First slide

Heat transfer

Question

The lengths and radii of two rods made of same materials are in the ratio of 1:2 and 2:3 respectively, if the temperature difference between the ends for the two rods be the same, then in steady state, the amount of heat flowing through them will be in the ratio

Moderate

A

1:3
B

4:3
C

8:9
D

3:2

Solution

given $\frac{l_{1}}{l_{2}} = \frac{1}{2}$
K=constant= coefficient of thermal conductivity
$\frac{r_{1}}{r_{2}} = \frac{2}{3}$ ( here r is radius of the cross section)
$Δθ$ =constant= temperature difference
$\frac{Q_{1}}{Q_{2}} =?$
$P o w e r P = \frac{Q}{t} = \frac{{KA(θ}_{1} {-θ}_{2})}{l} = \frac{K A ∆ θ}{l} = \frac{K π r^{2} ∆ θ}{l}$
$\frac{Q}{t} α \frac{r^{2}}{l}$
$Pα \frac{r^{2}}{l}$
$\frac{P_{1}}{P_{2}} = {(\frac{r_{1}}{r_{2}})}^{2} \times \frac{l_{2}}{l_{1}}$
$\frac{P_{1}}{P_{2}} = {(\frac{2}{3})}^{2} \times \frac{2}{1}$
$\frac{P_{1}}{P_{2}} = \frac{4}{9} \times \frac{2}{1} = \frac{8}{9} \frac{P_{1}}{P_{2}} = \frac{8}{9} i s r a t i o o f a m o u n t o f h e a t f l o w p e r s e c o n d$

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