Answer

(a) Length of edge = 1cm = 1/100 m
Volume of the cube = side³
Putting the value of side, we get
Volume of the cube = (1/100 m)³

The volume of a cube of side 1 cm is equal to 10^-6 m³

(b) Given,
Radius, r = 2.0 cm = 20 mm (convert cm to mm)
Height, h = 10.0 cm =100 mm
The formula of total surface area of a cylinder S = 2πr (r + h)
Putting the values in this formula, we get
Surface area of a cylinder S = 2πr (r + h = 2 x 3.14 x 20 (20+100)
= 15072 = 1.5 × 10⁴ mm²
The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to 1.5 × 10⁴ mm²

(c) Using the conversion,
Given,
Time, t = 1 sec
speed = 18 km h^-1 = 18 km / hour
1 km = 1000 m and 1hour = 3600 sec
Speed = 18 × 1000 /3600 sec = 5 m /sec
Use formula
Speed = distance / time
Cross multiply it, we get
Distance = Speed × Time = 5 × 1 = 5 m
A vehicle moving with a speed of 18 km h^–1covers 5 m in 1 s.

(d) Density of lead = Relative density of lead × Density of water
Density of water = 1 g/cm³
Putting the values, we get
Density of lead = 11.3 × 1 g/ cm³
= 11.3 g cm^-3
1 cm = (1/100 m) =10^–2 m3
1 g = 1/1000 kg = 10^-3 kg
Density of lead = 11.3 g cm^-3 = 11.3
Putting the value of 1 cm and 1 gram
11.3 g/cm³ = 11.3 × 10^-3 kg (10^-2m)^-3 = 11.3 ×10^–3 × 10⁶ kg m^-3 =1.13 × 10³ kg m^–3
The relative density of lead is 11.3. Its density is 11.3 g cm^-3 g cm^–3 or 1.13 × 10³ kg m^–3.

2.2. Fill in the blanks by suitable conversion of units:
(a) 1 kg m²s^–2= ____ g cm² s^–2
(b) 1 m =____ ly
(c) 3.0 m s^–2=____ km h^–2
(d) G= 6.67 × 10^–11 N m² (kg)^–2=____ (cm)³s^–2 g^–1.

Answer

(a) 1 kg = 10³ g
1 m² = 10⁴ cm²
1 kg m² s^–2 = 1 kg × 1 m² × 1 s^–2
=10³ g × 10⁴ cm² × 1 s^–2 = 10⁷ g cm² s^–2
1 kg m²s^–2= 10⁷ g cm² s^–2

(b) Distance = Speed × Time
Speed of light = 3 × 10⁸ m/s
Time = 1 year = 365 days = 365 × 24 hours = 365 × 24 × 60 × 60 sec
Putting these values in above formula we get
1 light year distance = (3 × 10⁸ m/s) × (365 × 24 × 60 × 60 s) = 9.46 × 10¹⁵ m
9.46 × 10¹⁵ m = 1 ly
So that 1 m = 1/ 9.46 × 10¹⁵ ly = 1.06 × 10^-16 ly

(c) 1 hour = 3600 sec so that 1 sec = 1/3600 hour
1 km = 1000 m so that 1 m = 1/1000 km
3.0 m s^–2 = 3.0 (1/1000 km)( 1/3600 hour)^-2 = 3.0 × 10^–3 km × ((1/3600)^-2 h–2)
= 3.0 × 10^–3 km × (3600)² h^–2 = 3.88 × 10⁴ km h^–2
3.0 m s^–2= 3.88 × 10⁴ km h^–2

(d) Given,
G= 6.67 × 10^–11 N m² (kg)^–2
We know that
1 N = 1 kg m s^–2
1 kg = 10³ g
1 m = 100 cm = 10² cm
Putting above values, we get
6.67 × 10^–11 N m² kg^–2 = 6.67 × 10^–11 × (1 kg m s^–2) (1 m²) (1Kg^–2)
Solve and cancel out the units we get
⇒ 6.67 × 10^–11 × (1 kg^–1 × 1 m³ × 1 s^–2)
Putting above values to convert Kg to g and m to cm
⇒ 6.67 × 10^–11 × (10³ g)^-1 × (10² cm)³ × (1 s^–2)
⇒ 6.67 × 10^–11 × 10^-3 g^-1 × 10⁶ cm³ × (1 s^–2)
⇒ 6.67 × 10^–8 cm³ s^–2 g^–1
G= 6.67 × 10^–11 N m² (kg)^–2= 6.67 × 10^–8 (cm)³s^–2 g^–1.

2.3. A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m²s^–2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α^–1 β^–2 γ² in terms of the new units.

Answer

The given statement is true because a dimensionless quantity may be large or small in comparision to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.
(a) An atom is a very small object in comparison to a soccer ball.
(b) A jet plane moves with a speed greater than that of a bicycle.
(c) Mass of Jupiter is very large as compared to the mass of a cricket ball.
(d) The air inside this room contains a large number of molecules as compared to that present in a geometry box.
(e) A proton is more massive than an electron.
(f) Speed of sound is less than the speed of light.

2.5. A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

Answer

(b) Least count of screw gauge = Pitch/Number of divisions = 1/1000 = 0.001 cm.

(c) Wavelength of light, λ ≈ 10^-5 cm = 0.00001 cm

Hence, it can be inferred that an optical instrument is the most suitable device to measure length.

2.7. A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair?

(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?

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 = V_Sodium
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 / V_Sodium
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 A^1/3
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 A^1/3
r₀ = 1.2 f = 1.2 × 10^-15 m
Volume of nucleus, V = (4 / 3) π r³
= (4 / 3) π (r₀ A^1/3)³ = (4 / 3) π r₀ A    ….. (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,
ρ_Sodium = 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.

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 × (1 / m_pm_e² c³G)

Where,
t = Age of Universe
e = Charge of electrons = 1.6 ×10^–19 C
ε₀ = Absolute permittivity
m_p = Mass of protons = 1.67 × 10^–27 kg
m_e = 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 × 10⁹ Nm²/C²
Substituting these values in the equation, we get
t = (1.6 × 10^-19)⁴ × (9 × 10⁹)² / (9.1 × 10^-31)² × 1.67 × 10^-27 × (3 × 10⁸)³ × 6.67 × 10^-11