Chapter 2 Units and Measurements Solutions
Question - 21 : - Precise measurements of physical quantities are a need ofscience. For example, to ascertain the speed of an aircraft, one must have anaccurate method to find its positions at closely separated instants of time.This was the actual motivation behind the discovery of radar in World War II.Think of different examples in modern science where precise measurements oflength, time, mass etc. are needed. Also, wherever you can, give a quantitativeidea of the precision needed.
Answer - 21 : -
It is indeed very true that precise measurements of physicalquantities are essential for the development of science. For example,ultra-shot laser pulses (time interval ∼ 10–15 s)are used to measure time intervals in several physical and chemical processes.
X-ray spectroscopy is used to determine the inter-atomicseparation or inter-planer spacing.
The development of mass spectrometer makes it possible to measurethe mass of atoms precisely.
Question - 22 : - Just as precise measurements are necessary in science, it isequally important to be able to make rough estimates of quantities usingrudimentary ideas and common observations. Think of ways by which you canestimate the following (where an estimate is difficult to obtain, try to get anupper bound on the quantity):
(a) the total mass of rain-bearing clouds over India during theMonsoon
(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.
Answer - 22 : -
(a) During monsoons, a metrologist records about 215 cm ofrainfall in India i.e., the height of water column, h = 215 cm= 2.15 m
Area of country, A = 3.3 × 1012 m2
Hence, volume of rain water, V = A × h =7.09 × 1012 m3
Density of water, ρ = 1 × 103 kgm–3
Hence, mass of rain water = ρ × V =7.09 × 1015 kg
Hence, the total mass of rain-bearing clouds over India isapproximately 7.09 × 1015 kg.
(b) 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 theship (d2) in this case.
Volume of water displaced by the ship with the elephant onboard, Vbe= Ad2
Volume of water displaced by the elephant = Ad2 – Ad1
Density of water = D
Mass of elephant = AD (d2 – d1)
(c) Wind speed during a storm can be measured by an anemometer. Aswind blows, it rotates. The rotation made by the anemometer in one second givesthe value of wind speed.
(d) Area of the head surface carrying hair = A
With the help of a screw gauge, the diameter and hence, the radiusof a hair can be determined. Let it be r.
∴Area of one hair = πr2
Numberof strands of hair 
(e) Let the volume of the room be V.
One mole of air at NTP occupies 22.4 l i.e., 22.4 × 10–3 m3 volume.
Number of molecules in one mole = 6.023 × 1023
∴Number of molecules in room ofvolume V
=
= 134.915 × 1026 V
= 1.35 × 1028 V
The Sun is a hot plasma (ionized matter) with its inner core at atemperature exceeding 107 K, and its outer surface at atemperature of about 6000 K. At these high temperatures, no substance remainsin a solid or liquid phase. In what range do you expect the mass density of theSun to be, in the range of densities of solids and liquids or gases? Check ifyour guess is correct from the following data: mass of the Sun = 2.0 × 1030 kg, radiusof the Sun = 7.0 × 108 m.
Answer - 23 : -
Mass of the Sun, M = 2.0 × 1030 kg
Radius of the Sun, R = 7.0 × 108 m
Volumeof the Sun, V =
Density of the Sun =
The density of the Sun is in the density range of solids andliquids. This high density is attributed to the intense gravitationalattraction of the inner layers on the outer layer of the Sun.
When the planet Jupiter is at a distance of 824.7 millionkilometers from the Earth, its angular diameter is measured to be 
of arc. Calculate thediameter of Jupiter.
Answer - 24 : -
Distance of Jupiter from the Earth, D = 824.7 ×106 km = 824.7 × 109 m
Angulardiameter = 
Diameter of Jupiter = d
Using the relation,
A man walking briskly in rain with speed v mustslant his umbrella forward making an angle θ with the vertical. A studentderives the following relation between θ and v: tan θ = v andchecks that the relation has a correct limit: as v →0, θ→ 0, as expected. (We are assuming there is no strong wind and that the rainfalls vertically for a stationary man). Do you think this relation can becorrect? If not, guess the correct relation.
Answer - 25 : -
Answer: Incorrect;on dimensional ground
Therelation is
Dimension of R.H.S = M0 L1 T–1
Dimension of L.H.S = M0 L0 T0
(
The trigonometric functionis considered to be a dimensionless quantity)
Dimension of R.H.S is not equal to the dimension of L.H.S. Hence,the given relation is not correct dimensionally.
To make the given relation correct, the R.H.S should also bedimensionless. One way to achieve this is by dividing the R.H.S by the speed ofrainfall
.
Therefore, the relation reduces to
. This relation is dimensionally correct.
It is claimed that two cesium clocks, if allowed to run for 100years, free from any disturbance, may differ by only about 0.02 s. What doesthis imply for the accuracy of the standard cesium clock in measuring atime-interval of 1 s?
Answer - 26 : -
Difference in time of caesium clocks = 0.02 s
Time required for this difference = 100 years
= 100 × 365 × 24 × 60 × 60 = 3.15 × 109 s
In 3.15 × 109 s, the caesium clock shows a timedifference of 0.02 s.
In1s, the clock will show a time difference of
Hence, the accuracy of a standard caesium clock inmeasuring a time interval of 1 s is
Question - 27 : - Estimate the average mass density of a sodium atom assuming itssize to be about 2.5
. (Use the known values ofAvogadro’s number and the atomic mass of sodium). Compare it with the densityof sodium in its crystalline phase: 970 kg m–3. Are the twodensities of the same order of magnitude? If so, why?
Answer - 27 : - Diameter of sodium atom = Size of sodium atom =2.5
Radius of sodium atom, r =
= 1.25 × 10–10 m
Volumeof sodium atom, V =
According to the Avogadro hypothesis, one mole of sodium contains6.023 × 1023 atoms and has a mass of 23 g or 23 × 10–3 kg.
∴ Mass of one atom =
Density of sodium atom, ρ =
It is given that the density of sodium in crystalline phase is 970kg m–3.
Hence, the density of sodium atom and the density of sodium in itscrystalline phase are not in the same order. This is because in solid phase,atoms are closely packed. Thus, the inter-atomic separation is very small inthe crystalline phase.
where r is the radius of the nucleus, A itsmass number, and r0 is a constant equal to about,1.2 f. Show that the rule implies that nuclear mass density is nearly constantfor different nuclei. Estimate the mass density of sodium nucleus. Compare itwith the average mass density of a sodium atom obtained in Exercise. 2.27.
Answer - 28 : -
Radius of nucleus r is given by the relation,
… (i)
= 1.2 f = 1.2 × 10–15 m
Volumeof nucleus, V =
Now, the mass of a nuclei M is equal to its massnumber i.e.,
M = A amu= A × 1.66 × 10–27 kg
Density of nucleus,
ρ =
This relation shows that nuclear mass depends only onconstant. Hence, the nuclear mass densities of all nuclei are nearly thesame. Density of sodium nucleus is given by,
Question - 29 : - A LASER is a source of very intense, monochromatic, andunidirectional beam of light. These properties of a laser light can beexploited to measure long distances. The distance of the Moon from the Earthhas been already determined very precisely using a laser as a source of light.A laser light beamed at the Moon takes 2.56 s to return after reflection at theMoon’s surface. How much is the radius of the lunar orbit around the Earth?
Answer - 29 : -
Time taken by the laser beam to return to Earth after reflectionfrom the Moon = 2.56 s
Speed of light = 3 × 108 m/s
Timetaken by the laser beam to reach Moon =
Radius of the lunar orbit = Distance between the Earth and theMoon = 1.28 × 3 × 108 = 3.84 × 108 m = 3.84 ×105 km
A SONAR (sound navigation and ranging) uses ultrasonic waves todetect and locate objects under water. In a submarine equipped with a SONAR thetime delay between generation of a probe wave and the reception of its echoafter reflection from an enemy submarine is found to be 77.0 s. What is thedistance of the enemy submarine? (Speed of sound in water = 1450 m s–1).
Answer - 30 : -
Let the distance between the ship and the enemy submarine be ‘S’.
Speed of sound in water = 1450 m/s
Time lag between transmission and reception of Sonar waves = 77 s
In this time lag, sound waves travel a distance which is twice thedistance between the ship and the submarine (2S).
Timetaken for the sound to reach the submarine
∴ Distance between the ship andthe submarine (S)= 1450 × 38.5 = 55825 m = 55.8 km