A) rate of cooling is same in both
B) rate of cooling of A is four times that of B
C) rate of cooling of A is twice that of B
D) rate of cooling of A is \[\frac{1}{4}\] times that of B
Correct Answer: C
Solution :
Rate of cooling \[{{R}_{c}}=\frac{A\varepsilon \sigma ({{T}^{4}}-T_{0}^{4})}{mc}\] \[=\frac{A\varepsilon \sigma ({{T}^{4}}-T_{0}^{4})}{V\rho c}\] \[\Rightarrow \] \[{{R}_{c}}=\frac{A}{V}\propto \frac{1}{r}\propto \frac{1}{(Diameter)}\](\[\because \]\[m=\rho V\]) Since, diameter of A is half that of B, so its rate of cooling will be doubled that of B.
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