$A$ body cools from $50.0^{\circ}C$ to $49.9^{\circ}C$ in $5 \, s$. How long will it take to cool from $40.0^{\circ}C$ to $39.9^{\circ}C$ (in $, s$)? The temperature of the surroundings is $30^{\circ}C$. Assume Newton's Law of Cooling applies.

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:

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 :

For a small temperature difference between the body and the surroundings,the relation between the rate of loss of heat $R$ and the temperature of the body $\theta$ is depicted by:

$A$ hot body is allowed to cool. The surrounding temperature is constant at $30^{\circ} C$. It takes time $t_{1}$ to cool from $70^{\circ} C$ to $68^{\circ} C$ and time $t_{2}$ to cool from $60^{\circ} C$ to $59.5^{\circ} C$. Then:

Hot water cools from $60\,^{\circ}C$ to $50\,^{\circ}C$ in the first $10\, min$ and to $42\,^{\circ}C$ in the next $10\, min$. The temperature of the surrounding is ........ $^{\circ}C$.