Question -
Answer -
(a) During monsoons, a metrologist records about 215 cm ofrainfall in India i.e., the height of water column,┬аh┬а= 215 cm= 2.15 m
Area of country,┬аA┬а= 3.3 ├Ч 1012┬аm2
Hence, volume of rain water,┬аV┬а=┬аA┬а├Ч┬аh┬а=7.09 ├Ч 1012┬аm3
Density of water,┬а╧Б┬а= 1 ├Ч 103┬аkgmтАУ3
Hence, mass of rain water =┬а╧Б┬а├Ч┬аV┬а=7.09 ├Ч 1015┬аkg
Hence, the total mass of rain-bearing clouds over India isapproximately 7.09 ├Ч 1015┬аkg.
(b) Consider a ship of known base area floating in the sea.Measure its depth in sea (say┬аd1).
Volume of water displaced by the ship,┬аVb┬а=┬аA┬аd1
Now, move an elephant on the ship and measure the depth of theship (d2) in this case.
Volume of water displaced by the ship with the elephant onboard,┬аVbe=┬аAd2
Volume of water displaced by the elephant =┬аAd2┬атАУ┬аAd1
Density of water =┬аD
Mass of elephant =┬аAD┬а(d2┬атАУ┬аd1)
(c) Wind speed during a storm can be measured by an anemometer. Aswind blows, it rotates. The rotation made by the anemometer in one second givesthe value of wind speed.
(d) Area of the head surface carrying hair =┬аA
With the help of a screw gauge, the diameter and hence, the radiusof a hair can be determined. Let it be┬аr.
тИ┤Area of one hair = ╧Аr2
Numberof strands of hair┬а
(e) Let the volume of the room be┬аV.
One mole of air at NTP occupies 22.4 l i.e., 22.4 ├Ч 10тАУ3┬аm3┬аvolume.
Number of molecules in one mole = 6.023 ├Ч 1023
тИ┤Number of molecules in room ofvolume┬аV
=
= 134.915 ├Ч 1026┬аV
= 1.35 ├Ч 1028┬аV