Q140 / 160 · AP EAMCET 2006 · Physics

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 :

  • A
    $\frac{r_A}{r_B}$
  • Correct answer
    B
    $\frac{r_B}{r_A}$
  • C
    $\frac{r_A^2}{r_B^2}$
  • D
    $\frac{r_B^2}{r_A^2}$

Solution

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

Hence, the correct answer is option: 'B'.

Relevant Questions