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A cylindrical rod having temperature `T_(1)` and `T_(2)` at its ends. The rate of flow of heat is `Q_(1) cal//sec`. If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat `Q_(2)` will beA. `4Q_1`B. `2Q_2`C. `Q_1/4`D. `Q_1/2` |
Answer» Correct Answer - B Heat flow rate =`(KA(T_1-T_2))/L=Q` when linear dimensions are doubled `A_1 prop r_1^2 " " L_1=L` `A_2 prop 4r_1^2 " " L_1=2L_1` so `Q_2=2Q_1` |
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