If a liquid cools from $95^{\circ}C$ to $90^{\circ}C$ in $30 \, s$ and from $55^{\circ}C$ to $50^{\circ}C$ in $70 \, s$,then the temperature of the surroundings is ...... $^{\circ}C$. (in $.5$)

$A$ cup of tea cools from $80\,^{\circ}C$ to $60\,^{\circ}C$ in $1$ minute. The ambient temperature is $30\,^{\circ}C$. How long will it take to cool from $60\,^{\circ}C$ to $50\,^{\circ}C$?

Hot water cools from $60\,^{\circ}C$ to $50\,^{\circ}C$ in the first $10\, min$ and to $42\,^{\circ}C$ in the next $10\, min$. The temperature of the surrounding is ........ $^{\circ}C$.

$A$ body cools from $62^{\circ}C$ to $50^{\circ}C$ in $10$ minutes and to $42^{\circ}C$ in the next $10$ minutes. The temperature of the surrounding is ........ $^{\circ}C$.

$A$ hot liquid kept in a beaker cools from $80\,^{\circ}C$ to $70\,^{\circ}C$ in $2\, \text{min}$. If the surrounding temperature is $30\,^{\circ}C$,then the time taken for the same liquid to cool from $60\,^{\circ}C$ to $50\,^{\circ}C$ is ........ $\text{sec}$.

$A$ body cools in a surrounding which is at a constant temperature of $\theta_0$. Assuming 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 $A(\theta = \theta_1)$ and $B(\theta = \theta_2)$. These tangents meet the time-axis at angles $\alpha_1$ and $\alpha_2$ as shown in the graph. Then: