NCERT Solutions - Class 11 Physics Chapter 1: Units and Measurements
Exercises
Q1.1 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 ..... (mm)2
(c) A vehicle moving with a speed of 18 km h–1 covers .... m in 1 s
(d) The relative density of lead is 11.3. Its density is .... g cm–3 or .... kg m–3
(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 ..... (mm)2
(c) A vehicle moving with a speed of 18 km h–1 covers .... m in 1 s
(d) The relative density of lead is 11.3. Its density is .... g cm–3 or .... kg m–3
Answer -
a. 10–6 m3
Explanation:
Length of each side
= 1 cm
= 1/100 m
= 10–2 m
Volume of cube
= (side)3
= (10–2 m)3
= 10–6 m3
b. 1.5 × 104 (mm)2
Explanation:
Radius, r
= 2.0 cm
= 2.0 × 10 mm
= 20 mm
Height, h
= 10.0 cm
= 10.0 × 10 mm
= 100 mm
Surface area of a cylinder
= 2πr(h + r)
= 2 × 3.14 × 20 × (100 + 20)
= 15072 mm2
= 1.5 × 104 mm2
c. 5 m
Explanation:
Time, t = 1 s
Speed, v
= 18 km/h
= (18 × 1000 m)/(3600 s)
= 18 × (5/18) m/s
= 5 m/s
Therefore, distance covered
= Speed × Time
= 5 m/s × 1 s
= 5 m
d. 11.3 g/cm3, 1.13 × 104 kg/m3
Explanation:
Relative density of lead = 11.3
We know, density of water = 1 g/cm3
Relative density of a substance
= Density of the substance / Density of water
⇒ Density of a substance
= Relative density of the substance × Density of water
⇒ Density of lead
= Relative density of lead × Density of water
⇒ Density of lead
= 11.3 × 1 g/cm3
= 11.3 g/cm3
= (11.3 / 103 kg)/(1 / 106 m3)
= 1.13 × 104 kg/m3
a. 10–6 m3
Explanation:
Length of each side
= 1 cm
= 1/100 m
= 10–2 m
Volume of cube
= (side)3
= (10–2 m)3
= 10–6 m3
b. 1.5 × 104 (mm)2
Explanation:
Radius, r
= 2.0 cm
= 2.0 × 10 mm
= 20 mm
Height, h
= 10.0 cm
= 10.0 × 10 mm
= 100 mm
Surface area of a cylinder
= 2πr(h + r)
= 2 × 3.14 × 20 × (100 + 20)
= 15072 mm2
= 1.5 × 104 mm2
c. 5 m
Explanation:
Time, t = 1 s
Speed, v
= 18 km/h
= (18 × 1000 m)/(3600 s)
= 18 × (5/18) m/s
= 5 m/s
Therefore, distance covered
= Speed × Time
= 5 m/s × 1 s
= 5 m
d. 11.3 g/cm3, 1.13 × 104 kg/m3
Explanation:
Relative density of lead = 11.3
We know, density of water = 1 g/cm3
Relative density of a substance
= Density of the substance / Density of water
⇒ Density of a substance
= Relative density of the substance × Density of water
⇒ Density of lead
= Relative density of lead × Density of water
⇒ Density of lead
= 11.3 × 1 g/cm3
= 11.3 g/cm3
= (11.3 / 103 kg)/(1 / 106 m3)
= 1.13 × 104 kg/m3
Q1.2 Fill in the blanks by suitable conversion of units:
(a) 1 kg m2 s–2 = .... g cm2 s–2
(b) 1 m = ..... ly
(c) 3.0 m s–2 = .... km h–2
(d) G = 6.67 × 10–11 N m2 (kg)–2 = .... (cm)3 s–2 g–1
(a) 1 kg m2 s–2 = .... g cm2 s–2
(b) 1 m = ..... ly
(c) 3.0 m s–2 = .... km h–2
(d) G = 6.67 × 10–11 N m2 (kg)–2 = .... (cm)3 s–2 g–1
Answer -
a. 107 g cm2 s–2
Explanation:
1 kg = 103 g
1 m = 102 cm
1 kg m2 s–2
= 103 g × (102 cm)2 × s–2
= 103 g × 104 cm2 × s–2
= 107 g cm2 s–2
b. 1.057 × 10–16 ly
Explanation:
Speed of light = 3 × 108 m/s
1 light year = distance travelled by light in 1 year
= 3 × 108 m/s × 365 × 24 × 60 × 60 s
= 9.46 × 1015 m
So, 1 m = 1 / (9.46 × 1015) ly
= 1.057 × 10–16 ly
c. 3.89 × 104 km h–2
Explanation:
1 m = 1 / 1000 km
1 s = 1 / 3600 h
3.0 m/s2
= 3.0 × (1 / 1000) km / (1 / 3600)2 h2
= 3.0 × 10–3 km / (7.77 × 10–7) h2
= 3.89 × 104 km h–2
d. 6.67 × 10–8 cm3 s–2 g–1
Explanation:
G = 6.67 × 10–11 N m2 kg–2
1 N = 1 kg m/s2
So,
G = 6.67 × 10–11 kg–1 m3 s–2
1 kg = 103 g
1 m = 102 cm
G = 6.67 × 10–11 × (102)3 cm3 × s–2 × (103)–1 g–1
= 6.67 × 10–11 × 106 / 103
= 6.67 × 10–8 cm3 s–2 g–1
a. 107 g cm2 s–2
Explanation:
1 kg = 103 g
1 m = 102 cm
1 kg m2 s–2
= 103 g × (102 cm)2 × s–2
= 103 g × 104 cm2 × s–2
= 107 g cm2 s–2
b. 1.057 × 10–16 ly
Explanation:
Speed of light = 3 × 108 m/s
1 light year = distance travelled by light in 1 year
= 3 × 108 m/s × 365 × 24 × 60 × 60 s
= 9.46 × 1015 m
So, 1 m = 1 / (9.46 × 1015) ly
= 1.057 × 10–16 ly
c. 3.89 × 104 km h–2
Explanation:
1 m = 1 / 1000 km
1 s = 1 / 3600 h
3.0 m/s2
= 3.0 × (1 / 1000) km / (1 / 3600)2 h2
= 3.0 × 10–3 km / (7.77 × 10–7) h2
= 3.89 × 104 km h–2
d. 6.67 × 10–8 cm3 s–2 g–1
Explanation:
G = 6.67 × 10–11 N m2 kg–2
1 N = 1 kg m/s2
So,
G = 6.67 × 10–11 kg–1 m3 s–2
1 kg = 103 g
1 m = 102 cm
G = 6.67 × 10–11 × (102)3 cm3 × s–2 × (103)–1 g–1
= 6.67 × 10–11 × 106 / 103
= 6.67 × 10–8 cm3 s–2 g–1
Q1.3 A calorie is a unit of heat (energy in transit) 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 α kg, the unit of length equals β m, and 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.
Answer -
We are given:
1 calorie = 4.2 J
1 J = 1 kg·m2/s2
This means:
1 calorie = 4.2 kg·m2/s2
Now let us convert this into the new units:
New unit of mass = α kg
New unit of length = β m
New unit of time = γ s
So,
1 new unit of mass = α kg ⇒ 1 kg = 1/α new mass unit
1 new unit of length = β m ⇒ 1 m = 1/β new length unit
1 new unit of time = γ s ⇒ 1 s = 1/γ new time unit
Now write the expression for energy:
1 calorie = 4.2 kg·m2/s2
Substitute:
= 4.2 × (1/α) new mass unit × (1/β)2 new length unit2 / (1/γ)2 new time unit2
= 4.2 × (1/α) × (1/β2) × (γ2)
= 4.2 × α–1 × β–2 × γ2
Hence, in the new system of units, the magnitude of 1 calorie is:
= 4.2 α–1 β–2 γ2
We are given:
1 calorie = 4.2 J
1 J = 1 kg·m2/s2
This means:
1 calorie = 4.2 kg·m2/s2
Now let us convert this into the new units:
New unit of mass = α kg
New unit of length = β m
New unit of time = γ s
So,
1 new unit of mass = α kg ⇒ 1 kg = 1/α new mass unit
1 new unit of length = β m ⇒ 1 m = 1/β new length unit
1 new unit of time = γ s ⇒ 1 s = 1/γ new time unit
Now write the expression for energy:
1 calorie = 4.2 kg·m2/s2
Substitute:
= 4.2 × (1/α) new mass unit × (1/β)2 new length unit2 / (1/γ)2 new time unit2
= 4.2 × (1/α) × (1/β2) × (γ2)
= 4.2 × α–1 × β–2 × γ2
Hence, in the new system of units, the magnitude of 1 calorie is:
= 4.2 α–1 β–2 γ2
Q1.4 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
(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.
(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 -
A physical quantity like length, mass, or speed cannot be called large or small unless it is compared to a reference or standard value. Without such a comparison, the terms 'large' or 'small' are not meaningful.
Reframed statements:
A physical quantity like length, mass, or speed cannot be called large or small unless it is compared to a reference or standard value. Without such a comparison, the terms 'large' or 'small' are not meaningful.
Reframed statements:
- Atoms are very small compared to a pinhead or a grain of sand.
- A jet plane moves with great speed compared to a car or a train.
- The mass of Jupiter is very large compared to the mass of Earth.
- The air inside this room contains a large number of molecules compared to the air in a balloon.
- A proton is about 1836 times more massive than an electron.
- The speed of sound (≈ 343 m/s) is much smaller than the speed of light (≈ 3 × 108 m/s).
Q1.5 How many times does a clock’s second hand rotate in 1 hour? A new unit of length is defined 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?
Answer - According to the question, the speed of light in vacuum is unity for the new unit of length. In other words, the length of the new unit is equivalent to the distance covered by the light in one second.
We know, speed of light
= 3 × 108 m/s
So, 1 new unit = 3 × 108 m/s
Time, t
= 8 min 20 s
= (8 × 60 s) + 20 s
= 480 + 20 s
= 500 s
Required distance
= Speed × Time
= 3 × 108 m/s × 500 s
= 1 new unit/s × 500 s
= 500 new units
We know, speed of light
= 3 × 108 m/s
So, 1 new unit = 3 × 108 m/s
Time, t
= 8 min 20 s
= (8 × 60 s) + 20 s
= 480 + 20 s
= 500 s
Required distance
= Speed × Time
= 3 × 108 m/s × 500 s
= 1 new unit/s × 500 s
= 500 new units
Q1.6 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?
(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 -
Precision depends on the least count (smallest value measurable).
Therefore, an optical instrument is the most precise.
Precision depends on the least count (smallest value measurable).
- Vernier callipers: Least count = 1 mm / 20 = 0.05 mm
- Screw gauge: Least count = Pitch / Number of divisions = 1 mm / 100 = 0.01 mm
- Optical instrument: Can measure up to the order of light wavelength ≈ 10–6 m = 0.000001 mm
Therefore, an optical instrument is the most precise.
Q1.7 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?
Answer -
Magnified width = 3.5 mm
Magnification = 100
Actual thickness = Measured value / Magnification
=> Thickness = 3.5 mm / 100 = 0.035 mm = 35 µm
Therefore, estimated thickness = 35 micrometres (µm)
Magnified width = 3.5 mm
Magnification = 100
Actual thickness = Measured value / Magnification
=> Thickness = 3.5 mm / 100 = 0.035 mm = 35 µm
Therefore, estimated thickness = 35 micrometres (µm)
Q1.8 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?
(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 -
- To estimate the diameter of the thread, wind it tightly around a pencil in a single layer, ensuring no gaps. Measure the total length covered by, say, 50 turns of the thread using a metre scale. Then, divide the total length by the number of turns to get the diameter of the thread.
- No, increasing the number of divisions on the circular scale cannot increase accuracy arbitrarily. Practical limits such as precision in engraving, readability, and thermal or mechanical limitations restrict how finely the scale can be divided. Beyond a point, such subdivisions will not translate to real improvements in accuracy.
- A larger number of measurements reduces the effect of random errors. When we take 100 measurements and calculate the mean, the influence of outliers and random fluctuations gets averaged out, giving a more reliable and accurate estimate than just 5 measurements.
Q1.9 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 -
Image area = 1.55 m² = 1.55 × 10⁴ cm²
Object area = 1.75 cm²
Area magnification
= Image area / Object area
= (1.55 × 10⁴) / 1.75
≈ 8857.14
Linear magnification = √(Area magnification)
=> m = √8857.14 ≈ 94.1
Therefore, linear magnification ≈ 94.1
Image area = 1.55 m² = 1.55 × 10⁴ cm²
Object area = 1.75 cm²
Area magnification
= Image area / Object area
= (1.55 × 10⁴) / 1.75
≈ 8857.14
Linear magnification = √(Area magnification)
=> m = √8857.14 ≈ 94.1
Therefore, linear magnification ≈ 94.1
Q1.10 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
(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
Answer -
Rules for counting significant figures:
Rules for counting significant figures:
- All non-zero digits are always significant.
- Zeros between non-zero digits are significant.
- Leading zeros (zeros before the first non-zero digit) are not significant.
- Trailing zeros after a decimal point are significant.
- Trailing zeros in a whole number without a decimal point are not significant.
- 0.007 → 1 significant figure (only '7')
- 2.64 × 1024 → 3 significant figures (2, 6, and 4)
- 0.2370 → 4 significant figures ('0' after decimal is significant)
- 6.320 → 4 significant figures (includes trailing zero after decimal)
- 6.032 → 4 significant figures
- 0.0006032 → 4 significant figures (leading zeros not counted)
Q1.11 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 -
Rules for Significant Figures:
Length = 4.234 m
Breadth = 1.005 m
Thickness = 2.01 cm = 0.0201 m
Total surface area of the cuboid:
Total surface area
= 2(lb + bh + hl)
= 2 × (4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)
= 2 × (4.25517 + 0.0202005 + 0.0851034)
= 2 × 4.3604739
= 8.7209478 m2
Correct to 3 significant figures (least precise: 2.01 → 3 sig. figs)
⇒ Area = 8.72 m2
Volume of the cuboid:
Volume
= l × b × h
= 4.234 × 1.005 × 0.0201
= 0.085535917 m3
Correct to 3 significant figures
⇒ Volume = 0.0855 m3
Rules for Significant Figures:
- For multiplication and division: The final answer should have as many significant figures as the value with the least number of significant figures.
- For addition and subtraction: The result should be rounded off to the least number of decimal places among the values.
Length = 4.234 m
Breadth = 1.005 m
Thickness = 2.01 cm = 0.0201 m
Total surface area of the cuboid:
Total surface area
= 2(lb + bh + hl)
= 2 × (4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)
= 2 × (4.25517 + 0.0202005 + 0.0851034)
= 2 × 4.3604739
= 8.7209478 m2
Correct to 3 significant figures (least precise: 2.01 → 3 sig. figs)
⇒ Area = 8.72 m2
Volume of the cuboid:
Volume
= l × b × h
= 4.234 × 1.005 × 0.0201
= 0.085535917 m3
Correct to 3 significant figures
⇒ Volume = 0.0855 m3
Q1.12 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, (b) the difference in the masses of the pieces to correct significant figures?
Answer -
Rules for Significant Figures:
Note: The box mass is in kg while the gold pieces are in g. So we convert everything to the same unit.
Rules for Significant Figures:
- For multiplication and division: The final answer should have as many significant figures as the value with the least number of significant figures.
- For addition and subtraction: The result should be rounded off to the least number of decimal places among the values.
Note: The box mass is in kg while the gold pieces are in g. So we convert everything to the same unit.
- Total mass of the box:
Mass of box = 2.30 kg
Gold piece 1 = 20.15 g = 0.02015 kg
Gold piece 2 = 20.17 g = 0.02017 kg
Total mass
= 2.30 + 0.02015 + 0.02017
= 2.34032 kg
Now, limiting to 2 decimal places (as in 2.30):
⇒ Total mass = 2.34 kg - Difference in masses of the two gold pieces:
Difference
= 20.17 g – 20.15 g
= 0.02 g
Both values have 2 decimal places, so answer should also have 2 decimal places:
⇒ Difference = 0.02 g
Q1.13 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 = m₀ / (1 − v²)1/2
Guess where to put the missing c.
m = m₀ / (1 − v²)1/2
Guess where to put the missing c.
Answer -
We apply dimensional analysis to find the correct position of the missing constant c.
The given formula is:
m = m₀ / √(1 − v²)
Since mass (m and m₀) has dimension [M], the right-hand side must also have the same dimension as mass. So the denominator √(1 − v²) must be dimensionless.
But v² has the dimension of velocity squared:
[v²] = [L² T–2]
So, (1 − v²) is not dimensionless.
To make it dimensionless, we must divide v² by a quantity with the same dimension, which is c² (where c = speed of light):
⇒ v² / c² is dimensionless
Therefore, the correct relation is:
m = m₀ / √(1 − v² / c²)
So, the missing c must be placed as:
m = m₀ / √(1 − v² / c²)
We apply dimensional analysis to find the correct position of the missing constant c.
The given formula is:
m = m₀ / √(1 − v²)
Since mass (m and m₀) has dimension [M], the right-hand side must also have the same dimension as mass. So the denominator √(1 − v²) must be dimensionless.
But v² has the dimension of velocity squared:
[v²] = [L² T–2]
So, (1 − v²) is not dimensionless.
To make it dimensionless, we must divide v² by a quantity with the same dimension, which is c² (where c = speed of light):
⇒ v² / c² is dimensionless
Therefore, the correct relation is:
m = m₀ / √(1 − v² / c²)
So, the missing c must be placed as:
m = m₀ / √(1 − v² / c²)
Q1.14 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?
Answer -
Radius of a hydrogen atom
= 0.5 Å
= 0.5 × 10–10 m
Volume of one hydrogen atom (assuming spherical shape)
= (4/3)πr3
= (4/3) × 3.14 × (0.5 × 10–10)3 m3
= (4/3) × 3.14 × 0.125 × 10–30 m3
= 0.5233 × 10–30 m3
= 5.233 × 10–31 m3
Total number of atoms in 1 mole = 6.022 × 1023
Total volume of 1 mole of hydrogen atoms
= Volume of 1 atom × Number of atoms
= 5.233 × 10–31 m3 × 6.022 × 1023
= 3.15 × 10–7 m3
Radius of a hydrogen atom
= 0.5 Å
= 0.5 × 10–10 m
Volume of one hydrogen atom (assuming spherical shape)
= (4/3)πr3
= (4/3) × 3.14 × (0.5 × 10–10)3 m3
= (4/3) × 3.14 × 0.125 × 10–30 m3
= 0.5233 × 10–30 m3
= 5.233 × 10–31 m3
Total number of atoms in 1 mole = 6.022 × 1023
Total volume of 1 mole of hydrogen atoms
= Volume of 1 atom × Number of atoms
= 5.233 × 10–31 m3 × 6.022 × 1023
= 3.15 × 10–7 m3
Q1.15 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 hydrogen molecule to be about 1 Å). Why is this ratio so large?
Answer -
Molar volume of gas at STP
= 22.4 L
= 22.4 × 10–3 m3
Radius of hydrogen molecule
= 0.5 Å
= 0.5 × 10–10 m
Volume of one hydrogen molecule (spherical shape)
= (4/3)πr3
= (4/3) × 3.14 × (0.5 × 10–10)3
= 0.523 × 10–30 m3
Volume of 1 mole of hydrogen molecules
= 0.523 × 10–30 × 6.022 × 1023
= 3.15 × 10–7 m3
Ratio of molar volume to atomic volume
= (22.4 × 10–3) / (3.15 × 10–7)
= 7.11 × 104
The ratio is very large because gas molecules are widely separated from each other. Most of the volume occupied by a gas is actually empty space.
Molar volume of gas at STP
= 22.4 L
= 22.4 × 10–3 m3
Radius of hydrogen molecule
= 0.5 Å
= 0.5 × 10–10 m
Volume of one hydrogen molecule (spherical shape)
= (4/3)πr3
= (4/3) × 3.14 × (0.5 × 10–10)3
= 0.523 × 10–30 m3
Volume of 1 mole of hydrogen molecules
= 0.523 × 10–30 × 6.022 × 1023
= 3.15 × 10–7 m3
Ratio of molar volume to atomic volume
= (22.4 × 10–3) / (3.15 × 10–7)
= 7.11 × 104
The ratio is very large because gas molecules are widely separated from each other. Most of the volume occupied by a gas is actually empty space.
Q1.16 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).
Answer -
This is due to the phenomenon of parallax, which is the apparent shift in the position of an object when viewed from different positions.
Nearby objects appear to shift more against the background in response to your movement because the angle subtended by them at your eyes changes rapidly. This gives the illusion that they are moving very fast in the opposite direction.
Distant objects, like the Moon or hilltops, are so far away that the angle they subtend changes very little as you move. So they appear almost stationary or even seem to move along with you.
This difference in apparent motion is a direct consequence of the geometry of vision and how relative motion is perceived based on distance.
This is due to the phenomenon of parallax, which is the apparent shift in the position of an object when viewed from different positions.
Nearby objects appear to shift more against the background in response to your movement because the angle subtended by them at your eyes changes rapidly. This gives the illusion that they are moving very fast in the opposite direction.
Distant objects, like the Moon or hilltops, are so far away that the angle they subtend changes very little as you move. So they appear almost stationary or even seem to move along with you.
This difference in apparent motion is a direct consequence of the geometry of vision and how relative motion is perceived based on distance.
Q1.17 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 -
Even though the Sun is made up of high-temperature plasma, its total mass is extremely large and compressed under strong gravitational forces. So we can expect its average density to be much higher than typical gases and possibly comparable to liquids or even solids.
Volume of the Sun (assuming spherical shape)
= (4/3)πR3
= (4/3) × 3.14 × (7.0 × 108 m)3
= (4/3) × 3.14 × 343 × 1024 m3
= 1.436 × 1027 m3
Density
= Mass / Volume
= (2.0 × 1030 kg) / (1.436 × 1027 m3)
= 1.39 × 103 kg/m3
This value is around 1400 kg/m3, which is comparable to the density of liquids like water.
So, even though the Sun is in a plasma state, its average mass density lies in the range of liquids, not gases. Our guess is correct.
Even though the Sun is made up of high-temperature plasma, its total mass is extremely large and compressed under strong gravitational forces. So we can expect its average density to be much higher than typical gases and possibly comparable to liquids or even solids.
Volume of the Sun (assuming spherical shape)
= (4/3)πR3
= (4/3) × 3.14 × (7.0 × 108 m)3
= (4/3) × 3.14 × 343 × 1024 m3
= 1.436 × 1027 m3
Density
= Mass / Volume
= (2.0 × 1030 kg) / (1.436 × 1027 m3)
= 1.39 × 103 kg/m3
This value is around 1400 kg/m3, which is comparable to the density of liquids like water.
So, even though the Sun is in a plasma state, its average mass density lies in the range of liquids, not gases. Our guess is correct.