First slide

Conduction

Question

Two rods (one semi-circular and other straight) of same material and of same cross-sectional area are joined as shown in fig. The point A and B are maintained at different temperatures. The ratio of the heat transferred through a cross-section of a semicircular rod to the heat transferred through a cross-section of the straight rod in a given time is

Easy

A

$2 : π$
B

1 : 2
C

$π : 2$
D

3 : 2

Solution

We know that $\frac{dQ}{dt} = \frac{KA (T_{1} - T_{2})}{d}$
$∴ \frac{dQ}{dt} \propto \frac{1}{d}$
So, $\frac{(dQ / dt)_{1}}{(dQ / dt)_{2}} = \frac{1}{(πr / 2 r)} = \frac{2 r}{πr} = \frac{2}{π}$

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