Units and Measurements Class 11 NCERT Solutions: Download Topic-wise Questions & Answers PDF

• Introduction to Units and Measurement
• The international system of units
• Measurement of length
• Measurement of mass
• Measurement of time
• Accuracy, precision of instruments and errors in measurement Significant figures
• Dimensions of physical quantities
• Dimensional formulae and dimensional equations
• Dimensional analysis and its applications

Units and Measurement is an important chapter for the students because it forms the foundation of various other concepts. We are providing the solutions for all the NCERT questions given in this chapter for the students. The students can download the NCERT solutions for Units and Measurement Chapter here for free.

Download Here: NCERT Solution for Class XI Physics Units and Measurements Chapter

Q.2.1 Fill in the blanks

Ans.2.1

(a) The volume of a cube of side 1  ${cm}^3$ is equal to  ${10}^{-6}{m}^3$

The volume of the cube = 1 cm × 1 cm × 1 cm = 1 cm³

1 cm3 = 1 × 10^-6 m³

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to1.25 × 10⁴(mm)²

The surface area of a solid cylinder = 2ԯr(r+)h, where r = 2.0 cm = 20 mm, h = 10 cm = 100 mmSurface area = 2 × 22/7 × 20 ×(100 +20)  ${mm}^2$ = 1.51 ×  ${10}^4$  ${mm}^2$

(c) A vehicle moving with a speed of 18 km h^–1 covers 5 m in 1 s

Vehicle speed = 18 km/h = 18000/3600 m/s = 5 m/s

(d) The relative density of lead is 11.3. Its density is 11.3 g cm^–3 or 11.3 × 10³....kg m^–3.

Relative density or Specific gravity = Density of the subject / density of water

Density of water = 1 gm/cc= 10^-3 / 10^-6 

Q.2.2 Fill in the blanks by suitable conversion of units

Ans.2.2(a) 1 kg m² s^–2 = 10⁷g cm² s^–2

kg m²s^-2 = 10³ × 10⁴ = 10⁷

(b) 1 m = 10^-16..... ly

1 m = 1.057× 10^-16 light year

(c) 3.0 m s^–2 = .... km h^–2

= (3/10³) × (3.6 x 10³)² = 3.9 × 10⁴

(d) G = 6.67 × 10^–11 N m² (kg) ^–2 = 6.67 × 10^-8 (cm) ³ s^–2 g^–1.

1N =1 kg = 1000 gm = 10³ gm

G = 6.67 × 10^-11 Nm²(kg) ^-2 = 6.67 × 10^-11 x (1 kg m/s² ) × ( 1 m²) × ( 1kg ^-2)

= 6.67 × 10^-11 × ( 1 kg^-1 × 1m³ × 1s^-2)

= 6.67 × 10^-11 × (103g)^-1 × (102 cm)³× (1s^-2)

= 6.67 × 10^-8 cm³s^-2g^-1

Q.2.3 A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1 kg m² s^–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 γ ² in terms of the new units.

Ans.2.3

1 Calorie = 4.2 J = 4.2 kgm²s^-2

Standard formula for conversion

Given unit / New unit = (M1/M2)^x(L1/L2)^y(T1/T2)^z

Formula for energy = M¹L²T^-2

Here x = 1, y = 2, z = -2

In this problem

M1 = 1 kg, L1 = 1 m, T1 = 1 s and

M2 = α kg, L2 = β m, T2 = γ s

So 1 Calorie = 4.2 (1/α)¹(1/β)²(1/γ)^-2

                        = 4.2α^-1β^-2γ²

Q.2.4 Explain this statement clearly:

“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary:

(a) atoms are very small objects

(b) a jet plane moves with great speed

(c) the mass of Jupiter is very large

(d) the air inside this room contains a large number of molecules

(e) a proton is much more massive than an electron

(f) the speed of sound is much smaller than the speed of light.

Ans.2.4

• Atoms are very small compared to a cricket ball
• Compared to a car on a busy city road, a jet plane moves with great speed
• When compared to the average mass of a human being, mass of Jupiter is very large.
• When compared to the air inside a football, the air inside this room contains a large number of molecules
• A proton has higher mass compared to that of an electron
• The speed of sound is much smaller when calling your friend from your balcony than signaling him with a torch light

Q.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?

Ans.2.5

Distance between the Sun and the Moon = Speed of light × time taken by light to cover the distance

Speed of light = 1 unit

Time taken = 8 min 20 s = 480 + 20 s = 500 s

The distance between the Sun and the Moon = 1 × 500 = 500 units

Q.2.6 Which of the following is the most precise device for measuring length?

(a) a vernier calipers with 20 divisions on the sliding scale

(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale

(c) an optical instrument that can measure length to within a wavelength of light ?

Ans.2.6

• The least count of a vernier calliper = 0.1 mm
• The least count of the screw gauge = 1/100 mm = 0.01 mm
• The least count of the optical instrument = wave length of the light = 10^-5cm = 10^-3 mm = 0.001 mm

Hence the answer is (c)

Q.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?

Ans.2.7

Microscope magnification = 100

Average width of the hair = 3.5 mm

Hence, actual width of the hair = 3.5/100 = 0.035 mm

Q.2.8 Answer the following:

(a)You are given a thread and a meter scale. How will you estimate the diameter of the thread?

(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?

(c) The mean diameter of a thin brass rod is to be measured by vernier calipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?

Ans.2.8

• We require a small cylindrical object on which the thread can be wrapped. Wind the thread in a circular fashion on to the cylinder so that the thread forms a coil. If L is the length of the thread and n is the number of turns, then the diameter D of the thread can be expressed as D = L/n

The least count of the screw gauge = Pitch / number of divisions on the circular scale 1/200 = 5  $\times {10}^{-3}$

By increasing the number of turns on the circular scale, when denominator increases, the least count will increase and the accuracy will increase. However, when the number of turns increases, there may the problem of readability of the scale.

• Taking 100 measurements, instead of 5 will reduce the probability of error and the observation will be more accurate.

Q.2.9 The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector-screen arrangement?

Ans.2.9

The magnification of the projected image can be expressed as

$m_a$ = Area of the image / Area of the object = 1.55 m² / 1.75 cm² = 1.55 × 10⁴ / 1.75 = 8.857 × ${10}^3$

So linear magnification is  $m_l$ =  $\surd m_a$ = 94.11

Q.2.10 State the number of significant figures in the following:

(a) 0.007 m²

(b) 2.64 × 10²⁴ kg

(c) 0.2370 g cm^–3

(d) 6.320 J

(e) 6.032 N m^–2

(f) 0.0006032 m²

 Ans.2.10

• The only significant digit in this number is 7, hence significant digit number is 1
• For significant value determination, the power of 10 is irrelevant. The relevant is 2.64. Hence the number of significant digit is 3
• In the expression of 0.2370, the 0 before decimal is insignificant, the numbers 2,3,7,0 are significant. The number of significant digit is 4
• All the digits in this number 6.320 are significant. Hence number of significant digit is 4
• 032, in this number all digits 6,0,3,2 are significant. Hence number of significant digit is 4
• 0006032 – here the digits 6,0,3,2 are significant. Hence the number of significant digit is 4

Q.2.11 The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Ans.2.11

Length ,l = 4.234 m

Bredth, b = 1.005 m

Thickness, t = 2.01 cm = 0.0201 m

The area of the rectangular sheet=  $2lb+2lt+2bt$ = 2  $\times (4.234\times 1.005+4.234\times 0.0201+1.005\times 0.0201)m^2$ = 8.72  $m^2$

The volume of the rectangular sheet =  $l\times b\times t$ = 4.234  $\times 1.005\times 0.0201m^3$ = 85.53  $\times {10}^{-3}{m}^3$

Q.2.12 The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?

Ans.2.12

The mass of the box = 2.30 kg = 2300 g

Mass of the 1st gold piece = 20.15 g

Mass of the 2nd gold piece = 20.17 g

Total mass of the box with gold pieces = 2300 + 20.15 + 20.17 = 2340.32 g = 2.3 kg

The mass difference between 2 gold pieces = 20.17 – 20.15 = 0.02 g

Q.2.13 A physical quantity P is related to four observables a, b, c and d as follows:

P =a3b2/(√c d)

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?

Ans.2.13

From the given equation,

ΔP = (3Δa/a + 2Δb/b+0.5Δc/c +Δd/d)

Given value Δa/a = 1%, Δb/b = 3%, Δc/c = 4% and Δd/d = 2 %

Hence,

(ΔP/P) × 100% = 3 × 1 + 2 × 3 + 0.5 × 4 + 2 = 13

So ΔP = 13% of P

P = 3.763, so ΔP = 13 × 3.763 = 0.489 = 0.5

Hence error lies in the 1st decimal place. So P = 3.8

Q.2.14 A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:

(a) y = a sin 2π t/T

(b) y = a sin vt

(c) y = (a/T) sin t/a

(d) y =( a√2) (sin 2ԯt / T + cos 2ԯt / T )

(a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.

Ans.2.14

The notation for a dimension can be expressed in the form of M^x L^y T^z

 For (a):

1. LHS y = M⁰ L¹ T⁰
2. RHS a = M⁰ L¹ T⁰ , sin 2π t/T = M⁰ L⁰ T⁰

Both LHS and RHS have contained similar dimension, hence correct.

For (b):

1. LHS y = M⁰ L¹ T⁰
2. RHS a = M⁰ L¹ T^0, sin vt = M⁰ L¹ T⁰

Since LHS and RHS dimensions are not matching, the equation is incorrect

For (c):

1. LHS y = M⁰ L¹ T⁰
2. RHS (a/T) = M⁰ L¹ T^-1 , sin t/a = M⁰ L^-1 T¹

Since LHS and RHS dimensions are not matching, the equation is incorrect

For (d):

1. LHS = M⁰ L¹ T⁰
2. RHS ( a√2) = M⁰ L¹ T⁰ , (sin 2ԯt / T + cos 2ԯt / T ) = M⁰ L¹ T⁰

Both LHS and RHS have contained similar dimension, hence correct.

Q.2.15 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo o f 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 =        m_0/(1-v²)^1/2

 Guess where to put the missing c.

Ans.2.15

From the given equation, we can write m₀/m = (1-v²)^1/2

Now LHS is dimension less, hence RHS also needs to be dimension less. Since v and c have the same dimension i.e. speed, the RHS expression needs to be (1-v² /c²)^1/2 to be dimension less.

Q.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?

Ans.2.16

The radius of the hydrogen atom, r = 0.5 × 10^-10 m

The volume of the hydrogen atom = (4/3) πr³ = 5.238 × 10^-31 m³

1 hydrogen mole contains 6.023 × 10²³ hydrogen atom.

So the volume of 1 mole of hydrogen atom = 6.023 × 10²³ × 5.238 × 10^-31 m³

                                                                                                            = 3.15 × 10 ^-07 m³

Q.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?

Ans.2.17 From 2.16 we have derived, the volume of  1 mole of hydrogen atom = 3.15 × 10 ^{-07 $m^3$}

1 mole of ideal gas volume = 22.4 L = 22.4 × 10^{-3 $m^3$}

Ratio   = (22.4 × 10^-3)/ (3.15 × 10^-7)

          = 7.1 × 10⁴

The molar volume is 7.1 x 10⁴ times more than the atomic volume. So the inter-atomic space in hydrogen atom

is larger than the size of the hydrogen atom.

Q.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).

Ans.2.18

When we look out of the window of a fast moving train, a line of sight is created (imaginary) between our eyes and the object outside. The line of sight of the distant object does not change whereas the line of sight of the objects closer to the train changes rapidly as the train moves. So the nearby objects appear to move rapidly in the opposite direction but the distant object remained stationary.

Q.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 × 10¹¹m. 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 metres?

Ans.2.19

Diameter of the Earth’s orbit = 3 × 10¹¹m, Radius = 1.5 × 10¹¹m

Parallax angle ϴ  = 1” (sec), from the equation π/180 deg = 1 rad and 1 degree = 60 × 60 sec we get Parallax

angle ϴ  = 1” (sec) = 4.848 × 10^-6 rad

Let the distance of the star from the Earth be D

Parsec is defined as the distance at which the average radius of the Earth’s orbit subtends an angle, which

same as Parallax angle.

Therefore D = r/ ϴ = 1.5×10¹¹/4.848×10^-6 = 3.094×10¹⁶ m

Q.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?

Ans.2.20

1 light year = 9.46 × 10¹⁵m

So the distance of the nearest star = 4.29 light year = 4.29 × 9.46 × 10¹⁵m = 4.058 × 10¹⁶m

1 parsec = 3.08 × 10¹⁶m

So the distance of the star = (4.058 × 10¹⁶) / 3.08 × 10¹⁶m = 1.3175 parsec = 1.32 parsec

We know ϴ = d/D

Where d is the diameter of the Earth orbit = 3 × 10¹¹m

D is the distance of the star from the Earth, D = 4.29 light year = 4.058 × 10¹⁶m

Then ϴ = 3 × 10¹¹m / 4.058 ×10¹⁶m = 7.39 × 10^-6 rad = 7.39 × 10^-6 /  4.848 × 10^-6=  1.52 “

Q 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.

Ans.2.21

Precise measurement is a must for any scientific experiment. For precise measurement of length, optical measurement is used. For time, ultra short pulse is used. For mass, mass spectrometer is used for measurement of mass of atoms.

Q.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.

Ans.2.22

• During monsoons, a metrologist records about 215 cm of rainfall in India. So

Height of water column, h = 215 cm = 2.15 m

Area of the country, A = 3.3

$\times {10}^{12}$  $m^2$

Hence, volume of rain water, V = A  $\times h$ = 7.09  $\times {10}^{12}$  $m^3$

Density of water,  $\rho$ = 1  $\times {10}^3$ kg/  $m^3$

Hence, mass of rain water =  $\rho \times V$ = 7.09  $\times {10}^{15}$ kg

Hence, the total mass of rain-bearing clouds over India is approximately 7.09  $\times {10}^{15}$ kg

Consider a ship of known base area floating in the sea.

Let the depth of the ship be  $d_1$

Volume of water displaced by the ship,  $V_b$ = A  $d_1$

Now, move an elephant on the ship and let the depth of the ship be  $d_2$

Volume of water displaced by the ship with elephant on board = A  $d_2$

Volume of water displaced by the elephant = A  $d_2-{Ad}_1$

If the density of the water is  $\rho .\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}$ of the elephant is = A  $\rho (d_2-d_1)$

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 the wind speed.

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, so that area of one hair =  $\pi r^2$

Number of strands of hair =  $\frac{Totalsurfacearea}{Areaofonehair}$ =  $\frac{A}{\pi r^2}$

Let the volume of the room be V

One mole of air at NTP occupies 22.4 lit = 22.4  $\times {10}^{-3}$  $m^3$ volume

Number of molecules in one mole = 6.023  $\times {10}^{23}$

Number of molecules in room of volume V =  $\frac{6.023\mathrm{}\times {10}^{23}}{22.4\mathrm{}\times {10}^{-3}}\times V$ = 2.69  $\times {10}^{25}$ V

Q.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.

Ans.2.23

Mass of the Sun = 2.0 × 10³⁰ kg

Radius of the Sun = 7.0 × 10⁸ m

Volume of the Sun = (4/3)πr³ =  4/3 x π x (7.0 × 10⁸)³ = 1.436 × 10²⁷

Density of the Sun = Mass / volume = 2.0 × 10³⁰ kg / 1.436 × 10²⁷ kg/ m³ = 1.39 × 10³ kg/ m³

The density is in the range of solids and liquids.

Q.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.

Ans.2.24

 Distance of the planet Jupiter = 824.7 million Km = 824.7 x 10⁶ km

 ϴ = 35.72" = 35.72 × 4.85 × 10^-6 rad = 173.242 × 10^-6 rad

Diameter of Jupiter d = ϴ × D = 173.242 × 10^-6 × 824.7 × 10⁶ km = 142872.67 km = 1.43 × 10⁵ km

Q.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 $\to$ 0, ϴ $\mathit{}\mathit{}\to$

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.

Ans.2.25

In the equation v = tan ϴ, the dimension v = L¹T^-1. Since tan ϴ is dimensionless, the equation v = tanϴ is incorrect.

To make it correct v must be divided by the velocity of rain u and the correct expression should be v/u = tan ϴ

Q.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?

Ans.2.26

Total time in 100 years = 100 × 365 × 24 × 60  × 60 s

Error in 100 years = 0.02 sec

Error in 1 sec= 0.02 / (100 × 365 × 24 × 60  × 60) s = 6.34 × 10^-12

Q.2.27Estimate 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 mass density of sodium in its crystalline phase: 970 kg/m³. Are the two densities of the same order of magnitude? If so, why?

Ans.2.27

The diameter of the sodium atom = 2.5 Å = 2.5 × 10^-10 m

Radius of the sodium atom = 1.25 × 10^-10 m

Volume of the sodium atom = (4/3)π r³ = (4/3) × (22/7) × (1.25 × 10^-10)  m³ = 8.18 × 10^-30

Mass of 1 mole atom of sodium = 23 g = 23 × 10^-3 kg

1 mole of sodium contains 6.023 × 10²³ atoms.

Hence mass of 1 sodium atom = (23 × 10^-3)/ (6.023 × 10²³) = 3.818 × 10^-26 kg

Atomic mass density of sodium = M/V = (3.818 × 10^-26)/ (8.18 × 10^-30) kg/m³ = 4667.48  kg/m³

The density of sodium in its solid state is 4667.48 kg/m³ but in the crystalline phase is 970 kg/m^3.

In crystalline phase there are voids in between atoms, hence the density is less.

Q.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.

Ans.2.28

Radius of the nucleus  r = r₀ A^1/3

r₀ = 1.2f = 1.2 × 10^–15 m

Volume of the nucleus = (4/3)πr³ = (4/3)π(r₀ A^1/3)³ = (4/3)π(r₀)³ A

Mass of the nucleus = mA, where m = average mass of the nucleus and A is the number of nucleus

Nucleus mass density = Mass of the nucleus / Volume of the nucleus

                        = mA/(4/3πr³) = 3mA/4πr³A = 3m/4πr₀³

Using m = 1.66 × 10^-27 kg, r₀ = 1.2 ×x 10^–15 m

Nucleus mass density = (3 × 1.66 × 10^-27)/4π(1.2 × 10^–15)³ kg/m³

            = 4.98 × 10^-27/4 × 3.1316 × 1.728 × 10^-45 kg/m³

            = 2.3 × 10¹⁷ kg/m³

The nuclear mass density is much larger than the atomic mass density of sodium, obtained in 2.27

Q.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?

Ans.2.29

Time taken by the laser beam after reflection = 2.56 s

The speed of laser = 3 × 10⁸ m/s       

If the distance between Earth and the Moon is d, the distance covered by laser beam is 2d

From the relation distance = speed × time, we get

d = (2.56 × 3 × 10⁸)/2 = 3.84 × 10⁸ m

Q.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).

Ans.2.30

Speed of the sound in water = 1450 m s^–1

Time taken from generation + reflection = 2t = 77 s

From the relation s = ut, where s is the distance of the enemy ship, we get

s = 1450 × 77/2 m = 55825 m = 55.8 km

Q.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?

Ans.2.31

Speed of light = 3 × 10⁸ m/s, Time taken = 3 billion years = 3 × 10⁹ years = 3 × 365 × 24 × 60 × 60 × 10⁹ s

The distance = speed x time = 3 × 10⁸ × 3 × 365 × 24 × 60 × 60 × 10⁹ m

                                    = 2.838 × 10²⁵ m

Q.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.

Ans.2.32

Distance of the moon from Earth = 3.84 × 10⁸ m

Distance of the Sun from Earth = 1.496 × 10¹¹ m

Sun’s diameter = 1.39 × 10⁹ m

Sun’s angular diameter = 1920° = 1920 × 4.85 × 10^-6 rad = 9.3 × 10^-3 rad

 During total Solar eclipse, the moon completely covers the Sun, then the angular diameter of both Sun and the

Moon will be equal

So the angular diameter of moon, ϴ = 9.3 × 10^-3 rad, distance d of moon from Earth = 3.84 × 10⁸ m and

diameter of the moon = angular diameter × distance

So the approximate diameter of the Moon = 9.3 × 10^-3 × 3.84 × 10⁸ m = 35.712 × 10⁵ m