The new resistance of a wire of R , Omega,who | vedclass

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(A) The resistance of a wire is given by $R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$.Since the volume of the wire remains constant during stretchi

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$16 \, R$

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(A) The resistance of a wire is given by $R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$.Since the volume of the wire remains constant during stretching,$V = A_1 L_1 = A_2 L_2$,which implies $L_1 r_1^2 = L_2 r_2^2$.Therefore,$\frac{L_2}{L_1} = \left( \frac{r_1}{r_2} \right)^2$.The ratio of the new resistance $R_2$ to the original resistance $R_1$ is $\frac{R_2}{R_1} = \frac{\rho L_2 / A_2}{\rho L_1 / A_1} = \left( \frac{L_2}{L_1} \right) \left( \frac{A_1}{A_2} \right) = \left( \frac{r_1}{r_2} \right)^2 \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{r_1}{r_2} \right)^4$.Given that the radius is reduced to half,$r_2 = \frac{r_1}{2}$,so $\frac{r_1}{r_2} = 2$.Substituting this into the formula,$\frac{R_2}{R} = (2)^4 = 16$.Thus,the new resistance is $R_2 = 16 \, R$.

The new resistance of a wire of $R \, \Omega$,whose radius is reduced to half,is

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