Question -
Answer -
(a) During monsoons, a metrologist records about 215 cm ofrainfall in India i.e., the height of water column, h = 215 cm= 2.15 m
Area of country, A = 3.3 × 1012 m2
Hence, volume of rain water, V = A × h =7.09 × 1012 m3
Density of water, ρ = 1 × 103 kgm–3
Hence, mass of rain water = ρ × V =7.09 × 1015 kg
Hence, the total mass of rain-bearing clouds over India isapproximately 7.09 × 1015 kg.
(b) Consider a ship of known base area floating in the sea.Measure its depth in sea (say d1).
Volume of water displaced by the ship, Vb = A d1
Now, move an elephant on the ship and measure the depth of theship (d2) in this case.
Volume of water displaced by the ship with the elephant onboard, Vbe= Ad2
Volume of water displaced by the elephant = Ad2 – Ad1
Density of water = D
Mass of elephant = AD (d2 – d1)
(c) Wind speed during a storm can be measured by an anemometer. Aswind blows, it rotates. The rotation made by the anemometer in one second givesthe value of wind speed.
(d) Area of the head surface carrying hair = A
With the help of a screw gauge, the diameter and hence, the radiusof a hair can be determined. Let it be r.
∴Area of one hair = πr2
Numberof strands of hair 
(e) Let the volume of the room be V.
One mole of air at NTP occupies 22.4 l i.e., 22.4 × 10–3 m3 volume.
Number of molecules in one mole = 6.023 × 1023
∴Number of molecules in room ofvolume V
=
= 134.915 × 1026 V
= 1.35 × 1028 V