$A$ body cools from $70^{\circ} C$ to $40^{\circ} C$ in $5 \text{ min}$. Calculate the time it takes to cool from $60^{\circ} C$ to $40^{\circ} C$. The temperature of the surrounding is $20^{\circ} C$. (in $\text{ min}$)

Consider two hot bodies $B_1$ and $B_2$ which have temperatures $100^oC$ and $80^oC$ respectively at $t = 0$. The temperature of the surroundings is $40^oC$. The ratio of the respective rates of cooling $R_1$ and $R_2$ of these two bodies at $t = 0$ will be

$A$ vessel contains hot water at $100^{\circ}C$. If it cools to $80^{\circ}C$ in time $T_1$ and from $80^{\circ}C$ to $60^{\circ}C$ in time $T_2$,then:

$A$ body cools from $80\,^{\circ} C$ to $50\,^{\circ} C$ in $5$ minutes. Calculate the time (in $min$) it takes to cool from $60\,^{\circ} C$ to $30\,^{\circ} C$. The temperature of the surroundings is $20\,^{\circ} C$.

$A$ body takes $5$ minutes for cooling from $50^oC$ to $40^oC$. Its temperature comes down to $33.33^oC$ in the next $5$ minutes. The temperature of the surroundings is ....... $^oC$.

$A$ solid cube and a solid sphere of identical material and equal masses are heated to the same temperature and left to cool in the same surroundings. Then,