$A$ body takes $4\, \text{min}$ to cool from $61^{\circ} \text{C}$ to $59^{\circ} \text{C}$. If the temperature of the surroundings is $30^{\circ} \text{C}$,the time taken by the body to cool from $51^{\circ} \text{C}$ to $49^{\circ} \text{C}$ is $....\, \text{min}$.

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:

$A$ sphere,a cube,and a thin circular plate,all of the same material and same mass,are initially heated to the same high temperature and allowed to cool down under similar conditions. Then,the

$A$ liquid in a beaker has temperature $\theta(t)$ at time $t$ and $\theta_0$ is the temperature of the surroundings. According to Newton's law of cooling,the correct graph between $\log_e(\theta - \theta_0)$ and $t$ is:

$A$ body cools from $70^{\circ} C$ to $40^{\circ} C$ in $5$ minutes. Calculate the time it takes to cool from $60^{\circ} C$ to $30^{\circ} C$. The temperature of the surroundings is $20^{\circ} C$. (in $min.$)

The temperature of a body falls from $50^oC$ to $40^oC$ in $10$ minutes. If the temperature of the surroundings is $20^oC$,then the temperature of the body after another $10$ minutes will be ........ $^oC$.