A liquid in a beaker has temperature $\theta (t)$ at time $t$ and $\theta_0$ is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge $log_e(\theta - \theta_0) $ and $t$ is

A body cools from ${60^o}C$ to ${50^o}C$ in $10$ minutes when kept in air at ${30^o}C$. In the next $10$ minutes its temperature will be

Two identical beakers $A$ and $B$ contain equal volumes of two different liquids at $60\,^oC$ each and left to cool down. Liquid in $A$ has density of $8 \times10^2\, kg / m^3$ and specific heat of $2000\, Jkg^{-1}\,K^{-1}$ while liquid in $B$ has density of $10^3\,kgm^{-3}$ and specific heat of $4000\,JKg^{-1}\,K^{-1}$ . Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)

A solid sphere of radius $R$ and a hollow sphere of inner radius $r$ and outer radius $R$ made of copper are heated to the same temperature and are allowed to cool in the same environment. Then, choose the $CORRECT$ statement

Read the following statements:

$A.$ When small temperature difference between a liquid and its surrounding is doubled the rate of loss of heat of the liquid becomes twice.

$B.$ Two bodies $P$ and $Q$ having equal surface areas are maintained at temperature $10^{\circ}\,C$ and $20^{\circ}\,C$. The thermal radiation emitted in a given time by $P$ and $Q$ are in the ratio $1: 1.15$

$C.$ A carnot Engine working between $100\,K$ and $400\,K$ has an efficiency of $75 \%$

$D.$ When small temperature difference between a liquid and its surrounding is quadrupled, the rate of loss of heat of the liquid becomes twice.

Choose the correct answer from the options given below :

A body takes $4\, {min}$. to cool from $61^{\circ} {C}$ to $59^{\circ} {C}$. If the temperature of the surroundings is $30^{\circ} {C}$, the time taken by the body to cool from $51^{\circ} {C}$ to $49^{\circ} {C}$ is $....\,min$