Physics Ncert Solutions Class 12th

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New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

Q = CV

= ( 5 0 μ F ) * 2 V

= 1 0 0 * 1 0 6 C

Q = 1 0 0 μ C (on upper plate)

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

  R e q = 6 ?

i = 1 2 v 6 Ω 2 A (through battery)

v 1 5 Ω = 6 V

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2 months ago

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New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

d = 2 R h T + 2 R h R

d m = ( 2 * 6 4 0 0 * 1 0 3 * 3 2 0 + 2 * 6 4 0 0 * 1 0 3 * 2 0 0 0 )

= 224 km

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A
alok kumar singh

Contributor-Level 10

β = I C I E = I C I E I C = I C / I E 1 l C / I E = α 1 α

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V
Vishal Baghel

Contributor-Level 10

Active region of the CE transistor is linear region and is best suited for its use as an amplifier.

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A
alok kumar singh

Contributor-Level 10

For x < a,

B1 = μ 0 i 0 x 2 π a 2  

For a < x < b,

B 2 = μ 0 i 0 2 π x   

B 1 B 2 = x 2 a 2           

 

New answer posted

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

λ > h λ > 4 0 0 m

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V
Vishal Baghel

Contributor-Level 10

V A B = 5 5 * 4 = 4 V

Q = C e q V = 2 x = 8 μ c

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A
alok kumar singh

Contributor-Level 10

Current through zener diode,

l z = 2 4 1 0 1 1 0 5 m A = 1 2 m A              

Power across zener diode,

P z = V z . l z = 1 0 * 1 2 = 1 2 0 m W          

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