(C) According to Stefan's Law,the rate of cooling (rate of fall of temperature) is given by $\frac{d\theta}{dt} = \frac{\sigma A (T^4 - T_0^4)}{MS}$.S

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

$\frac{R_2 \rho_2 S_2}{R_1 \rho_1 S_1}$

(C) According to Stefan's Law,the rate of cooling (rate of fall of temperature) is given by $\frac{d\theta}{dt} = \frac{\sigma A (T^4 - T_0^4)}{MS}$.Since $M = V \rho = \frac{4}{3} \pi R^3 \rho$ and $A = 4 \pi R^2$,we have $\frac{d\theta}{dt} \propto \frac{A}{MS} = \frac{4 \pi R^2}{(\frac{4}{3} \pi R^3 \rho) S} = \frac{3}{R \rho S}$.Therefore,the ratio of the rates of cooling is $\frac{(d\theta/dt)_1}{(d\theta/dt)_2} = \frac{R_2 \rho_2 S_2}{R_1 \rho_1 S_1}$.

Class 11 Physics 10-2.Heat Transfer Newton's Law of Cooling

Medium

Two spheres of radii $R_1$ and $R_2$ have densities $\rho_1$ and $\rho_2$ and specific heats $S_1$ and $S_2$. If they are heated to the same temperature,what is the ratio of their rates of cooling?

A
$\frac{R_2}{R_1}$
B
$\frac{R_1 \rho_2 S_2}{R_2 \rho_1 S_1}$
C
$\frac{R_2 \rho_2 S_2}{R_1 \rho_1 S_1}$
D
$\left( \frac{R_1}{R_2} \right)^2$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

10-2.Heat Transfer

Subtopic

Newton's Law of Cooling

If a body cools from a temperature of $62^{\circ} C$ to $50^{\circ} C$ in $10 \, \text{minutes}$ and to $42^{\circ} C$ in the next $10 \, \text{minutes}$, then the temperature of the surroundings is (in $^{\circ} C$)

EasyAP EAMCET 2025

View Solution

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 :

DifficultAP EAMCET 2006

View Solution

$A$ body cools in $7$ minutes from $60^\circ C$ to $40^\circ C$. What time (in minutes) does it take to cool from $40^\circ C$ to $28^\circ C$, if its surrounding temperature is $10^\circ C$? (Newton's law of cooling holds good)

DifficultMHT CET 2026

View Solution

$A$ 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

Easy

View Solution

$A$ liquid in a beaker has temperature $\theta(t)$ at time $t$ and $\theta_0$ is the temperature of the surroundings. According to Newton's law of cooling,which of the following is the correct graph between $\log_e(\theta - \theta_0)$ and $t$?

Medium

View Solution

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

Two spheres of radii $R_1$ and $R_2$ have densities $\rho_1$ and $\rho_2$ and specific heats $S_1$ and $S_2$. If they are heated to the same temperature,what is the ratio of their rates of cooling?

Similar Questions

If a body cools from a temperature of $62^{\circ} C$ to $50^{\circ} C$ in $10 \, \text{minutes}$ and to $42^{\circ} C$ in the next $10 \, \text{minutes}$, then the temperature of the surroundings is (in $^{\circ} C$)

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 :

$A$ body cools in $7$ minutes from $60^\circ C$ to $40^\circ C$. What time (in minutes) does it take to cool from $40^\circ C$ to $28^\circ C$, if its surrounding temperature is $10^\circ C$? (Newton's law of cooling holds good)

$A$ 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

$A$ liquid in a beaker has temperature $\theta(t)$ at time $t$ and $\theta_0$ is the temperature of the surroundings. According to Newton's law of cooling,which of the following is the correct graph between $\log_e(\theta - \theta_0)$ and $t$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module