NCERT Solutions for Class 11th: Ch 2 Units and Measurements Physics Science

2.15. A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :
m = m0 / (1-v2)1/2
Guess where to put the missing c.AnswerGiven the relation,
m = m/ (1-v2)1/2
Dimension of m = M1 L0 T0
Dimension of m0 = M1 L0 T0
Dimension of v = M0 L1 T–1
Dimension of v2 = M0 L2 T–2
Dimension of c = M0 L1 T–1
The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S. This is only possible when the factor, (1-v2)1/2 is dimensionless i.e., (1 – v2) is dimensionless. This is only possible if v2 is divided by c2. Hence, the correct relation is
m = m0 / (1 – v2/c2)1/2

2.16. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10-10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms ?AnswerRadius of hydrogen atom, r = 0.5 Å = 0.5 × 10-10 m
Volume of hydrogen atom = (4/3) π r3
= (4/3) × (22/7) × (0.5 × 10-10)3
= 0.524 × 10-30 m3
1 mole of hydrogen contains 6.023 × 1023 hydrogen atoms.
∴ Volume of 1 mole of hydrogen atoms = 6.023 × 1023 × 0.524 × 10–30
= 3.16 × 10–7 m3

2.17. One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large ?AnswerRadius of hydrogen atom, r = 0.5 Å = 0.5 × 10-10 m
Volume of hydrogen atom = (4/3) π r3
= (4/3) × (22/7) × (0.5 × 10-10)3
= 0.524 × 10-30 m3
Now, 1 mole of hydrogen contains 6.023 × 1023 hydrogen atoms.
∴ Volume of 1 mole of hydrogen atoms, Va = 6.023 × 1023 × 0.524 × 10–30
= 3.16 × 10–7 m3
Molar volume of 1 mole of hydrogen atoms at STP,
Vm = 22.4 L = 22.4 × 10–3 m3

Hence, the molar volume is 7.08 × 104 times higher than the atomic volume. For this reason, the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom.2.18. Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).AnswerLine of sight is defined as an imaginary line joining an object and an observer’s eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly.
On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.2.19. The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 × 1011 m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1 (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1 (second) of arc from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of meters?

Answer

Diameter of Earth’s orbit = 3 × 1011 m
Radius of Earth’s orbit, r = 1.5 × 1011 m
Let the distance parallax angle be 1″ = 4.847 × 10–6 rad.
Let the distance of the star be D.
Parsec is defined as the distance at which the average radius of the Earth’s orbit subtends an angle of 1″

= 0.309 × 10-6 ≈ 3.09 × 1016 m
Hence, 1 parsec ≈ 3.09 × 1016 m

2.20. The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?

Answer

Distance of the star from the solar system = 4.29 ly
1 light year is the distance travelled by light in one year.
1 light year = Speed of light × 1 year
= 3 × 108 × 365 × 24 × 60 × 60 = 94608 × 1011 m
∴ 4.29 ly = 405868.32 × 1011 m
∵ 1 parsec = 3.08 × 1016 m
∴ 4.29 ly = 405868.32 × 1011 / 3.08 × 1016  =  1.32 parsec
Using the relation,
θ = d / D
where,
Diameter of Earth’s orbit, d = 3 × 1011 m
Distance of the star from the earth, D = 405868.32 × 1011 m
∴ θ = 3 × 1011 / 405868.32 × 1011  =  7.39 × 10-6 rad
But, 1 sec = 4.85 × 10–6 rad
∴ 7.39 × 10-6 rad = 7.39 × 10-6 / 4.85 × 10-6 =  1.52

2.21. Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate 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 of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

Answer

It is indeed very true that precise measurements of physical quantities 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-atomic separation or inter-planer spacing.

The development of mass spectrometer makes it possible to measure the mass of atoms precisely.

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.

Answer

(a) During monsoons, a Metrologist records about 215 cm of rainfall 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, A × h = 7.09 × 1012 m3
Density of water, ρ = 1 × 103 kg m–3
Hence, mass of rain water = ρ × V = 7.09 × 1015 kg
Hence, the total mass of rain-bearing clouds over India is approximately 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 the ship (d2) in this case.
Volume of water displaced by the ship with the elephant on board, VbeAd2
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. As wind blows, it rotates. The rotation made by the anemometer in one second gives the 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 radius of a hair can be determined. Let it be r.
∴Area of one hair = πr2
Number of strands of hair ≈ Total surface area / Area of one hair = A / πr2

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

2.23. The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 × 1030 kg, radius of the Sun = 7.0 × 108 m.

Answer

Mass of the Sun, M = 2.0 × 1030 kg
Radius of the Sun, R = 7.0 × 108 m
Density, ρ = ?

The density of the Sun is in the density range of solids and liquids. This high density is attributed to the intense gravitational attraction of the inner layers on the outer layer of the Sun.

2.24. When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72 of arc. Calculate the diameter of Jupiter.

Answer

Distance of Jupiter from the Earth, D = 824.7 × 106 km = 824.7 × 109 m
Angular diameter = 35.72 = 35.72 × 4.874 × 10-6 rad
Diameter of Jupiter = d
Using the relation,
θ = d/ D
d = θ D = 824.7 × 109 × 35.72 × 4.872 × 10-6
= 143520.76 × 103 m = 1.435 × 105 Km

Additional Exercises

2.25. A man walking briskly in rain with speed v must slant his umbrella forward making an angle θ with the vertical. A student derives the following relation between θ and v : tan θ = v and checks 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 rain falls vertically for a stationary man). Do you think this relation can be correct ? If not, guess the correct relation.

Answer

Incorrect; on dimensional ground
The relation is tan θ = ν
Dimension of R.H.S = M0 L1 T–1
Dimension of L.H.S = M0 L0 T0
(∵ The trigonometric function is 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 be dimensionless. One way to achieve this is by dividing the R.H.S by the speed of rainfall ν’
Therefore, the relation reduces to
tan θ = ν / ν’
This relation is dimensionally correct.

2.26. It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s?

Answer

Error in 100 years = 0.02 s

Hence, the accuracy of a standard caesium clock in measuring a time interval of 1 s is 10-12 s.

2.27. Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase : 970 kg m-3. Are the two densities of the same order of magnitude ? If so, why ?

Answer

Diameter of sodium atom = Size of sodium atom = 2.5 Å
Radius of sodium atom, r = (1/2) × 2.5 Å = 1.25 Å = 1.25 × 10-10 m
Volume of sodium atom, V = (4/3) π r3
= (4/3) × 3.14 × (1.25 × 10-10)3 = VSodium
According to the Avogadro hypothesis, one mole of sodium contains 6.023 × 1023 atoms and has a mass of 23 g or 23 × 10–3 kg.
∴ Mass of one atom = 23 × 10-3 / 6.023 × 1023  Kg = m1
Density of sodium atom, ρ = m1 / VSodium
Substituting the value from above, we get
Density of sodium atom, ρ =4.67 × 10-3 Kg m-3
It is given that the density of sodium in crystalline phase is 970 kg m–3.

Hence, the density of sodium atom and the density of sodium in its crystalline 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 in the crystalline phase.

2.28. The unit of length convenient on the nuclear scale is a Fermi : 1 f = 10-15 m. Nuclear sizes obey roughly the following empirical relation :
r = r0 A1/3
where r is the radius of the nucleus, A its mass number, and r0 is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

Answer

Radius of nucleus r is given by the relation,
r = r0 A1/3
r0 = 1.2 f = 1.2 × 10-15 m
Volume of nucleus, V = (/ 3) π r3
(/ 3) π (rA1/3)3 = (/ 3) π rA    ….. (i)

Now, the mass of a nuclei M is equal to its mass number i.e.,
M = A amu = A × 1.66 × 10–27 kg
Density of nucleus, ρ = Mass of nucleus / Volume of nucleus
= A X 1.66 × 10-27 / (4/3) π r03 A

= 3 X 1.66 × 10-27 / 4 π r03  Kg m-3
his relation shows that nuclear mass depends only on constant r0. Hence, the nuclear mass densities of all nuclei are nearly the same.
Density of sodium nucleus is given by,
ρSodium = 3 × 1.66 × 10-27 / 4 × 3.14 × (1.2 × 10-15)3
= 4.98 × 1018 / 21.71 = 2.29 × 1017 Kg m-3

2.29. A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has 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 the Moon’s surface. How much is the radius of the lunar orbit around the Earth?

Answer

Time taken by the laser beam to return to Earth after reflection from the Moon = 2.56 s
Speed of light = 3 × 108 m/s
Time taken by the laser beam to reach Moon  = 1 / 2 × 2.56 = 1.28 s
Radius of the lunar orbit = Distance between the Earth and the Moon = 1.28 × 3 × 108 = 3.84 × 108 m = 3.84 × 105 km

2.30. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine?
(Speed of sound in water = 1450 m s-1).

Answer

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 the distance between the ship and the submarine (2S).
Time taken for the sound to reach the submarine = 1/2 × 77 = 38.5 s
∴ Distance between the ship and the submarine (S) = 1450 × 38.5 = 55825 m = 55.8 km

2.31. The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us ?

Answer

Time taken by quasar light to reach Earth = 3 billion years
= 3 × 109 years
= 3 × 109 × 365 × 24 × 60 × 60 s
Speed of light = 3 × 108 m/s
Distance between the Earth and quasar
= (3 × 108) × (3 × 109 × 365 × 24 × 60 × 60)
= 283824 × 1020 m
= 2.8 × 1022 km

2.32. It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

Answer

The position of the Sun, Moon, and Earth during a lunar eclipse is shown in the given figure.
Position of sun moon during lunar eclipse
Distance of the Moon from the Earth = 3.84 × 108 m
Distance of the Sun from the Earth = 1.496 × 1011 m
Diameter of the Sun = 1.39 × 109 m

It can be observed that ΔTRS and ΔTPQ are similar. Hence, it can be written as:
PQ / RS = VT / UT
1.39 x 109 / RS = 1.496 × 1011 / 3.84 × 108
RS = (1.39 × 3.84 / 1.496) × 106 = 3.57 × 106 m
Hence, the diameter of the Moon is 3.57 × 106 m.

2.33. A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?

Answer

One relation consists of some fundamental constants that give the age of the Universe by:

t = (e2/4πε0)2 × (/ mpmec3G)

Where,
t = Age of Universe
e = Charge of electrons = 1.6 ×10–19 C
ε0 = Absolute permittivity
m= Mass of protons = 1.67 × 10–27 kg
m= Mass of electrons = 9.1 × 10–31 kg
c = Speed of light = 3 × 108 m/s
G = Universal gravitational constant = 6.67 × 1011 Nm2 kg–2
Also, 1 / 4πε0 = 9 × 10Nm2/C2
Substituting these values in the equation, we get
t = (1.6 × 10-19)4 × (× 109)2 / (9.1 × 10-31)2 × 1.67 × 10-27 × (× 108)3 × 6.67 × 10-11

=  [ (1.6)× 81 / 9.1 × 1.67 × 27 × 6.67 ] × 10-76+18-62+27-24+11 seconds
[(1.6)4 × 81 / 9.1 × 1.67 × 27 × 6.67 × 365 × 24 × 3600 ] × 10-76+18+62+27-24+11 years
≈  6 X 10-9 × 1018 years
= 6 billion years.