__Question 2.22__:

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.

__Solution__:

(a) During monsoons, a Metrologist records about 215 cm of rainfall in India i.e., the height of water column,

*h*= 215 cm = 2.15 m

Area of country,

*A*= 3.3 × 10

^{12}m

^{2}

Hence, volume of rain water,

*V*=

*A*×

*h*= 7.09 × 10

^{12}m

^{3}

Density of water,

*ρ*= 1 × 10

^{3}kg m

^{–3}

Hence, mass of rain water =

*ρ*×

*V*= 7.09 × 10

^{15}kg

Hence, the total mass of rain-bearing clouds over India is approximately 7.09 × 10

^{15}kg.

(b) Consider a ship of known base area floating in the sea. Measure its depth in sea (say

*d*

_{1}).

Volume of water displaced by the ship,

*V*

_{b}=

*A*

*d*

_{1}

Now, move an elephant on the ship and measure the depth of the ship (

*d*

_{2}) in this case.

Volume of water displaced by the ship with the elephant on board,

*V*

_{be}=

*Ad*

_{2}

Volume of water displaced by the elephant =

*Ad*

_{2}–

*Ad*

_{1}

Density of water =

*D*

Mass of elephant =

*AD*(

*d*

_{2}–

*d*

_{1})

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

*r*

^{2}

Number of strands of hair ≈ Total surface area / Area of one hair = A / π

*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 × 10

^{–3}m

^{3}volume.

Number of molecules in one mole = 6.023 × 10

^{23}

∴Number of molecules in room of volume V

= 6.023 X 10

^{23}X V / 22.4 X 10

^{-3}= 134.915 X 10

^{26}V = 1.35 X 10

^{28}V

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