Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):

A. the total mass of rain-bearing clouds over India during the Monsoon

B. the mass of an elephant

C. the wind speed during a storm

D. the number of strands of hair on your head

E. the number of air molecules in your classroom.

A. Height of water column during monsoon is recorded as 215 cm.

H = 200 cm = 2.0 m

Area of the country, A = 3.3 × 10¹² m²

Volume of water column, V = AH

⇒ V = 3.3 × 10¹² m² × 2.0 m

⇒ V = 6.6 × 10¹² m³

Density of water, ρ = 10³ kg m^-3

Mass of the rain-water, m = Vρ

⇒ m = 6.6 × 10¹² m³ × 10³ kg m^-3

⇒ m = 6.6 × 10¹⁵ kg

The mass of the rain-bearing clouds should be approximately equal to this mass.

B. Consider a boat of known base-area A. Measure the depth upto which it sinks in water when it is empty. Let this height be h.

Volume of water displaced by boat, v = Ah

Measure the depth again when elephant is kept on the boat. Let this depth be h’.

Volume of water displaced, V = Ah’

Volume of water displaced by elephant, V’ = V – v = A(h’-h)

The mass of this volume of water is equal to the mass of the elephant.

Density of water, ρ = 10³ kg m^-3

Mass of water displaced by elephant, m = V’ρ

⇒ m = A(h’-h)×10³ kg

This is approximately equal to the mass of the elephant.

C. Consider a balloon filled with air. When there is no wind, note the height of the balloon from a fixed point. When the wind blows, measure the distance of the balloon from the fixed point after a certain time. The displacement of balloon can be calculated from the angular displacement. Now, the ratio of the displacement to the time of flight is the speed of wind.

Also, an anemometer can be used to measure the speed of the wind. The number of rotations in one second gives the speed of wind.

D. Let A be the area of the head covered with hair.

If r is the radius of hair strand, the base area of hair strand,

a = πr²

So, number of hairs, n = A/a = A/πr²

This gives an approximation of the number of hair strands on a head.

E. If l, b and h are the length, breadth and height of the classroom, then its volume is v = lbh.

If r is the radius of an air molecule, then volume of the air molecule, v’ = (4/3)πr³

So, number of air molecules in the classroom, n = v/v’

⇒

⇒ n = 3lbh/4πr³

This gives an approximation of the number of air molecules in the classroom.