Question

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

Answer 1

(a) Rainfall is measured extensively covering the entire nation along with an approximate demarcation of the areas involved. This will give the data for computing the total volume of rainfall in the country. The projection of the trend for future can be forecast in general based on the statistical data collected over years. An average rainfall of about 100 cm during monsoon in India is already recorded by the meteorologist.

(b)Consider a ship of known base area floating in the sea. Measure its depth $d_1$ in the sea. The volume of water displaced by the ship:

$V_1=Ad$. Now, move an elephant on the ship and measure the depth of the ship, $d_2$ in this case. The volume of water displaced by the ship with the elephant on board: $V_2=A{d}_2$. The volume of water displaced by the elephant= $A{d}_2-A{d}_1$.The density of water$=D$. Mass of elephant = $D\times (A{d}_2-A{d}_1)$

(c) Wind speed during a storm can be measured by an anemometer. As the wind blows, it rotates. The rotation made by the anemometer in one second gives the value of wind speed.

or
Wind speed can be estimated by floating a gas-filled balloon at a known height $h$. At the onset of wind, the angle drift of balloon in one second can be estimated. Using simple trigonometry, we can estimate the wind speed.

(d) Area of the head surface carrying hair $=A$

With the help of a screw gauge, the diameter and hence, the radius of a hair can be determined. Let it be $r$.
Area of one hair = $r^2$
Number of strands of hair = Total surface area / Area of one hair =$\frac{A}{r^2}$

(e) Let the volume of the room be $V$

One mole of air at NTP occupies $22.4$ $l$ i.e., $22.4$ $\times$10${}^{-3}$ $m^3$ volume.
Number of molecules in one mole =$6.023$ $\times$10${}^{23}$
Number of molecules in room of volume $V$
=$6.023$ $\times$ 10${}^{23}$ $V/22.4$ $\times$10${}^{-3}$ = $134.915$ $\times$ 10${}^{26}$ $V=1.35$ $\times$ 10${}^{28}$ $V$