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New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

T = 2 π l g

2 = 2 π 2 g

g = 2 π 2 m / s 2

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

At constant pressure,

d U = n C V d T = n . 5 2 R d T

d Q = n C P d T = n . 7 2 R d T

dW = nRdT

dU : dQ : dW = 5 : 7 : 2

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

D1 is in forward bias and D2 is in reverse bias.

Current,   I = 5 0 . 7 1 0 = 0 . 4 3 A

New question posted

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New answer posted

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V
Vishal Baghel

Contributor-Level 10

Thermal stress is developed on heating when expansion of rod is hindered.

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Vishal Baghel

Contributor-Level 10

L C = 1 1 0 0 m m = 0 . 0 1 m m

Zero error = +0.08 mm.

Diameter = 1 + 72 * 0.01 – 0.08 = 1.64 mm

Radius = 0.82 mm

New answer posted

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Vishal Baghel

Contributor-Level 10

Zener break down occurs in p-n junction having p and n both : Heavily doped and have narrow depletion layer.

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Vishal Baghel

Contributor-Level 10

T = 2 π L g o r T 2 = 4 π 2 ( L g )

g = 4 π 2 ( L T 2 )

Δ g g % = ( Δ L L + 2 Δ T T ) % = [ 1 m m 1 m + 2 * 0 . 0 1 1 . 9 5 ] * 1 0 0 % = ( 0 . 0 0 1 + 0 . 0 1 0 2 ) * 1 0 0 = 1 . 1 3 %

New answer posted

2 months ago

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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

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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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