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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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The correct answer is:
$\frac{r_B}{r_A}$
$\frac{4 \pi}{3} r^3 \rho c\left(-\frac{d T}{d t}\right)=\sigma 4 \pi r^2\left(T^4-T_0^4\right)$
$\therefore\left(-\frac{d T}{d t}\right)=\frac{3 \sigma}{\rho r c}\left(T^4-T_0^4\right)=H$ (say)
Ratio of rates of fall of temperature
$\frac{H_A}{H_B}=\frac{r_B}{r_A}$
$\therefore\left(-\frac{d T}{d t}\right)=\frac{3 \sigma}{\rho r c}\left(T^4-T_0^4\right)=H$ (say)
Ratio of rates of fall of temperature
$\frac{H_A}{H_B}=\frac{r_B}{r_A}$
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