(a) The average rainfall of nearly $100 \mathrm{~cm}$ or $1 \mathrm{~m}$ is recorded by meteorologists, during Monsoon, in India. If $\mathrm{A}$ is the area of the country, then $\mathrm{A}=3.3$ million sq. $\mathrm{km}=3.3 \times 10^6(\mathrm{~km})^2$ $=3.3 \times 10^6 \times 10^6 \mathrm{~m}^2=3.3 \times 10^{12} \mathrm{~m}^2$
Mass of rain - bearing clouds
$=$ Volume of rain water $\times$ density of water $=3.3 \times 10^{12} \mathrm{~m}^2 \times 1 \mathrm{~m} \times 1000 \mathrm{~kg} / \mathrm{m}^3$ $=3.3 \times 10^{15} \mathrm{~kg}$.
(b) Measure the depth of an empty boat in water. Let it be $d_1$. If $A$ be the base area the boat, then volume of water displaced by boat, $V_1=A \mathrm{~d}_1$.
Let $d_2$ be the depth of boat in water when the elephant is moved into the boat. Volume of water displaced by (boat $+$ elephant), $V_2=A d_2$
Volume of water displaced by elephant,
$$
V=V_2-V_1=A\left(d_2-d_1\right)
$$
If $\rho$ be the density of water, then mass of elephant $=$ mass of water displaced by it $=A\left(d_2-d_1\right) \rho$.
(c) Wind speed during storms can be estimated by floating a gas-filled balloon in air at a known height $h$. When there is no wind, the balloon is at $\mathrm{A}$. Suppose the wind starts blowing to the right such that the balloon reaches position B in 1 second. Now, $A B=d=h \theta$.


The value of directly gives the wind speed.
(d) Let us assume that the hair on our head are uniformly distributed. Thickness of a human hair is $5 \times 10^{-5} \mathrm{~m}$.
Number of hair on the head
$=\frac{\text { area of the head }}{\text { area of cross-section of a hair }}$
Taking head to be a circle of radius $8 \mathrm{~cm}$, number of hair on head
$$
=\frac{\pi(0.08)^2}{\pi\left(5 \times 10^{-5}\right)^2}=\frac{64 \times 10^{-4}}{25 \times 10^{-10}}=2.56 \times 10^6
$$
The number of hair on the human head is of the order of one million.
(e) Consider a class room of size $10 \mathrm{~m} \times 8 \mathrm{~m} \times 4 \mathrm{~m}$. Volume of this room is $320 \mathrm{~m}^3$. We know that $22.4 \mathrm{~L}$ or $22.4 \times 10^{-3} \mathrm{~m}^3$ of air has $6.02 \times 10^{23}$ molecules (equal to' Avogadro's number).
$\therefore \quad$ Number of molecules of air in the class room
$$
=\frac{6.02 \times 10^{23}}{22.4 \times 10^{-3}} \times 320=8.6 \times 10^{27}
$$