1.

A cylindrical rod is having temperature T_(1) and T_(2) ate its ends. The rate of flow of heat is Q_(1) cal/sec. If all the linear dimensions are doubled keeping temperatures constant, then rate of flow of heat Q_(2) will be :

Answer»

`4Q_(1)`
`Q_(1)//4`
`2Q_(1)`
`Q_(1)//2`.

Solution :`(dQ)/(dt)="k A"((T_(1)-T_(2))t)/(d)`
When linear DIAMENSION are doubled then A becomes 4 times and d becomes 2 times.
`:.(dQ)/(dt)` becomes 2 times
Correct choice is (c ).


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