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

Hint : In order to solve this question, we are going to solve every part one by one depending on the things that have been asked in that part. The parts solution contain the rough measurement based on a random idea that we can take for the measurement of the mass of elephant, wind speed, number of strands, etc

Complete Step By Step Answer:
(a) For the measurement of the rainfall, the rainfall in all the areas covering the extensive part of the rainfall in the nation is considered and is summed up as a whole. Thus, the data of the total rainfall of the whole year will be computed by this. The projection of this trend of rainfall can be forecast for the future of the nation. An average rainfall of $100cm$ during the monsoon in India has been recorded by meteorologists.
(b) Consider a ship with the known surface area floating at the sea. Measure the depth of the sea, ${d_1}$
Volume of the water displaced by the sea can be calculated as:
${V_1} = A{d_1}$
If an elephant is moved at the ship and the depth, ${d_2}$ is measured
Then, volume of the water displaced is
${V_2} = A{d_2}$
Volume of the water displaced by the elephant is
$V = A{d_2} - A{d_1}$
If density of water is $D$
Mass of elephant is $M = \left( {A{d_2} - A{d_1}} \right)D$
(c) Weight of a floating air balloon at a height $h$ can be used to determine. At the onset of wind, the angle drift of the balloon in one second can be estimated. Using simple trigonometry, we can estimate the wind speed.
(d) Area of head surface carrying hair is equal to $A$
With the help of 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 $= \dfrac{{Total\,surface\,area}}{{Area\,of\,one\,hair}} = \dfrac{A}{{{r^2}}}$
(e) Let the volume of the room be $V$
One mole of air at NTP occupies $22.4l = 22.4 \times {10^{ - 3}}{m^3}$ volume
The number of molecules in the volume $V$ is:
$= \dfrac{{6.023 \times {{10}^{23}}}}{{22.4 \times {{10}^{ - 3}}}} = 134.915 \times {10^{26}}V = 1.35 \times {10^{28}}V$ .

Note :
Alternatively, in part (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.
It is important to note that in all other parts, simple mathematical and physical principles have been used.