Q.
The identical square rods of metal are welded end to end as shown in figure, $Q$ cal of heat flow through this combination in 4 min. If the rods were welded as shown in figure, the same amount of heat will flow through the combination in
A) 16 min
B) 12 min
C) 1 min
D) 4 min
Explanation
(c) From $\frac{\mathrm{\Delta}Q}{\mathrm{\Delta}t} = KA\left( \frac{\mathrm{\Delta}T}{\mathrm{\Delta}x} \right)$
$\mathrm{\Delta}t = \ \frac{\mathrm{\Delta}Q\mathrm{\Delta}x}{KA(\mathrm{\Delta}T)}$
In arrangement (b), $A$ is doubled and $\mathrm{\Delta}x\ $is halved.
$\therefore\ \mathrm{\Delta}t\ \rightarrow \ \frac{1/2}{2} \rightarrow \frac{1}{4}$ time
$ie\ ,\frac{1}{4} \times 4$ min= 1 min
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.