Two identical square rods of metal are welded end to end as shown in figure $(a)$. Assume that $10\, cal$ of heat flows through the rods in $2\, min$. Now the rods are welded as shown in figure, $(b)$. The time it would take for $10$ cal to flow through the rods now, is ........ $\min$
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$\frac{\mathrm{Q}_{1}}{\mathrm{t}_{1}}=\mathrm{i}_{\mathrm{H}_{1}}=\frac{100-0}{2 \mathrm{R}}=\frac{50}{\mathrm{R}}$
$\mathrm{i}_{\mathrm{H}_{2}}=\frac{100}{\mathrm{R} / 2}=\frac{200}{\mathrm{R}}=\frac{\mathrm{Q}_{2}}{\mathrm{t}_{2}}$
$\mathrm{Q}_{1}=\mathrm{Q}_{2}=10 \mathrm{cal}$
$\frac{50}{\mathrm{R}} \times(2)=\frac{200}{\mathrm{R}} \times \mathrm{t}_{2}$
$\mathrm{t}_{2}=\frac{1}{2} \mathrm{min}$
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