The resistance of a 5, cm long wire is 10, Om | vedclass

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(A) When a wire is stretched uniformly,its volume remains constant. Since $V = A 	imes l$,we have $A_1 l_1 = A_2 l_2$. Given $l_1 = 5\, cm$ and $l_2

Vedclass · Accepted Answer

$160$

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(A) When a wire is stretched uniformly,its volume remains constant. Since $V = A 	imes l$,we have $A_1 l_1 = A_2 l_2$. Given $l_1 = 5\, cm$ and $l_2 = 20\, cm$,the ratio of lengths is $\frac{l_2}{l_1} = \frac{20}{5} = 4$. The resistance $R$ is given by $R = ho \frac{l}{A}$. Since $A = \frac{V}{l}$,we get $R = ho \frac{l^2}{V}$. Since $ho$ and $V$ are constant,$R \propto l^2$. Therefore,$\frac{R_2}{R_1} = \left( \frac{l_2}{l_1} ight)^2 = (4)^2 = 16$. Given $R_1 = 10\, \Omega$,we find $R_2 = 16 	imes 10 = 160\, \Omega$.

The resistance of a $5\, cm$ long wire is $10\, \Omega$. It is uniformly stretched so that its length becomes $20\, cm$. The resistance of the wire is ............. $\Omega$.

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