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  • 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.

Q 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.

Answers (1)

best_answer

(a) Height of water column during monsoon is recorded as 215 cm.

H = 215 cm = 2.15 m

Area of the country, A = 3.3 \times 10^{12} m^2

Volume of water column, V = AH

 V = 3.3 \times 10^{12} m^2 \times 2.15 m = 7.1 \times 10^{12} m^3

Mass of the rain-bearing clouds over India during the Monsoon, m = Volume x Density

 m = 7.1 \times 10^{12} m^3 \times 10^3 kg m^{-3} =  7.1 \times 10^{15} kg       (Density of water = 103 kg m-3 )

b) Consider a large solid cube of known density having a density less than water.

Measure the volume of water displaced when it immersed in water = v

Measure the volume again when the elephant is kept on the cube = V

The volume of water displaced by elephant, V' = V – v

The mass of this volume of water is equal to the mass of the elephant.

Mass of water displaced by elephant, m = V' x Density of water

This gives the approximate mass of the elephant.

 

(c) A rotating device can be used to determine the speed of the wind. As the wind blows, the number of rotations per second will give the wind speed.

 

(d) Let A be the area of the head covered with hair.

If r is the radius of the root of the hair, the area of the hair strand, a = \pi r^2

So, the number of hair , n = A/a = A/\pi r^2 

 

(e) Let l, b and h be the length, breadth and height of the classroom, \therefore Volume of the room, v = lbh.

The volume of the air molecule,  v' = (4/3)\pi r^3  (r is the radius of an air molecule)

So, the number of air molecules in the classroom, n = v'/v = 4\pi r^3/3lbh

Posted by

Devendra Khairwa

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