Question

if 25% part of length of wire is stretched by 25% then percentage change in resistance of wire will be about.​

Answer 1

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Answer 2

Answer:

The percentage change in resistance of wire will be approximately 14%.

Explanation:

Given, 25% part of the wire will be stretched by 25%.

Consider that, l is the total length of wire. Then 25% of the wire length will be l/4.

So when we stretch only l/4 of the wire then new length will be l'.

$l'=\frac{l}{4} + 25\%(\frac{l}{4} )$

$l'=\frac{l}{4} +0.25\times\frac{l}{4}$

$l' = (1.25)\frac{l}{4}$

Now the area of that part will be A'.

$lA = l'A'$

$\frac{l}{4}A=1.25(\frac{l}{4})A'$

$A'=\frac{A}{1.25}$

Now we know that the resistance of the wire is given by:

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

Total resistance of the wire :

$R_{eq} =\rho (\frac{3}{4A} l)+ \rho \frac{l'}{A'}$

Substitute the value of l' and A' in above equation:

$R_{eq}=\frac{3}{4} R+\rho\frac{l}{A}\times \frac{1}{4}(1.25)^2$

$R_{eq}=\frac{3}{4}R+\frac{R}{4}(1.25)^2$

$R_{eq}=1.140R$

The percentage % change in resistance of a wire can be calculated as:

$\frac{R_{eq}-R}{R}\times 100=\frac{1.140R-R}{R}\times 100= 14\%$

Therefore, the percentage change in resistance of wire will be  approximately 14%.

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