$A$ 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,which of the following is the correct graph between $\log_e(\theta - \theta_0)$ and $t$?

$A$ very thin metallic shell of radius $r$ is heated to temperature $T$ and then allowed to cool. The rate of cooling of the shell is proportional to ........

$A$ 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 $:$

$A$ 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 ....... $sec$.

$A$ body cools in a surrounding which is at a constant temperature of $\theta_0$. Assume that it obeys Newton's law of cooling. Its temperature $\theta$ is plotted against time $t$. Tangents are drawn to the curve at the points $P(\theta = \theta_2)$ and $Q(\theta = \theta_1)$. These tangents meet the time axis at angles of $\varphi_2$ and $\varphi_1$,as shown in the figure.

$A$ bucket full of water cools from $75^{\circ}C$ to $70^{\circ}C$ in time $T_1$,from $70^{\circ}C$ to $65^{\circ}C$ in time $T_2$,and from $65^{\circ}C$ to $60^{\circ}C$ in time $T_3$. Which of the following is correct?