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New answer posted

11 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

For mirror

V l 1 M = m 2 V O 1 M

V l 1 M  Velocity of image w.r.t. mirror

V 0 M  Velocity of object w.r.t. mirror

m magnification

1 v + 1 1 . 9 = 1 0 . 1

1 v = 1 1 . 9 1 0 . 1

= 0 . 1 1 . 9 1 . 9 * 0 . 1

v = 1 . 9 * 0 . 1 1 . 8 = 0 . 1 9 1 . 8 = 0 . 1 0 5

m = v u = ( 0 . 1 1 . 9 )

| V l 1 M | = ( 1 1 9 ) 2 | V O 1 M |

= 1 ( 1 9 ) 2 * 4 0 = 0 . 1 m / s

New answer posted

11 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

h c λ 1 = W + e V 1

h c λ 2 = W + e V 2

= 1 2 4 0 0 ( 1 2 8 0 0 1 4 0 0 0 ) [ λ 1 & λ 2 a r e t a k e n i n A ° ]

= 4 . 4 3 . 1 = 1 . 3 V

New answer posted

11 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

As power delivered by the bulb is 500 W, potential difference across it should be 00 V. Thus,

  R = R b u l b = 1 0 0 2 5 0 0 = 2 0 Ω

 

            

New answer posted

11 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Inside a uniform spherical shell, electric field is zero every where & electric potential is constant but not zero, every where

New answer posted

11 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

    Y = ( A + B ¯ ¯ + A ) + ( A + B ¯ ¯ + B ) ¯ + ( A + B ¯ + A ) . ( A + B ¯ + B )           

              A            B            Y

              0            1

              0            1            0&nbs

...more

New answer posted

11 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Power is maximum at resonance

So, X L = X C = 1 ω C

C = 1 ω * X L

= 1 2 5 0 * 1 0 0 * π 2

= 4 μ F [ π 2 = 1 0 ]

New answer posted

11 months ago

0 Follower 7 Views

R
Raj Pandey

Contributor-Level 9

F. B. D of Beaker

By NLM2         

f = m a = m ω 2 R m ω 2 R μ N R μ g / ω 2

New answer posted

11 months ago

0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

t a n β = B s i n 6 0 ° A B c o s 6 0 °

β = t a n 1 ( 3 B 2 A B )

 

              

             

New answer posted

11 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

P = E L 2 M 5 G 2

= [ M L 2 T 2 ] [ M L 2 T 1 ] 2 [ M 5 ] [ M 1 L 3 T 2 ] 2

= M 0 L 0 T 0

New answer posted

11 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

U B = 1 2 L l 2 . . . . . . . . . . . . ( i )

PR = l2R

=> 640 = (8)2 R

R = 1 0 Ω

from (i) 64 = 1 2 * L * ( 8 ) 2

L = 2H

l = L R = 2 1 0 = 0 . 2 s e c o n d        

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