2018-10-01_Physics_For_You

rmin

()()()(. )

.

=

××

××

−

−

9109 2921610

51610

192

13

or rmin = 5.3 × 10–14 m or rmin = 5.3 × 10–12 cm

e distance of closest approach is of the order of

10 –12 cm.

21. (c) : At Vg = –1 V, IP = (0.125 VP – 7.5) × 10–3 A

?

dI

dV

P

P

=×0 125 10. −^3

A

V

or r

dV

p dI

p

p

==

× −

1

0 125 10.^3

V

A

or rP = 8 × 10^3 : ...(i)

Again, IP = (0.125 VP – 7.5)mA

IP = (0.125 × 300 – 7.5) at VP = 300 V, Vg = –1 V.

or IP = (37.5 – 7.5) mA or IP = 30 mA

In second case, Vg = –3 V, V = 300 V, I = 5 mA

? g

I

m V

P

g

=

∆

∆

at VP constant

or gm=

−×

−−−

()−

[()]

30 510

13

3

or gm=

×

=×

25 10− −−

2

12510

(^331)
.AV ...(ii)

22. (a) : ' N
    N

e t

0

= −λ

ere is a simultaneous emission of two particles.

∴ N −−

N

e

N

N

tte

0

( + ) 0

0

=or ( + )

4

λλ (^12) = λλ 12
or log 4 = (O 1 + O 2 ) t log e
Nowaλλ 1 ==0 693 nd
1620

0 693

(^2810)

..

∴ 







2.303[ 2 × 0.3]=0.693^1

1620

+ 1

810

t

or ..

.

tt= ×× or year

×

2 303 (^061620) =
0 693 3

1080

23. (d) : EMF developed due to motion = H
    ? H = vBl where v= ×
    ×

170 1000= −

60 60

47 .m 2 s^1

H = 47.2 × (0.2 × 10–4) = 9.44 × 10–4 = 0.944 mV

24. (b) : e mass of a wire length l, cross-sectional
    area A and density d is given by
    m = Ald or A

m

ld

=

? e resistance of wire of resistivity ρ is

R l

A

dl

m

==ρρ=kl

2

(^2) ...(i)

? I

V

R

=−[] 1 e−Rt/L , when current grows in L-R

circuit.

or 1

12

6

1 68410

3

=−−×

−

[]e t/(.)

or t = 0.97 × 10–3 s = 0.97 ms or t = 1 ms.

18. (a) : Refer the gure (a). Given i = 60°, G = 30° and
    A = 30°.

We h a v e

G = i + e – A ...(i)

From equation (i), we get 30° = 60° + e – 30° or

e = 0, i.e. the emergent ray is perpendicular to the

face AC through which it emerges. is is shown in

gure (b). Here also i = 60º and G = 30º

erefore, r 1 = i – G = 60° – 30° = 30°. Hence

μ= =

°

°

===

sin

sin

sin

sin

/

/

.

i

r 1

60

30

32

12

31732

19. (a) : Given u = –200 cm, fo = 50 cm. e distance
    of the image I formed by the objective is given by
    111
    vuf 0

−= or

1111

50

1

vfo u 200

=+=−

which gives v =

200

3

cm. erefore,

e image I serves as the object for the eye-piece.

us vc = –25 cm, fe = 5 cm. e object distance

uc is given by

1111

25

1

′ 5

=

′

−=−−

uvfe

which gives uc = −

25

6

cm.

Separation between objective and eye-piece

||vu=||′=+^200 ==..

3

25

6

425

6

70 8cm

20. (c) : Energy is conserved.
    Loss in kinetic energy = Gain in potential energy
    1
    4

(^251610)
0
13
πε

()() (. )

min

Ze e

r

=× × − J

∴=

××−

r

Ze

min

.

1

4

2

051610

2

πε^13