Explanation

(b) Given, that, the two resistances $R\ and\ 2R$ are connected in parallel so the potential drop is equal for both.
The thermal energy developed in a circuit is
$\ \ \ \ \ \ \ H = \frac{V^{2}t}{R}\ \ \Rightarrow \ H \propto \frac{1}{R}$
$\Rightarrow \frac{H_{1}}{H_{2}} = \frac{R_{2}}{R_{1}} \Rightarrow \frac{H_{1}}{H_{2}} = \frac{2R}{R}$
$\left\lbrack Here,\ R_{1} = R,\ R_{2} = 2R \right\rbrack$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ H_{1}:H_{2} = 2\ :1$