Physics NCERT Exemplar Solutions Class 12th Chapter Eight

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Payal Gupta

Contributor-Level 10

F.B.D. of point 'P'

T s i n θ = 3 0            - (1)

T c o s θ = 1 0 0          - (2)

( 1 ) ÷ ( 2 )

t a n θ = 3 1 0 θ = t a n 1 ( 3 * 1 0 1 )    

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Payal Gupta

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For high wavelength, size of antenna should be high.

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Payal Gupta

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For Transistor

to act as a switch Saturation & cut – off state

to act as an amplifier Active Region

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Payal Gupta

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E 2 > E 1

For statement 1

              hf = E1  - E2

Since, E2 > E1

So, it should be

hf = E2 – E1

Therefore statement 1 is wrong

For statement 2

For the jumping of electron from higher energy orbit

(E2) to lower energy orbit (E1)

hf = E2 – E1

  f = E 2 E 1 h              

Statement (2) is correct

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Payal Gupta

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  μ 2 > μ 1

μ 2 μ 1 > 1               

μ 2 = c v 2 , μ 1 = c v 1               

μ 2 μ 1 = v 1 v 2 > 1 v 1 > v 2               

Frequency remains constant while refraction since energy is constant

υ = v λ               

λ α v               

λ 1 > λ 2               

λ decreases

wavelength and speed decreases but frequency remains constant

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Payal Gupta

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R 1 = R  

R 2 = R       

P = 1 f = ( μ 1 ) ( 1 R 1 1 R 2 )               

P = 1 f = ( μ 1 ) ( 1 R ( 1 R ) )

P = ( μ 1 ) 2 R              

L2

R1 = R

R 2 =               

R 1 =               

R2 = ­­-R

Power of  L 1 = 1 f ' ' = ( μ 1 ) ( 1 R 1 1 R 2 )

Power of L 1 = ( μ 1 ) * 2 R = P  

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Payal Gupta

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Statement 1 is true

Speed of EM ware in vaccum is given by,

C = 1 μ 0 0                

Speed of EM wave in a material medium

v = 1 u r μ 0 r 0                

So, statement (2) is false.

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Payal Gupta

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Dimension

RC → [T]

L R [ T ]

L c [ T ]

L C dimensionless

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