2018-10-01_Physics_For_You

SOLUTIONS

1. (d) : i =

ε

R

e

Rt

1 − L

















−

; i =

12

6

1

6

V 84103

Ω









−

























−

×











−

e

t

.

|1 A| = ( 21.

6

A) −^84103

























−

×















−

e

t

Ÿ − ×

−

e

6 t

84. (^103) =^1
    2
    or

6

84 103

t

. × −

= ln 2 Ÿ t = (.0 693)(^84 .)^10

6

× −^3

= 0.00097 s = 0.97 ms | 1 ms.

2. (d) : Vrms =

20

2

V

XL = ZL = (5 × 10–3 H) 2000

rad

s









= 10 :

XC =

1

ωC

=

1

()50 10×^6 . 2000







−F rad

s

= 10 :

R = RR + RL Ÿ R = 6 : + 4 : = 10 :

Z = ()XXLC−+^22 R

= () 10 ΩΩ−+ 10 22 () 10 Ω = 10 :

Irms =

V

Z

rms =^20

210

V

()Ω

= ()^2 A = 1.4 A

Vvoltmeter = Irms ˜  ()XXLC−+RL







22

= () 21 A. () 01 ΩΩ−+ 0422 ()Ω 

= 42 V = 5.6 V

3. (d) : I =

V

R

=

220 100

20

sin( πt)

Ÿ I = 11 sin (100pt)

Here peak value = 11 A

It happens for the rst time when (100pt 1 ) =

π

2

Ÿ t 1 =

1

200

s

e rms value of I =

11

2

.

It happens when (100pt 2 ) =

3

4

π

t 2 =^3

400

(t 2 – t 1 ) =

3

400

–

1

200

=

1

400

s = 2.5 × 10 –3 s.

4. (a) : For terminal velocity net force on the rod must
   be zero

BIlm==gieI

×

×

,.., =

..

.

02 98.

06 1

98

3

A

Now if e is the emf induced in the rod,

e × I = P = P 1 + P 2

So, e=

+

=

(. .)

(./)

.

076120

983

06 V

Now as this e is generated due to motion of rod with

terminal velocity in the magnetic eld, i.e.,

eBvl v

e

TTBl

===

×

so = ms−

06

06 1

1 1

.

.

Power dissipated across a conductor P

V

R

=

2

i.e., R

V

P

=

2

e resistance are in parallel combination so,

V 1 = V 2 = e

So, R

e

(^1) P
2
1

062

076

== = 047

(.)

.

. Ω

And, R e

(^2) P
2
2

062

12

== = 03

(.)

.

. Ω.

5. (b) :

di

dt

= 10^3 A s–1

? Induced emf across inductance, |e| = L

di

dt

|e| = (5 × 10–3) (10^3 ) V = 5 V

Since , the current is decreasing , the polarity of this

emf would be so as to increase the existing current.

e circuit can be redrawn as shown in the gure.

1 W 15V 5 mH B

A I

15V

Now, VA – 5 + 15 + 5 = VB

? VA – VB = –15 V or VB – VA = 15 V

6. (d) : B =

μ 0 2

2 2232

iR

()RX+ /

B =

μ 0 2

232232

iR

()RR+ /

=

μ 0 2

24232

iR

()R /

=

μμ 0 2

3

0

22 16

iR

R

i

.. R

=

φ = NBA cos 45° =^2

16

1

2

μ 0 i 2

R

a

φ

μ

=^0

2

82

ia

R

A 1 W 15 V 5 mH B

I