Question 33:
A piece of wire of resistance 20 Ω is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.
Lakhmir Singh Physics Class 10
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R₁=20Ω r₂
l₁ l₂=2l₁
a₁ a₂=¹/₂a₁(since,length is doubled thickness gets half)
r₁/r₂=(ρl₁/a₁)/(ρl₂/a₂)
20/r₂=(l₁/a₁)/(2l₁/¹/₂a₁)
20/r₂=1/4
r₂=20*4
r₂=80Ω
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mark ans a brai....
l₁ l₂=2l₁
a₁ a₂=¹/₂a₁(since,length is doubled thickness gets half)
r₁/r₂=(ρl₁/a₁)/(ρl₂/a₂)
20/r₂=(l₁/a₁)/(2l₁/¹/₂a₁)
20/r₂=1/4
r₂=20*4
r₂=80Ω
Hop its helpful
mark ans a brai....
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