$A$ body takes $5 \ min$ to cool from $90^{\circ}C$ to $60^{\circ}C$. If the temperature of the surroundings is $20^{\circ}C$,how much time (in $min$) will it take to cool from $60^{\circ}C$ to $30^{\circ}C$?

$A$ body cools in a surrounding which is at a constant temperature of $\theta_0$. Assuming that it obeys Newton's law of cooling,its temperature $\theta$ is plotted against time $t$. Tangents are drawn to the curve at the points $A(\theta = \theta_1)$ and $B(\theta = \theta_2)$. These tangents meet the time-axis at angles $\alpha_1$ and $\alpha_2$ as shown in the graph. Then:

$A$ sphere and a cube made of the same material and having the same surface area are heated to the same temperature and kept in the same surroundings. The ratio of their initial rates of cooling will be

Two metallic spheres $S_1$ and $S_2$ are made of the same material and have identical surface finish. The mass of $S_1$ is three times that of $S_2$. Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. The ratio of the initial rate of cooling of $S_1$ to that of $S_2$ is

Two metal spheres,one solid and one hollow,are heated to $300^{\circ}C$ and allowed to cool in the same environment. The rate of heat loss will be:

An object kept in a large room having air temperature of $25^{\circ}C$ takes $12 \text{ minutes}$ to cool from $80^{\circ}C$ to $70^{\circ}C$. The time taken to cool for the same object from $70^{\circ}C$ to $60^{\circ}C$ would be nearly.....$min$