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