A) The density of the metal
B) The temperature gradient perpendicular to the area
C) The temperature to which the metal is heated
D) The area of the metal plate
Correct Answer: B
Solution :
\[\frac{dQ}{dt}=KA\frac{d\theta }{dl}\]\[\Rightarrow \] \[\frac{dQ}{dt}\propto \frac{d\theta }{dl}\] (Temperature gradient)
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