In the given figure, a battery of emf E is connected across a conductor PQ of length l

Current is constant in conductor

i = constant

Resistance of element $\mathrm{dR}=\frac{\rho \mathrm{dx}}{\pi \mathrm{r}^{2}}$

$\mathrm{dV}=\mathrm{idR}=\frac{i \rho \mathrm{dx}}{\pi \mathrm{r}^{2}}$

$\mathrm{E}=\frac{\mathrm{dV}}{\mathrm{dx}}=\frac{\mathrm{i \rho}}{\pi \mathrm{r}^{2}}$

$\& V_{d}=\frac{e E \tau}{m}$

$\therefore V_{d \propto} E$

$\rightarrow \quad \mathrm{E} \propto \frac{1}{\mathrm{r}^{2}}$

if $\mathrm{r}$ decreases, $\mathrm{E}$ will increase $\therefore \mathrm{V}_{\mathrm{d}}$ will increase