Answer :

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 × 10^{12} m^{2}

Volume of water column, V = AH

⇒ V = 3.3 × 10^{12} m^{2} × 2.0 m

⇒ V = 6.6 × 10^{12} m^{3}

Density of water, ρ = 10^{3} kg m^{-3}

Mass of the rain-water, m = Vρ

⇒ m = 6.6 × 10^{12} m^{3} × 10^{3} kg m^{-3}

⇒ m = 6.6 × 10^{15} 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 = 10^{3} kg m^{-3}

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

⇒ m = A(h’-h)×10^{3} 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 = πr^{2}

So, number of hairs, n = A/a = A/πr^{2}

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)πr^{3}

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

⇒

⇒ n = 3lbh/4πr^{3}

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

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