Question

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 monsoonb. The mass of an elephantc. The wind speed during a stormd. The number of strands on of hair on your head e. The number of molecules in your classroom

Answer 1

(a) During rainstorm, a Metrologist records around 215 cm of precipitation in India i.e., the tallness of water segment, h = 215 cm = 2.15 m

Territory of nation, A = 3.3 × 1012 m2

Thus, volume of rain water, V = A × h = 7.09 × 1012 m3

Thickness of water, ρ = 1 × 103 kg m– 3

Thus, mass of rain water = ρ × V = 7.09 × 1015 kg

Thus, the aggregate mass of rain-bearing mists over India is roughly 7.09 × 1015 kg.

(b) Consider a ship of realized base territory drifting in the ocean. Measure its profundity in ocean (say d1).

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

Presently, move an elephant on the ship and measure the profundity of the ship (d2) for this situation.

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

Volume of water dislodged by the elephant = Ad2 – Ad1

Thickness of water = D

Mass of elephant = AD (d2 – d1)

(c) Wind speed amid a tempest can be estimated by an anemometer. As wind blows, it turns. The pivot made by the anemometer in one moment gives the estimation of wind speed.

(d) Area of the head surface conveying hair = A

With the assistance of a screw measure, the distance across and thus, the range of a hair can be resolved. Give it a chance to be r.

∴Area of one hair = πr2

Number of strands of hair ≈ Total surface region/Area of one hair = A/πr2

(e) Let the volume of the room be V.

One mole of air at NTP involves 22.4 l i.e., 22.4 × 10– 3 m3 volume.

Number of atoms in a single mole = 6.023 × 1023

∴Number of atoms in room of volume V

= 6.023 × 1023 × V/22.4 × 10-3 = 134.915 × 1026 V = 1.35 × 1028 V

Answer 2

Hey dear,

# Ways for rough estimate-

(i) The total mass of rain-bearing clouds over India during the Monsoon-
Mass of rainwater over India = avg rain fall × area of india × density of water

(ii) The mass of an elephant-
By lever (mass around 3000 kg for adult one)

(iii) The wind speed during storm-
By couple of devices including balloons (80-300 kmph)

(iv) The number of strands of hair on our head-
No of human hair = area of hair / cross sectional area of a hair

(v) The number of air molecules in your classroom-
No of air molecules in given class room

Hope that is helpful...