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.

(N/A) During monsoons,the average rainfall in India is about $215 \ cm$,i.e.,the height of the water column $h = 2.15 \ m$. The area of India is $A \approx 3.3 \times 10^{12} \ m^2$. The volume of rainwater $V = A \times h \approx 7.09 \times 10^{12} \ m^3$. Given the density of water $\rho = 10^3 \ kg/m^3$,the mass of rain water is $M = \rho V \approx 7.09 \times 10^{15} \ kg$.
$(b)$ Consider a ship of known base area $A$ floating in the sea. Measure its depth $d_1$. When an elephant is placed on the ship,the depth increases to $d_2$. The volume of water displaced by the elephant is $V = A(d_2 - d_1)$. The mass of the elephant is $M = \rho_{water} \times A(d_2 - d_1)$.
$(c)$ Wind speed during a storm can be estimated using an anemometer,which measures the rate of rotation of cups driven by the wind. Alternatively,one can observe the displacement of objects over a known time interval.
$(d)$ Measure the surface area of the head $A$ and the diameter of a single strand of hair $d$ using a screw gauge. The number of strands $N \approx A / (\pi (d/2)^2)$.
$(e)$ Let the volume of the room be $V$. At $NTP$,one mole of air occupies $22.4 \times 10^{-3} \ m^3$. The number of molecules in one mole is $6.023 \times 10^{23}$. Thus,the number of molecules in the room is $N = (V / 22.4 \times 10^{-3}) \times 6.023 \times 10^{23} \approx 2.7 \times 10^{25} \times V$.