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Consider two rods of same length and different specific heats (`S_1` and `S_2`), conductivities `K_1` and `K_2` and area of cross section (`A_1` and `A_2`) and both having temperature `T_1` and `T_2` at their ends. If the rate of heat loss due to conduction is equal thenA. `K_1A_1=K_2A_2`B. `K_2A_1=K_1A_2`C. `(K_1A_1)/(S_1)=(K_2A_2)/(S_2)`D. `(K_2A_1)/(S_2)=(K_1A_2)/(S_1)`

Answer» Correct Answer - A
According to problem, rate of heat loss in both rods is
equal, i.e.,`((dQ)/(dt))_(1)=((dQ)/(dt))_(2)`
`implies(K_1A_1Deltatheta_1)/(l_1)=(K_2A_2Deltatheta_2)/(l_2)`
`K_1A_1=K_2A_2[AsDeltatheta_1=Deltatheta_2=(T_1-T_2)` and `l_1=l_2` given]


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