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

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

$A$ body cools from a temperature $3T$ to $2T$ in $10$ minutes. The room temperature is $T$. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of the next $10$ minutes will be

$A$ body initially at $80^\circ C$ cools to $64^\circ C$ in $5$ minutes and to $52^\circ C$ in $10$ minutes. The temperature of the body after $15$ minutes will be ...... $^\circ C$.

Consider two hot bodies $B_1$ and $B_2$ which have temperatures $100^oC$ and $80^oC$ respectively at $t = 0$. The temperature of the surroundings is $40^oC$. The ratio of the respective rates of cooling $R_1$ and $R_2$ of these two bodies at $t = 0$ 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)