Question

Three rods made of the same material and having the same cross-section,same length have been joined as shown in the figure. The extreme left and right ends are kept at $0^{\circ }C$ and ${90}^{\circ }C$ respectively. The temperature of the junction of the three rods will be

• A. ${45}^{\circ }C$
• B. ${60}^{\circ }C$
• C. ${30}^{\circ }C$
• D. ${20}^{\circ }C$

Answer 1

The correct option is B ${60}^{\circ }C$

Let ${\theta }^{\circ }$$C$ be the temperature at $B$.
$Q$ be the heat flowing per second from $A$ to $B$ on account of temperature difference by conductivity.

$\therefore Q=\frac{KA(90-\theta )}{l}$ .....(i)

Where, $k$ = thermal conductivity of the rod,
$A$ = Area of cross section of the rod,
$l$ = length of the rod.

By symmetry, heat flow from $C$ to $B$ will also be same as $Q$

$\therefore$ The heat flowing per second from $B$ to $D$ will be

$2Q=\frac{KA(\theta -0)}{l}$ .....(ii)

Dividing Eq. (ii) by Eq. (i)

$2=\frac{\theta }{90-\theta }\Rightarrow \theta ={60}^{\circ }C$