Physics NCERT Exemplar Solutions Class 12th Chapter Eight

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New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

C = 10-6 f

  f = 0.5 * 103 = 500

z = x R 2 + ( x L x C ) 2  

for minimum impedance :

xL = xC

ω 2 = 1 L C  

L = 1 π 2 = 1 1 0 * 1 0 0 0 = 1 0 0 m H

 

New question posted

a month ago

0 Follower 1 View

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

r = m v q B = 2 k m q B

2 * 5 * 1 0 3 * 1 . 6 * 1 0 1 9 * 2 4 * 1 . 6 6 * 1 0 2 7 1 . 6 * 1 0 1 9 * 1 2

= 9.975 * 10-2 cm

= 9.975 cm

r 1 0 cm

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

U = n c v T

= 2 * 3 2 * R * 3 0 0 = 7 4 7 9 J

New answer posted

a month ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

Using newton's law:

2mg – T = 4ma  - (1)

 T – mg = ma       - (2)

From (1) & (2) :

m g = 5 m a a = g 5 = 2 m / s 2  

T = m g + m g 5 = 6 m g 5  

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Using conservation of momentum:

60 * v = (60 + 120) * 2

⇒v = 6m/s

 

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

L ? = r ? * m v ?

= ( 3 i ^ ? j ^ ) * ( 3 j ^ + k ^ )

= 9 k ^ + 3 ( ? j ^ ) ? i ^

= ? i ^ ? 3 j ^ + 9 k ^

| L ? | = 1 + 9 + 8 1 = 9 1

New answer posted

a month ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

x R = 1 2 0 Ω

x L = ω L = 1 0 0 * 1 0 0 * 1 0 3 1 0 Ω

x C = 1 ω c = 1 1 0 0 * 1 0 0 * 1 0 6 = 1 0 0 Ω

z = 1 2 0 2 + ( 1 0 0 1 0 ) 2 = 1 2 0 2 + 9 0 2

= 150  Ω

3 2 = 1 6 0 7 5 * t t = 3 2 * 7 5 1 6 0 = 1 5

New answer posted

a month ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Using whetstone

R 3 = 4 6 R = 2 Ω

R e q u = 5 4 1 5 Ω

i = 3 6 5 4 1 5 = 1 0 A

New answer posted

a month ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

At null point:

1 5 4 3 + 2 = R 1 0 0 4 3

R = 1 9 Ω

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