Explanation:
B Given,
Same length of two identical wires.
length of wire $=l$
Let the resistance be $\mathrm{R}$ and current be I for each wire.
If the wire are first connected in series the equivalent resistance is, $\mathrm{R}_{1}=\mathrm{R}+\mathrm{R}$
$\mathrm{R}_{1}=2 \mathrm{R}$
If connected in parallel, the equivalent resistance is,
$\frac{1}{\mathrm{R}_{2}}=\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}$
$\mathrm{R}_{2}=\frac{\mathrm{R}}{2}$
The ratio of heat generated in two cases,
$\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{I}^{2} \mathrm{R}_{1} \mathrm{t}}{\mathrm{I}^{2} \mathrm{R}_{2} \mathrm{t}}$
$\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}$
$\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{2 \mathrm{R}}{\frac{\mathrm{R}}{2}}$
$\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{4}{1}$
$\therefore \quad \mathrm{H}_{1}: \mathrm{H}_{2}=4: 1$