Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

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alok kumar singh

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v = 1 ε μ = 1 2 ε 0 . 2 μ 0 = 1 2 ε 0 μ 0 = c 2 = 1 5 * 1 0 7 m / s .

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alok kumar singh

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  N p N s = V p V s N p = V p V s * N s = 2 2 0 1 2 * 2 4 = 4 4 0

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alok kumar singh

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l = l 0 c o s 2 θ = 1 0 0 * c o s 2 ( 3 0 ° ) = 7 5 L u m e n s  

Note : As the incident light is unpolarized, intensity of emerging light does not depend on the polarization axis of polarizer.

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alok kumar singh

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If the block does not slide,

mg sin    θ μ m g c o s θ

  t a n θ μ d y d x μ x 2 0 . 5 x 1 m .         

 Thus, y 1 2 4 = 0 . 2 5 m = 2 5 c m .  

 

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alok kumar singh

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Using conversation of momentum in direction perpendicular to the original direction of motion,

mv1 sin 30° = mv2 sin 30°

v 1 v 2 = 1

 

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alok kumar singh

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Quality factor = ω L R = 2 * 3 . 1 4 * 1 0 * 1 0 6 * 2 * 1 0 4 6 . 2 8 = 2 0 0 0

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alok kumar singh

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f = W = 0.5 * 10 = 5 N

N = F

For block not to slide,

  f μ N            

5 0 . 2 F F 2 5 N

 

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alok kumar singh

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Ratio of masses on two pistons of the hydraulic lift equals to that of their cross- section area.

A 1 A 2 = 1 0 0 m            

Now,   4 2 A 1 A 2 / 4 2 = M m M = 2 5 6 A 1 A 2 . m = 2 5 6 0 0 k g .  

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alok kumar singh

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Percentage Modulation =   A m A c * 1 0 0 = 2 0 8 0 * 1 0 0 = 2 5 %

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alok kumar singh

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Potential difference across 2k Ω  is 5V, thus current through it,

i = 5 2 * 1 0 3 = 2 5 * 1 0 4 A .           

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