A) \[\frac{M\alpha }{4\pi {{R}^{2}}\sigma }\ln \left( \frac{3}{2} \right)\]
B) \[\frac{M\alpha }{4\pi {{R}^{2}}\sigma }\ln \left( \frac{16}{3} \right)\]
C) \[\frac{M\alpha }{16\pi {{R}^{2}}\sigma }\ln \left( \frac{16}{3} \right)\]
D) \[\frac{M\alpha }{16\pi {{R}^{2}}\sigma }\ln \left( \frac{3}{2} \right)\]
Correct Answer: C
Solution :
In the given problem, fall in temperature of sphere, \[dT=(3{{T}_{0}}-2{{T}_{0}})={{T}_{0}}\] Temperature of surrounding, \[{{T}_{surr}}={{T}_{0}}\] Initial temperature of sphere, \[{{T}_{initial}}=3{{T}_{0}}\] Specific heat of the material of the sphere varies as, \[c=\alpha {{T}^{3}}\]per unit mass (\[\alpha =a\] constant) Applying formula, \[\frac{dT}{dt}=\frac{\sigma A}{McJ}\left( {{T}^{4}}-T_{surr}^{4} \right)\] \[\Rightarrow \]\[\frac{{{T}_{0}}}{dt}=\frac{\sigma 4\pi {{R}^{2}}}{M\alpha {{\left( 3{{T}_{0}} \right)}^{3}}J}\left[ {{\left( 3{{T}_{0}} \right)}^{4}}-{{\left( {{T}_{0}} \right)}^{4}} \right]\] \[\Rightarrow \]\[dt=\frac{M\alpha 27T_{0}^{4}J}{\sigma 4\pi {{R}^{2}}\times 80T_{0}^{4}}\]Solving we get, Time taken for the sphere to cool down temperature \[2{{T}_{0}}\],\[t=\frac{M\alpha }{16\pi {{R}^{2}}\sigma }\ln \left( \frac{16}{3} \right)\]
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