A hot body, obeying Newton's law of cooling is cooling down from its peak value 80°C to an ambient temperature of 30°C. It takes 5 minutes in cooling down from 80°C to 40°C. How much time will it take to cool down from 62°C to 32°C? (Given ln 2 = 0.693, ln 5 = 1.609)

Newton's law of cooling states that the rate of cooling of a body is proportional to the difference between its temperature and the ambient temperature. The ambient temperature is given as 30°C. The differential equation is:

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

where $$ T $$ is the temperature of the body at time $$ t $$, $$ T_s = 30^\circ \text{C} $$ is the ambient temperature, and $$ k $$ is a positive constant. Separating variables and integrating:

$$\int \frac{dT}{T - T_s} = -k \int dt$$

$$\ln|T - T_s| = -kt + C$$

where $$ C $$ is the integration constant. Since the body's temperature is above the ambient temperature, we can write:

$$T - T_s = e^{-kt + C} = e^C e^{-kt}$$

Let $$ A = e^C $$, so:

$$T - T_s = A e^{-kt}$$

$$T = T_s + A e^{-kt} = 30 + A e^{-kt}$$

At time $$ t = 0 $$, the temperature is 80°C:

$$80 = 30 + A e^{0} \Rightarrow 80 = 30 + A \Rightarrow A = 50$$

Thus, the temperature equation is:

$$T = 30 + 50 e^{-kt}$$

It takes 5 minutes for the temperature to drop from 80°C to 40°C. At $$ t = 5 $$ minutes, $$ T = 40^\circ \text{C} $$:

$$40 = 30 + 50 e^{-k \cdot 5}$$

$$10 = 50 e^{-5k}$$

$$\frac{10}{50} = e^{-5k}$$

$$0.2 = e^{-5k}$$

Taking natural logarithm on both sides:

$$\ln(0.2) = \ln(e^{-5k})$$

$$\ln(0.2) = -5k$$

Since $$ \ln(0.2) = \ln\left(\frac{1}{5}\right) = -\ln(5) $$ and given $$ \ln(5) = 1.609 $$:

$$-1.609 = -5k$$

$$k = \frac{1.609}{5} = 0.3218$$

Now, we need to find the time taken to cool from 62°C to 32°C. Let $$ t_1 $$ be the time when the temperature is 62°C and $$ t_2 $$ be the time when it is 32°C. The time required is $$ t_2 - t_1 $$. Using the temperature equation:

At $$ T = 62^\circ \text{C} $$:

$$62 = 30 + 50 e^{-k t_1}$$

$$32 = 50 e^{-k t_1}$$

$$e^{-k t_1} = \frac{32}{50} = 0.64$$

At $$ T = 32^\circ \text{C} $$:

$$32 = 30 + 50 e^{-k t_2}$$

$$2 = 50 e^{-k t_2}$$

$$e^{-k t_2} = \frac{2}{50} = 0.04$$

Dividing the two expressions:

$$\frac{e^{-k t_1}}{e^{-k t_2}} = \frac{0.64}{0.04} = 16$$

$$\frac{e^{-k t_1}}{e^{-k t_2}} = e^{-k(t_1 - t_2)} = e^{k(t_2 - t_1)}$$

So:

$$e^{k(t_2 - t_1)} = 16$$

Taking natural logarithm:

$$k(t_2 - t_1) = \ln(16)$$

$$\ln(16) = \ln(2^4) = 4 \ln 2 = 4 \times 0.693 = 2.772$$

Substituting $$ k = \frac{\ln(5)}{5} = \frac{1.609}{5} $$:

$$t_2 - t_1 = \frac{\ln(16)}{k} = \frac{2.772}{0.3218} \approx 8.614$$

Alternatively, using the exact expression:

$$t_2 - t_1 = \frac{\ln(16)}{k} = \frac{4 \ln 2}{\frac{\ln 5}{5}} = \frac{4 \ln 2 \times 5}{\ln 5} = \frac{20 \ln 2}{\ln 5}$$

Substituting the given values:

$$t_2 - t_1 = \frac{20 \times 0.693}{1.609} = \frac{13.86}{1.609} \approx 8.614$$

Rounding to one decimal place, the time is 8.6 minutes.

Hence, the correct answer is Option B.