Two liquids of equal volume are cooled under identical conditions from $60^{\circ}C$ to $50^{\circ}C$ in $324 \; sec$ and $810 \; sec$ respectively. If the ratio of their specific heats is $3:4$,find the ratio of their densities. (The water equivalent of the calorimeter is negligible.)

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 $min$)

In $5\, \text{minutes}$,a body cools from $75^{\circ} \text{C}$ to $65^{\circ} \text{C}$ at a room temperature of $25^{\circ} \text{C}$. The temperature of the body at the end of the next $5\, \text{minutes}$ is $......\,^{\circ} \text{C}$.

$A$ container with $1 \ kg$ of water is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received from sunlight is $700 \ W \ m^{-2}$ and it is absorbed by the water over an effective area of $0.05 \ m^2$. Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in $^{\circ}C$) in the temperature of water and the surroundings after a long time will be. . . . . . . . (Ignore the effect of the container, and take the constant for Newton's law of cooling $k = 0.001 \ s^{-1}$, Heat capacity of water $s = 4200 \ J \ kg^{-1} \ K^{-1}$)

$A$ sphere,a cube,and a thin circular plate,all of the same material and same mass,are initially heated to the same high temperature and allowed to cool down under similar conditions. Then,the

$A$ 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