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New answer posted

5 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

In given circuit inductor behave as a simple wire so resultant circuit will be

Ref = 2 + 1 = 3 Ω  

V = IR


l = 3 0 3 = 1 0 A  

 

New answer posted

5 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

l = V 0 B l R = V 0 * 5 * 2 0 * 1 0 2 4 + 1         

  2 * 1 0 3 = V 0 * 2 0 * 1 0 2

V 0 = 2 * 1 0 3 2 0 * 1 0 2 = 2 2 * 1 0 2 = 1 * 1 0 2 m / s

V0 = 1 cm/s

New answer posted

5 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Object is moving in upward direction with constant velocity so in upward motion (+2N) and for downward motion (-2N) So option (1) is correct representation.

 

New answer posted

5 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Adiabatic equation : P 1 V γ = P 2 V γ ( 2 0 0 ) ( 1 2 0 0 ) γ = P ( 3 0 0 ) γ

P = ( 2 0 0 ) ( 4 ) 3 2 = 1 6 0 0 K P a

W = P 1 V 1 P 2 V 2 γ 1 = 2 4 0 4 8 0 1 . 5 1 = 4 8 0 J  

New answer posted

5 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Battery is connected while dielectric is inserted so potential difference will be  remains same.

U i = 1 2 c V 2

U f = 1 2 K c V 2

Δ U = 1 2 ( K 1 ) c V 2 = 1 2 * 1 * 2 0 0 * 1 0 9 * 2 0 0 2 = 4

New answer posted

5 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

from given information : I = A = 60°, r = A 2 = 3 0 ° =>  μ = 3 = c v

v = c 3 t i m e = ( 5 3 1 0 0 ) ( 3 * 1 0 8 3 ) = 5 * 1 0 1 0 s

New answer posted

5 months ago

0 Follower 15 Views

A
alok kumar singh

Contributor-Level 10

Maximum energy = 10 J

1 2 K x 2 = 1 0              

K = 5

Given Tpendulum = Tspring

2 π l g = 2 π m K             

4 g = 5 5           

g = 4m/s2

New answer posted

5 months ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

X c = 1 ω c , X L = ω L so at very high frequencies capacitor behaves as conduct and inductor behaves as open circuit. The effective impedance will be 2 Ω .

 

New answer posted

5 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Distance of 4th bright fringe from central maxima = 4 λ D d  

S o , 2 . 4 1 0 0 = 8 λ D d λ = 2 . 4 * 0 . 3 * 1 0 3 1 0 0 * 8 * 1 . 5 = 6 * 1 0 7 m

f = c λ = 3 * 1 0 8 6 * 1 0 7 = 5 * 1 0 1 4 H z

New answer posted

5 months ago

0 Follower 21 Views

V
Vishal Baghel

Contributor-Level 10

Δ M M = Δ μ μ = 2 5 0 5 0 0 = 1 2

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