The total mass of rain-bearing clouds over India during the monsoon
If meteorologist record 10 cm of average rainfall during monsoon then
Height of  average rainfall (h) = 10 cm = 0.1 m
Area of India (A) = 3.3 million square km
= 3.3 x 10⁶ square km  [\(\because \) 1 million = 10⁶]
= 3.3 x 10⁶(10³m)²
= 3.3 x 10⁶ x 10⁶m² = 3.3 x 10¹²m²   [\(\because \) 1 km = 10³]
Volume of rain water (V) = Area x height
= Axh = 3.3 x 10¹²m² x 0.1m
=3.3x10¹¹m³
Density of water (\(\rho \) ) = 10³ kg/m³
Mass of rain water (m) = Volume x Density
= m = vx 
= 3.3 x 10¹¹m³ x 10³ kg/m³
[Density = [\(\because \) mass/Volume]
= 3.3 x 10¹⁴ kg
Therefore, total mass of rain bearing clouds over India during the monsoon is 3.3 x 10¹⁴ kg.

We can measure the mass of an elephant using the law of flotation. According to which, in floating condition the mass of an object dipped in a liquid is equal to the mass of the liquid displaced by the submerged part of the body in liquid. We take a boat of known base area A. Now, empty submerged part of the boat in water (d₁ ) is measured. 
The volume of water displaced by the boat (V₁) = Ad₁
\(\therefore \) Now move the elephant into this boat and again depth of the submerged part of the boat in water (d₂) is measured.
The volume of water displaced by the boat elephant 
V₂ = Ad₂
Volume of water displaced by the elephant
\(={ V }_{ 2 }-{ V }_{ 1 }={ Ad }_{ 2 }-{ Ad }_{ 1 }=A({ d }_{ 2 }-{ d }_{ 1 })\)
If is the density of water then, mass of the elephant
= mass of the water displaced by the elephant
= A(d₂-d₂)\(\rho \)

The number of strands of hair on our head
If we assume a uniform distribution of strands of hair on head then, the number of strands of hair
= Area of the head/Area of cross-section of hair
The thickness of a strand of hair is measured by an appropriate instrument, if it is obtained
d = 5 x 10^-5 = 5 x 10^-3 cm
Then, area of cross-section of human hair
 \(=\pi \left( \frac { d }{ 2 } \right) ^{ 2 }=\frac { \pi d^{ 2 } }{ 4 } \)
\(\\ =\frac { 3.14\times (5\times 10^{ -3 })^{ 2 } }{ 4 } =\frac { 3.14\times 25 }{ 4 } \times 10^{ -6 }cm^{ 2 }\)
Average radius of human head (r)=8 cm
 \(\because \) Area of human head \(=\pi r^{ 2 }=3.14\times (8)^{ 2 }\)
3.14 x 64 cm²
\(\therefore \)The number of strands of hair \(=\frac { 3.14\times 64 }{ 3.14\times \frac { 25 }{ 4 } \times 10^{ -6 } } \)
 \(\approx 10\times 10^{ 6 }=10^{ 7 }\)

The number of air molecules in your classroom
We measure the length(l), breadth (b) and height (h) of our classroom.
\(\therefore \) The volume of classroom = Volume of air in classroom
 \(V=l\times b\times h\)
We know that at NTP, one mole of air occupies a volume of 22.4 L i.e., 22.4 x 10^-3m³
Number of air molecules in one mole of air
= Avogadro number(N) = 6.023 x 10²³
\(\therefore \) Number of air molecules in our classroom
\(=\frac { 6.023\times 10^{ 23 } }{ 22.4\times 10^{ -3 } } \times (l\times b\times h)\)