Updated By SaraNextGen

On April 21, 2024, 11:35 AM

Question 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 =  = 1.32 parsec

Using the relation,

But, 1 sec = 4.85 × 10^–6 rad

∴

Question 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.

Question 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, V = 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, V_b = 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, V_be= 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. 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 

(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

= = 134.915 × 10²⁶ V

= 1.35 × 10²⁸ V

Question 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

Volume of the Sun, V =

Density of the Sun =

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.

Question 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   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 = 

Diameter of Jupiter = d

Using the relation,

Question 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:

Answer: Incorrect; on dimensional ground

The relation is .

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 .

Therefore, the relation reduces to

. This relation is dimensionally correct.

Question 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:

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 time difference of 0.02 s.

In 1s, the clock will show a time difference of .

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

Question 2.27:

Estimate the average mass density of a sodium atom assuming its size to be about 2.5 . (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 

Radius of sodium atom, r =

= 1.25 × 10^–10 m

Volume of sodium atom, V =

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 =

Density of sodium atom, ρ =

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.

Question 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 : 

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,

 … (i)

= 1.2 f = 1.2 × 10^–15 m

Volume of nucleus, V =

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 =

This relation shows that nuclear mass depends only on constant . Hence, the nuclear mass densities of all nuclei are nearly the same.

Density of sodium nucleus is given by,

Question 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 =

Radius of the lunar orbit = Distance between the Earth and the Moon = 1.28 × 3 × 10⁸ = 3.84 × 10⁸ m = 3.84 × 10⁵ km