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NCERT Solutions Class 11 Physics Chapter 2 Units and Measurements have a one-step solution to all your questions about Chapter 2 Units and Measurements. Students who are looking for study material for preparation for their CBSE boards examination then they are at the right place. Our NCERT Solutions will not only help students learn the basic concepts of Chapter 2 Units and Measurement but it will also help them prepare for competitive exams like JEE Mains, JEE Advance, NTSE, and Olympiad. 

In this NCERT Solutions Class 11 Physics Chapter 2 Units and Measurements we have discussed all kinds of queries which include NCERT intext questions and exercise questions. Experts of the subject matter have made the solutions with utmost concern that it is easy to understand for the students. All the important topics related to the chapter on Units and Measurements are discussed here, such as

  • Physical Quantities – Such physical properties of the substance or any system which can be quantified by measurement and can be expressed as numerical values are called Physical quantities. Example – Mass, amount of substance, length, time, temperature, electric current, light intensity, force, velocity, density, etc.
  • Units – a definite magnitude of a quantity is called the unit of measurement. A unit of measurement is defined and adopted by the convention of laws.
    • Properties of Unit
      • It must be well defined
      • Must be of suitable size
      • Can be easily reproducible as per need
      • Must not change with time
      • Should remain constant under any influence like temperature, pressure, etc.
      • Must be easy to compare experimentally similar kinds of physical quantities.
    • Types of Units
      • Fundamental Units – A base unit that is used for the measurement of base quantities are called a fundamental unit. Fundamental units of measurement are
      • Name Symbol Quantity
        meter m Length
        kilogram kg Mass
        second s Time
        kelvin K Temperature
        ampere A Electric current
        mole mol Number of particles
        candela cd Luminous intensity

      • Derived Units – derived units used for measurement of seven base units adopted by the international system of units are called SI derived units. Some examples of derived units are 

        Name Symbol Quantity
        Hertz Hz Frequency
        Radian rad angle
        newton N force, weight
        farad F electrical capacitance
        ohm Ω electrical resistance, impedance, reactance
        weber Wb magnetic flux
        degree Celsius ֯C temperature relative to 273.15 K
        Becquerel Bq radioactivity (decays per unit time)

    • System of Units
      •  FPS System
      • CGS System
      • MKA System
      • SI System
  • Dimensional formula
  • Units and Dimensions of some derived quantities
  • Principal of Homogeneity

These solutions are based on the latest syllabus approved by the CBSE. NCERT Solutions Class 11 Physics will help students learn tough concepts with ease. Our solution is designed in such a way that it will assist students in their last-minute preparation. Our expert on the subject advises that one must go through all the topics mentioned here to score good marks in their exam.

All the solutions are at your fingertip, get started now.

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NCERT Solutions Class 11 Physics Chapter 2 Units and Measurements

1. Fill in the blanks:

(i) The volume of a cube of side 1 cm is equal to.....m

(ii) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ... (mm)

(iii) A vehicle moving with a speed of 18 km h–1 covers....m in 1 s 

(iv) The relative density of lead is 11.3. Its density is ....g cm–3 or . ...kg m–3.

Solution:

(i) 1 cm = \(\frac 1{100}m\)

Volume of the cube = 1 cm3

But, 1 cm3 = 1 cm × 1 cm × 1 cm = (\(\frac 1{100}\) ) m x  (\(\frac 1{100}\) ) m x  (\(\frac 1{100}\) )m

∴ 1 cm3 = 10–6 m3

Hence, the volume of a cube of side 1 cm is equal to 10–6 m3

(ii) The total surface area of a cylinder of radius r and height h is S = 2πr (r + h). 

Given that, r = 2 cm = 2 × 1 cm = 2 × 10 mm = 20 mm 

h = 10 cm = 10 × 10 mm = 100 mm = 15072 = 1.5 × 104 mm2

∴ S = 2 x3.14 x 20 x (20 +100)

(iii) Using the conversion,

1 km/h = \(\frac 5{18}\)m/s

18 km/h = 18 x \(\frac 5{18}\) = 5 m/s

Therefore, distance can be obtained using the relation: 

Distance = Speed × Time = 5 × 1 = 5 m 

Hence, the vehicle covers 5 m in 1 s. 

(iv) Relative density of a substance is given by the relation,

Relative density = \(\frac{Density \,of\,substance}{Density\,of\,water}\)

Density of water = 1 g/cm3

Density of lead = Relative density of lead x Density of water

= 11.3 x 1 = 11.3 g/cm3

Again, 1g = \(\frac {1}{1000}\) kg

1 cm3 = 10–6 m3

1 g/cm3 = \(\frac {10^{-3}}{10^{-6}}\)kg/m3 = 103 kg/m3

∴ 11.3 g/cm3 = 11.3 × 103 kg/m3

2. Fill in the blanks by suitable conversion of units:

(i) 1 kg m2 s–2 = ....g cm2 s–2 

(ii) 1 m =..... ly

(iii) 3.0 m s–2 =.... km h–2 

(iv) G= 6.67 × 10–11 N m2 (kg)–2 =.... (cm)3 s–2 g–1.

Solution:

(i) 1 kg = 103

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

(ii) Light year is the total distance travelled by light in one year. 

1 ly = Speed of light × One year 

= (3 × 108 m/s) × (365 × 24 × 60 × 60 s) 

= 9.46 × 1015 m

∴ 1m = \(\frac {1}{9.46\times10^{15}}\) = 1.057 x 10-16 ly

(iii) 1 m = 10–3 km

Again, 1 s = \(\frac {1}{3600}\)h

1 s–1 = 3600 h–1

1 s–2 = (3600) 2 h–2 

∴ 3 m s–2 = (3 × 10–3 km) × ((3600)2 h–2 )= 3.88 × 10–4 km h–2

(iv) 1 N = 1 kg m s–2

1 kg = 10–3 g–1 

1 m3 = 106 cm3 

∴ 6.67 × 10–11 N m2 kg–2 

= 6.67 × 10–11 × (1 kg m s–2 ) (1 m2 ) (1 s–2

= 6.67 × 10–11 × (1 kg × 1 m3 × 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

3. A calorie is a unit of heat or energy 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 γ2 in terms of the new units.

Solution:

Given that, 1 calorie = 4.2 (1 kg) (1 m2) (1 s–2

New unit of mass = α kg

Hence, in terms of the new unit, 1 kg = 1/α = α-1

In terms of the new unit of length,

1 m = 1/β = β-1 or 1 m2 = β-2

And, in terms of the new unit of time,

1 s = \(\frac 1γ =γ ^{-1}\)

1 s2 = γ-2

1 s-2 = γ2

∴ 1 calorie = 4.2 (1 α–1) (1 β–2) (1 γ2) = 4.2 α–1 β–2 γ2

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: 

Atoms are very small objects a jet plane moves with great speed the mass of Jupiter is very large the air inside this room contains a large number of molecules a proton is much more massive than an electron the speed of sound is much smaller than the speed of light.

Solution:

The given statement is true because a dimensionless quantity may be large or small in comparison 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. 

  • An atom is a very small object in comparison to a soccer ball. 
  • A jet plane moves with a speed greater than that of a bicycle. 
  • Mass of Jupiter is very large as compared to the mass of a cricket ball.
  • The air inside this room contains a large number of molecules as compared to that present in a geometry box. 
  • A proton is more massive than an electron. 
  • Speed of sound is less than the speed of light.
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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? 

Solution:

Distance between the Sun and the Earth: 

= Speed of light × Time taken by light to cover the distance 

Given that in the new unit, speed of light = 1 unit 

Time taken, t = 8 min 20 s = 500 s 

Distance between the Sun and the Earth = 1 × 500 = 500 units

6. Which of the following is the most precise device for measuring length: a vernier callipers with 20 divisions on the sliding scale a screw gauge of pitch 1 mm and 100 divisions on the circular scale an optical instrument that can measure length to within a wavelength of light? 

Solution:

A device with minimum count is the most suitable to measure length. 

Least count of vernier callipers 

= 1 standard division (SD) – 1 vernier division (VD)

= 1 - \(\frac {9}{10} = \frac {1}{10}\) = 0.01 cm

Least count of screw gauge = \(\frac {Pitch}{Number\,of\,divisions} = \frac {1}{1000}\) = 0.001 cm

Least count of an optical device = Wavelength of light ∼ 10–5cm = 0.00001 cm 

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

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? 

Solution: 

Magnification of the microscope = 100 

Average width of the hair in the field of view of the microscope = 3.5 mm 

∴Actual thickness of the hair is \(\frac {3.5}{100}\) = 0.035 mm.

8. Answer the following: 

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

(ii) 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? 

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

Solution: 

(i) Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a meter scale. The diameter of the thread is given by the relation.

Diameter = \(\frac {Length\,of\,thread}{Number\,of\,turns}\)

(ii) It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only. 

(iii) A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.

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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? 

Solution: 

Area of the house on the slide = 1.75 cm2 

Area of the image of the house formed on the screen = 1.55 m2 

= 1.55 × 104 cm2

Arial magnification, ma = \(\frac {Area\, of\, image}{Area\,of\,object} = \frac {1.55}{1.75}\times 10^4\)

Linear magnifications, ml = \(\sqrt {m_a}\)

\(\sqrt {\frac {1.55}{1.75}} \times10^4\) = 94.11

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 

Solution:

a). The given quantity is 0.007 m2. If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) are insignificant. This means that here, two zeros after the decimal are not significant. Hence, only 7 is a significant figure in this quantity. 

b). The given quantity is 2.64 × 1024 kg. Here, the power of 10 is irrelevant for the determination of significant figures. Hence, all digits i.e., 2, 6 and 4 are significant figures. 

c). The given quantity is 0.2370 g cm–3. For a number with decimals, the trailing zeroes are significant. Hence, besides digits 2, 3 and 7, 0 that appears after the decimal point is also a significant figure. 

d). The given quantity is 6.320 J. For a number with decimals, the trailing zeroes are significant. Hence, all four digits appearing in the given quantity are significant figures. 

e). The given quantity is 6.032 Nm–2 . All zeroes between two non-zero digits are always significant. 

f). The given quantity is 0.0006032 m2. If the number is less than one, then the zeroes on the right of the decimal point (but left to the first non-zero) are insignificant. Hence, all three zeroes appearing before 6 are not significant figures. All zeros between two non-zero digits are always significant. Hence, the remaining four digits are significant figures.

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. 

Solution:

Length of sheet, l = 4.234 m 

Breadth of sheet, b = 1.005 m 

Thickness of sheet, h = 2.01 cm = 0.0201 m 

The given table lists the respective significant figures:

Quantity  Number Significant Figure
l 4.234 4
b 1.005 4
h 2.01 3

Hence, area and volume both must have least significant figures i.e., 3. 

Surface area of the sheet = 2 (l × b + b × h + h × l) 

= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)

= 2 (4.25517 + 0.02620 + 0.08510) 

= 2 × 4.360 

= 8.72 m2 

Volume of the sheet = l × b × h 

= 4.234 × 1.005 × 0.0201 

= 0.0855 m3 

This number has only 3 significant figures i.e., 8, 5, and 5

12. The mass of a box measured by a grocer’s balance is 2.300 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?

Solution:

(a) Mass of grocer’s box = 2.300 kg 

Mass of gold piece I = 20.15g = 0.02015 kg 

Mass of gold piece II = 20.17 g = 0.02017 kg 

Total mass of the box = 2.3 + 0.02015 + 0.02017 = 2.34032 kg 

In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is 2.3 kg. 

(b) Difference in masses = 20.17 – 20.15 = 0.02 g

In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

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

P = \(\frac {a^3b^2}{(\sqrt{cd})}\)

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?

Solution:

 P = \(\frac {a^3b^2}{(\sqrt{cd})}\)

\(\frac {\triangle P}{P} = \frac {3\triangle a}{a}+ \frac {2\triangle b}{b} + \frac 12 \frac {\triangle c}{c}+ \frac {\triangle d}{d}\)

\((\frac {\triangle P}{P} \times 100) \% = (3\times \frac {\triangle a}{a} \times 100+2\times \frac {\triangle b}{b} \times 100 + \frac 12 \times \frac {\triangle c}{c}\times 100 + \frac {\triangle d}{d}\times 100)\%\)

= 3 x 1 + 2 x 3 + 1/2 x 4 + 2

= 3 + 6 + 2 + 2 = 13 %

Percentage error in P = 13 % 

Value of P is given as 3.763. 

By rounding off the given value to the first decimal place, we get P = 3.8.

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

y = a sin \((\frac {2 \pi t}{T})\)

y = a sin vt

y = \((\frac aT) sin \frac ta\)

y = (a√2) \((sin\frac {2 \pi t}{T} + cos\frac {2 \pi t}{T})\)

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

Solution:

Correct  

y = a sin \(\frac {2\pi t}{T}\)

Dimension of y = M0 L1 T0 

Dimension of a = M0 L1 T0 

Dimension of  sin \(\frac {2\pi t}{T}\)= M0 L0 T0 

∵ Dimension of L.H.S = Dimension of R.H.S 

Hence, the given formula is dimensionally correct. 

Incorrect  

y = a sin vt 

Dimension of y = M0 L1 T0  

Dimension of a = M0 L1 T0  

Dimension of vt = M0 L1 T–1 × M0 L0 T1 = M0 L1 T0

But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the given formula is dimensionally incorrect.

Incorrect y = \((\frac {a}{T}) sin (\frac ta)\)

Dimension of y = M0 L1T0

Dimension of \(\frac aT\)= M0 L1T-1

Dimension of \(\frac ta\)= M0 L-1T1

But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the formula is dimensionally incorrect.

Correct

y =  (a√2) \((sin\,2 \pi\frac { t}{T} + cos\,2 \pi\frac { t}{T})\) 

Dimension of y = M0 L1 T0 

Dimension of a = M0 L1 T0 

Dimension of \(\frac tT\)= M0 L0 T0 

Since the argument of the trigonometric function must be dimensionless (which is true in the given case), the dimensions of y and a are the same. Hence, the given formula is dimensionally correct.

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:

\(m = \frac {m_0}{(1-v^2)^{\frac 12}}\)

Solution:

Given the relation, \(m = \frac {m_0}{(1-v^2)^{\frac 12}}\)

Dimension of m = M1 L0 T

Dimension of = 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 − 2)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

\(m = \frac {m_0}{(1-\frac {v^2}{c^2})^{\frac 12}}\)

16. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å ∶ 1Å = 10−10 The size of a hydrogen atom is about 0.5Å, what is the total atomic volume in m3 of a mole of hydrogen atoms? 

Solution:

Radius of hydrogen atom, r = 0.5 A = 0.5 × 10–10 m

Volume of hydrogen atom = \(\frac 43 \pi r^3\)

\(\frac 43 \times \frac {22}7 \times (0.5\times 10^{-10})^3\)

= 0.524 x 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

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 A). Why is this ratio so large? 

Solution:

Radius of hydrogen atom, r = 0.5 A= 0.5 × 10–10

Volume of hydrogen atom = \(\frac 43\pi r^3\)

\(\frac 43 \times \frac {22}7 \times (0.5\times 10^{-10})^3\)

= 0.524 x 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 m

Molar volume of 1 mole of hydrogen atoms at STP, 

Vm = 22.4 L = 22.4 × 10–3 m3

∴ \(\frac {V_m}{V_a} = \frac {22.4 \times 10^{-3}}{3.16\times10^{-7}} = 7.08 \times 10^4\)

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.

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). 

Solution:

Line 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.

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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 meters? 

Solution: 

Diameter of Earth’s orbit = 3 × 1011

Radius of Earth’s orbit, r = 1.5 × 1011

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''.

∴ We have \(\theta = \frac rD\)

D = \(\frac rD\) = \(\frac {1.5 x 10^{11}}{4.847\times10^{-6}}\) = 0.309 x 10-6 ≈ 3.09 x 1016 m

Hence, 1 parsec ≈ 3.09 x 1016 m

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? 

Solution: 

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

∴ 4.29 ly = 405868.32 × 1011

∵ 1 parsec = 3.08 × 1016 m

∴ 4.29 ly = \(\frac {405868.32\times 10^{11}}{3.08\times 10^{16}}\) = 1.32 parsec

Using the relation, \(\theta = \frac dD\)

where.

Diameter of Earth's orbit, d = 3 x 1011m

Distance of the star from the Earth, D = 405868.32 × 1011

∴ \(\theta = \frac {3\times 10^{11}}{405868.32 \times 10^{11}}\) = 7.39 x 10-6 rad

But, 1 sec = 4.85 x 10-6 rad

7.39 x 10-6 rad = \(\frac {7.39\times 10^{-4}}{4.85\times10^{-6}}\) = 1.52''

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. 

Solution:

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.

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): the total mass of rain-bearing clouds over India during the Monsoon the mass of an elephant the wind speed during a storm the number of strands of hair on your head the number of air molecules in your classroom. 

Solution:

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. 

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, Vbe= Ad2 

Volume of water displaced by the elephant = Ad2 – Ad

Density of water = D 

Mass of elephant = AD (d2 – d1

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. 

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 = \(\frac {Total\,surface\, area}{Area\, of\, one\, hair\, \pi} = \frac {A}{r^2}\)

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

\(\frac {6.023 \times 10^{23}}{22.4\times10^{-3}}\times V\) = 134.915 x 1026 V = 1.35 x 1028 V

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. 

Solution:

Mass of the Sun, M = 2.0 × 1030 kg 

Radius of the Sun, R = 7.0 × 108

Volume of the Sun, V =  \(\frac 43\pi r^3\)

\(\frac 43 \times \frac {22}{7} \times (7.0 \times 10^8)^3\)

\(\frac {88}{21} \) x 343 x 1024 = 1437.3 x 1024 m3

Density of the Sun = \(\frac {Mass}{Volume}=\frac {2.0 \times 10^{30}}{1437.3\times 10^{24}}\) = 1.4 x 103 kg/m5

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.

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. 

Solution: 

Distance of Jupiter from the Earth, D = 824.7 × 106 km = 824.7 × 109

Angular diameter = 35.72'' = 35.72 x 4.874 x 10-6 rad

Diameter of Jupiter = d 

Using the relation,

\(\theta = \frac dD\)

d = θD = 824.7 x 109 x 35.27 x 4.872 x 10-6

= 143520.76 x 103 = 1.435 x 105 km

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 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.

Solution:

Answer is Incorrect: 

On dimensional ground, the relation is tan =  

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 tan = v/v′ 

This relation is dimensionally correct.

 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? 

Solution:

Difference in time of cesium clocks = 0.02 s 

Time required for this difference = 100 years 

= 100 × 365 × 24 × 60 × 60 = 3.15 × 109

In 3.15 × 109 s, the cesium clock shows a time difference of 0.02 s. 

In 1s, the clock will show a time difference of \(\frac {0.02}{3.15\times10^9}s\)

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

\(=\frac {3.15\times10^9} {0.02}\) 

= 157.5 x 109

= 1.5 x 1011 s

 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? 

Solution:

Diameter of sodium atom = Size of sodium atom = 2.5 A 

Radius of sodium atom, r = 1/2 x 2.5 A = 1.25 A  = 1.25 × 10–10

Volume of sodium atom, V = \(\frac 43\pi r^3\)

\(\frac 43\times 3.14 \times (1.25 \times 10^{10})^3\)

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 = \(\frac {23\times10^{-3}}{6.023\times 10^{23}}kg\)

Density of sodium atom, ρ = \(\frac {\frac {23\times10^{-3}}{6.023\times10^{23}}}{\frac 43\times 3.14 \times (1.25\times 10^{-10})^3}\) = 4.67 x 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.

 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: = 01/3 

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. 

Solution:

Radius of nucleus r is given by the relation,

= 01/3……………. (i)

r0 = 1.2 f = 1.2 x 10-15m

Volume of nucleus, V = \(\frac 43\pi r^3\)

\(\frac 43\pi (r_0A^{\frac 13})^3 = \frac 43\pi r^3_0A\)

Now, the mass of a nuclei M is equal to its mass number i.e. 

M = A 

a mu = A × 1.66 × 10–27 kg 

Density of nucleus,

p = \(\frac {Mass\,of\,nucleus}{Volume\,of\,nucleus}\)

\(\frac {A\times1.66\times10^{-27}}{\frac 43\pi r^3_0A} = \frac {3\times 1.66\times 10^{-27}}{4\pi r^3_0}\) kg/m3

This 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,

psodium \(\frac {3\times1.66\times10^{-27}}{4\times3.14\times(1.2\times10^{-15})^3}\)

\(\frac {4.98}{21.71}\times 10^{18} = 2.29\times 10^{17} kg\, m^{-3}\)

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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?

Solution:

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 = \(\frac 12 \times 2.56 = 1.28 s\)

Radius of the lunar orbit 

= Distance between the Earth and the Moon 

= 1.28 × 3 × 108 

= 3.84 × 108 m = 3.84 × 105 km 

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).

Solution:

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 = \(\frac 12\) x 77 = 38.5 s

Distance between the ship and the submarine (S) = 1450 × 38.5 = 55825 m = 55.8 km

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?

Solution:

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

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. 

Solution:

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

Distance of the Sun from the Earth = 1.496 × 1011

Diameter of the Sun = 1.39 × 109

It can be observed that ΔTRS and ΔTPQ are similar. Hence, it can be written as:

\(\frac {PQ}{RS} = \frac {VT}{UT} \)

\(\frac {1.39 \times10^9}{RS} = \frac {1.496 \times10^{11}}{3.84 \times10^8}\)

RS = \(\frac {1.39\times3.84}{1.496}\times10^6\) = 3.57 x 106m

Hence, the diameter of the Moon is 3.57× 106 m.

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? 

Solution:

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

\(t = (\frac {e^2}{4\pi ɛ_0 })^2 \times \frac {1}{m_pm_e^2c^3G}\)

Where, 

t = Age of Universe 

e = Charge of electrons = 1.6 ×10–19 

ɛ0  = Absolute permittivity 

mp = Mass of protons = 1.67 × 10–27 kg 

me = 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, \(\frac {1}{4\pi ɛ_0 } \) = 9 x 109 Nm2/C2

Substituting these values in the equation, we get

t = \(\frac {(1.6\times10^{-19})^4 \times (9\times10^9)^2}{(9.1 \times10^{31})^2 \times 1.67 \times 10^{-27}\times (3\times10^8)^3 \times 6.67\times10^{-11}}\)

\(\frac {(1.6)^4 \times 81}{9.1 \times 1.67 \times 27\times 6.67} \times 10 ^{-76+18+62+27-24+11}\) s

=  \(\frac {(1.6)^4\times 81}{9.1\times1.67\times27\times6.67\times365\times24\times3600} \times 10^{-76+18+62+27-24+11}\) years

= 6 x 10-9 x 1018 years

= 6 billion years

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