A wire of resistance 160,Omega is melted and | vedclass

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(A) The volume of the wire remains constant during the process of melting and redrawing.$V_1 = V_2 \implies A_1 L_1 = A_2 L_2$Given $L_2 = \frac{L_1}{

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$10$

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(A) The volume of the wire remains constant during the process of melting and redrawing.$V_1 = V_2 \implies A_1 L_1 = A_2 L_2$Given $L_2 = \frac{L_1}{4}$,we have $A_1 L_1 = A_2 (\frac{L_1}{4})$,which simplifies to $A_2 = 4 A_1$.The resistance of a wire is given by $R = ho \frac{L}{A}$.Thus,the ratio of the new resistance $R_2$ to the original resistance $R_1$ is:$\frac{R_2}{R_1} = \frac{ho L_2 / A_2}{ho L_1 / A_1} = \frac{L_2}{L_1} 	imes \frac{A_1}{A_2}$Substituting the values: $\frac{R_2}{R_1} = (\frac{1}{4}) 	imes (\frac{A_1}{4 A_1}) = \frac{1}{16}$.Therefore,$R_2 = \frac{R_1}{16} = \frac{160}{16} = 10\,\Omega$.

$A$ wire of resistance $160\,\Omega$ is melted and drawn into a wire of one-fourth of its original length. The new resistance of the wire will be $......\Omega$.

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