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(B) Let the temperature of the junction be $	heta$.Since the rods are made of the same material and have the same length and cross-section,they have

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$60$

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(B) Let the temperature of the junction be $	heta$.Since the rods are made of the same material and have the same length and cross-section,they have the same thermal resistance $R$.According to the principle of conservation of energy (steady state),the heat flowing into the junction must equal the heat flowing out of the junction.Let the heat flow from the $90^{\circ}C$ ends towards the junction and then towards the $0^{\circ}C$ end.$Q_{in} = Q_{out}$$\frac{90 - 	heta}{R} + \frac{90 - 	heta}{R} = \frac{	heta - 0}{R}$$2(90 - 	heta) = 	heta$$180 - 2	heta = 	heta$$3	heta = 180$$	heta = 60^{\circ}C$.

Three rods made of the same material and having the same cross-section are joined as shown in the figure. Each rod is of the same length. The 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 ...... $^{\circ}C$.

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