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

Two bodies $A$ and $B$ of equal masses,surface area,and emissivity are cooling under Newton's law of cooling from the same initial temperature. Their cooling curves are represented by the graph. If $\theta$ is the instantaneous temperature of the body and $\theta_0$ is the temperature of the surroundings,then the relationship between their specific heats $S_A$ and $S_B$ is:

Two hot bodies $A$ and $B$ have temperatures $100^{\circ}C$ and $80^{\circ}C$ respectively. The surrounding temperature is $40^{\circ}C$. The ratio of their rates of cooling $R_1 : R_2$ at $t = 0$ is:

Two metallic spheres $S_1$ and $S_2$ are made of the same material and have identical surface finish. The mass of $S_1$ is three times that of $S_2$. Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. The ratio of the initial rate of cooling of $S_1$ to that of $S_2$ is

$A$ solid sphere and a hollow sphere of the same material and same radius are heated to the same temperature. They are placed in an environment of the same temperature. If the temperature difference between both and the surroundings is $T$,then .......

$A$ body cools from a temperature $3\theta$ to $2\theta$ in $10 \text{ minutes}$. The room temperature is $\theta$. The temperature of the body at the end of the next $10 \text{ minutes}$ is '$x$'. Assuming that Newton's law of cooling is applicable, the value of '$x$' will be