Question

Two metallic rods $P Q$ and $Q R$ of different materials are joined together at the junction $Q$ as shown in figure. It is observed that if the ends $P$ and $R$ are kept at $100^{\circ} C$ and $0^{\circ} C$ respectively, the temperature of the junction $Q$ is $60^{\circ} C$. There is no loss of heat to the surroundings. The rod $Q R$ is replaced by another rod $Q R^{\prime}$ of the same material and length $\left(Q R=Q R^{\prime}\right)$. If the area of cross-section of $Q R^{\prime}$ is twice that of $Q R$ and the ends $P$ and $R^{\prime}$ are maintained at $100^{\circ} C$ and $0^{\circ} C$ respectively, the temperature of the junction $Q$ will be nearly

• A. $29^{\circ} C$
• B. $33^{\circ} C$
• C. $60^{\circ} C$
• D. $43^{\circ} C$

Answer 1

Answer: (D)

In the first case,

Since the rods are connected in series, therefore rate of heat
i.e., heat current is same.
$\therefore H_{PQ} = H_{QR}$
$\therefore \frac{K_{PQ} A_{PQ}(100 - 60)}{L_{PQ}}$
$ = \frac{K_{QR}A_{QR}(60 - 0}{L_{QR}}$
or $\frac{K_{PQ}}{K_{QR}} = \frac{A_{QR}L_{PQ}(60 - 0)}{A_{PQ}L_{QR}(100 - 60)} \dots$(i)
In second case,

Let $T$ be the temperature of the junction.
As $H_{P Q}=H_{Q R'}$ in steady state
$\therefore \frac{K_{P Q} A_{P Q}(100-T)}{L_{P Q}}$
$=\frac{K_{Q R'} A_{Q R'}(T-0)}{L_{Q R'}} $
or $ \frac{K_{P Q}}{K_{Q R'}}=\frac{A_{Q R'} L_{P Q}(T-0)}{A_{P Q} L_{Q R'}(100-T)}\dots$(ii)
Divide (i) by (ii), we get
$\frac{K_{Q R'}}{K_{Q R}}=\frac{A_{Q R} L_{Q R'}(60-0)(100-T)}{A_{Q R'} L_{Q R}(100-60)(T-0)}$
Since both rods $Q R$ and $Q R'$ are made of same material
$\therefore K_{Q R'}=K_{Q R}$
Also $ A_{Q R'}=2 A_{Q R}$ and $ L_{Q R'}=L_{Q R} $ (Given)
Substituting these values in eqn (iii), we get
$2(100-60)(T-0)=(60-0)(100-T)$
or $80(T)=60(100-T)$ or
$8\, T=600-6 \,T$ or $14\, T=600$
or $T=\frac{600}{14} \approx 43^{\circ} C$