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(B) Let the initial length of the wire be $l$. After stretching,the new length $l^{\prime}$ is given by: $l^{\prime} = l + 10\% 	ext{ of } l = l + 0.

Vedclass · Accepted Answer

$1.21$ times,same

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(B) Let the initial length of the wire be $l$. After stretching,the new length $l^{\prime}$ is given by: $l^{\prime} = l + 10\% 	ext{ of } l = l + 0.1l = 1.1l$. Since the volume of the wire remains constant,$A \cdot l = A^{\prime} \cdot l^{\prime}$,which implies $A^{\prime} = A / 1.1$. The new resistance $R^{\prime}$ is given by $R^{\prime} = ho \frac{l^{\prime}}{A^{\prime}} = ho \frac{1.1l}{A/1.1} = (1.1)^2 ho \frac{l}{A} = 1.21R$. Thus,the resistance becomes $1.21$ times the original resistance. Specific resistance (resistivity) $ho$ is an intrinsic property of the material and depends only on the nature of the material and temperature,not on the dimensions of the wire. Therefore,the specific resistance remains the same.

$A$ wire of a certain material is stretched slowly by $10 \%$. Its new resistance and specific resistance become respectively:

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