$A$ 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:

$A$ liquid cools from $50^{\circ}C$ to $45^{\circ}C$ in $5$ minutes and from $45^{\circ}C$ to $41.5^{\circ}C$ in the next $5$ minutes. The temperature of the surrounding is......... $^{\circ}C$.

$A$ body cools from $60^{\circ} C$ to $40^{\circ} C$ in $6$ minutes. After the next $6$ minutes,its temperature will be (Temperature of the surroundings is $10^{\circ} C$). (in $^{\circ} C$)

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

The temperature of a liquid drops from $365 \ K$ to $361 \ K$ in $2$ minutes. Find the time during which the temperature of the liquid drops from $344 \ K$ to $342 \ K$. The temperature of the room is $293 \ K$.

$A$ solid sphere of radius $R$ and a hollow sphere of inner radius $r$ and outer radius $R$ made of copper are heated to the same temperature and are allowed to cool in the same environment. Then,choose the $CORRECT$ statement.