According to Newton's law of cooling,the rate of cooling of a body is proportional to $(\Delta \theta )^n$,where $\Delta \theta$ is the difference of the temperature of the body and the surroundings,and $n$ is equal to

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:

Two bottles $A$ and $B$ have radii $R_{A}$ and $R_{B}$ and heights $h_{A}$ and $h_{B}$ respectively,with $R_{B}=2 R_{A}$ and $h_{B}=2 h_{A}$. These are filled with hot water at $60^{\circ} C$. Consider that heat loss for the bottles takes place only from side surfaces. If the time the water takes to cool down to $50^{\circ} C$ is $t_{A}$ and $t_{B}$ for bottles $A$ and $B$,respectively,then $t_{A}$ and $t_{B}$ are best related as:

Hot water kept in a beaker placed in a room cools from $70^{\circ}C$ to $60^{\circ}C$ in $4$ minutes. The time taken by it to cool from $69^{\circ}C$ to $59^{\circ}C$ will be

$A$ bowl filled with very hot water cools from $98^\circ C$ to $86^\circ C$ in $2$ minutes when the room temperature is $22^\circ C$. How long will it take to cool from $75^\circ C$ to $69^\circ C$?

$A$ 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 ........ $\text{sec}$.