Four spheres A, B, C and D are of the same ra | vedclass

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

Question

(C) The rate of cooling $R_F = \frac{d	heta}{dt}$ is given by Newton's Law of Cooling as $R_F \propto \frac{1}{ms}$,where $m$ is the mass and $s$ is

Vedclass · Accepted Answer

$D$

Vedclass · Answer

(C) The rate of cooling $R_F = \frac{d	heta}{dt}$ is given by Newton's Law of Cooling as $R_F \propto \frac{1}{ms}$,where $m$ is the mass and $s$ is the specific heat capacity.Since the spheres have the same radius,their volumes $V$ are equal. Mass $m = ho V$,where $ho$ is the density.Thus,$R_F \propto \frac{1}{(ho V)s} \propto \frac{1}{ho s}$.We need to calculate the product $ho s$ for each sphere:For $A$: $ho s = 6 	imes 2 = 12$For $B$: $ho s = 3 	imes 5 = 15$For $C$: $ho s = 4 	imes 4 = 16$For $D$: $ho s = 5 	imes 6 = 30$The rate of cooling is slowest when the product $ho s$ is maximum.Comparing the values $12, 15, 16, 30$,the maximum value is $30$,which corresponds to sphere $D$.

Similar Questions

Two hot bodies $A$ and $B$ have temperatures $100^{\circ}C$ and $80^{\circ}C$ respectively. The surrounding temperature is $40^{\circ}C$. The ratio of their rates of cooling $R_1 : R_2$ at $t = 0$ is:

$A$ body takes $5$ minutes to cool from $90^oC$ to $60^oC$. If the temperature of the surroundings is $20^oC$,the time taken by it to cool from $60^oC$ to $30^oC$ will be ...... $\min.$

It takes $10$ minutes to cool a liquid from $61^{\circ} C$ to $59^{\circ} C$. If the room temperature is $30^{\circ} C$,then the time taken in cooling from $51^{\circ} C$ to $49^{\circ} C$ is .......... $min$.

$A$ body cools from a temperature of $60^{\circ} C$ to $50^{\circ} C$ in $10 \text{ minutes}$ and $50^{\circ} C$ to $40^{\circ} C$ in $15 \text{ minutes}$. The time taken in minutes for the body to cool from $40^{\circ} C$ to $30^{\circ} C$ is

In a room where the temperature is $30^{\circ} C$, a body cools from $61^{\circ} C$ to $59^{\circ} C$ in $4$ minutes. The time taken by the body to cool from $51^{\circ} C$ to $49^{\circ} C$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module