(a) The metrologist records 325 cm of rain, which is the height of the water column. h = 325 cm = 3.25 m

Area = 3.3 x 10¹² m²

Volume of water = A x h = 3.25 x 3.3 x 10¹² = 10.725 x 10¹² m³

Density of water = 1 x 10³ kg m^-3

Therefore, the mass of rain water = $\rho$ x V = 1 x 10³ x 10.725 x 10¹² = 10.725 x 10¹⁵ kg

Thus, the total mass of rain-bearing clouds is 10.725 x 10¹⁵ kg

(b) 

Let a known base area be floating in the sea. Let the depth of the sea be d₁.

The volume of water displaced = A d₁

Now measure the depth of the ship with an elephant onboard.

The volume of water displaced = A d₂

From the above equations, the volume of water displaced by the elephant = A d₁ – A d₂

Water density = D

Elephant’s mass = AD(d₂ – d₁)

(c) Anemometer is used to measure the speed of the wind. As the wind blows, it rotates the anemometer and the number of rotations per second gives the wind speed.

(d) The surface area of the head = A

Let r be the radius

Area of one hair = $\pi$r²

Number of strands of hair = $\frac{Total<mspace\; width="1em"></mspace>surface<mspace\; width="1em"></mspace>area}{Area<mspace\; width="1em"></mspace>of<mspace\; width="1em"></mspace>one<mspace\; width="1em"></mspace>hair}=\frac{A}{r^2}$$$

(e) Let V be the volume of the room

In mole, the number of molecules = 6.023 x 10²³

One mole of air = 22.4 x 10^-3 m³ volume

Number of molecules in room = $\frac{6.023\times {10}^{23}}{22.4\times {10}^{-3}}=134.915\times {10}^{26}$$$