A meter bridge is used to determine the resistance of an unknown wire by measuring the balance point length l. If the wire is replaced by another wire of same material but with double the length and half the thickness, the balancing point is expected to be

In a meter bridge the ratio of two resistance is $\frac{R}{R^{\text{'}}}=\frac{l}{l^{\text{'}}}$ (where l and l' are balancing lengths)
Resistance $R=\frac{\rho l}{A}=\frac{\rho l}{\pi r^2}$
If material remains same $\rho ={\rho }^{\text{'}}$
Given $l^{\text{'}}=2l r^{\text{'}}=\frac{r}{2}$
$\therefore R^{\text{'}}=\frac{\rho l^{\text{'}}}{A^{\text{'}}}=\frac{\rho 21}{\pi {\left(\frac{r}{2}\right)}^2}=\frac{8\rho l}{\pi r^2}=R^{\text{'}}=8R$
Therefore the new balancing point is expected to be 8l. Hence the correct answer is option (c).