Question

a wire of resistance r and of radius r and compressed back to another wire of radius 2r, the new resistance of wire is????​ Answer is r/16

Answer 1

Answer:

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Answer 2

Answer:

new resistance of wire is R/16.

Given:

a wire of resistance R and of radius r and compressed back to another wire of radius 2r.

To find:

The new resistance of wire.

Step-by-step by explanation:

It is given that, the a wire of resistance R and of radius r and compressed back to another wire of radius 2r.

So,

In this case when wire is compressed the volume will remain constant.

volume of wire before compressing = πr²l

volume of wire before compressing = π(2r)²L

∴ πr²l = π(2r)²L

⇒ πr²l = π4r²L

⇒ L = l/4

Hence, new length of wire will be l/4.

Resistance is given by ρl/Α

where ρ is resistivity, l is length and A is cross sectional area.

so, resistance of wire before compressing will be

R = ρl/ πr²

and resistance of wire after compressing will be

R' = ρL/Α'

R' = ρ(l/4)/π4r² = ρl/16πr²

as R = ρl/ πr² so R' = R/16.

Hence, the new resistance of wire is R/16.

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