A metal wire has a resistance of 35 ,Omega. I | vedclass

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(B) Given,initial resistance $R_{1} = 35 \,\Omega$ and final length $l_{2} = 2l_{1}$.Since the volume of the wire remains constant during stretching,$

Vedclass · Accepted Answer

$140$

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(B) Given,initial resistance $R_{1} = 35 \,\Omega$ and final length $l_{2} = 2l_{1}$.Since the volume of the wire remains constant during stretching,$V_{1} = V_{2}$.$A_{1}l_{1} = A_{2}l_{2} \implies A_{2} = A_{1} \left(\frac{l_{1}}{l_{2}}ight) = A_{1} \left(\frac{l_{1}}{2l_{1}}ight) = \frac{A_{1}}{2}$.The resistance of a wire is given by $R = ho \frac{l}{A}$.Therefore,the new resistance $R_{2} = ho \frac{l_{2}}{A_{2}} = ho \frac{2l_{1}}{A_{1}/2} = 4 \left(ho \frac{l_{1}}{A_{1}}ight) = 4R_{1}$.Substituting the value of $R_{1}$,we get $R_{2} = 4 	imes 35 = 140 \,\Omega$.

$A$ metal wire has a resistance of $35 \,\Omega$. If its length is increased to double by drawing it,then its new resistance will be (in $\Omega$)

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