This MCQ module is based on: NCERT Exercises and Solutions: Units and Measurements
TOPIC 4 OF 33
NCERT Exercises and Solutions: Units and Measurements
🎓 Class 11
Physics
CBSE
Theory
Ch 1 – Units and Measurements
⏱ ~8 min
🌐 Language: [gtranslate]
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NCERT Exercises and Solutions: Units and Measurements
Chapter Summary
- Physics is built on measurement — expressing a physical quantity as a numerical value multiplied by a unit.
- The SI system has 7 base units: metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), and candela (cd). All other units are derived from these.
- Two supplementary units exist: radian (rad) for plane angle and steradian (sr) for solid angle.
- Significant figures indicate the precision of a measurement. Rules govern counting sig figs, arithmetic operations, and rounding.
- Errors in measurement are classified as systematic (instrumental, personal, environmental) and random. Systematic errors shift all readings consistently; random errors fluctuate.
- Absolute error: \(|\Delta a_i| = |a_i - \bar{a}|\). Percentage error: \(\frac{\Delta\bar{a}}{\bar{a}} \times 100\%\).
- Error propagation: for sum/difference, absolute errors add; for product/quotient, relative errors add; for powers, relative error is multiplied by the power.
- Accuracy = closeness to the true value. Precision = closeness of repeated measurements to each other.
- Every physical quantity has a dimensional formula expressed as powers of [M], [L], [T], etc.
- Dimensional analysis can check equations, derive relations, and convert units — but cannot determine dimensionless constants or handle trigonometric/exponential functions.
Key Terms
Physical QuantityA quantity that can be measured and expressed as number × unit.
SI SystemInternational System of Units with 7 base units, adopted in 1971.
Significant FiguresMeaningful digits in a measurement (known + one estimated).
Absolute Error\(|\Delta a_i| = |a_i - \bar{a}|\)
Relative Error\(\Delta\bar{a}\,/\,\bar{a}\) (dimensionless)
Percentage Error\((\Delta\bar{a}\,/\,\bar{a}) \times 100\%\)
Systematic ErrorConsistent shift in one direction; not reduced by averaging.
Random ErrorStatistical fluctuations; reduced by taking many readings.
Dimensional FormulaPowers of base quantities: e.g. Force = \([M\,L\,T^{-2}]\).
Dimensional AnalysisTechnique using dimensions to verify, derive, or convert.
AccuracyCloseness of measurement to the true value.
PrecisionCloseness of repeated measurements to each other.
NCERT Exercises
Exercise 1.1
Fill in the blanks:
(a) The volume of a cube of side 1 cm is equal to _____ m³.
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to _____ mm².
(c) A vehicle moving with a speed of 18 km/h covers _____ m in 1 s.
(d) The relative density of lead is 11.3. Its density is _____ g/cm³ or _____ kg/m³.
(a) Side = 1 cm = \(10^{-2}\) m
Volume = \((10^{-2})^3 = \boxed{10^{-6} \text{ m}^3}\)
(b) Total surface area of a solid cylinder = \(2\pi r(r + h)\)
\(r = 2.0\) cm, \(h = 10.0\) cm
\(= 2\pi \times 2.0 \times (2.0 + 10.0) = 2\pi \times 2.0 \times 12.0 = 48\pi \approx 150.8\) cm²
Converting to mm²: \(1\text{ cm}^2 = 100\text{ mm}^2\)
\(= 150.8 \times 100 = \boxed{1.508 \times 10^4 \text{ mm}^2} \approx 15080\text{ mm}^2\)
(c) \(18\text{ km/h} = 18 \times \frac{1000}{3600}\text{ m/s} = 18 \times \frac{5}{18} = \boxed{5\text{ m/s}}\)
Distance in 1 s = 5 m.
(d) Relative density = density of substance / density of water.
Density of water = 1 g/cm³ = 1000 kg/m³.
Density of lead = \(11.3 \times 1 = \boxed{11.3\text{ g/cm}^3}\)
In SI: \(11.3 \times 1000 = \boxed{11300\text{ kg/m}^3} = 1.13 \times 10^4\text{ kg/m}^3\)
Volume = \((10^{-2})^3 = \boxed{10^{-6} \text{ m}^3}\)
(b) Total surface area of a solid cylinder = \(2\pi r(r + h)\)
\(r = 2.0\) cm, \(h = 10.0\) cm
\(= 2\pi \times 2.0 \times (2.0 + 10.0) = 2\pi \times 2.0 \times 12.0 = 48\pi \approx 150.8\) cm²
Converting to mm²: \(1\text{ cm}^2 = 100\text{ mm}^2\)
\(= 150.8 \times 100 = \boxed{1.508 \times 10^4 \text{ mm}^2} \approx 15080\text{ mm}^2\)
(c) \(18\text{ km/h} = 18 \times \frac{1000}{3600}\text{ m/s} = 18 \times \frac{5}{18} = \boxed{5\text{ m/s}}\)
Distance in 1 s = 5 m.
(d) Relative density = density of substance / density of water.
Density of water = 1 g/cm³ = 1000 kg/m³.
Density of lead = \(11.3 \times 1 = \boxed{11.3\text{ g/cm}^3}\)
In SI: \(11.3 \times 1000 = \boxed{11300\text{ kg/m}^3} = 1.13 \times 10^4\text{ kg/m}^3\)
Exercise 1.2
Fill in the blanks by suitable conversion of units:
(a) 1 kg m² s² = _____ g cm² s²
(b) 1 m = _____ ly (light year)
(c) 3.0 m s² = _____ km h²
(d) \(G = 6.67 \times 10^{-11}\) N m² kg² = _____ cm³ s² g¹
(a) \(1\text{ kg m}^2\text{s}^{-2} = (10^3\text{ g})(10^2\text{ cm})^2(1\text{ s})^{-2}\)
\(= 10^3 \times 10^4 = \boxed{10^7\text{ g cm}^2\text{ s}^{-2}}\)
(b) 1 light year = distance light travels in 1 year
\(= 3 \times 10^8 \times 365.25 \times 24 \times 3600 \approx 9.46 \times 10^{15}\text{ m}\)
\(1\text{ m} = \frac{1}{9.46 \times 10^{15}}\text{ ly} = \boxed{1.057 \times 10^{-16}\text{ ly}}\)
(c) \(3.0\text{ m s}^{-2}\)
\(1\text{ m} = 10^{-3}\text{ km}\), \(1\text{ s} = \frac{1}{3600}\text{ h}\)
\(= 3.0 \times 10^{-3}\text{ km} \times (3600)^2\text{ h}^{-2}\)
\(= 3.0 \times 10^{-3} \times 1.296 \times 10^7 = \boxed{3.888 \times 10^4\text{ km h}^{-2}}\)
(d) \(G = 6.67 \times 10^{-11}\text{ N m}^2\text{kg}^{-2}\)
\([G] = [M^{-1}\,L^3\,T^{-2}]\)
\(= 6.67 \times 10^{-11} \times (10^3)^{-1} \times (10^2)^3 \times 1\)
\(= 6.67 \times 10^{-11} \times 10^{-3} \times 10^6 = \boxed{6.67 \times 10^{-8}\text{ cm}^3\text{ s}^{-2}\text{ g}^{-1}}\)
\(= 10^3 \times 10^4 = \boxed{10^7\text{ g cm}^2\text{ s}^{-2}}\)
(b) 1 light year = distance light travels in 1 year
\(= 3 \times 10^8 \times 365.25 \times 24 \times 3600 \approx 9.46 \times 10^{15}\text{ m}\)
\(1\text{ m} = \frac{1}{9.46 \times 10^{15}}\text{ ly} = \boxed{1.057 \times 10^{-16}\text{ ly}}\)
(c) \(3.0\text{ m s}^{-2}\)
\(1\text{ m} = 10^{-3}\text{ km}\), \(1\text{ s} = \frac{1}{3600}\text{ h}\)
\(= 3.0 \times 10^{-3}\text{ km} \times (3600)^2\text{ h}^{-2}\)
\(= 3.0 \times 10^{-3} \times 1.296 \times 10^7 = \boxed{3.888 \times 10^4\text{ km h}^{-2}}\)
(d) \(G = 6.67 \times 10^{-11}\text{ N m}^2\text{kg}^{-2}\)
\([G] = [M^{-1}\,L^3\,T^{-2}]\)
\(= 6.67 \times 10^{-11} \times (10^3)^{-1} \times (10^2)^3 \times 1\)
\(= 6.67 \times 10^{-11} \times 10^{-3} \times 10^6 = \boxed{6.67 \times 10^{-8}\text{ cm}^3\text{ s}^{-2}\text{ g}^{-1}}\)
Exercise 1.3
A calorie is a unit of heat energy and it equals about 4.2 J, where 1 J = 1 kg m² s². Suppose we employ a system of units in which the unit of mass equals \(\alpha\) kg, the unit of length equals \(\beta\) m, the unit of time is \(\gamma\) s. Show that a calorie has a magnitude \(4.2\,\alpha^{-1}\,\beta^{-2}\,\gamma^2\) in terms of the new units.
Solution:
Energy has dimensions \([M\,L^2\,T^{-2}]\).
In SI: 1 calorie = 4.2 J = 4.2 kg m² s²
In the new system: unit of mass = \(\alpha\) kg, unit of length = \(\beta\) m, unit of time = \(\gamma\) s.
\[n_2 = n_1 \left(\frac{M_1}{M_2}\right)^1 \left(\frac{L_1}{L_2}\right)^2 \left(\frac{T_1}{T_2}\right)^{-2}\] \[= 4.2 \left(\frac{1\text{ kg}}{\alpha\text{ kg}}\right) \left(\frac{1\text{ m}}{\beta\text{ m}}\right)^2 \left(\frac{1\text{ s}}{\gamma\text{ s}}\right)^{-2}\] \[= 4.2 \times \frac{1}{\alpha} \times \frac{1}{\beta^2} \times \gamma^2\] \[\boxed{= 4.2\,\alpha^{-1}\,\beta^{-2}\,\gamma^2 \text{ (in new units)}}\]
Energy has dimensions \([M\,L^2\,T^{-2}]\).
In SI: 1 calorie = 4.2 J = 4.2 kg m² s²
In the new system: unit of mass = \(\alpha\) kg, unit of length = \(\beta\) m, unit of time = \(\gamma\) s.
\[n_2 = n_1 \left(\frac{M_1}{M_2}\right)^1 \left(\frac{L_1}{L_2}\right)^2 \left(\frac{T_1}{T_2}\right)^{-2}\] \[= 4.2 \left(\frac{1\text{ kg}}{\alpha\text{ kg}}\right) \left(\frac{1\text{ m}}{\beta\text{ m}}\right)^2 \left(\frac{1\text{ s}}{\gamma\text{ s}}\right)^{-2}\] \[= 4.2 \times \frac{1}{\alpha} \times \frac{1}{\beta^2} \times \gamma^2\] \[\boxed{= 4.2\,\alpha^{-1}\,\beta^{-2}\,\gamma^2 \text{ (in new units)}}\]
Exercise 1.4
Explain this statement clearly: "To call a dimensional quantity 'large' or 'small' is meaningless without specifying a standard for comparison."
Solution:
A dimensional quantity has both a number and a unit. The numerical value changes when the unit changes. For example, the distance between two cities might be "small" if measured in light-years (0.000000... ly) but "large" in millimetres (hundreds of millions of mm). The same physical quantity appears large or small depending entirely on the unit (standard) chosen for comparison. Therefore, calling a quantity "large" or "small" only makes sense when we compare it to some reference standard of the same kind.
A dimensional quantity has both a number and a unit. The numerical value changes when the unit changes. For example, the distance between two cities might be "small" if measured in light-years (0.000000... ly) but "large" in millimetres (hundreds of millions of mm). The same physical quantity appears large or small depending entirely on the unit (standard) chosen for comparison. Therefore, calling a quantity "large" or "small" only makes sense when we compare it to some reference standard of the same kind.
Exercise 1.5
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 20 s to cover this distance?
Solution:
In the new system, \(c = 1\) new unit of length per second.
Time taken = 8 min 20 s = \(8 \times 60 + 20 = 500\) s.
Distance = speed × time = \(1 \times 500 = \boxed{500 \text{ new units of length}}\)
(In conventional units, this is \(3 \times 10^8 \times 500 = 1.5 \times 10^{11}\) m.)
In the new system, \(c = 1\) new unit of length per second.
Time taken = 8 min 20 s = \(8 \times 60 + 20 = 500\) s.
Distance = speed × time = \(1 \times 500 = \boxed{500 \text{ new units of length}}\)
(In conventional units, this is \(3 \times 10^8 \times 500 = 1.5 \times 10^{11}\) m.)
Exercise 1.6
Which of the following is the most precise device for measuring length:
(a) a vernier caliper 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?
Solution:
(a) Vernier caliper: Least count = \(\frac{1\text{ MSD}}{20} = \frac{1\text{ mm}}{20} = 0.05\text{ mm} = 5 \times 10^{-5}\text{ m}\)
(b) Screw gauge: Least count = \(\frac{\text{pitch}}{\text{no. of divisions}} = \frac{1\text{ mm}}{100} = 0.01\text{ mm} = 10^{-5}\text{ m}\)
(c) Optical instrument: wavelength of visible light \(\approx 500\text{ nm} = 5 \times 10^{-7}\text{ m}\)
The smallest least count gives the highest precision.
\(\boxed{\text{The optical instrument (c) is the most precise.}}\)
(a) Vernier caliper: Least count = \(\frac{1\text{ MSD}}{20} = \frac{1\text{ mm}}{20} = 0.05\text{ mm} = 5 \times 10^{-5}\text{ m}\)
(b) Screw gauge: Least count = \(\frac{\text{pitch}}{\text{no. of divisions}} = \frac{1\text{ mm}}{100} = 0.01\text{ mm} = 10^{-5}\text{ m}\)
(c) Optical instrument: wavelength of visible light \(\approx 500\text{ nm} = 5 \times 10^{-7}\text{ m}\)
The smallest least count gives the highest precision.
\(\boxed{\text{The optical instrument (c) is the most precise.}}\)
Exercise 1.7
State the number of significant figures in the following:
(a) 0.007 m² (b) 2.64 × 10²&sup4; kg (c) 0.2370 g/cm³ (d) 6.320 J (e) 6.032 N m² (f) 0.0006032 m²
Solution:
(a) 0.007: Leading zeros not significant. Only 7 counts. 1 sig fig.
(b) \(2.64 \times 10^{24}\): Digits 2, 6, 4 in the coefficient. 3 sig figs.
(c) 0.2370: Leading zero not significant, but trailing zero after decimal is. Digits: 2, 3, 7, 0. 4 sig figs.
(d) 6.320: All digits significant (trailing zero after decimal). 4 sig figs.
(e) 6.032: All digits significant (zero between 6 and 3 counts). 4 sig figs.
(f) 0.0006032: Leading zeros not significant. Digits: 6, 0, 3, 2. 4 sig figs.
(a) 0.007: Leading zeros not significant. Only 7 counts. 1 sig fig.
(b) \(2.64 \times 10^{24}\): Digits 2, 6, 4 in the coefficient. 3 sig figs.
(c) 0.2370: Leading zero not significant, but trailing zero after decimal is. Digits: 2, 3, 7, 0. 4 sig figs.
(d) 6.320: All digits significant (trailing zero after decimal). 4 sig figs.
(e) 6.032: All digits significant (zero between 6 and 3 counts). 4 sig figs.
(f) 0.0006032: Leading zeros not significant. Digits: 6, 0, 3, 2. 4 sig figs.
Exercise 1.8
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:
\(l = 4.234\) m (4 sig figs), \(b = 1.005\) m (4 sig figs), \(t = 2.01\) cm = 0.0201 m (3 sig figs).
Surface area (of the largest face):
\(A = l \times b = 4.234 \times 1.005 = 4.255170\text{ m}^2\)
Both have 4 sig figs, so result: \(\boxed{A = 4.255\text{ m}^2}\) (4 sig figs).
Volume:
\(V = l \times b \times t = 4.234 \times 1.005 \times 0.0201 = 0.08552...\text{ m}^3\)
The factor with fewest sig figs is \(t\) (3 sig figs), so:
\(\boxed{V = 0.0855\text{ m}^3}\) (3 sig figs).
\(l = 4.234\) m (4 sig figs), \(b = 1.005\) m (4 sig figs), \(t = 2.01\) cm = 0.0201 m (3 sig figs).
Surface area (of the largest face):
\(A = l \times b = 4.234 \times 1.005 = 4.255170\text{ m}^2\)
Both have 4 sig figs, so result: \(\boxed{A = 4.255\text{ m}^2}\) (4 sig figs).
Volume:
\(V = l \times b \times t = 4.234 \times 1.005 \times 0.0201 = 0.08552...\text{ m}^3\)
The factor with fewest sig figs is \(t\) (3 sig figs), so:
\(\boxed{V = 0.0855\text{ m}^3}\) (3 sig figs).
Exercise 1.9
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 placed in the box. What is (a) the total mass of the box, (b) the difference in the masses of the two gold pieces to correct significant figures?
Solution:
(a) Total mass:
Convert gold pieces to kg: 20.15 g = 0.02015 kg, 20.17 g = 0.02017 kg.
Total = 2.300 + 0.02015 + 0.02017 = 2.34032 kg.
The least precise measurement is 2.300 kg (3 decimal places in kg).
Rounding to 3 decimal places: \(\boxed{2.340\text{ kg}}\)
(b) Difference:
\(20.17 - 20.15 = 0.02\) g.
Each has 4 sig figs but 2 decimal places. Result: \(\boxed{0.02\text{ g}}\)
(a) Total mass:
Convert gold pieces to kg: 20.15 g = 0.02015 kg, 20.17 g = 0.02017 kg.
Total = 2.300 + 0.02015 + 0.02017 = 2.34032 kg.
The least precise measurement is 2.300 kg (3 decimal places in kg).
Rounding to 3 decimal places: \(\boxed{2.340\text{ kg}}\)
(b) Difference:
\(20.17 - 20.15 = 0.02\) g.
Each has 4 sig figs but 2 decimal places. Result: \(\boxed{0.02\text{ g}}\)
Exercise 1.10
Each side of a cube is measured to be 7.203 m. What is the total surface area and volume of the cube to appropriate significant figures?
Solution:
Side \(a = 7.203\) m (4 sig figs).
Surface area: \(S = 6a^2 = 6 \times (7.203)^2 = 6 \times 51.883 = 311.3\text{ m}^2\)
\((7.203)^2 = 51.8832\), keeping 4 sig figs: 51.88.
\(S = 6 \times 51.88 = 311.3\text{ m}^2\) (4 sig figs).
\(\boxed{S = 311.3\text{ m}^2}\)
Volume: \(V = a^3 = (7.203)^3 = 7.203 \times 7.203 \times 7.203 = 373.7\text{ m}^3\)
Keeping 4 sig figs: \(\boxed{V = 373.7\text{ m}^3}\)
Side \(a = 7.203\) m (4 sig figs).
Surface area: \(S = 6a^2 = 6 \times (7.203)^2 = 6 \times 51.883 = 311.3\text{ m}^2\)
\((7.203)^2 = 51.8832\), keeping 4 sig figs: 51.88.
\(S = 6 \times 51.88 = 311.3\text{ m}^2\) (4 sig figs).
\(\boxed{S = 311.3\text{ m}^2}\)
Volume: \(V = a^3 = (7.203)^3 = 7.203 \times 7.203 \times 7.203 = 373.7\text{ m}^3\)
Keeping 4 sig figs: \(\boxed{V = 373.7\text{ m}^3}\)
Exercise 1.11
5.74 g of a substance occupies 1.2 cm³. Express its density by keeping the significant figures in view.
Solution:
\(\rho = \frac{m}{V} = \frac{5.74}{1.2} = 4.7833...\text{ g/cm}^3\)
\(m = 5.74\) has 3 sig figs, \(V = 1.2\) has 2 sig figs.
Result should have 2 sig figs (the fewest).
\(\boxed{\rho = 4.8\text{ g/cm}^3}\)
\(\rho = \frac{m}{V} = \frac{5.74}{1.2} = 4.7833...\text{ g/cm}^3\)
\(m = 5.74\) has 3 sig figs, \(V = 1.2\) has 2 sig figs.
Result should have 2 sig figs (the fewest).
\(\boxed{\rho = 4.8\text{ g/cm}^3}\)
Exercise 1.12
The unit of length convenient on the nuclear scale is a fermi: 1 f = 10¹&sup5; m. Nuclear sizes obey roughly the empirical formula \(r = r_0 A^{1/3}\), where \(r_0 = 1.2\) f and \(A\) is the mass number. What is the size of a nucleus of &sup5;&sup5;Mn (mass number 55)?
Solution:
\(r = r_0 A^{1/3} = 1.2 \times (55)^{1/3}\) f
\((55)^{1/3} = (55)^{0.333} \approx 3.803\)
\(r = 1.2 \times 3.803 = 4.6\) f
Converting to metres: \(r = 4.6 \times 10^{-15}\) m
\(\boxed{r \approx 4.6\text{ f} = 4.6 \times 10^{-15}\text{ m}}\)
\(r = r_0 A^{1/3} = 1.2 \times (55)^{1/3}\) f
\((55)^{1/3} = (55)^{0.333} \approx 3.803\)
\(r = 1.2 \times 3.803 = 4.6\) f
Converting to metres: \(r = 4.6 \times 10^{-15}\) m
\(\boxed{r \approx 4.6\text{ f} = 4.6 \times 10^{-15}\text{ m}}\)
Exercise 1.13
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)
Solution:
The sound wave travels to the submarine and back, so the total distance is \(2d\).
\[2d = v \times t = 1450 \times 77.0 = 111650\text{ m}\] \[d = \frac{111650}{2} = 55825\text{ m}\] \[\boxed{d \approx 55.8\text{ km} = 5.58 \times 10^4\text{ m}}\] (To 3 sig figs, since 77.0 has 3 sig figs.)
The sound wave travels to the submarine and back, so the total distance is \(2d\).
\[2d = v \times t = 1450 \times 77.0 = 111650\text{ m}\] \[d = \frac{111650}{2} = 55825\text{ m}\] \[\boxed{d \approx 55.8\text{ km} = 5.58 \times 10^4\text{ m}}\] (To 3 sig figs, since 77.0 has 3 sig figs.)
Exercise 1.14
The unit of length convenient on the atomic scale is known as an angstrom: 1 A = 10¹° m. The size of a hydrogen atom is about 0.5 A. What is the total atomic volume in m³ of a mole of hydrogen atoms?
Solution:
Radius of H atom: \(r = 0.5\) A \(= 0.5 \times 10^{-10}\) m \(= 5 \times 10^{-11}\) m
Volume of one atom: \(V_1 = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (5 \times 10^{-11})^3\)
\(= \frac{4}{3}\times 3.14159 \times 1.25 \times 10^{-31}\)
\(= 5.236 \times 10^{-31}\text{ m}^3\)
1 mole = \(6.022 \times 10^{23}\) atoms.
Total volume = \(6.022 \times 10^{23} \times 5.236 \times 10^{-31}\)
\(= 3.154 \times 10^{-7}\text{ m}^3\)
\(\boxed{V \approx 3.15 \times 10^{-7}\text{ m}^3}\)
Radius of H atom: \(r = 0.5\) A \(= 0.5 \times 10^{-10}\) m \(= 5 \times 10^{-11}\) m
Volume of one atom: \(V_1 = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (5 \times 10^{-11})^3\)
\(= \frac{4}{3}\times 3.14159 \times 1.25 \times 10^{-31}\)
\(= 5.236 \times 10^{-31}\text{ m}^3\)
1 mole = \(6.022 \times 10^{23}\) atoms.
Total volume = \(6.022 \times 10^{23} \times 5.236 \times 10^{-31}\)
\(= 3.154 \times 10^{-7}\text{ m}^3\)
\(\boxed{V \approx 3.15 \times 10^{-7}\text{ m}^3}\)
Exercise 1.15
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? (Radius of hydrogen molecule = 1 A). Why is this ratio so large?
Solution:
Molar volume at STP: \(V_m = 22.4\text{ L} = 22.4 \times 10^{-3}\text{ m}^3\)
Radius of H&sub2; molecule: \(r = 1\text{ A} = 10^{-10}\text{ m}\)
Volume of one molecule: \(V_1 = \frac{4}{3}\pi (10^{-10})^3 = 4.189 \times 10^{-30}\text{ m}^3\)
Volume of 1 mole of molecules: \(V_a = 6.022 \times 10^{23} \times 4.189 \times 10^{-30} = 2.522 \times 10^{-6}\text{ m}^3\)
Ratio: \(\frac{V_m}{V_a} = \frac{22.4 \times 10^{-3}}{2.522 \times 10^{-6}} \approx \boxed{8886 \approx 10^4}\)
Why so large? In a gas at STP, the molecules are very far apart compared to their size. The intermolecular spacing is much larger than the molecular diameter. The actual volume occupied by molecules is a tiny fraction of the total gas volume — most of the space is empty.
Molar volume at STP: \(V_m = 22.4\text{ L} = 22.4 \times 10^{-3}\text{ m}^3\)
Radius of H&sub2; molecule: \(r = 1\text{ A} = 10^{-10}\text{ m}\)
Volume of one molecule: \(V_1 = \frac{4}{3}\pi (10^{-10})^3 = 4.189 \times 10^{-30}\text{ m}^3\)
Volume of 1 mole of molecules: \(V_a = 6.022 \times 10^{23} \times 4.189 \times 10^{-30} = 2.522 \times 10^{-6}\text{ m}^3\)
Ratio: \(\frac{V_m}{V_a} = \frac{22.4 \times 10^{-3}}{2.522 \times 10^{-6}} \approx \boxed{8886 \approx 10^4}\)
Why so large? In a gas at STP, the molecules are very far apart compared to their size. The intermolecular spacing is much larger than the molecular diameter. The actual volume occupied by molecules is a tiny fraction of the total gas volume — most of the space is empty.
Exercise 1.16
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 last motivation behind the high-accuracy atomic clocks. Comment on the accuracy of different types of clocks.
Solution:
Different clocks offer vastly different levels of accuracy:
Sundial: Accuracy ~ 15 minutes. Depends on sunshine and observer's judgement.
Mechanical clock (pendulum): Accuracy ~ 1 second per day. Affected by temperature, humidity, and friction.
Quartz clock: Accuracy ~ 1 second in a few years. Uses vibrations of a quartz crystal at ~32,768 Hz.
Caesium atomic clock: Accuracy ~ 1 second in \(10^{13}\) seconds (about 300,000 years). Based on the hyperfine transition of caesium-133 at 9,192,631,770 Hz. This is used to define the SI second.
Optical/lattice clocks: Even more precise, losing about 1 second in \(10^{18}\) seconds (the age of the universe).
The need for high precision in measuring positions at closely separated time intervals (for aircraft tracking, GPS, space navigation) drove the development of these increasingly accurate timepieces.
Different clocks offer vastly different levels of accuracy:
Sundial: Accuracy ~ 15 minutes. Depends on sunshine and observer's judgement.
Mechanical clock (pendulum): Accuracy ~ 1 second per day. Affected by temperature, humidity, and friction.
Quartz clock: Accuracy ~ 1 second in a few years. Uses vibrations of a quartz crystal at ~32,768 Hz.
Caesium atomic clock: Accuracy ~ 1 second in \(10^{13}\) seconds (about 300,000 years). Based on the hyperfine transition of caesium-133 at 9,192,631,770 Hz. This is used to define the SI second.
Optical/lattice clocks: Even more precise, losing about 1 second in \(10^{18}\) seconds (the age of the universe).
The need for high precision in measuring positions at closely separated time intervals (for aircraft tracking, GPS, space navigation) drove the development of these increasingly accurate timepieces.
Frequently Asked Questions - NCERT Exercises and Solutions: Units and Measurements
What are the key NCERT exercise types in Chapter 1 Units and Measurements?
NCERT Class 11 Physics Chapter 1 Units and Measurements exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question in the chapter, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Units and Measurements?
For numerical problems in NCERT Class 11 Physics Chapter 1 Units and Measurements: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 1?
From NCERT Class 11 Physics Chapter 1 (Units and Measurements), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 11 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula (M^a L^b T^c). For Chapter 1 Units and Measurements problems, write the dimensions of each quantity, substitute, and simplify. If dimensions match the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 1 exercises?
Common mistakes in NCERT Class 11 Physics Chapter 1 Units and Measurements exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for vectors, (3) forgetting to consider all forces in free-body diagrams, (4) algebra errors during derivations, and (5) misreading the problem statement. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 11 Physics Chapter 1 Units and Measurements solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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