Hot water is kept in a thermally insulated closed container. It takes $T_1 \, min$ for the temperature to drop from $75^{\circ}C$ to $70^{\circ}C$,$T_2 \, min$ to drop from $70^{\circ}C$ to $65^{\circ}C$,and $T_3 \, min$ to drop from $65^{\circ}C$ to $60^{\circ}C$. Then:

The temperature of a body falls from $50^oC$ to $40^oC$ in $10$ minutes. If the temperature of the surroundings is $20^oC$,then the temperature of the body after another $10$ minutes will be ........ $^oC$.

$A$ 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:

$A$ solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature $120\ ^oC$,then:

The temperature of a body falls from $62^\circ C$ to $50^\circ C$ in $10$ minutes. If the temperature of the surroundings is $26^\circ C$,the temperature in the next $10$ minutes will become ...... $^\circ C$.

$A$ body takes $4$ minutes to cool from $100^{\circ}C$ to $70^{\circ}C$. To cool from $70^{\circ}C$ to $40^{\circ}C$ it will take ........ $\text{min.}$ (room temperature is $15^{\circ}C$)