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(D) According to Newton's Law of Cooling,the rate of heat loss is given by $\frac{dQ}{dt} = \sigma A (T - T_0)$,where $\sigma$ is a constant related t

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$\frac{1}{r \rho c}$

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(D) According to Newton's Law of Cooling,the rate of heat loss is given by $\frac{dQ}{dt} = \sigma A (T - T_0)$,where $\sigma$ is a constant related to the surface nature,$A$ is the surface area,and $(T - T_0)$ is the temperature difference.Since $dQ = mc \cdot dT$,we have $mc \frac{dT}{dt} = -\sigma A (T - T_0)$.The rate of cooling is $\frac{-dT}{dt} = \frac{\sigma A}{mc} (T - T_0)$.Substituting the values for a sphere: $A = 4 \pi r^2$ and $m = \rho \cdot V = \rho \cdot \frac{4}{3} \pi r^3$.Thus,$\frac{-dT}{dt} = \frac{\sigma (4 \pi r^2)}{\rho (\frac{4}{3} \pi r^3) c} (T - T_0)$.Simplifying this,we get $\frac{-dT}{dt} \propto \frac{1}{\rho r c}$.

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