Question

Two metallic rods $PQ$ and $QR$ 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 $QR$ is replaced by another rod $Q{R}^{\prime }$ of the same material and length $\left(QR=Q{R}^{\prime }\right)$. If the area of cross-section of $Q{R}^{\prime }$ is twice that of $QR$ 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 $H_{PQ}=H_{Q{R}^{\prime }}$ in steady state
$\therefore \frac{K_{PQ}{A}_{PQ}(100-T)}{L_{PQ}}$
$=\frac{K_{Q{R}^{\prime }}{A}_{Q{R}^{\prime }}(T-0)}{L_{Q{R}^{\prime }}}$
or $\frac{K_{PQ}}{K_{Q{R}^{\prime }}}=\frac{A_{Q{R}^{\prime }}{L}_{PQ}(T-0)}{A_{PQ}{L}_{Q{R}^{\prime }}(100-T)}\dots$(ii)
Divide (i) by (ii), we get
$\frac{K_{Q{R}^{\prime }}}{K_{QR}}=\frac{A_{QR}{L}_{Q{R}^{\prime }}(60-0)(100-T)}{A_{Q{R}^{\prime }}{L}_{QR}(100-60)(T-0)}$
Since both rods $QR$ and $Q{R}^{\prime }$ are made of same material
$\therefore K_{Q{R}^{\prime }}=K_{QR}$
Also $A_{Q{R}^{\prime }}=2A_{QR}$ and $L_{Q{R}^{\prime }}=L_{QR}$ (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
$8T=600-6T$ or $14T=600$
or $T=\frac{600}{14}\approx 43^{\circ }C$