Question:
In the given figure, a battery of emf $E$ is connected across a conductor PQ of length $l '$ and different area of cross-sections having radii $r_{1}$ and $r_{2}\left(r_{2} Choose the correct option as one moves from P to Q :
Correct Option: 1
Solution:
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