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A wire of length $L$ is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were $10 \Omega$, its new resistance would be
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Verified Answer
The correct answer is:
$160 \Omega$.
Let the original diameter of the wire be $D$. Therefore the new diameter is $D / 2$.
Original area of cross-section is $\frac{\pi D^2}{4}$ and the final area of cross-section is $\frac{\pi D^2}{16}$. The new length of the wire is given by
$$
L \times \frac{\pi D^2}{4}=L^{\prime} \times \frac{\pi D^2}{16} \Rightarrow L^{\prime}=\frac{16}{4} L=4 L
$$
Now, we know that the resistance is given by $R=\rho \frac{L}{A}$.
$$
\begin{aligned}
\therefore \quad & R^{\prime}=\rho \frac{L^{\prime}}{A^{\prime}}=\rho \frac{4 L}{A / 4}=16 R . \\
& {\left[\because A^{\prime}=\frac{\pi D^2}{16}=\frac{A}{4}\right] } \\
\therefore \quad & R^{\prime}=16 \times 10=160 \Omega .
\end{aligned}
$$
Original area of cross-section is $\frac{\pi D^2}{4}$ and the final area of cross-section is $\frac{\pi D^2}{16}$. The new length of the wire is given by
$$
L \times \frac{\pi D^2}{4}=L^{\prime} \times \frac{\pi D^2}{16} \Rightarrow L^{\prime}=\frac{16}{4} L=4 L
$$
Now, we know that the resistance is given by $R=\rho \frac{L}{A}$.
$$
\begin{aligned}
\therefore \quad & R^{\prime}=\rho \frac{L^{\prime}}{A^{\prime}}=\rho \frac{4 L}{A / 4}=16 R . \\
& {\left[\because A^{\prime}=\frac{\pi D^2}{16}=\frac{A}{4}\right] } \\
\therefore \quad & R^{\prime}=16 \times 10=160 \Omega .
\end{aligned}
$$
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