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.
a)
During the monsoon, the average rainfall recorded by Meteorologist in India is,
h = 100 cm = 1 m
The area of India is,
A = 3.3 × 10 12 m 2
We know that the density of water is 1000 kg / m 3 .
The mass of rain bearing cloud is given as,
m = ρ A h
Where, m is the mass of rain bearing cloud, ρ is the density of water, A is the area of India and h is the average rainfall recorded by meteorologist in India.
By substituting the values in the above expression, we get
m = 1000 kg / m 3 × 3.3 × 10 12 m 2 × 1 m = 3.3 × 10 15 kg
Hence, the total mass of the rain bearing cloud over India during monsoon is 3.3 × 10 15 kg .
b)
Consider an empty boat floating in water.
The volume of water displaced by empty boat is given as,
V 1 = A d 1
Where, V 1 is the volume of water displaced by empty boat, A is the area of the base of boat and d 1 is the depth of the boat immersed in water.
Now, moved the elephant into the boat and measure the depth of boat in water.
The volume of water displaced by boat when elephant is moved into the boat is,
V 2 = A d 2
Where, V 2 is the volume of water displaced by boat when elephant is moved in the boat and d 2 is the depth of the boat immersed in the water when elephant is moved into the boat.
The volume of water displaced by elephant is given as,
V = V 2 − V 1 = A d 2 − A d 1 = A ( d 2 − d 1 )
The mass of the elephant is given as,
m = ρ V = ρ A ( d 2 − d 1 )
Thus, the mass of the element is ρ A ( d 2 − d 1 ) .
c)
The wind speed during a storm can be estimated by using a gas-filled balloon. The gas must be lighter than air, so that balloon floats in air.
The figure below shows the required apparatus to measure the speed of wind during wind storm.
The position A of the balloon is the position when there is no wind and position B of the balloon is the position when the wind is blowing and displacing the balloon from its original position.
By knowing the value of d , the speed of the wind storm can be calculated.
The speed of wind can also be calculated by using an anemometer. The blowing wind rotates the anemometer and the rotation of anemometer in one minute gives the information about the wind speed during the wind storm.
d)
The radius of human head is given as,
R = 8 × 10 − 2 m
The thickness of human hair measured by screw gauge is given as,
t = 5 × 10 − 5 m
The number of strands of hair on human is given as,
n = A H a h = π R 2 π t 2 = ( R t ) 2
Where, n is the number of strands of hair on human, R is the radius of human head, t is the thickness of human hair.
By substituting the values in the above expression, we get
n = ( 8 × 10 − 2 5 × 10 − 5 ) 2 = 2.56 × 10 6
Thus, the number of strands of hair on human is 2.56 × 10 6 .
e)
Let us assume the length of the classroom is 10 m, the width of the classroom is 8 m and the height of the classroom is 4 m.
The volume of the class room is,
V = l b h
Where, V is the volume of classroom, l is the length of the class room, b is the width of the classroom and h is the height of the classroom.
By substituting the values in the above expression, we get
V = 10 m × 8 m × 4 m = 320 m 2
We know that 22.4 × 10 − 3 m 3 of air have 6.02 × 10 23 molecules of air.
The number of molecules present in the classroom is given as,
n = 6.02 × 10 23 22.4 × 10 − 3 × V
By substituting the values in the above expression, we get
n = 6.02 × 10 23 22.4 × 10 − 3 × 320 = 8.6 × 10 27
Thus, the number of air molecules in the classroom is 8.6 × 10 27 .
a)
During the monsoon, the average rainfall recorded by Meteorologist in India is,
The area of India is,
We know that the density of water is
The mass of rain bearing cloud is given as,
Where,
By substituting the values in the above expression, we get
Hence, the total mass of the rain bearing cloud over India during monsoon is
b)
Consider an empty boat floating in water.
The volume of water displaced by empty boat is given as,
Where,
Now, moved the elephant into the boat and measure the depth of boat in water.
The volume of water displaced by boat when elephant is moved into the boat is,
Where,
The volume of water displaced by elephant is given as,
The mass of the elephant is given as,
Thus, the mass of the element is
c)
The wind speed during a storm can be estimated by using a gas-filled balloon. The gas must be lighter than air, so that balloon floats in air.
The figure below shows the required apparatus to measure the speed of wind during wind storm.
The position A of the balloon is the position when there is no wind and position B of the balloon is the position when the wind is blowing and displacing the balloon from its original position.
By knowing the value of
The speed of wind can also be calculated by using an anemometer. The blowing wind rotates the anemometer and the rotation of anemometer in one minute gives the information about the wind speed during the wind storm.
d)
The radius of human head is given as,
The thickness of human hair measured by screw gauge is given as,
The number of strands of hair on human is given as,
Where,
By substituting the values in the above expression, we get
Thus, the number of strands of hair on human is
e)
Let us assume the length of the classroom is 10 m, the width of the classroom is 8 m and the height of the classroom is 4 m.
The volume of the class room is,
Where,
By substituting the values in the above expression, we get
We know that
The number of molecules present in the classroom is given as,
By substituting the values in the above expression, we get
Thus, the number of air molecules in the classroom is
