Question

A metallic conductor of irregular cross-section is shown in the figure. A constant potential difference is applied across the ends $\left(1\right)$ and $\left(2\right).$ Then

• A. the current at cross-section $P$ equals the currrent at the cross-section $Q$.
• B. the electric field intensity at $P$ is less than that at $Q$.
• C. the number of electrons crossing per unit area of cross-section at $P$ is less than that at $Q$ .
• D. the rate of heat generated per unit time at $Q$ is greater than that at $P$.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A the current at cross-section $P$ equals the currrent at the cross-section $Q$.
B the electric field intensity at $P$ is less than that at $Q$.
C the number of electrons crossing per unit area of cross-section at $P$ is less than that at $Q$ .
D the rate of heat generated per unit time at $Q$ is greater than that at $P$.


We know that current is constant across cross-section. Thus, $I$ is the same at both $P$ and $Q$. Option A is correct.

Also, $R=\frac{\rho l}{A}$

Crosssectional area at $P$ is greater that crosssectional area at $Q$. So, option C is also correct.

The electric field intensity $E_1$ can be found by
$E_1=\frac{dV}{dx}\Rightarrow \frac{iR}{dx}=\frac{i\rho dx}{A\cdot dx}$

(where $dV$ is the potential difference across small section $dx$)

$\therefore E_1=\frac{i\rho }{A}\Rightarrow E_1\propto \frac{1}{A}$

Option B is correct.

Rate of heat generated =
$P=i^{2}R=i^2\frac{\rho dx}{A}$

$\Rightarrow P\propto \frac{1}{A}$. Option D is correct.