[XII – Physics] 188

SAMPLE PAPER – I

1. Because magnetic monopole does not exist.

2. A device that allows the passage of a number of frequencies (Wc ± Wm)along with carrier frequency Wc

3. Magnifying power is independent of aperture.

4. (i) IR radiation (ii) Radiowaves

5. Because emission is followed by antiparticles and .

6. Plane Wavefront.

9. A device consisting of a transmitter and a receiver. This is used to increasecommunication Range as

1,2,3 and 4 etc are the repeaters arranged between source of informationand the use of information.

10. Drift speed

deVV am

Relaxation time

V potential diff. across the conductor

1dV

(i) Drift speed is halved.

(ii)´´´ 2 2

V V EEl

Electric field is halved.

11. B B Y

0 0 0

1 0 1

0 1 1

1 1 1

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189 [XII – Physics]

OR + NOT NOR

12. centreBBaxial

8

r

x P

20 0

3 22 2

.8 22

Ir r Irr x

3 .x r

13. BN = BP – BQ

BP = BQ

2 4 2 3 4

o p o QI Ir r

IQ = 3IP.

14. Ey = Eoy cos (wt – kx)

22

83 102

Hz

(i) Speed of em waves 8

83 10 2 10 /1.5

V m s

(ii)3

–28

4 10 2 10 .2 10

oo

EB T

V

15. WAB = qo V = qo (VB – VA)

Akq kQVa b

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[XII – Physics] 190

Bkq kQVb b

1 1 .

AB oW q kqb a

Aq

Q

B

16. In S.I. system the unit of radioactivity is becquerel.

1 becquerel = 1 disintegration/second.

1 curie = 3.7 × 1010 bq.

T½ = mean life × ln 2

mean life1.44

24001.44

= 1667 years

17. Definition of modulation N.C.E.R.T. Pg. 517 Vol-II.

18. Linear fringe width

Dd

Angular fringe width,

D d

in some medium

d

(i)1 decreases d

(ii) is undependent of D, no effect.

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191 [XII – Physics]

(iii) angular width increases

(iv) 1/M angular width decreases.

19. Statement of Gauss’s law

E E d A

= 1 + 2 + 3 + 4 + 5 + 6

1 0 0 0 0 0 E d A

20 E ai a i

= E0a3

(ii) E = q/E0 q = E E0 = E0a3 E0

20.2 1

hcE E

22 21 2

1 113.6

hc

zn n

For shortest n1 = 1, n2 =

shortest –34 8

86.624 10 3 10 2.25 101 113.6 41

m

For longest 21 1, 2 n n

For longest 43shortest

–84 2.25 103 m

3 × 10–8m

t Re

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[XII – Physics] 192

21. Principle, working 1, 2

OR

(i) = MB Sin

max = MB sun 90° = MB

(ii) W = U2 – U1 U = MB Cos

MB – (–MB) U = – MB (stable equilib)

= 2 MB U = + MB (Unstable equilib)

22. (i) C = k Co = 5 x 4.8 = 24.0 f

(ii) 2

2/ 2

5/ 2

o

o

q C CCq C

(iii) 0

1

6 1.2 .5

V

V voltk

23. NCERT (Part I) Solved Example

24. Principle of potentiometer when a steady current is passed through apotentiometer wire AB of length L, and V is the potential difference acrosswire AB, then

VPotentialgradient kL

–3

PQ

E 4I= 8×10 AR+R 480 20

VPQ = I RPQ = 8 × 10–3 × 20 = 0.16 Volt.

Potential gradient 0.16 /400

PQ

PQ

Vk V cm

L

–340×10AJ= 400 100 .0.16

cm

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193 [XII – Physics]

25. Define photoelectric effect

(i) Since Intensity is same, number of electrons emitted per secondremain same

(ii) K. Emax a ha

violet > blue

(E.Emax)violet > (K. Emax)blue

27. Refer NCERT

OR

28. Refer NCERT

P = Vrms Irms Cos

rms

PCos =I

rmsV

Ploss = I2rms R. R Þ resistance of transmission line.

Ploss a I2rms

Low power factor (cos) high

high power loss.

29. Refer NCERT

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