Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image 

 

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R
Raj Pandey

Contributor-Level 9

I A B = m ( 0 ) 2 + 2 m a 2 2 + 3 m a 2 2

I A B = m a 2 2 + 3 m a 2 4 = 5 m a 2 4 I A B = 25 m a 2 20 = N 20 m a 2 N = 25

 

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R
Raj Pandey

Contributor-Level 9

Acceleration of the combined system along the incline is given by

a = g s i n ? 37 ? a = 3 g 5

With respect to block M :

N = m g - m a s i n ? 37 ? = m g - 3 m g 5 3 5 N = 16 25 m g μ N = m a c o s ? 37 ? μ 16 25 m g = 3 m g 5 4 5 μ = 3 4 = 750 1000 I = 750

 

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Raj Pandey

Contributor-Level 9

Δ E 1 = 12400 e V ? 1085 ? = 11.43 e V

Δ E 2 = 12400 304 e V = 40.80 e V

Δ E 1 + Δ E 2 = 54.4 e V 1 n 1 2 - 1 n 2 2 52.23 = 54.4 1 n 1 2 - 1 n 2 2

If n 1 = 1 n 2 = 5

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R
Raj Pandey

Contributor-Level 9

1 20 = ( 1.55 - 1 ) 1 R 1 - 1 R 2

1 f = ( 1.50 - 1 ) 1 R 1 - 1 R 2

Divide (1) by (2), f 20 = 0.55 0.50

f = 22 c m

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A
alok kumar singh

Contributor-Level 10

90 ?

Sol. | P ? + Q ? | = | P ? |

P 2 + Q 2 + 2 P Q c o s ? θ = P 2

Q + 2 P c o s ? θ = 0

c o s ? θ = - Q 2 P

t a n ? α = 2 P s i n ? θ 2 P c o s ? θ + Q =

? [ 2 P c o s ? θ + Q = 0 ]

α = 90 ?

 

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alok kumar singh

Contributor-Level 10

r in   = l 1 - l 2 l 2 R ext   = 60 500 * 10

r = 6 5 = 1.2 Ω

n = 12

 

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A
alok kumar singh

Contributor-Level 10

6

Sol. Common potential after connection.

V common   = C 1 V 1 + C 2 V 2 C 1 + C 2 = 60 * 20 + 0 120 = 10 V o l t

 

Loss of energy = 1 2 C V 2 - 1 2 ( 2 C ) * V Common   2

= 1 2 * 60 * 10 - 12 * ( 20 ) 2 - 60 * 10 - 12 * ( 10 ) 2 = 60 * 10 - 12 ( 200 - 100 ) = 6000 * 10 - 12 = 6 n J

 

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alok kumar singh

Contributor-Level 10

M ice   L f + m ice   ( 40 - 0 ) C w = m steam   L v + m steam   ( 100 - 40 ) C w

200 [ 80 + 40 ( 1 ) ] = M [ 540 + 60 ( 1 ) ]

200 ( 120 ) = M ( 600 )

M = 40 g m

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