An ordinary body cools from $4 \theta$ to $3 \theta$ in $t$ minutes. The temperature of that body after the next $t$ minutes is (Assume Newton's law of cooling and room temperature is $\theta$)

$A$ block of metal is heated to a temperature much higher than the room temperature and placed in an evacuated cavity. Which curve correctly represents the rate of cooling? ($T$ is the temperature of the block and $t$ is the time.)

$A$ 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 $:$

$A$ body cools from $80\,^{\circ} C$ to $50\,^{\circ} C$ in $5$ minutes. Calculate the time (in $min$) it takes to cool from $60\,^{\circ} C$ to $30\,^{\circ} C$. The temperature of the surroundings is $20\,^{\circ} C$.

$A$ body takes $5 \text{ min}$ to cool from $80^{\circ}C$ to $50^{\circ}C$. How many minutes will it take to cool from $60^{\circ}C$ to $30^{\circ}C$,if the room temperature is $20^{\circ}C$?

$A$ 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.