$A$ 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 $30^{\circ}C$?

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 $t$?

In a room where the temperature is $30^{\circ} C$, a body cools from $61^{\circ} C$ to $59^{\circ} C$ in $4$ minutes. The time taken by the body to cool from $51^{\circ} C$ to $49^{\circ} C$ will be:

$A$ pan filled with hot food cools from $94\,^{\circ}C$ to $86\,^{\circ}C$ in $2$ minutes when the room temperature is at $20\,^{\circ}C$. How long (in $s$) will it take to cool from $71\,^{\circ}C$ to $69\,^{\circ}C$?

$A$ body cools in $7$ minutes from $60^\circ C$ to $40^\circ C$. What time (in minutes) does it take to cool from $40^\circ C$ to $28^\circ C$, if its surrounding temperature is $10^\circ C$? (Newton's law of cooling holds good)

$A$ hot body,obeying Newton's law of cooling,is cooling down from its peak value $80\,^oC$ to an ambient temperature of $30\,^oC$. It takes $5\,minutes$ to cool down from $80\,^oC$ to $40\,^oC$. How many minutes will it take to cool down from $62\,^oC$ to $32\,^oC$? (Given $\ln 2 = 0.693, \ln 5 = 1.609$)