(D) $1$. Since the conductor is in a steady state,the current $I$ flowing through any cross-section must be the same. Therefore,$I_P = I_Q$.$2$. Curre

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(D) $1$. Since the conductor is in a steady state,the current $I$ flowing through any cross-section must be the same. Therefore,$I_P = I_Q$.$2$. Current density $J = I/A$. Since the area $A_P > A_Q$,the current density $J_P $3$. The rate of heat generated per unit time is power $H = I^2 R$. For a small segment of length $dx$,$dR = \rho \frac{dx}{A}$. Since $A_P > A_Q$,the resistance $R_Q > R_P$. Therefore,$H_Q > H_P$.$4$. All statements are correct.

Class 12 Physics Current Electricity Mix Examples-Current Electricity

Difficult

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

A
the current at the cross-section $P$ equals the current at the cross-section $Q$
B
the electric field intensity at $P$ is less than that at $Q$
C
the rate of heat generated per unit time at $Q$ is greater than that at $P$
D
All of the above