Two solid spheres A and B made of the same ma | vedclass

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(B) According to the Stefan-Boltzmann law,the rate of heat loss is given by $\frac{dQ}{dt} = e \sigma A (T^4 - T_0^4)$.Since $Q = mc\Delta T = (\rho V

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$r_B / r_A$

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(B) According to the Stefan-Boltzmann law,the rate of heat loss is given by $\frac{dQ}{dt} = e \sigma A (T^4 - T_0^4)$.Since $Q = mc\Delta T = (\rho V c) \Delta T$,the rate of change of temperature is $\frac{dT}{dt} = \frac{1}{mc} \frac{dQ}{dt}$.For a sphere,$V = \frac{4}{3} \pi r^3$ and $A = 4 \pi r^2$.Substituting these,we get $\frac{dT}{dt} = \frac{\sigma (4 \pi r^2) (T^4 - T_0^4)}{\rho (\frac{4}{3} \pi r^3) c} = \frac{3 \sigma (T^4 - T_0^4)}{\rho r c}$.Thus,the rate of change of temperature is inversely proportional to the radius: $\frac{dT}{dt} \propto \frac{1}{r}$.Therefore,the ratio of the rate of change of temperature for spheres $A$ and $B$ is $\frac{(dT/dt)_A}{(dT/dt)_B} = \frac{r_B}{r_A}$.

Two solid spheres $A$ and $B$ made of the same material have radii $r_A$ and $r_B$ respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature of $A$ and $B$ is :

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