MHTCET 11 Apr 2026 Shift 1 Paper

Concept:In steady state, the heat current through each rod in series is the same because no heat is lost from the insulated sides.Explanation:Let each rod have length $L$ and area of cross-section $A$.For rods in series, heat current $I$ is equal in all three rods:$$I = \frac{2KA(3T - T_1)}{L} = \frac{KA(T_1 - T_2)}{L} = \frac{2KA(T_2 - T)}{L}$$Cancelling the common factor $\frac{KA}{L}$, we get:$$2(3T - T_1) = T_1 - T_2 = 2(T_2 - T)$$Equating the first and third parts:$$2(3T - T_1) = 2(T_2 - T)$$$$3T - T_1 = T_2 - T$$$$T_1 + T_2 = 4T \quad \text{(1)}$$Equating the middle and third parts:$$T_1 - T_2 = 2(T_2 - T)$$$$T_1 - 3T_2 = -2T \quad \text{(2)}$$Subtract equation (2) from equation (1):$$(T_1 + T_2) - (T_1 - 3T_2) = 4T - (-2T)$$$$4T_2 = 6T$$$$T_2 = \frac{3T}{2}$$Substitute $T_2 = \frac{3T}{2}$ into equation (1):$$T_1 + \frac{3T}{2} = 4T$$$$T_1 = \frac{5T}{2}$$Therefore,$$\frac{T_1}{T_2} = \frac{\frac{5T}{2}}{\frac{3T}{2}} = \frac{5}{3}$$Answer:$$\frac{T_1}{T_2} = \frac{5}{3}$$So, the correct option is C.