Units and Measurement
Based on NCERT Class 11th Physics Chapter 2
Excercise
Q2.1 Fill in the blanks
(a) The volume of a cube of side 1 cm is equal to ..... m3
Answer: (a) 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 × \(\frac{1}{100}\)m × \(\frac{1}{100}\)m
1 cm3 = 10–6 m3
Hence, the volume of a cube of side 1 cm is equal to 10–6 m3..
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to
...(mm)2
Answer: (b) The total surface area of a cylinder of radius r and height h is S = 2πrh + 2πr2 = 2πr (r + h). Given that, r = 2 cm = 2 × 10 mm = 20 mm ∴S = 2 × 3.14 × 20(20+100) = 15072 = 1.5 × 104 mm2
h = 10 cm = 10 × 10 mm = 100 mm
(c) A vehicle moving with a speed of 18 km h–1 covers .... m in 1 s
Answer: Using Conversion 1 km/h = \(\frac{1000}{3600}\)m/s = \(\frac{5}{18}\)m/s 18 km/h = 18 × \(\frac{5}{18}\) = 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.
(d) The relative density of lead is 11.3. Its density is ....g cm–3 or ....kg m–3.
Answer: 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 × Density of water . = 11.3 × 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 11.3 g/cm3 = 11.3 × 103 kg/m3
Fill in the blanks by suitable conversion of units.
(a) 1 kg m2 s–2 = ....g cm2 s–2
Answer: 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
(b) 1 m = ..... ly
Answer: 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 ∴ 1 m = \(\frac{1}{9.46 × 10^{15}}\) = 1.057 x 10-16 ly
(c) 3.0 m s–2 = .... km h–2
Answer: 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
(d) G = 6.67 × 10–11 N m2 (kg)–2 = .... (cm)3 s–2 g–1.
Answer: 1 N = 1 kg m s–2 1 kg-1 = (103)-1 g-1 = 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 kg–2) = 6.67 × 10–11 × (1 kg-1 × 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
1 kg = 103 g
Q 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 γ 2 in terms of the new units.
Answer: 
Q 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
Answer: Physical quantities are called large or small depending on the unit (standard) of measurement. For example, the distance between two cities on earth is measured in kilometres but the distance between stars or inter —galactic distances are measured in parsec. The later standard parsec is equal to 3.08 x 1016m or 3.08 x 1012 km is certainly larger than metre or kilometre. Therefore, the inter-stellar or intergalactic distances are certainly larger than the distances between two cities on earth. (a) The size of an atom is much smaller than even the sharp tip of a pin. (b) A Jet plane moves with a speed greater than that of a super fast train. (c) The mass of Jupiter is very large compared to that of the earth. (d) The air inside this room contains more number of molecules than in one mole of air. (e) This is a correct statement. – a proton is much more massive than an electron (f) This is a correct statement -the speed of sound is much smaller than the speed of light
Q 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: Distance between Sun and Earth = Speed of light in vacuum x time taken by light to travel from Sum to Earth = 3 x 108 m/ s x 8 min 20 s = 3 x 108 m/s x 500 s = 500 x 3 x 108 m. In the new system, the speed of light in vacuum is unity. So, the new unit of length is 3 x 108 m. .•. distance between Sun and Earth = 500 new units.
Q Which of the following is the most precise device for measuring length :
(a) a vernier callipers 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 ?
Answer: (a) Least count of vernier callipers = 1/20 = 0.05 mm = 5 x 10-5 m (b) Least count of screw gauge =Pitch/No. of divisions on circular scale = 1 x 10-3/100 = 1 x 10-5 m (c) Least count of optical instrument = 6000 Å (average wavelength of visible light as 6000 Å) = 6 x 10-7m As the least count of optical instrument is least, it is the most precise device out of three instruments given to us.
Q 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 ?
Answer: As magnification, m =thickness of image of hair/ real thickness of hair = 100 .•. Thickness of hair =3.5 mm/100 = 0.035 mm
and average width of the image of hair as seen by microscope = 3.5 mm
Q2.8 Answer the following :
(a)You are given a thread and a metre 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 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 ?
Answer: (a) 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 metre scale. The diameter of the thread is given by the relation, Diameter of thread = \(\frac{Length\space of \space thread}{Number\space of\space turns}\) (b) We know that least count = Pitch/number of divisions on circular scale When number of divisions on circular scale is increased, least count is decreased. Hence the accuracy is increased. However, this is only a theoretical idea.Practically speaking, increasing the number of ‘turns would create many difficulties. As an example, the low resolution of the human eye would make observations difficult. The nearest divisions would not clearly be distinguished as separate. Moreover, it would be technically difficult to maintain uniformity of the pitch of the screw throughout its length. (c) Due to random errors, a large number of observation will give a more reliable result than smaller number of observations. This is due to the fact that the probability (chance) of making a positive random error of a given magnitude is equal to that of making a negative random error of the same magnitude. Thus in a large number of observations, positive and negative errors are likely to cancel each other. Hence more reliable result can be obtained.
Q2.9 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.
Answer: Here area of the house on slide = 1.75 cm2 = 1.75 x 10-4 m2 and area of the house of projector-screen = 1.55 m2 .•. Areal magnification =Area on screen/Area on slide = 1.55 m2 / 1.75 x 10-4 m2 = 8.857 x 103 .•. Linear magnification = \(\sqrt{\text Areal, magnification}\) = \(\sqrt{8.857 \times 10^3}\) =94.1
Q2.10 State the number of significant figures in the following :
(a) 0.007 m2
Answer: The given quantity is 0.007 m2. This means that here, two zeros after the decimal are not significant. Hence, only 7 is a significant figure in this quantity
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.
(b) 2.64 × 1024 kg
Answer: The given quantity is 2.64 × 1024 kg. Hence, all digits i.e., 2, 6 and 4 are significant figures
Here, the power of 10 is irrelevant for the determination of significant figures.
c) 0.2370 g cm–3
Answer: 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) 6.320 J
Answer: 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) 6.032 N m–2
Answer: The given quantity is 6.032 Nm–2. All zeroes between two non-zero digits are always significant.
(f) 0.0006032 m2
Answer: The given quantity is 0.0006032 m2. 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.
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.
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.
Answer: Length of sheet, l = 4.234 m – ( significant figure 4 ) 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.
Breadth of sheet, b = 1.005 m – ( significant figure 4 )
Thickness of sheet, h = 2.01 cm = 0.0201 m – ( significant figure 3 )
Q 2. 12. The mass of a box measured by a grocer’s balance is 2.3 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?
Answer: Mass of grocer’s box = 2.300 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. Since the least number of decimal places is 1, Hence, the total mass of the box is 2.3 kg. Difference in masses = 20.17 – 20.15 = 0.02 g Since the least number of decimal places is 2 so the difference in masses to the correct significant figures is 0.02 g.
Mass of gold piece I = 20.15g = 0.02015 kg
Mass of gold piece II = 20.17 g = 0.02017 kg
Q2.13: A physical quantity P is related to four observables a, b, c and d as follows:
P = \(a^3 b^2 \over \sqrt {cd} \)
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?
Answer: As the error lies in first decimal place, the answer should be rounded off to first decimal place. Hence, we shall express the value of P after rounding it off as P = 3.8.
Q2.14 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 = \(a\sqrt 2\)( sin \(\frac{2πt}{T}\) + cos \(\frac{2πt}{T}\) )
(a = maximum displacement of the particle, v = speed of the particle. T = time-period
of motion). Rule out the wrong formulas on dimensional grounds.
Answer :
Q2.15 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo 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= \(m_0\over (1-v^2)^{1/2}\)
Guess where to put the missing c.
Answer: 
Q2.16 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 ?
Answer: Radius of hydrogen atom, r = 0.5 Å = 0.5 × 10–10 m Volume of hydrogen atom = \(\frac{4}{3}πr^3\) = \(\frac{4}{3} × \frac{22}{7}\) (0.5 × 10-10)3 = 5.23 x 10-31 m3 According to Avagadro’s hypothesis, one mole of hydrogen contains 6.023 x 1023 atoms. Therefore Atomic volume of 1 mole of hydrogen atoms = 6.023 x 1023 x 5.23 x 10-31 = 3.15 x 10-7m3.
Q2.17 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 ?
Answer: Volume of one mole of ideal gas, Vg Radius of hydrogen molecule = 1Å/2 Volume of hydrogen atom = \(\frac{4}{3}πr^3\) = \(\frac{4}{3} × \frac{22}{7}\) (0.5 × 10-10)3 = 0.523 x 10-30 m3 One mole contains 6.023 x 1023 molecules. Volume of one mole of hydrogen, Vh = 0.5238 x 10-30 x 6.023 x 1023 m3 = 3.1548 x 10-7 m3 Now Vg/VH=22.4 x 10-3/3.1548 x 10-7 =7.1 x 104 The ratio is very large. This is because the inter atomic separation in the gas is very large compared to the size of a hydrogen molecule.
= 22.4 litre = 22.4 x 10-3 m3
= 0.5 Å = 0.5 x 10-10 m
Q2.18 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).
Answer: 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.
Q2.19 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 ?
Answer: Diameter of Earth’s orbit = 3 × 1011 m Let the distance parallax angle be = 4.847 × 10–6 rad. Let the distance of the star be D. we can say θ=r /D , D = distance of distant object or star Since, θ = 1″(s) = 4.847 × 10–6 rad. and r = 1.5 × 1011 m D = \(\frac{r}{θ}\) = 1.5 x 1011/ 4.85 x 10-6 m = 3.09 x 1016 m = 3 x 1016 m Hence, 1 parsec ≈ 3 x 1016 m
∴ Radius of Earth’s orbit, r = 1.5 × 1011 m
Parsec is defined as the distance at which the average radius of the Earth’s orbit subtends an angle of 1″
Q2.20 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 traveled by light in one year. 1 light year = Speed of light × 1 year = 3 × 108 × 365 × 24 × 60 × 60 = 94608 × 1011 m ……………….( Speed of light = 3 × 108 m/s) 1 parsec = 3.08 × 1016 m 4.29 ly = \(405868.32 × 10^{11} \over 3.08 × 10^{16}\) = 1.32 parsec Using the relation, θ = d/D Where diameter’s of earth orbit = 3 x 1011 m Distance of the star from the earth = 405868.32 × 1011 m θ = \( (3 × 10^{11}) \over (405868.32 × 10^{11})\) = 7.39 x 10-6 rad But, 1 sec = 4.85 × 10–6 rad 7.39 x 10-6 rad = 7.39 x 10-6 rad / 4.85 × 10–6 rad = 1.52″
4.29 ly = 405868.32 × 1011 m
Q2.21 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. OR – As an example, while launching a satellite using a space launch rocket system we must measure time to a precision of 1 micro second. Again working with lasers we require length measurements to an angstrom unit (1 A° = 10-10m) or even a fraction of it. For estimating nuclear sizes we require a precision of 10-15 m. To measure atomic masses using mass spectrograph we require a precision of 10-30kg and so on.
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.
Answer: (a) During monsoons, a metrologist records about 215 cm of rainfall in India i.e., the height Density of water, ρ = 1 × 103 kg m–3 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, Vbe= 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 Area of one hair = πr2 Number of strands of hair = \( Total \space surface \space area \over Area \space of \space one \space hair \) = \(\frac{A}{r^2}\) (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 = \(\frac{6.023 × 10^{23}}{22.4 × 10^{-3}}\)
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
Hence, mass of rain water = ρ × V = 7.09 × 1015 kg
With the help of a screw gauge, the diameter and hence, the radius of a hair can be determined. Let it be r.
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.
Answer : Given M = 2 x 1030 kg, r = 7 x 108 m ∴ Volume of Sun = \(\frac{4}{3}πr^3\) = \(\frac{4}{3} ×\frac{22}{7} × (7 × 10^8)^3 \) = 1.437 x 1027 m3 As p = M/V, ∴ p = 2 x 1030/1.437 x 1027= 1391.8 kg m-3 = 1.4 x 103 kg m-3 Mass density of Sun is in the range of mass densities of solids/liquids and not gases.
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.
Answer : Given angular diameter θ = 35.72= 35.72 x 4.85 x 10-6 rad = 173.242 x 10-6 = 1.73 x 10-4 rad Diameter of Jupiter D = θ x d = 1.73 x 10-4 x 824.7 x 109 m =1426.731 x 103 = 1.43 x 108 m