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(C) When a wire is stretched,its volume remains constant. Since $Volume = Area 	imes Length$,if the length becomes $n$ times,the area becomes $1/n$ t

Vedclass · Accepted Answer

$3:1$

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(C) When a wire is stretched,its volume remains constant. Since $Volume = Area 	imes Length$,if the length becomes $n$ times,the area becomes $1/n$ times.Resistance $R = ho \frac{l}{A}$. Therefore,$R \propto l^2$.Given that the length becomes $2$ times $(l_2 = 2l_1)$,the new resistance $R_2$ is:$R_2 = R_1 	imes (2)^2 = 4R_1$.The change in resistance is $\Delta R = R_2 - R_1 = 4R_1 - R_1 = 3R_1$.The ratio of the change in resistance to the initial resistance is $\frac{\Delta R}{R_1} = \frac{3R_1}{R_1} = 3:1$.

$A$ thick wire is stretched so that its length becomes two times. Assuming that there is no change in its density,what is the ratio of the change in resistance of the wire to the initial resistance of the wire?

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