A) (3/16)H
B) (16/3)H
C) (9/27)H
D) (1/16)H
Correct Answer: A
Solution :
Rate of cooling \[\propto ({{T}^{4}}-T_{0}^{4})\] Þ \[\frac{H}{H'}=\frac{(T_{1}^{4}-T_{0}^{4})}{(T_{2}^{4}-T_{0}^{4})}=\]\[\frac{{{400}^{4}}-{{200}^{4}}}{{{600}^{4}}-{{200}^{4}}}\] or \[H'=\frac{(16+4)(16-4)H}{(36+4)(36-4)}\]\[=\frac{3}{16}H\]
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