$A$ body cools in $7$ minutes from $60^{\circ}C$ to $40^{\circ}C$. What time (in minutes) does it take to cool from $40^{\circ}C$ to $28^{\circ}C$ if the surrounding temperature is $10^{\circ}C$? Assume Newton's Law of cooling holds.

$A$ solid cube and a solid sphere of identical material and equal masses are heated to the same temperature and left to cool in the same surroundings. Then,

$A$ certain quantity of water cools from $70\,^{\circ}C$ to $60\,^{\circ}C$ in the first $10\, minutes$ and to $54\,^{\circ}C$ in the next $10\, minutes$. The temperature of the surrounding is ......... $^{\circ}C$.

$A$ metal sphere cools at a rate of $1.5^{\circ} C / min$ when its temperature is $80^{\circ} C$. When the temperature of the sphere is $40^{\circ} C$,its rate of cooling is $0.3^{\circ} C / min$. The temperature of the surrounding $\left(\theta_0\right)$ is (in $^{\circ} C$)

Two spheres have a radius ratio of $1:2$ and a density ratio of $2:1$. They have the same specific heat. If they are heated to the same temperature and placed in the same environment,what is the ratio of their rate of cooling?

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