If the wire is replaced by another wire of same material but with double the length and half the thickness, the resistance of the wire as a whole will change. Let us calculate this change.
Initial resistance, $\mathrm{R}=\rho \frac{\mathrm{L}}{\mathrm{A}}=\rho \frac{\mathrm{L}}{\pi \mathrm{r}^{2}}$
Fubak resistance, $\mathrm{R}^{\prime}=\rho \frac{2 \mathrm{~L}}{\mathrm{~A}^{\prime}}=\rho \frac{2 \mathrm{~L}}{\pi\left(\frac{\mathrm{r}}{2}\right)^{2}}$
$=\rho \frac{8 \mathrm{~L}}{\pi \mathrm{r}^{2}}=8\left(\rho \frac{\mathrm{L}}{\pi \mathrm{r}^{2}}\right)=8 \mathrm{R}$
Therefore, the resistance will increase by eight times.
In a meter-bridge, we have
$\begin{array}{l}
\frac{\mathrm{R}}{\mathrm{S}}=\frac{l}{100-l} \\
\text { where } \quad \mathrm{R}=\text { unknown resistor } \\
\quad \mathrm{S}=\text { known resistor } \\
l=\text { balancing length } \\
\therefore l=\frac{100 \mathrm{R}}{\mathrm{S}+\mathrm{R}}
\end{array}$
When the resistance of the new wire is $8 \mathrm{R}$ then the new balancing length will be
$\begin{aligned}
l &=\frac{100 \times 8 \mathrm{R}}{\mathrm{S}+8 \mathrm{R}}=\frac{100 \times 8 \mathrm{R}}{(\mathrm{S}+\mathrm{R})+7 \mathrm{R}} \\
&=\frac{100 \times 8 \mathrm{R}}{\left(\frac{100 \mathrm{R}}{l}\right)+7 \mathrm{R}} \quad \therefore l^{\prime}=\frac{800 l}{100+7 l}
\end{aligned}$
No option is correct.