A given piece of wire of length l and radius | vedclass

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(C) The volume of the wire remains constant during stretching. Therefore,$V = A_1 l_1 = A_2 l_2$.Since $A = \pi r^2$,we have $\pi r^2 l_1 = \pi (r/2)^

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) The volume of the wire remains constant during stretching. Therefore,$V = A_1 l_1 = A_2 l_2$.Since $A = \pi r^2$,we have $\pi r^2 l_1 = \pi (r/2)^2 l_2$.This simplifies to $r^2 l_1 = \frac{r^2}{4} l_2$,which gives $l_2 = 4 l_1$.The resistance of a wire is given by $R = ho \frac{l}{A} = ho \frac{l}{\pi r^2}$.Thus,$R \propto \frac{l}{r^2}$.Let $R_1 = R$ and $R_2$ be the new resistance.$\frac{R_2}{R_1} = \frac{l_2}{l_1} 	imes (\frac{r_1}{r_2})^2$.Substituting the values: $\frac{R_2}{R} = (4) 	imes (\frac{r}{r/2})^2 = 4 	imes 2^2 = 4 	imes 4 = 16$.Therefore,the new resistance is $16R$.

$A$ given piece of wire of length $l$ and radius $r$ and resistance $R$ is stretched uniformly to a wire of radius $\frac{r}{2}$. Its new resistance is .................. $R$.

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