ = - = -u uor m

OR (a) Two sources are said to be coherent if they emit

wavelength in same frequency or wavelength of light with a constant phase difference. The phase difference between two sodium vapor lamps is not constant so the light emitted by the source is not constant and therefore the interference pattern does not appear.

(b) Let two slits are separated by a distance d. Screeen is placed from at a distance D.

The slits S1 and S2 then behave like two coherent sources because light waves coming out from S1 and S2 are derived from the same original source and any abrupt phase change in S will manifest in exactly similar phase changes in the light coming out from S1 and S2. Thus, the two sources S1 and S2 will be locked in phase;

The constructive interference happen when the wavelets emitting from both sources are in phase

i.e S2P – S1P = nl; n = 0, 1, 2 . . . Then

S P D xd

Dx

d

D22

212

2

21

12

2= + +

≈ +

+



Similarly

S P Dx

d

D1

2

112

2= +-

S1

d

G

P

OZ

G

S2 D

x x

zy

Therefore, path difference

S P S Px

dx

d

DxdD2 1

2 2

2 22

- =+

- -

=

Therefore, the point P will be maximum intensity if

xdD

n= λ

or xn D

d= λ

The condition for destructive interference is

path difference = +

n12

λ

Therefore,

xdD

n= +

12

λ

or xn D

d= +( )2 1

2λ

The fringe width is the separation between the con-secutive bright or dark fringes

Taking bright fringes

β λ λ λ= - = + - =+x xn D

dn D

dD

dn n1

1( )

30. An a.c source generating a voltage V = Vmsinωt is connected to a capacitor of capacitance C. Find the expression for the current i, flowing through it, plot a

graph of v and i versus ωt to show that the current is π2

, ahead of the voltage.

A resistor of 200 Ω and a capacitor of 15 μF are connected in series to a 220 V, 50 Hz a.c source. Calculate the current in the circuit and the rms voltage across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox. [5]

OR

Explain briefly, with the help of a labeled diagram, the basic principle of the working of an a.c generator.

In an a.c generator, coil of N turns and area A is rotated at V revolutions per second in a uniform magnetic field B.

Write the expression for the emf produced. A 100 turn coil of area 0.1 m2 rotates at half a revolution per second. It is placed in a magnetic field 0.01 T perpendicular to the axis of rotation of the coil. Calculate the maximum voltage generated in the coil. [5]

Solution A The charge in the capacitor at time t is

q CV CV tm= = sinω

Therefore the current in the circuit idqdt

CV tm= = ω ωcos

iVm

C

t tm= = +

1 2

ω

ω ω πcos sinI

1

0

−1

−2

1 2 3 4 5 6 7 8 9

The average current in the circuit

IV

ZV

R Xrms

rms

c

= =+ ( )2 2

Z RC

= +

= +

× × × ×

-

22 1 2

26

21200

12 3 14 50 15 10ω

/

.

=

1 2

291 5

/

. Ω

Therefore

Irms = =220291 5

0 75.

. A

V I RR rms= = × =0 75 200 150. V

V I Xc rms c= = × =0 75 212 159. V

V V Vc R m+ = + >150 159

This because of the phase difference between the voltages is not in phase. The voltage in the circuit should be

V V VR c= + = + =2 2 2 2150 159 218 6. V

OR



A.C. generator

Sliprings

Carbonbrushes

Alternating emf

Coil Axle

SN

The device used to generate ac voltage (or current ) using the principle of electromagnetic induction is called a.c. generator. An a.c. generator consists of a metallic coil called armature rotating between the poles of the field magnet as shown in the figure. The axis of rotation of the coil is perpendicular to the direction of the magnetic field. The ends of the armature coil are welded to two metallic rings called slip rings which rotate along with the armature. Two carbon brushes rub against the slip rings and the wires which carries the power to the external circuit is connected to the carbon brushes. As the armature rotates in the magnetic field, The flux linked to the coil changes and hence induced currents generated in the coil.

When the coil rotated with constant angular velocity w, then the angle q between the magneti field vector and the area vecotor is

θ ω= t (Assuming q = 0 at t = 0)

There for the flux at time t is

φ θ ω( ) cost BA BA t= =cos

There for the emf induced in the coil

ε φ ω ω= - =Nddt

NBA tsin

The maximum emf induced

ε ω0 = NBA

Therefore

ε ε ω= 0 sin t

We have

ε ω0 = NBA

And it is given

ω π π π= = =2 22T

rad/ sec

N B A= = = =100 0 01 0 1 2; . ; . / secT m and radω π

Therefore

ε π0 100 0 01 0 1= × × ×. . volts

ε0 0 314= . volts

The average voltage

Vrms = = =ε0

20 314

20 22

.. .V


physics (cbse 2008)jayroy.in/cbse/2008 qp with solns.pdf · mass, and so the line b represents the...

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