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$(A)$ Let the initial length be $l_1 = l$. Since the volume of the wire remains constant, $V = A_1 l_1 = A_2 l_2$.Given that the length is increased b

Vedclass · Accepted Answer

$300$

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$(A)$ Let the initial length be $l_1 = l$. Since the volume of the wire remains constant, $V = A_1 l_1 = A_2 l_2$.Given that the length is increased by $100 \%$, the new length $l_2 = l_1 + 1.00 l_1 = 2l_1$.From the volume conservation, $A_1 l_1 = A_2 (2l_1) \Rightarrow A_2 = \frac{A_1}{2}$.The resistance of a wire is given by $R = ho \frac{l}{A}$.Thus, the initial resistance $R_1 = ho \frac{l_1}{A_1}$ and the final resistance $R_2 = ho \frac{l_2}{A_2} = ho \frac{2l_1}{A_1/2} = 4 ho \frac{l_1}{A_1} = 4R_1$.The percentage change in resistance is $\frac{\Delta R}{R_1} 	imes 100 = \frac{R_2 - R_1}{R_1} 	imes 100 = \frac{4R_1 - R_1}{R_1} 	imes 100 = 300 \%$.

The length of a given cylindrical wire is increased by $100 \%$. Due to the consequent decrease in diameter, the change in the resistance of the wire will be .................. $\%$

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