$A$ 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 ...... $\min.$

If the initial temperatures of a metallic sphere and a disc,of the same mass,radius,and material,are equal,then the ratio of their rate of cooling in the same environment will be:

$A$ cup of tea cools from $65.5 ^\circ C$ to $62.5 ^\circ C$ in $1$ minute in a room at $22.5 ^\circ C$. How long will the same cup of tea take,in minutes,to cool from $46.5 ^\circ C$ to $40.5 ^\circ C$ in the same room? (Choose the nearest value)

Two spheres of the same metal have radii $r$ and $2r$. They are heated to the same temperature and placed in the same environment. The ratio of their rates of cooling will be:

$A$ 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:

$A$ mass of $50\,g$ of water in a closed vessel,with surroundings at a constant temperature,takes $2\,minutes$ to cool from $30\,^oC$ to $25\,^oC$. $A$ mass of $100\,g$ of another liquid in an identical vessel with identical surroundings takes the same time to cool from $30\,^oC$ to $25\,^oC$. The specific heat of the liquid is .......... $kcal/(kg \cdot ^oC)$ (The water equivalent of the vessel is $30\,g$).