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Exercises 1.1 to 1.5 from NCERT Class 11 Physics Chapter 1: Units and Measurement, step by step with explanations and appropriate unit conversions.
1.1 Fill in the blanks
(a) The volume of a cube of side 1 cm is equal to:
\( V = (1 \text{ cm})^3 = 1 \text{ cm}^3 = 1 \times 10^{-6} \text{ m}^3 \)
✅ Answer: \( 1 \times 10^{-6} \text{ m}^3 \)
(b) Surface area of a solid cylinder:
Formula:
\( A = 2\pi r^2 + 2\pi rh \)
Given:
\( r = 2.0 \text{ cm} = 20 \text{ mm}, \) \( h = 10.0 \text{ cm} = 100 \text{ mm} \)
\( A = 2\pi r(r + h) = 2\pi (20)(20 + 100) \) \(= 2\pi (20)(120) \) \( = 4800\pi \text{ mm}^2 \approx 15080 \text{ mm}^2 \)
✅ Answer: \( \approx 1.51 \times 10^4 \text{ mm}^2 \)
(c) Speed = 18 km/h ; Convert to m/s:
\( \dfrac{18 \times 1000}{3600} = 5 \text{ m/s} \)
So, in 1 s, it covers 5 m.
✅ Answer: 5 m
(d) Relative density of lead, \( \rho_r = 11.3 \)
That means:
\( \text{Density} = 11.3 \text{ g/cm}^3 \)
Convert to kg/m³:
\( 1 \text{ g/cm}^3 = 1000 \text{ kg/m}^3 \) \(\Rightarrow 11.3 \text{ g/cm}^3 = 11300 \text{ kg/m}^3 \)
✅ Answer:
\( 11.3 \text{ g/cm}^3 \) or \( 11300 \text{ kg/m}^3 \)
1.2 Fill in the blanks by suitable conversion of units
(a) \( 1 \text{ kg} = 1000 \text{ g},\quad 1 \text{ m} = 100 \text{ cm} \)
\( 1 \text{ kg m}^2 \text{ s}^{-2} \) \(= 1000 \text{ g} \times (100 \text{ cm})^2 \times \text{ s}^{-2} \)
\( = 1000 \times 10^4 \text{ g cm}^2 \text{ s}^{-2} \)
\(= 10^7 \text{ g cm}^2 \text{s}^{-2} \)
✅ Answer:\( 10^7 \text{ g cm}^2 \text{s}^{-2} \)
(b)1 ly (light year) = distance light travels in 1 year
Speed of light c = 3 \( \times 10^8 \text{ m/s} \)
\( 1 \text{ year} = 365.25 \times 24 \times 3600 \)
\( \approx 3.156 \times 10^7 \text{ s} \)
\( 1 \text{ ly} = 3 \times 10^8 \times 3.156 \times 10^7 \)
\(= 9.47 \times 10^{15} \text{ m} \)
So,
\( 1 \text{ m} = \dfrac{1}{9.47 \times 10^{15}} \text{ ly} \)
✅ Answer:\( \approx 1.06 \times 10^{-16} \text{ ly} \)
(c) Convert 3.0 m/s² to km/h²
\( 1 \text{m/s}^2 = \left(\dfrac{1}{1000}\, \text{km}\right) / \left(\dfrac{1}{3600}\, \text{h}\right)^2 \)
\( = \dfrac{1}{1000} \times (3600)^2 \)
\( = \dfrac{1}{1000} \times 1.296 \times 10^7 = 12960\, \text{km/h}^2 \)
\( 3.0\, \text{m/s}^2 = 3.0 \times 12960 = 38880\, \text{km/h}^2 \)
✅ Answer: \( 3.89 \times 10^4\, \text{km/h}^2 \)
(d) Given: \( G = 6.67 \times 10^{-11}\, \text{N m}^2 \text{kg}^{-2} \)
We know: \( 1\, \text{N} = 10^5\, \text{dyne},\quad 1\, \text{m} = 100\, \text{cm},\) \( \quad 1\, \text{kg} = 1000\, \text{g} \)
\( G = 6.67 \times 10^{-11} \times 10^5\, \text{dyne} \) \( \times (100)^2\, \text{cm}^2 / (1000)^2\, \text{g}^2 \)
\( = 6.67 \times 10^{-11} \times 10^5 \times \dfrac{10^4}{10^6} \)
\( = 6.67 \times 10^{-8} \)
✅ Answer: \( 6.67 \times 10^{-8}\, \text{cm}^3 \text{s}^{-2} \text{g}^{-1} \)
1.3 Calorie in new units
Given:
-
\( 1\, \text{cal} = 4.2\, \text{J} = 4.2\, \text{kg m}^2 \text{s}^{-2} \)
In new units:
\( 1\, \text{kg} = \alpha,\quad 1\, \text{m} = \beta,\quad 1\, \text{s} = \gamma \)
So,
\( 1\, \text{J} \) \( = \dfrac{1}{\alpha}\, \text{mass unit} \times \dfrac{1}{\beta^2}\, \text{length unit}^2 \times \) \( \gamma^2\, \text{time unit}^{-2} \) \( = \dfrac{\gamma^2}{\alpha \beta^2}\)
So,
✅ Answer:\( \text{Magnitude of 1 calorie} = 4.2 \alpha^{-1} \beta^{-2} \gamma^2 \)
1.4 Rewriting Statements (Explanation-based)
The term “large” or “small” is relative. Without a reference standard, these terms are meaningless.
Refined statements:
(a) Atoms are very small compared to a grain of sand.
(b) A jet plane moves with great speed compared to a bicycle.
(c) The mass of Jupiter is very large compared to Earth's mass.
(d) The air inside this room contains a large number of molecules compared to the number in an evacuated container.
(e) A proton is about 1836 times more massive than an electron.
(f) The speed of sound (~343 m/s) is much smaller than the speed of light (~3 × 10^8 m/s).
1.5 Sun-Earth Distance in New Units
Let c = 1 in new units.
Time light takes = 8 min 20 s = \( 8 \times 60 + 20 = 500\, \text{s} \)
In new units where c = 1:
\( \text{Distance} = c \times t = 1 \times 500 = 500\, \text{new units} \)
✅ Answer:
Distance between Sun and Earth = 500 units (in new units where speed of
light = 1)
Detailed solutions to questions 1.6 to 1.17 from Chapter 1: Units and Measurement (Class 11 Physics - NCERT):
1.6 Most precise device:
-
(a) Vernier Callipers: Least count = 0.01 cm = 0.1 mm
-
(b) Screw Gauge: Least count = pitch / number of divisions = 1 mm / 100 = 0.01 mm
-
(c) Optical Instrument: Precision ≈ wavelength of light ≈ 0.0005 mm
✅ Answer: (c) An optical instrument is the most precise.
1.7 Estimate thickness of hair:
Given:
-
Magnified image width = 3.5 mm
-
Magnification = 100
So, actual thickness = \( \frac{3.5\ \text{mm}}{100} = 0.035\ \text{mm} = 35\ \mu m \)
✅ Answer: Thickness of hair = 35 micrometers
1.8
(a) Estimate thread diameter:
Wrap the thread around a pencil multiple times (say, 50 turns), measure
total length using a scale, and divide by number of turns.
\( \text{Diameter} = \frac{\text{Total length}}{\text{Number of turns}} \)
(b) Can accuracy of screw gauge be increased arbitrarily?
No. Mechanical limitations like backlash error and wear & tear limit
practical accuracy despite increasing scale divisions.
(c) Why 100 measurements more reliable than 5?
Because more data minimizes random errors through averaging → improves
reliability and accuracy.
1.9
Given:
-
Slide area = 1.75 cm²
-
Screen area = 1.55 m² = 15,500 cm²
\( \text{Area magnification} = \frac{15,500}{1.75} = 8857.14 \)
\( \text{Linear magnification} = \sqrt{8857.14} \approx 94.1 \)
✅ Answer: Linear magnification ≈ 94.1
1.10 Significant Figures:
(a) 0.007 → 1 sig. fig.
(b) 2.64 × 10²⁴ → 3 sig. figs.
(c) 0.2370 → 4 sig. figs.
(d) 6.320 → 4 sig. figs.
(e) 6.032 → 4 sig. figs.
(f) 0.0006032 → 4 sig. figs.
1.11
-
Length = 4.234 m (4 sig. figs.)
-
Breadth = 1.005 m (4 sig. figs.)
-
Thickness = 2.01 cm = 0.0201 m (3 sig. figs.)
Area:
\( A = 4.234 \times 1.005 = 4.25517 \) \( \Rightarrow 4.255\ \text{m}^2\ (\text{3 sig. figs.}) \)
Volume:
\( V = A \times \text{thickness} = 4.255 \times 0.0201 \) \(= 0.08561 \Rightarrow 0.0856\ \text{m}^3\ (\text{3 sig. figs.}) \)
1.12
(a) Total mass =
\( 2.30\ \text{kg} + 0.02015\ \text{kg} + 0.02017\ \text{kg} \) \( = 2.34032\ \text{kg} \Rightarrow 2.34\ \text{kg} \)
(b) Difference =
\( |20.17 - 20.15| = 0.02\ \text{g} \ (\text{2 sig. figs.}) \)
✅ Answer:
(a) Total mass = 2.34 kg
(b) Difference = 0.02 g
1.14
1 mole of H = 6.022 × 10²³ atoms
Size of one H atom ≈ 0.5 Å → radius
Volume of one atom ≈ \( \frac{4}{3} \pi r^3 \)
\(r = 0.5 \times 10^{-10} \text{ m},\) \(\quad V_{\text{atom}} = \frac{4}{3}\pi (0.5 \times 10^{-10})^3 \) \( \approx 5.24 \times 10^{-31} \text{ m}^3 \)
Total atomic volume = \( 5.24 \times 10^{-31} \times 6.022 \times 10^{23} \) \( \approx 3.15 \times 10^{-7} \text{ m}^3 \)
✅ Answer: Total atomic volume ≈ \(3.15 \times 10^{-7}\ \text{m}^3 \)
1.15
Molar volume = 22.4 L = 22.4 × 10⁻³ m³
Atomic volume from 1.14 ≈ \( 3.15 \times 10^{-7} \text{ m}^3 \)
\( \text{Ratio} = \frac{22.4 \times 10^{-3}}{3.15 \times 10^{-7}} \approx 7.1 \times 10^{4} \)
✅ Answer:
Ratio ≈ \( 7.1 \times 10^4 \)
This is large because atoms are point-like and gases have vast empty
space between them.
1.16
This is due to parallax. Nearby objects subtend larger angles and shift more rapidly relative to your line of sight, while distant objects subtend very small angles and appear almost stationary.
1.17
\( \text{Volume of Sun} \) \( = \frac{4}{3}\pi R^3 = \frac{4}{3} \pi (7 \times 10^8)^3 \) \( \approx 1.43 \times 10^{27} \text{ m}^3 \) \( \text{Density} = \dfrac{2.0 \times 10^{30}}{1.43 \times 10^{27}} \) \(\approx 1400\, \text{kg/m}^3 \)
✅ Answer: Density of Sun ≈ 1400 kg/m³ → closer to liquids (e.g., water = 1000 kg/m³), not gases.
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