e solution of this equation is Q = Q 0 sin (Zt + φ),

where t = 0, Q = Q 0

QCV

CC

CC

0 eq^12 V

12

==

+ =

×× ×

×

−

−

8210 20

10 10

12

6

=32 × 10–6C ∵ (^) φ=π
2
? Q = (32 PC) cos (2.5 × 10^4 × 125 × 10–6) 32 PC

9. (b) : For power to be consumed at the rate of
   1100
   5

= 220 W, we have P = EvIv cos q

220

220 220

222222

=

×

+

×

RL +

R

ωωRL

where R =

V

P

2 2202

1100

== 44 Ω

220 220 44

44

44 220 44

2

22

= ×^22

+

() ⇒+ =×

()

()

L

L

ω

ω

Lω= 220 44 44×−^2 = 88 Ω

L

f

=

×

=

×

=

×

×=

88

2

88

250

88

222

7

50

028

ππ

.H

10. (c) : Let f 1 and f 2 be the focal length in water.

en,^111112

1

11

fRwwRR

=−









+







=−

















μ

μ

μ

μ

...(i)

1

1

11

1

2

2

22

fRwwRR

=−







−−=−







− 

μ

μ

μ

μ

...(ii)

Adding (i) and (ii), we get

112

12

12

ff wR

+=

()μμ−

μ

or^1

30

(^212)

()μμ−
μwR
? (μ 1 – μ 2 ) =
μwR
60

1

3

=

11. (a) : e detector will
    receive the maximum light
    when the image of point
    source of light coincides
    with the position of
    detector.
    Let at any time t, image of S and detector coincides
    then,
    for u = – (100 – 10t)cm, v = 20t cm, f = 10 cm
    e detector receives maximum light.
    From,^111
    vuf

−=^ Ÿ^

1

20

1

100 10

1

tt 10

−

−−()

=

Ÿ 2 t^2 – 19t + 10 = 0 Ÿ t = 0.56 s and 8.94 s

12. (c) : Omin =

hc

eV

==

12400

40000

03 1. Å

At 40 kV : Omin =

12400

40000 = 0.31 Å

S D

100 –1 0 t 20 t

Wavelength of KD is independent of applied potential.

For KD , X-ray :

3

4

() 13. 61 ()ZE−^2 ==

hc

λKα

Given that λKα = 3 Omin Ÿ

1216

()Z− 12

= 3 × 0.31

⇒−()ZZ 1 =⇒−= ⇒=Z

1216

093

(^2130813637)
.


13. (c)


14. (c) : We know that N = N 0

1

2









n

For A, NN N

N

A

nA

=







 =







(^00)  =
4
(^10)
2

1

216

(^) ∵n t
A TA

 ===











80

20

4

For B, NNB N N

nB

= (^00)   =   =
2
(^10)
2

1

24

(^) ∵ n t
B TB

===













80

40

2

?

N

N

A

B

=

1

4

or NA : NB = 1:4

15. (a)