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  • A wire of resistance is melted and drawn in a wire of one-fourth of its length.

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

Option: 1

640 \, \, \Omega


Option: 2

40 \, \, \Omega


Option: 3

10 \, \, \Omega


Option: 4

16 \, \, \Omega


Answers (1)

best_answer

Voulme remain same

\begin{aligned} & \mathrm{A} \ell=\mathrm{A}^{\prime} \ell^{\prime} \\ & \mathrm{A}^{\prime}=\frac{\mathrm{A} \ell}{\ell^{\prime}}=\frac{\mathrm{A} \ell}{\ell / 4} \\ & \mathrm{~A}^{\prime}=4 \mathrm{~A} \\ & L^{\prime}=\frac{L}{4} \\ & \frac{R^{\prime}}{R}=\frac{\rho \frac{L^{\prime}}{A^{\prime}}}{\rho \frac{L}{A}} \\ & R^{\prime}=R\left[\frac{L^{\prime}}{\mathrm{A}^{\prime}} \times \frac{A}{L}\right] \\ & R^{\prime}=160\left[\frac{\mathrm{L}}{4 \mathrm{~L}} \times \frac{\mathrm{A}}{4 \mathrm{~A}}\right] \\ & \mathrm{R}^{\prime}=10 \Omega \\ & \end{aligned}

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Devendra Khairwa

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