Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

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2 months ago

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R
Raj Pandey

Contributor-Level 9

μ = 9 * 1 0 4 k g / m

T = 900 N

f 0 = 5 0 0 H z resonance frequency

f 0 ' = 5 5 0 H z Next higher frequency

v = T μ = 9 0 0 N 9 * 1 0 4 = 1 0 0 0 m / s

5 0 0 H z = 1 0 * 1 0 0 0 2 * l

l = 1 0 m

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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)

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R
Raj Pandey

Contributor-Level 9

θ = 6 0 ° 2 = 3 0 °

 

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R
Raj Pandey

Contributor-Level 9

R e q = 3 Ω

 

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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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P
Payal Gupta

Contributor-Level 10

In amplitude modulation, the amplitude of the career signal is varied in according with the modulating signal.

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A
alok kumar singh

Contributor-Level 10

v = f λ

f = v λ = 3 * 1 0 8 1 0 3 * 1 0 9 = 3 * 1 0 1 4 H z

Channels =  2 % o f 3 * 1 0 1 4 H z 8 * 1 0 3

2 % o f 3 * 1 0 1 4 H z 8 * 1 0 3

2 % o f 3 * 1 0 1 4 H z 8 * 1 0 3

= 2 x 1 0 0 3 * 1 0 1 4 8 * 1 0 3 = 7 5 * 1 0 7

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A
alok kumar singh

Contributor-Level 10

              x = 10   2 ? ( n t ? x ? ) c m

              V (wave velocity) =   ? ? 2 ?

              vmax = 10 * 2p n

              10 * 2pn = 4 *   ? ? 2 ?

              10 * 2pn =   4 2 ? ? n ? 2 ?

               x = 5?

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A
alok kumar singh

Contributor-Level 10

Maximum particle velocity = Aω

Wave velocity = ωR

ωk=Aω

k=1A=12=cm

λ=2πk=4πcm

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