A hot liquid is kept in a big room. The logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time. Then the plot will be

A body cools at the rate of $0.75°\mathrm{C} {\mathrm{s}}^{-1}$, when it is $50^{\circ} \mathrm{C}$ above the surrounding. Its rate of cooling, when it is $30^{\circ} \mathrm{C}$ above the same surrounding, will be

A bucket full of hot water cools from $75 °\mathrm{C}$ to $70 °\mathrm{C}$ in time $T_1$, from $70 °\mathrm{C}$ to $65 °\mathrm{C}$ in time $T_2$ and from $65 °\mathrm{C}$ to $60 °\mathrm{C}$ in time $T_3$, then

A liquid cools from $70 °\mathrm{C}$ to $60 °\mathrm{C}$ in $5$ minutes. If the temperature of the surrounding is constant at $30 °\mathrm{C}$, then the time taken by the liquid to cool from $60 °\mathrm{C}$ to $50 °\mathrm{C}$ is 

An object is. cooled from $75 °\mathrm{C}$ to $65 °\mathrm{C}$ in $2 \mathrm{minutes}$ in a room at $30 °\mathrm{C}$. The time taken to cool another object from $55 °\mathrm{C}$ to $45 °\mathrm{C}$ in the same room in $\mathrm{minutes}$ is

A body cools from $100 °\mathrm{C}$ to $70 °\mathrm{C}$ in $8$ minutes. If temperature of surrounding is $15 °\mathrm{C}$, then time required for body to cool from$70 °\mathrm{C}$  to $40 °\mathrm{C}$ is

Certain quantity of water cools from $70 °\mathrm{C}$ to $60 °\mathrm{C}$ in the first $5$ minutes and to $54 °\mathrm{C}$ in the next $5$ minutes. The temperature of the surroundings is

A liquid in a beaker has temperature $\theta$ at time $t$ and ${\theta }_0$ is temperature of surroundings, then according to Newton's law of cooling the correct graph between $\log \left(\theta -{\theta }_0\right)$ and $t$ is

If a piece of metal is heated to temperature $\theta$ and then allowed to cool in a room which is at temperature ${\theta }_0$ the graph between the temperature $T$ of the metal and time $t$ will be closest to: