1.

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

Answer»

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

Solution :`(DQ)/(dt)=(KA(T_(1)-T_(2)))/(l)=(Kpir^(2)(T_(1)-T_(2)))/(l)=Q_(1)`
When all dimensions are DOUBLED then `(dQ)/(dt)` BECOMES two TIMES i.e. `2Q_(1)`
Correct choice is (c ).


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