Physics NCERT Exemplar Solutions Class 11th Chapter Eight

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2 months ago

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alok kumar singh

Contributor-Level 10

Elastic potential energy stored in the spring

U = 1 2 k x 2

= 1 2 F x = F 2 2 k

H e n c e U 1 U 2 = k 2 k 1 F i s s a m e f o r b o t h s p r i n g s

= 500 1200 = 5 12

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2 months ago

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alok kumar singh

Contributor-Level 10

From work energy theorem

W s p + W E = ? K

- 1 2 k x 0 2 + Q E . x 0 = 0

? x 0 = 2 Q E k

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alok kumar singh

Contributor-Level 10

Flux through a surface ? = E ? . A ?  

? = 6 i ? - 4 j ? + 7 k ? . 200 i ?

= 1200 u n i t

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alok kumar singh

Contributor-Level 10

Resistance of whole wire is

R = r ' * l

20 = 1 * l

l = 20 m

Let length of the shorter section = a

1 R = 1 a + 1 20 - a

1 1.8 = 20 - a + a a 20 - a ? 20 a - a 2 = 36

a = 2 a n d 18

? a = 2 m

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alok kumar singh

Contributor-Level 10

d T d t = k ( T T 0 )  

d T ( T T 0 ) = k d t  

k = 1 6 l n ( 2 3 )             - (1)

Now d T d t = k ( T T 0 )

t = 10.257

t2 = 6 + 10.257 = 16.257 minutes

16.25 and or 16

 

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alok kumar singh

Contributor-Level 10

Q = Δ υ + w n C Δ T = n C v Δ T = n C Δ T 4 = 3 4 n C Δ T = n C v Δ T

C = 4 3 C v = 4 3 * 3 2 R = 2 R

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alok kumar singh

Contributor-Level 10

given N = 1000 turns

A = 1m2

ω = 1 r e v / s e c

= 1 * 2 π r a d / s e c

V = d ? B d t = d d t ( N B A c o s ω t )  

1 4 0 * 2 2 7  

= 20 * 22

= 440 volts

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2 months ago

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alok kumar singh

Contributor-Level 10

d = 4 3 cm (Lateral shift)

By Snell's law

μ a i r s i n 6 0 ° = μ g s i n θ

θ = 30

1*32=3sinθ  

s i n θ = 1 2  

t = 12 cm

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New answer posted

2 months ago

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alok kumar singh

Contributor-Level 10

V P = V q + V - q

= K q 5 - 3 + K ( - q ) 5 + 3 * 10 2 = K q 2 - K q 8 * 10 2 = 3 K q 8 * 10 2

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