The temperature of a body in air falls from $$ 40^{o}C $$ to $$ 24^{o}C $$ in 4 minutes. The temperature of the air is $$ 16^{o}C $$. The temperature of the body in the next 4 minutes will be:

Newton’s law of cooling: the rate of fall of temperature of a body is proportional to the temperature difference between the body and the surroundings.

Write it as $$\frac{dT}{dt} = -k\,(T - T_s)$$, where $$T_s$$ is the surrounding temperature and $$k$$ is a positive constant.

Integrating, we obtain the exponential form
$$T - T_s = (T_0 - T_s)\,e^{-kt}$$ $$-(1)$$, where $$T_0$$ is the initial temperature at $$t = 0$$.

Given: $$T_s = 16^{\circ}C$$ and $$T_0 = 40^{\circ}C$$.
Initial temperature difference: $$\Delta T_0 = 40 - 16 = 24^{\circ}C$$.

After $$4$$ min the body cools to $$24^{\circ}C$$.
Temperature difference then: $$\Delta T_4 = 24 - 16 = 8^{\circ}C$$.

Insert these values into $$(1)$$ to determine $$k$$:
$$8 = 24\,e^{-k \times 4}$$

$$e^{-4k} = \frac{8}{24} = \frac{1}{3}$$

Taking natural logarithm on both sides:
$$-4k = \ln\!\left(\frac{1}{3}\right) = -\ln 3$$

Thus $$k = \frac{\ln 3}{4}$$.

We want the temperature after another 4 min, i.e. at $$t = 8$$ min.
From $$(1)$$:

$$T_8 - 16 = 24\,e^{-k \times 8}$$

Substitute $$k = \frac{\ln 3}{4}$$:
$$T_8 - 16 = 24\,e^{-2\ln 3} = 24\,(e^{\ln 3})^{-2} = 24 \times \frac{1}{3^2} = \frac{24}{9} = \frac{8}{3}$$

Hence $$T_8 = 16 + \frac{8}{3} = \frac{48}{3} + \frac{8}{3} = \frac{56}{3}\,^{\circ}C$$.

Therefore the temperature of the body after the next 4 minutes is $$\frac{56}{3}^{\circ}C$$. This corresponds to Option D.