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New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

  I = E R 2

U I n d u c t o r = H e a t l o s s i n R 1 = 1 2 L ( E R 2 ) 2              

New answer posted

4 months ago

0 Follower 14 Views

A
alok kumar singh

Contributor-Level 10

The time period of LC oscillations,

  T = 2 π L C          

The time at which charge on the capacitor will be zero is  T 4 .  

So t    = π 2 L C

New question posted

4 months ago

0 Follower 1 View

New answer posted

4 months ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Vertical component of velocity just after collision = u 2 2

k i = m u 2 2 k f = 1 2 m u 2 2 + 1 2 m u 2 8 = 5 m u 2 16

Fraction  = k i - k f k i = 1 - 5 m u 2 16 m u 2 2 = 1 - 5 8 = 3 8

 

New answer posted

4 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Heat lost = =  Heat gained

m * 540 + m * s ω * ( 100 - 5 ) = 10 * 80 + ( 10 + 74 + 10 ) * s ω * 5

m ( 540 + 95 ) = 800 + 94 * 5

m 635 = 1270

m = 2 g m

New answer posted

4 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

If the multimeter shows zero or low resistance reading for forward bias and does not change even on reversing the connection then the diode is defective. It is short.

If the diode shows a high resistance under both forward and reverse biased conditions, it is defective. It is open.

If the multimeter shows zero or low resistance reading for forward bias and shows high resistance reading on reversing the connection then the diode is not defective.

New answer posted

4 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

For gas A : P A (  initial ) = v f v i γ P = ( 2 ) 3 / 2 P

P A = 2 2 P

For gas B: P B (  initial ) = P

For gas C : P C (  initial ) = 2 V 0 V 0 P = 2 P  Ratio = 2 2 : 1 : 2

New answer posted

4 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  η = W . D . b y g a s H e a t a b s o r b

η = H e a t ( N e t ) H e a t ( A b s o r b )

η = A r e a u n d e r t h e c u r v e H e a t A b s o r b

η = 1 2 * S 0 T 0 1 2 ( 3 T 0 ) S 0 = 1 3

New answer posted

4 months ago

0 Follower 17 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

4 months ago

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R
Raj Pandey

Contributor-Level 9

Voltmeter readings

  20 = I 1 R 20 = R v R + R v 4 R R v R = 5 R + 5 R v R = 5 R v R v - 5 R = 5 1 - 5 R v

For 5 < R v < , R > 5 Ω

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