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Rate of heat flow through a cylindrical ...

Rate of heat flow through a cylindrical rod is `H_(1)`. Temperatures of ends of rod are `T_(1)` and `T_(2)`. If all the dimensions of rod become double and temperature difference remains same and rate of heat flow becomes `H_(2)`. Then `(H_(1))/(H_(2))` is `0.x`. Find value of x.

A

`H_(2) = 2H_(1)`

B

`H_(2) = H_(1)`

C

`H_(2) = H_(1)/4`

D

`H_(2) = 4H_(1)`

Text Solution

Verified by Experts

The correct Answer is:
A
Rate of heat flow, `H = Q/t = (KAT_(1)-T_(2))/l`
or `H alphaA/l` or `H_(2)=2H_(1)`
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