$A$ body takes $10 \, \text{minutes}$ to cool from $60 \, ^\circ\text{C}$ to $50 \, ^\circ\text{C}$. The temperature of the surroundings is constant at $25 \, ^\circ\text{C}$. Then, the temperature of the body after the next $10 \, \text{minutes}$ will be approximately ....... $^\circ\text{C}$

$A$ cup of coffee cools from $90^{\circ} C$ to $80^{\circ} C$ in $t$ minutes when the room temperature is $20^{\circ} C$. The time taken by the similar cup of coffee to cool from $80^{\circ} C$ to $60^{\circ} C$ at the same room temperature is $:$

There are two spherical balls $A$ and $B$ of the same material with the same surface,but the diameter of $A$ is half that of $B$. If $A$ and $B$ are heated to the same temperature and then allowed to cool,then

$A$ cup of tea cools from $80\,^{\circ}C$ to $60\,^{\circ}C$ in $1\,min$. The ambient temperature is $30\,^{\circ}C$. In cooling from $60\,^{\circ}C$ to $50\,^{\circ}C$,it will take ....... $sec$.

In $5\, \text{minutes}$,a body cools from $75^{\circ} \text{C}$ to $65^{\circ} \text{C}$ at a room temperature of $25^{\circ} \text{C}$. The temperature of the body at the end of the next $5\, \text{minutes}$ is $......\,^{\circ} \text{C}$.

$A$ body cools from $80^{\circ} C$ to $50^{\circ} C$ in $5 \text{ min}$. In the next time of $t \text{ min}$, the body continues to cool from $50^{\circ} C$ to $30^{\circ} C$. The total time taken by the body to cool from $80^{\circ} C$ to $30^{\circ} C$ is
[The temperature of the surroundings is $20^{\circ} C$.] (in $\text{ min}$)