The average mass of rain drops is $$3.0 \times 10^{-5}$$ kg and their average terminal velocity is 9 m/s. Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.

The average mass of each rain drop is $$ m = 3.0 \times 10^{-5} $$ kg, and the average terminal velocity is $$ v = 9 $$ m/s. The energy transferred by each rain drop when it hits the surface is due to its kinetic energy. The kinetic energy (KE) of one rain drop is given by:

$$ KE = \frac{1}{2} m v^2 $$

Substituting the values:

$$ KE = \frac{1}{2} \times (3.0 \times 10^{-5}) \times (9)^2 $$

First, calculate $$ 9^2 $$:

$$ 9^2 = 81 $$

Then multiply by the mass:

$$ m \times 81 = 3.0 \times 10^{-5} \times 81 = 243 \times 10^{-5} = 2.43 \times 10^{-3} $$

Now multiply by $$ \frac{1}{2} $$:

$$ KE = \frac{1}{2} \times 2.43 \times 10^{-3} = 1.215 \times 10^{-3} \text{ J} $$

So, each rain drop transfers $$ 1.215 \times 10^{-3} $$ J of energy.

The place receives 100 cm of rain in a year. Convert this to meters:

$$ 100 \text{ cm} = 1 \text{ m} $$

This is the height of the water column over an area of 1 square meter. The volume of rain falling on 1 m² area is:

$$ \text{Volume} = \text{area} \times \text{height} = 1 \times 1 = 1 \text{ m}^3 $$

The density of water is 1000 kg/m³, so the total mass of rain falling on 1 m² area is:

$$ \text{Mass} = \text{volume} \times \text{density} = 1 \times 1000 = 1000 \text{ kg} $$

The mass of one rain drop is $$ m = 3.0 \times 10^{-5} $$ kg. The number of rain drops falling on 1 m² area is:

$$ \text{Number of drops} = \frac{\text{Total mass}}{\text{Mass per drop}} = \frac{1000}{3.0 \times 10^{-5}} $$

Calculate:

$$ \frac{1000}{3.0 \times 10^{-5}} = \frac{1000}{0.00003} = \frac{1000 \times 10^5}{3} = \frac{10^8}{3} = 3.333 \times 10^7 $$

The total energy transferred to 1 m² area is the number of drops multiplied by the energy per drop:

$$ E_{\text{total}} = \text{Number of drops} \times KE = \left( \frac{1000}{3.0 \times 10^{-5}} \right) \times (1.215 \times 10^{-3}) $$

Alternatively, we can use the formula for total kinetic energy since all drops have the same terminal velocity:

$$ E_{\text{total}} = \frac{1}{2} \times \text{Total mass} \times v^2 = \frac{1}{2} \times M \times v^2 $$

Substituting $$ M = 1000 $$ kg and $$ v = 9 $$ m/s:

$$ E_{\text{total}} = \frac{1}{2} \times 1000 \times (9)^2 = \frac{1}{2} \times 1000 \times 81 $$

First, $$ \frac{1}{2} \times 1000 = 500 $$:

$$ 500 \times 81 = 40500 \text{ J} $$

Which is $$ 4.05 \times 10^4 $$ J.

Hence, the energy transferred by rain to each square metre of the surface is $$ 4.05 \times 10^4 $$ J.

Comparing with the options:

A. $$ 3.5 \times 10^5 $$ J

B. $$ 4.05 \times 10^4 $$ J

C. $$ 3.0 \times 10^5 $$ J

D. $$ 9.0 \times 10^4 $$ J

So, the answer is Option B.