GATE 2024 IN Q.36: solution and answer

Official answer: ASource: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. The current flows along the axis, so $\mathbf{J}=k{s}^2{\hat{a}}_z$ is normal to the circular cross-section. The total current is the flux of $\mathbf{J}$ through that cross-section: $I={\int }_S\mathbf{J}\cdot d\mathbf{S}$.

2. The integrand depends on $s$ alone, so split the disc into thin rings of radius $s$ and width $ds$. Each ring has area $dS=2\pi sds$ (in cylindrical coordinates, $dS=sdsd\varphi$ integrated over $\varphi$ from $0$ to $2\pi$).

3. $$I={\int }_0^{2\pi }{\int }_0^{a}k{s}^{2}sdsd\varphi =2\pi k{\int }_0^a{s}^{3}ds$$
4. $$I=2\pi k\frac{a^4}{4}=\frac{\pi k{a}^4}{2}$$

5. Check the powers: $J\propto s^2$ and the area element brings one more $s$, so $s^3$ integrates to $a^4$. That rules out options (B) and (C), which carry $a^3$. Option (D) is half the right value. The answer is (A).