Explanation

(c) Here, 2$\Omega$, 3$\Omega$ and 6$\Omega$ are in parallel. So potential drop across them will be the same. As heat produced, $= \frac{V^{2}}{R}t$ $ie,\ H \propto \frac{1}{R}$, so maximum heat will be generated across 2$\Omega$ resistance. Similarly 4$\Omega$ and $5\Omega$ are also in parallel, so more heat will e generated across 4$\Omega$. Now the effective circuit will become
Total resistance, $= 1 + \frac{20}{9} = \frac{29}{9}\Omega$
Current, $I = \frac{V}{29/9} = \frac{9V}{29}A$
$V_{1} = \frac{9V}{29} \times 1 = \frac{9}{29}V$
and $V_{2} = \frac{9V}{29} \times \frac{20}{9} = \frac{20V}{29}$
Power spent across 2 $\ \Omega$,
$P_{1} = \frac{V_{1}^{2}}{2} = \frac{\left( \frac{9V}{29} \right)^{2}}{2} = \frac{40.5V^{2}}{(29)^{2}}$
Power spent across 4 $\Omega$,
$P_{2} = \frac{V_{2}^{2}}{4} = \frac{\left( \ \frac{20V}{29} \right)^{2}}{4} = \frac{50V^{2}}{(29)^{2}}$
$\therefore P_{2} > P_{1}$. Hence maximum heat is produced in 4$\Omega$ resistance.