$A$ sphere and a cube made of the same material and having the same surface area are heated to the same temperature and kept in the same surroundings. The ratio of their initial rates of cooling will be

$A$ certain quantity of water cools from $70^\circ C$ to $60^\circ C$ in the first $5$ minutes and to $54^\circ C$ in the next $5$ minutes. The temperature of the surroundings is ..... $^\circ C$

An object cools from $100^{\circ} C$ to $40^{\circ} C$ in $10$ minutes, when the surrounding temperature is $10^{\circ} C$. Then the time taken by the object to cool from $70^{\circ} C$ to $20^{\circ} C$ is
$ [\text{Take } \ln 2=0.7, \ln 3=1.1, \ln 6=1.8 ]$ (in $min$)

$A$ 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:

$A$ body cools from a temperature of $60^{\circ} C$ to $50^{\circ} C$ in $10 \text{ minutes}$ and $50^{\circ} C$ to $40^{\circ} C$ in $15 \text{ minutes}$. The time taken in minutes for the body to cool from $40^{\circ} C$ to $30^{\circ} C$ is

$A$ heating element of mass $100 \,g$ and having specific heat of $1 \,J/(g^{\circ}C)$ is exposed to surrounding air at $27^{\circ}C$. The element attains a steady state temperature of $127^{\circ}C$, while absorbing $100 \,W$ of electric power. If the power is switched off, then the approximate time taken by the element to cool down to $126^{\circ}C$ will be (neglect radiation). (in $\,s$)