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Q.
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
Let the original diameter of the wirc 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
$L \times \frac{\pi D^{2}}{4}=L' \times \frac{\pi D^{2}}{16}$
$ \Rightarrow L'=\frac{16}{4} L=4 \,L $
Now, we know that the resistance is given by
$R=\rho \frac{L}{A}$.
$\therefore R'=\rho \frac{L'}{A'}=\rho \frac{4 L}{A / 4}=16 R$
${\left[\because A'=\frac{\pi D^{2}}{16}=\frac{A}{4}\right]} $
$\therefore R'=16 \times 10=160 \,\Omega$