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New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

R = 2 Ω

L = 2 mH

E = 9V

i = ε 2 R = 9 v 4 Ω = 2 . 2 5 A

Just after the switch 'S' is closed, the inductor acts as open circuit.

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

According to KTG, the gas exerts pressure because its molecule :

suffer change in momentum when impinge on the walls of container.

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5 months ago

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Vishal Baghel

Contributor-Level 10

? 1 = 3 5 E 0 ( 0 . 2 ) N m 2 C 1 , a n d ? 2 = 4 5 E 0 ( 0 . 3 ) N m 2 C 1

? 1 ? 2 = 3 * 0 . 2 * 5 5 * 0 . 3 * 4 = 1 2

New answer posted

5 months ago

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Vishal Baghel

Contributor-Level 10

Let v is velocity at the highest position.

T m a x = 5 T m i n m g + m ( v 2 + 4 g l ) l = 5 ( m v 2 l m g ) 4 . v 2 l = 1 0 g

v = 5 2 g l = 5 2 * 1 0 * 1 = 5 m / s

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5 months ago

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Vishal Baghel

Contributor-Level 10

Δ U + Δ K E = 0

f 2 n R Δ T = 1 2 m v 2 Δ T = m n v 2 f R = 4 * 1 0 3 * 3 0 2 3 R = 3 . 6 3 R K .

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5 months ago

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Vishal Baghel

Contributor-Level 10

For equilibrium,

d U d r = 0 1 0 α r 1 1 + 5 β r 6 = 0 r = ( 2 α β ) 1 5

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Vishal Baghel

Contributor-Level 10

L d i d t = V 2 d i = 3 t d t 2 0 i d i = 3 0 4 t d t i = 1 2 A
U = 1 2 L i 2 = 1 2 * 2 * 1 2 2 = 1 4 4 J

New answer posted

5 months ago

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Vishal Baghel

Contributor-Level 10

Let the charge of on each drop be q and radius of each drop be r.

k q r = 2

When all drops are joined, radius,

r ' = ( 5 1 2 ) 1 3 r = 8 r

Potential of the new drop,

V = k . 5 1 2 q 8 r = 6 4 k q r = 1 2 8 V

New answer posted

5 months ago

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Vishal Baghel

Contributor-Level 10

2 π f 1 L C L = λ 2 4 π 2 C c 2 = 9 6 0 2 4 * 3 . 1 4 2 * 2 . 5 6 * 1 0 6 * 9 * 1 0 1 6 =10-7 = 10 * 10-8 H

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5 months ago

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V
Vishal Baghel

Contributor-Level 10

Image is virtual, when an object is placed at distance 10 cm from the lens and image is real when object is placed at 20 cm from the lens.

Thus, m1 = -m2 => f f 1 0 = f f 2 0 f = 1 5 c m  

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