Question

Two rods of same material have same length and area. The heat $\Delta Q$ flows through them for $12$ minutes when they are joint side by side. If now both the rods are joined in parallel, then the same amount of heat $\triangle O$ will flow in :

• A. 24 min
• B. 3 min
• C. 12 min
• D. 6 min

Answer 1

Answer: (B)

3 min

Answer 2

When rods are joined in series then heat flow

$\Delta Q_{1}=\frac{A\left(T_{1}-T_{2}\right) t_{1}}{\frac{l_{1}}{K_{1}}+\frac{l_{1}}{K_{2}}}$
$=\frac{A\left(T_{1}-T_{2}\right) t_{1}}{\frac{l}{K_{1}}+\frac{l}{K_{2}}}=\frac{A\left(T_{1}-T_{2}\right) t_{1}}{l} \frac{K}{2}$
When rods are joined in parallel
$\Delta Q_{2} =\left(K_{1} A+K_{2} A\right) \frac{\left(T_{1}-T_{2}\right) t_{2}}{l}$
$=\frac{2 K A\left(T_{1}-T_{2}\right) t_{2}}{l}$
$\because Q_{1} = \Delta Q_{2}$ (given)

$\therefore t_{2}=\frac{t_{1}}{4}=\frac{12}{4}=3\, \min$