The temperature difference across two cylindrical rods $A$ and $B$ of same material and same mass are…
The temperature difference across two cylindrical rods $A$ and $B$ of same material and same mass are $40^{\circ} \mathrm{C}$ and $60^{\circ} \mathrm{C}$ respectively. In steady state, if the rates of flow of heat through the rods A and B are in the ratio $3: 8$, the ratio of the lengths of the rods $A$ and $B$ is
$1: 3$
$5: 3$
$4: 3$
$2: 3$
Solution
For $\operatorname{rod} \mathrm{A}, \Delta \mathrm{T}_1=40^{\circ} \mathrm{C}$
For rod $B, \Delta T_2=60^{\circ} \mathrm{C}$
$\therefore \frac{\mathrm{Q}_{\mathrm{A}}}{\mathrm{Q}_{\mathrm{B}}}=\frac{3}{8} \Rightarrow \frac{\mathrm{~A}_1 \frac{\Delta \mathrm{~T}_1}{l_1}}{\mathrm{~A}_2 \frac{\Delta \mathrm{~T}_2}{l_2}}=\frac{3}{8}$ ....(i)
Also, $\mathrm{m}_1=\mathrm{m}_2 \Rightarrow \delta_1^{\mathrm{V}_1=\delta_2} \Rightarrow \mathrm{~V}_1=\mathrm{V}_2$
$\Rightarrow \mathrm{A}_1 l_1=\mathrm{A}_2 l_2 \frac{\mathrm{~A}_1}{\mathrm{~A}_2}=\frac{l_2}{l_1}$ .....(ii)
From eq(i) and (ii), we get
$\left(\frac{\Delta \mathrm{T}_1}{\Delta \mathrm{~T}_2}\right)\left(\frac{l_2}{l_1}\right)^2=\frac{3}{8} \Rightarrow\left(\frac{40}{60}\right)\left(\frac{l_2}{l_1}\right)^2=\frac{3}{8}$
$\therefore \quad l_1: l_2=4: 3$