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NCERT Class 11 Physics Chapter 1 Units of Measurement
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Units of Measurement
Chapter: 1
| Part – I |
EXERCISE
Note: In stating numerical answers, take care of significant figures.
1. Fill in the blanks:
(a) The volume of a cube of 1 cm is equal to …..m3.
Ans: 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.
Ans: 1.26 × 104 mm2.
(c) A vehicle moving with a speed of 18 km h–1 covers….m in 1 s.
Ans: 5 ms-1.
(d) The relative density of lead is 11.3. Its density is ….g cm–3 or ….kg m–3.
Ans: 1.13 × 104 kgm-3.
2. Fill in the blanks by suitable conversion of units.
(a) 1 kg m2 s–2 = ….g cm2 s –2.
Ans:
(b) 1 m = ….. ly.
Ans:
(c) 3.0 m s–2 = …. km h-2
Ans:
= 3 × 3600 × 10-3 kmh-2
= 3.888 × 104 km h-2
(d) G = 6.67 × 10–11 N m2 (kg)–2 = …. (cm)3 s–2 g–1.
Ans: G = 6.67 × 10-11 Nm2 kh-2
= 6.67 × 10-11 × 1 kg-1 m3 s-2
= 6.67 × 10-8 cm-3 s-3 g-1.
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: 1 cal. = 4.2 kg m2S-2
| S.I. system | New System |
| N1 = 4.2 | N1 = ? |
| M1 = 1 kg | M2 = 𝞪 kg |
| L1 = 1m | L2 = β m |
| T1 = 1 Seconds | T2 = 𝝲 second |
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.
Ans: Atoms are very small objects compared to chalk.
(b) A jet plane moves with great speed.
Ans: A jet plane moves with higher speed as compared to an aeroplane.
(c) The mass of Jupiter is very large.
Ans: The mass of Jupiter is very large compared to Mercury.
(d) The air inside this room contains a large number of molecules.
Ans: The air inside this room contains a large number of molecules as compared to a car tire.
(e) A proton is much more massive than an electron.
Ans: These statements need not be reframed.
(f) The speed of sound is much smaller than the speed of light.
Ans: These statements need not be reframed.
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: The time sun lights taken to reach the earth t = 8 min 20s
Now, convert 8 min 20 second into seconds
t = 8 × 60 + 20
= 500.
Velocity of light in vacuum, c = 1 new unit of length s-1
Distance = speed x time ( c × t)
Distance = 1 new unit of length s-1 × 500.
= 500 new units of length.
6. 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?
Ans: An optical instrument which can measure length within a wavelength of light is capable of extremely high precision. Therefore, option (c) is the correct answer.
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: Magnification = 100
Average width of the hair in the field of view = 3.5 mm
The estimate on the thickness of hair = Average width in field of view/Magnification
= 3.5/100
=0.035mm.
8. Answer the following:
(a) You are given a thread and a metre scale. How will you estimate the diameter of the thread?
Ans: Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other.
The diameter of the thread is given by the relation,
Diameter = Length of rod/Number of turns.
(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?
Ans: Increasing the number of divisions of the circular scale will increase its accuracy to a certain extent only.
(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?
Ans: Because random errors involved in the former are very less as compared to the latter.
9. The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected onto 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: Area of the house on the screen A = 1.75 cm²
= 1.55.cm2 = 1.55 × 104 cm2.
Area of the house on the screen A′= 1.55 m²
= 1.55 × 104 / 1.75
= 8.857 × 104 cm2.
Linear magnification
= √areal magnification
= √ 8.857 × 103
= 94.1 cm2.
10. State the number of significant figures in the following:
(a) 0.007 m2.
Ans: 1.
(b) 2.64 × 104 kg.
Ans: 3.
(c) 0.2370 g cm–3.
Ans: 4.
(d) 6.320 J.
Ans: 4.
(e) 6.032 N m–2.
Ans: 4.
(f) 0.0006032 m2.
Ans: 4.
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: The area (AAA) of the rectangular sheet is given by:
A= l × b
Area = (4.234 × 1.005) x 2
= 8.51034 = 8.5 m2.
Volume The volume (VVV) of the rectangular sheet is given by:
V = l × b × t
Volume = (4.234 × 1.005) × (2.01 × 10-2)
= 8.55289 × 10-2
= 0.0855m3.
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.
Ans: Total mass of the box.
Mass of the box: 2. 30 kg.
First convert the masses of the gold pieces to kilograms:
20.15 g = 0.02015 kg.
20.17 g = 0.02017 kg.
Next, we add the masses:
(2.3kg + 0.02015 kg + 0.02017 kg)
= 2.3432 kg.
(b) the difference in the masses of the pieces to correct significant figures?
Ans: Difference in the Masses of the Pieces.
The masses of two gold pieces are:
First piece: 20.15g
Second piece: 20.17g
The difference is:
20.17g – 20.15 g
= 0.02 g.
13. 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:
Guess where to put the missing c.
Ans: On rearranging, we have
Since, the left hand side is dimensionless, so the right hand side should also be dimensionless. This will be so, if
14. 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: Given that:
r = 0.5 A = 0.5 × 10-10 m
Volume of hydrogen atom = (4/3) 𝜋r3
= 0.524 × 10-30 m3
1 mole of hydrogen contains 6.023 × 1023 hydrogen atoms.
Volume of 1 mole of hydrogen atom = 6.023 × 1023 × 0.524 × 10-30
= 3.16 × 10-7 m3.
15. 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 a hydrogen molecule to be about 1 Å). Why is this ratio so large?
Ans: Molar volume of ideal Gas:
1 mole of an ideal gas occupies 22.4 litres.
Size of hydrogen molecules: 1 Å = 1 × 10-10 metres.
V = 4/3 𝜋r3 (r = 0.5 × 10-10 metres)
V = 4/3 𝜋 (0.5 × 10-10)3
= 4/3 𝜋 (1.25 × 10-31)
= 5 × 10 -31 m3
(convert to litres 1 m3 = 1000 litres)
5 × 10-31 m3 = 5 × 10 -28 litres
Avogadro’s number = 6.022 × 1023
Volume = 6.022 × 1023 × 5 × 10-28 litres
= 0.030 litres
Ratio = 22.4 / 0.030
= 747.
This high ratio is because of intermolecular spaces in gas being much larger than the size of the molecule. Thus, even though individual hydrogen molecules are very small, the gas volume reflects the much larger space between molecules in the gas phase.
16. 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: Due to the motion of the train the nearby objects appear to move quickly in the opposite direction to the train’s motion. This is because as the train moves forward, your line of sight to these nearby objects changes rapidly. On the other hand the sight of a distant star or a cliff does not change its direction because they are so far away that their relative motion due to the train’s movement is imperceptible during the short duration of observation. Due to this the distant cliff or the star appears stationary.
17. The Sun is a hot plasma (ionised 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: Given that:
M = 2 × 1030 kg
r = 7 × 108 kg
Volume of sun = 4/3 𝜋r3
= 4/3 × 22/7 × ( 7 × 108)3
= 1.437 × 1027 m3.
Now V = 1.437 × 1027.
As, r = M/V
= 2 × 1030/ 1.437 × 1027.
= 1391.8 kg m-3.
= 1.4 × 103 kg m-3.
This is higher than the density of gases under standard conditions. Therefore, the mass density of the Sun is in the range of densities of solids and liquids, not gases.
