Units and Measurement
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Question : 22
Total: 33
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) 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 Monsoon in India, the average rain fall is about 100 cm
i.e. 1 m over the area of the country, which is about
A = 3.3 × 10 6 k m 2
= 3.3 × 10 6 × 10 6
= 3.3 × 10 12 m 2
Therefore, volume of the rain water,
V = A
h = 3.3 × 10 12 × 1
= 3.3 × 10 12 m 3
Now, density of water,ρ = 103 k g m – 3
Hence, the total mass of rain-bearing clouds over India,
m = V
ρ = 3.3 × 10 12 × 10 3
= 3.3 × 10 15 k g
(b) To estimate the mass of an elephant, consider a boat having base areaA in a river. Mark a point on the boat upto which it is inside the water.
Now, move the elephant into the boat and again mark a point on theboat upto which it is inside the water. If h is the distance between the twomarks, then
Volume of the water displaced by the elephant,V = A h
According to Archimedes’ principle, mass of the elephant,
M = mass of the water displaced by the elephant
Ifρ ( = 10 3 k g m – 3 ) M = V ρ = A h ρ
(c) The wind speed during a storm can be found by measuring the angleof drift of an air balloon in a known time.
During the storm, suppose that the balloon moves tothe point B in an extremely small time t as shown infigure.
Ifθ Δ O A B
tan θ =
=
( ∵ A B = x )
orx = h tan θ
Hence the wind speed during the storm,v =
=
(d) Let the area of the hair-bearing head be equal to A. With a finemicrometer, measure the thickness d (diameter) of the hair. Then,
area of cross-section of the hair,a = π d 2 ∕ 4
If we ignore the interspacing between the hair, then the number of strands of hair on the head,n =
=
=
(e) At N.T.P., one mole of air occupies a volume of 22.4 litres i.e.22.4 × 10 – 3 m 3 ( = 6.023 × 10 23 )
Therefore, number of air molecules per m 3 , n =
= 2.69 × 10 25
Suppose that the dimensions of the classroom are7 m × 5 m × 4 m
Therefore, volume of the classroom,
V = 7 m × 5 m × 4 m
= 140 m 3
Therefore, number of air molecules in the classroom,
N = V . n
= 140 × 2.69 × 10 25
= 3.77 × 10 27
i.e. 1 m over the area of the country, which is about
Therefore, volume of the rain water,
Now, density of water,
Hence, the total mass of rain-bearing clouds over India,
(b) To estimate the mass of an elephant, consider a boat having base areaA in a river. Mark a point on the boat upto which it is inside the water.
Now, move the elephant into the boat and again mark a point on theboat upto which it is inside the water. If h is the distance between the twomarks, then
Volume of the water displaced by the elephant,
According to Archimedes’ principle, mass of the elephant,
M = mass of the water displaced by the elephant
If
(c) The wind speed during a storm can be found by measuring the angleof drift of an air balloon in a known time.
Consider that an air balloon is at the point A at avertical height h above the observation point O on theground, when there is no wind storm.
During the storm, suppose that the balloon moves tothe point B in an extremely small time t as shown infigure.
If
or
Hence the wind speed during the storm,
(d) Let the area of the hair-bearing head be equal to A. With a finemicrometer, measure the thickness d (diameter) of the hair. Then,
area of cross-section of the hair,
If we ignore the interspacing between the hair, then the number of strands of hair on the head,
(e) At N.T.P., one mole of air occupies a volume of 22.4 litres i.e.
Therefore, number of air molecules per
Suppose that the dimensions of the classroom are
Therefore, volume of the classroom,
Therefore, number of air molecules in the classroom,
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