If the length of a copper wire is increased by 5% due to stretching. What will b

Question

If the length of a copper wire is increased by

5 %

due to stretching. What will be the percentage increase in its resistance?

Options:

A . 12%

B . 11.5%

C . 20.25%

D . 10.25%

Answer: Option D
:
D
Resistance,

R = ρ \frac{L}{A}

where:

ρ

: Resistivity,

L

: Length of the wire

A

: Cross-sectional area of the wire.
The length of the copper wire is increased by

5 %

.
So, new length is:

L^{'} = L + \frac{5}{100} L = 1.05 L

Let the new cross-sectional area be

A^{'}

.
On stretching the wire, its volume will remainconstant.

⟹ L A = L^{'} A^{'}

A^{'} = \frac{A}{1.05}

New resistance will become:

R^{'} = ρ \frac{L^{'}}{A^{'}}

R^{'} = ρ (\frac{1.05 L}{A / 1.05})

R^{'} = {1.05}^{2} R = 1.1025 R

Percentage increase in its resistance is,

p = \frac{R^{'} - R}{R} \times 100

p = \frac{(1.1025 - 1) R}{R} \times 100

p = 10.25 %

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