$A$ body takes $T$ minutes to cool from $62^{\circ}C$ to $61^{\circ}C$ when the surrounding temperature is $30^{\circ}C$. The time taken by the body to cool from $46^{\circ}C$ to $45^{\circ}C$ is:

The instantaneous temperature difference between a cooling body and its surroundings,which obeys Newton's law of cooling,is $\theta$. Which of the following graphs represents the variation of $\ln \theta$ with time $t$?

$A$ body cools from $60^{\circ}C$ to $50^{\circ}C$ in $10$ minutes when kept in air at $30^{\circ}C$. In the next $10$ minutes,its temperature 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:

$A$ hot liquid kept in a beaker cools from $80\,^{\circ}C$ to $70\,^{\circ}C$ in $2\, \text{min}$. If the surrounding temperature is $30\,^{\circ}C$,then the time taken for the same liquid to cool from $60\,^{\circ}C$ to $50\,^{\circ}C$ is ........ $\text{sec}$.

$A$ 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: