Two solid spheres of radii $R_1$ and $R_2$ are made of the same material and have similar surfaces. The spheres are raised to the same temperature and then allowed to cool under identical conditions. Assuming the spheres to be perfect conductors of heat,the ratio of their initial rates of cooling is:

  • A
    $R_1^2 / R_2^2$
  • B
    $R_1^4 / R_2^4$
  • C
    $R_2^3 / R_1^3$
  • D
    $R_2 / R_1$

Explore More

Similar Questions

Hot water is kept in a thermally insulated closed container. It takes $T_1 \, min$ for the temperature to drop from $75^{\circ}C$ to $70^{\circ}C$,$T_2 \, min$ to drop from $70^{\circ}C$ to $65^{\circ}C$,and $T_3 \, min$ to drop from $65^{\circ}C$ to $60^{\circ}C$. Then:

$A$ body takes $4$ minutes to cool from $100^{\circ}C$ to $70^{\circ}C$. To cool from $70^{\circ}C$ to $40^{\circ}C$ it will take ........ $\text{min.}$ (room temperature is $15^{\circ}C$)

Hot water cools from $60\,^oC$ to $50\,^oC$ in the first $10\,minutes$ and to $42\,^oC$ in the next $10\,minutes.$ The temperature of the surroundings is ...... $^oC$

$A$ body cools from a temperature $3\theta$ to $2\theta$ in $10 \text{ minutes}$. The room temperature is $\theta$. The temperature of the body at the end of the next $10 \text{ minutes}$ is '$x$'. Assuming that Newton's law of cooling is applicable, the value of '$x$' will be

$A$ liquid cools down from $70^{\circ}C$ to $60^{\circ}C$ in $5$ minutes. The time taken to cool it from $60^{\circ}C$ to $50^{\circ}C$ will be

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo