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 ...... $^\circ C$.

$A$ liquid cools from $50^oC$ to $45^oC$ in $5$ minutes and from $45^oC$ to $41.5^oC$ in the next $5$ minutes. The temperature of the surrounding is ...... $^oC$.

An object cools from $100^{\circ} C$ to $40^{\circ} C$ in $10$ minutes, when the surrounding temperature is $10^{\circ} C$. Then the time taken by the object to cool from $70^{\circ} C$ to $20^{\circ} C$ is
$ [\text{Take } \ln 2=0.7, \ln 3=1.1, \ln 6=1.8 ]$ (in $min$)

$A$ hot body placed in air cools down to a lower temperature. The rate of decrease of temperature is proportional to the temperature difference from the surrounding. The body loses $60 \%$ and $80 \%$ of the maximum heat it can lose in time $t_1$ and $t_2$ respectively. The ratio $t_2 / t_1$ will be

Two solid spheres $A$ and $B$ made of the same material have radii $r_A$ and $r_B$ respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature of $A$ and $B$ is :

An object kept in a large room having air temperature of $25^{\circ}C$ takes $12 \text{ minutes}$ to cool from $80^{\circ}C$ to $70^{\circ}C$. The time taken to cool for the same object from $70^{\circ}C$ to $60^{\circ}C$ would be nearly.....$min$