Question

The wire is stretched so that its length is 20% more than its initial length, the Percentage increase in the resistance of the wire

Answer 1

The percentage increases resistance of the wire is 40%.

Explanation:

Given that,

The wire is stretched so that its length is 20% more than its initial length

We need to calculate the mass of the wire

Using formula of density

$d=\dfrac{m}{V}$

$d=\dfrac{m}{A\times l}$

$A=\dfrac{m}{d\times l}$

Where, d = density

l = length

m = mass

A = area

We need to calculate the percentage increases resistance of the wire

Using formula of resistance

$R=\dfrac{\rho l}{A}$

Put the value of A in equation

$R=\dfrac{\rho I}{\dfrac{m}{d\times l}}$

$R=\dfrac{\rho\times l^2\times d}{m}$

Here, $\dfrac{\rho d}{m} = k$

k is a constant.

On differentiating

$\dfrac{dR}{R}\times100=\dfrac{2\times20}{100}\times100$

$\dfrac{dR}{R}\times100=40\%$

Hence, The percentage increases resistance of the wire is 40%.

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Topic : resistance

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