A body takes 4 min to cool from 61°C to 59°C. If the temperature of the surroundings is 30°C, the time taken by the body to cool from 51°C to 49°C is:

According to Newton’s Law of Cooling, the rate at which the temperature $$T$$ of a body changes is proportional to the excess temperature over that of the surroundings. If $$T_s$$ is the constant temperature of the surroundings and $$k$$ is the proportionality constant, the law is written as

$$\frac{dT}{dt}=-k\,(T-T_s).$$

Integrating between an initial temperature $$T_1$$ and a final temperature $$T_2$$ over a time interval $$t$$ gives the well-known working formula

$$t=\frac{1}{k}\,\ln\!\left(\frac{T_1-T_s}{\,T_2-T_s}\right).$$

We first use the given cooling from 61 °C to 59 °C in 4 min to determine the constant $$k$$. Here

$$T_1 = 61^\circ\text{C},\qquad T_2 = 59^\circ\text{C},\qquad T_s = 30^\circ\text{C},\qquad t = 4\ \text{min}.$$

Substituting in the formula, we have

$$4 = \frac{1}{k}\,\ln\!\left(\frac{61-30}{59-30}\right) = \frac{1}{k}\,\ln\!\left(\frac{31}{29}\right).$$

Solving for $$k$$,

$$k = \frac{1}{4}\,\ln\!\left(\frac{31}{29}\right).$$

Now we need the time required for the body to cool from 51 °C to 49 °C. For this second stage we put

$$T_1' = 51^\circ\text{C},\qquad T_2' = 49^\circ\text{C}.$$

Using the same surroundings temperature $$T_s = 30^\circ\text{C}$$ and the same $$k$$, the time $$t'$$ for this cooling is

$$t' = \frac{1}{k}\,\ln\!\left(\frac{T_1'-T_s}{T_2'-T_s}\right) = \frac{1}{k}\,\ln\!\left(\frac{51-30}{49-30}\right) = \frac{1}{k}\,\ln\!\left(\frac{21}{19}\right).$$

We substitute the value of $$k$$ obtained earlier:

$$t' = \left[\frac{1}{\dfrac{1}{4}\,\ln\!\left(\dfrac{31}{29}\right)}\right] \,\ln\!\left(\frac{21}{19}\right) = 4\;\frac{\ln\!\left(\dfrac{21}{19}\right)} {\ln\!\left(\dfrac{31}{29}\right)}.$$

Evaluating the logarithms (natural logarithms):

$$\ln\!\left(\frac{31}{29}\right)\approx 0.0665,\qquad \ln\!\left(\frac{21}{19}\right)\approx 0.1003.$$

So

$$t' \approx 4 \times \frac{0.1003}{0.0665} \approx 4 \times 1.508 \approx 6.03\ \text{min}.$$

The calculated time is practically 6 minutes.

Hence, the correct answer is Option D.