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): the total mass of rain-bearing clouds over India during the Monsoon the mass of an elephant the wind speed during a storm the number of strands of hair on your head the number of air molecules in your classroom.

Asked by Abhisek | 1 year ago |  59

Solution :-

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 $$7.09 × 10^{15} kg$$.

Consider a ship of known base area floating in the sea.

Measure its depth in sea (say d1).

Volume of water displaced by the ship, Vb = A d1

Now, move an elephant on the ship and measure the depth of the ship (d2) in this case.

Volume of water displaced by the ship with the elephant on board, Vbe= Ad2

Volume of water displaced by the elephant = Ad2 – Ad1

Density of water = D

Mass of elephant = AD (d2 – d1)

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.

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 = πr2

Number of strands of hair = $$\dfrac{A}{r^2}$$

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

$$\dfrac{ 6.023 × 10^{23}}{ 22.4 × 10^{–3}} \times v$$

$$134.915 × 10^{26} V = 1.35 × 10^{28} V$$

Answered by Pragya Singh | 1 year ago

Related Questions

It is a well-known fact that during a total solar eclipse-the disk of the moon almost completely

It is a well-known fact that during a total solar eclipse-the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

The farthest objects in our Universe discovered by modern astronomers are so distant that light

The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?

A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects

A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects underwater. In a submarine equipped with a SONAR, the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450 m s–1).