UNITS AND MEASUREMENT-Exercise

Q1. Fill in the blanks
(a) The volume of a cube of side 1 cm is equal to .....\(\mathrm{m^3}\)
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...\(\mathrm{(mm)^2}\)
(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 ....\(\mathrm{g cm^{–3}}\) or ....\(\mathrm{kg m^{–3}}\).

Solution

(a)

Side \(= 1\ \text{cm} = 10^{-2}\ \text{m}\).

Volume of cube \[\begin{aligned}&= (10^{-2})^3 \\&= 10^{-6}\ \text{m}^3\end{aligned}\] (b)

Radius \(r = 2.0\ \text{cm} = 20\ \text{mm}\),
height \(h = 10.0\ \text{cm} = 100\ \text{mm}\).
Total surface area

\[\begin{aligned} A &= 2\pi r(h + r)\\ &= 2\pi \times 20 \times (100 + 20)\ \text{mm}^2\\ &\approx 1.5\times 10^{4}\ \text{mm}^2 \end{aligned}\] (c) \[\begin{aligned} &18\ \text{km h}^{-1}\\ &= 18 \times \frac{1000}{3600}\ \text{m s}^{-1}\\\\ &= 5\ \text{m s}^{-1}. \end{aligned}\]

Distance covered in \(1\ \text{s} = 5\ \text{m}\).

(d)

Relative density of lead \(= 11.3\).
Density of water \(= 1\ \text{g cm}^{-3}\).
So density of lead

\[\begin{aligned} &= 11.3\ \text{g cm}^{-3}\\ &= 11.3 \times 10^{3}\ \text{kg m}^{-3}\\ &= 1.13\times 10^{4}\ \text{kg m}^{-3}.\\ \end{aligned}\]

Q2. Fill in the blanks by suitable conversion of units
(a) \(\mathrm{1\ kg\ m^2\ s^{–2}}\) = ....\(\mathrm{g\ cm^2\ s^{–2}}\)
(b) 1 m = ..... ly
(c) \(\mathrm{3.0\ m\ s^{–2} = ....\ km\ h^{–2}}\)
(d) \(\mathrm{G = 6.67 \times 10^{–11}\ N\ m^2\ (kg)^{–2} = ....\ (cm)^3\ s^{–2}\ g^{–1}}\).

Solution

(a) \[\begin{aligned} &1\ \text{kg m}^2 \text{s}^{-2}\\ &= (10^3\ \text{g})(10^2\ \text{cm})^2 \text{s}^{-2}\\ &= 10^3 \times 10^4\ \text{g cm}^2 \text{s}^{-2}\\ &= 10^7\ \text{g cm}^2 \text{s}^{-2} \end{aligned}\] (b) \[\begin{aligned} &1\ \text{ly} \approx 9.46\times 10^{15}\ \text{m}\\\\ &\Rightarrow 1\ \text{m} \approx \frac{1}{9.46\times 10^{15}}\ \text{ly}\\\\ &\approx 1.06\times 10^{-16}\ \text{ly}. \end{aligned}\] (c) \[\begin{aligned} &3.0\ \text{m s}^{-2}\\\\ &= 3.0 \times \frac{1\ \text{km}}{1000\ \text{m}} \times \left(\frac{3600\ \text{s}}{1\ \text{h}}\right)^{2}\\\\ &= 0.003 \times 3600^2\ \text{km h}^{-2}\\ &\approx 3.89\times 10^{4}\ \text{km h}^{-2}. \end{aligned}\] (d) \[\begin{aligned} G &= 6.67\times 10^{-11}\ \text{N m}^2 \text{kg}^{-2}\\\\ &= 6.67\times 10^{-11}\ \frac{\text{m}^3}{\text{s}^2\ \text{kg}}. \end{aligned}\]

Using \(1\ \text{m}^3 = 10^{6}\ \text{cm}^3\), \(1\ \text{kg} = 10^{3}\ \text{g}\),

\[\begin{aligned} G &= 6.67\times 10^{-11} \times \frac{10^{6}\ \text{cm}^3}{\text{s}^2 \times 10^{3}\ \text{g}}\\\\ &= 6.67\times 10^{-8}\ \text{cm}^3\ \text{s}^{-2}\ \text{g}^{-1}. \end{aligned}\]

Q3. A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where \(\mathrm{1J = 1\ kg\ m^2\ s^{–2}}\). 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 \(\mathrm{α^{–1}\ β^{–2}\ γ^2}\) in terms of the new units.

Solution

Let the new fundamental units be:
Unit of mass \(M' = \alpha\ \text{kg}\)
Unit of length \(L' = \beta\ \text{m}\)
Unit of time \(T' = \gamma\ \text{s}\)

Also, \[\begin{aligned} 1\ \text{calorie} &= 4.2\ \text{J} \\&= 4.2\ \text{kg m}^2 \text{s}^{-2}. \end{aligned}\] Express old units in terms of new units \[\begin{aligned} 1\ \text{kg} &= \frac{1}{\alpha}\, M',\\ 1\ \text{m} &= \frac{1}{\beta}\, L',\\ 1\ \text{s} &= \frac{1}{\gamma}\, T'. \end{aligned}\] Therefore, \[\begin{aligned} 1\ \text{J} &= 1\ \text{kg m}^2 \text{s}^{-2}\\ &= \left(\frac{1}{\alpha} M'\right) \left(\frac{1}{\beta} L'\right)^2 \left(\frac{1}{\gamma} T'\right)^{-2} \end{aligned}\] Now, \[ \left(\frac{1}{\gamma} T'\right)^{-2} = \gamma^{2} {T'}^{-2}, \] so \[\begin{aligned} 1\ \text{J} &= \frac{1}{\alpha}\cdot\frac{1}{\beta^{2}}\cdot \gamma^{2}\; M' {L'}^{2} {T'}^{-2}\\\\ &= \alpha^{-1}\beta^{-2}\gamma^{2}\; (M' {L'}^{2} {T'}^{-2}) \end{aligned}\]

In the new system, the unit of energy is \(M'{L'}^{2}{T'}^{-2}\), so the numerical value of \(1\ \text{J}\) is \(\alpha^{-1}\beta^{-2}\gamma^{2}\)

Calorie in new units

\[ 1\ \text{calorie} = 4.2\ \text{J} = 4.2\ \alpha^{-1}\beta^{-2}\gamma^{2}\; (\text{new energy unit}) \] Hence, in the new units, a calorie has magnitude \[ \boxed{4.2\,\alpha^{-1}\,\beta^{-2}\,\gamma^{2}} \]

Q4. 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.

Solution

(a) Atoms are very small objects
Atoms have sizes of the order of \(10^{−10}\) m, which is extremely small compared to everyday length scales like a millimetre or a metre.

​ (b) A jet plane moves with great speed
A jet plane typically moves at a speed of a few hundred metres per second, which is much greater than the speed of ordinary road vehicles.

(c) The mass of Jupiter is very large
The mass of Jupiter (about \(10^{27}\) kg) is extremely large compared to the mass of Earth or typical laboratory-scale objects.

​ (d) The air inside this room contains a large number of molecules
The air inside a typical room contains of the order of \(10^{25}\) molecules, which is enormous compared to the number of macroscopic objects we usually deal with.

​ (e) A proton is much more massive than an electron
A proton has a mass about \(1.67×10^{−27}\) kg, which is roughly 1836 times the mass of an electron, so it is much more massive relative to an electron.

​ (f) The speed of sound is much smaller than the speed of light
In air, the speed of sound (about \(\mathrm{3.4×10^2\ m\ s^{−1}}\)) is negligible compared to the speed of light in vacuum (about \(\mathrm{3.0×10^8\ m s^{−1}}\)).

Q5. 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?

Solution

Given:
In the new system, the speed of light \(c' = 1\) (new length unit per second)
Time taken by light from Sun to Earth: \(t = 8\ \text{min} + 20\ \text{s}\)

Convert time to seconds:

\[\begin{aligned} t &= 8 \times 60\ \text{s} + 20\ \text{s}\\ &= 480\ \text{s} + 20\ \text{s}\\ &= 500\ \text{s} \end{aligned}\]

With \(c' = 1\), distance in new units is simply

\[\begin{aligned} d' &= c' \times t \\ &= 1 \times 500 \\&= 500 \end{aligned}\] So the distance between the Sun and the Earth is \[ \boxed{500\ \text{(new length units)}} \]

Q6. 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 ?

Solution

The most precise device is the one with the smallest least count. (a) Vernier callipers with 20 divisions on the sliding (Vernier) scale typically have a least count of about \(10^{-2}\) cm=0.01 cm.

​ (b) Screw gauge: pitch = 1 mm, circular scale=100 divisions
\[\begin{aligned}\text{Least count}&= \dfrac{\text{pitch}}{\text{no. of divisions}}\\\\ &=\dfrac{1}{100}\\\\ &=0.01\ \mathrm{mm}\\\\ &=0.001\ \mathrm{cm} \end{aligned}\]
(c) Optical instrument measuring down to a wavelength of light \(∼10^{−5}\) cm has least count \(\approx 10^{−5}\) cm, which is smaller than both 0.01 cm and 0.001 cm.
Therefore, (c) an optical instrument that can measure length to within a wavelength of light is the most precise device.

Q7. 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 ?

Solution

Given:
Magnification of microscope \(M = 100\)
Average observed width of hair in microscope \(w_{\text{obs}} = 3.5\ \text{mm}\)

Magnification is defined as \[ M = \frac{\text{observed size}}{\text{actual size}} \] So, the actual thickness \(w_{\text{actual}}\) of the hair is \[\begin{aligned} w_{\text{actual}} &= \frac{w_{\text{obs}}}{M}\\\\ &= \frac{3.5\ \text{mm}}{100}\\\\ &= 0.035\ \text{mm}\\\\ \end{aligned}\] Thus, the estimated thickness of the hair is \[ \boxed{0.035\ \text{mm}} \]

Q8. 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 ?

Solution

(a) Wrap the thread tightly around a smooth ruler (or pencil) in closely packed turns, count the number of turns \(n\), and measure the total length \(L\) covered by these turns along the scale. Then the estimated diameter of the thread is \[d\approx \dfrac{n}{L}\]
(b) Increasing the number of circular‑scale divisions reduces the least count and in principle improves precision, but accuracy cannot be increased arbitrarily this way. Practical limits such as finite eye resolution, mechanical imperfections, backlash, zero errors, and thermal or elastic effects stop the accuracy from improving beyond a certain point even if more divisions are added.
​ (c) With 100 measurements, random errors (which are equally likely positive or negative) tend to cancel better when averaged, so the mean diameter is closer to the true value. A larger sample also gives a more reliable estimate of the spread (scatter) of readings, making the final reported value statistically more trustworthy than one based on only 5 measurements.

Q9. 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.

Solution

Given:
Area of house on slide (object): \(A_o = 1.75\ \text{cm}^2\)
Area of house on screen (image): \(A_i = 1.55\ \text{m}^2\)

First, convert \(A_o\) from \(\text{cm}^2\) to \(\text{m}^2\):

\[ 1\ \text{cm}^2 = 10^{-4}\ \text{m}^2 \] \[\begin{aligned} A_o &= 1.75\ \text{cm}^2 \\&= 1.75 \times 10^{-4}\ \text{m}^2 \end{aligned}\] Areal magnification: \[\begin{aligned} m_A &= \frac{A_i}{A_o}\\\\ &= \frac{1.55}{1.75 \times 10^{-4}}\\\\ &\approx 8.857 \times 10^{3} \end{aligned}\]

Since areal magnification is the square of linear magnification \(m\):

\[\begin{aligned} m_A &= m^2 \\\\\Rightarrow m &= \sqrt{m_A}\\\\ &= \sqrt{8.857 \times 10^{3}}\\\\ &\approx 9.41 \times 10^{1}\\\\ &\approx 94 \end{aligned}\] So, the linear magnification of the projector–screen arrangement is \[ \boxed{m \approx 94} \]

Q10. State the number of significant figures in the following :
(a) \(\mathrm{0.007\ m^2}\)
(b) \(\mathrm{2.64 \times 10^24\ kg}\)
(c) \(\mathrm{0.2370\ g\ cm^{–3}}\)
(d) \(\mathrm{6.320\ J}\)
(e) \(\mathrm{6.032\ N\ m{–2}}\)
(f) \(\mathrm{0.0006032\ m^2}\)

Solution

Q11. 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 surface area and volume of the sheet to correct significant figures.

Solution

Given:
Length \(l = 4.234\ \text{m}\) (4 significant figures)
Breadth \(b = 1.005\ \text{m}\) (4 significant figures)
Thickness \(t = 2.01\ \text{cm} = 0.0201\ \text{m}\) (3 significant figures)

Surface Area \[\begin{aligned} SA &= 2(lb+bt+ta)\\ &= \scriptsize 2\Bigl[(4.234 \times 1.005) + (1.005 \times 2.01\times10^{-2}) + (2.01\times10^{-2}\times 4.234)\Bigr]\ \text{m}^2\\ &=\scriptsize 2\Bigl[4.25517+0.0202005+0.0851034\Bigr]\ \text{m}^2\\ &=2\times 4.3604739\ \text{m}^2\\ &=8.7209478\ \text{m}^2 \end{aligned}\]

For multiplication, the result should have the same number of significant figures as the factor with the least significant figures. Here, among \(4.234\), \(1.005\) and \(2.01\), \(2.01\) have 3 significant figures, so:

\[\begin{aligned} SA \approx 8.72\ \text{m}^2 \\ \scriptsize(\text{3 significant figures}) \end{aligned}\] Volume \[\begin{aligned} V &= l \times b \times t\\ &= (4.234)(1.005)(0.0201)\ \text{m}^3\\ &= 4.25517 \times 0.0201\ \text{m}^3\\ &\approx 0.085536\ \text{m}^3 \end{aligned}\]

Among the factors, thickness \(t = 0.0201\ \text{m}\) has 3 significant figures, which is the least. So the volume should be given to 3 significant figures:

\[\begin{aligned} V \approx 0.0855\ \text{m}^3 \\ \scriptsize(\text{3 significant figures}) \end{aligned}\] Answers (with correct significant figures): \[\begin{aligned} \boxed{SA = 8.72\ \text{m}^2}\\\\ \boxed{V = 0.0855\ \text{m}^3} \end{aligned}\]

Q12. 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 ?

Solution

Given:
Mass of box: \(m_b = 2.30\ \text{kg}\)
Masses of gold pieces: \(m_1 = 20.15\ \text{g},\ m_2 = 20.17\ \text{g}\)

Convert grams to kilograms: \[\begin{aligned} m_1 &= 20.15\ \text{g} \\&= 0.02015\ \text{kg},\\\\ m_2 &= 20.17\ \text{g} \\&= 0.02017\ \text{kg} \end{aligned}\] (a) Total mass \[\begin{aligned} m_{\text{total}} &= m_b + m_1 + m_2\\ &= 2.30 + 0.02015 + 0.02017\ \text{kg}\\ &= 2.34032\ \text{kg} \end{aligned}\]

For addition, the result is limited by the least precise decimal place
\(2.30\ \text{kg}\) is known to \(0.01\ \text{kg}\) (2 decimal places)
\(0.02015\) and \(0.02017\ \text{kg}\) are more precise

So round to 2 decimal places: \[ m_{\text{total}} \approx 2.34\ \text{kg} \] (b) Difference in masses of the pieces \[\begin{aligned} \Delta m &= m_2 - m_1 \\ &= 20.17\ \text{g} - 20.15\ \text{g} \\ &= 0.02\ \text{g} \end{aligned}\]

Both \(20.17\ \text{g}\) and \(20.15\ \text{g}\) are given to 2 decimal places in grams, so the difference should also be given to 2 decimal places:

\[ \Delta m = 0.02\ \text{g} \] Answers: \[\begin{aligned} \boxed{m_{\text{total}} = 2.34\ \text{kg}}\\\\ \boxed{\Delta m = 0.02\ \text{g}} \end{aligned}\]

Q13. A famous relation in physics relates ‘moving mass’ \(\mathrm{m}\) to the ‘rest mass’ \(\mathrm{m_o}\) of a particle in terms of its speed \(\mathrm{v}\) and the speed of light, \(\mathrm{c}\). (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 \(\mathrm{c}\). He writes : \[m=\dfrac{m_0}{(1-v^2)^{\frac{1}{2}}}\] Guess where to put the missing \(\mathrm{c}\).

Solution

The boy has written \[ m = \frac{m_0}{\sqrt{1 - v^2}} \]

For this to be a valid physical relation:
1. The left-hand side and right-hand side must have the same dimensions.
2. The expression inside the square root, \(1 - v^2\), must be **dimensionless**, because you cannot subtract a quantity with dimensions from the pure number 1.

Here, \(v^2\) has dimensions of \([L^2 T^{-2}]\), so \(1 - v^2\) is not dimensionless.

To make the denominator dimensionless, we divide \(v^2\) by \(c^2\), where \(c\) is the speed of light and has the same dimensions as \(v\). Then

\[ \frac{v^2}{c^2} \]

is dimensionless, and so is \(1 - \dfrac{v^2}{c^2}\)

Thus, the correct form of the relation is \[ \boxed{m = \frac{m_0}{\sqrt{1 - \dfrac{v^2}{c^2}}}} \]

Q14. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by \(\mathrm{Å: 1\ Å = 10^{–10}\ m}\). 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 ?

Solution

Given:
\(1\ \text{Å} = 10^{-10}\ \text{m}\)
Size (diameter) of hydrogen atom \(\approx 0.5\ \text{Å}\)

Assume the hydrogen atom is spherical.
Radius of hydrogen atom
Diameter \(d = 0.5\ \text{Å} = 0.5 \times 10^{-10}\ \text{m}\)
So radius

\[\begin{aligned} r &= \frac{d}{2}\\\\ &= \frac{0.5 \times 10^{-10}}{2}\\\\ &= 0.25 \times 10^{-10}\ \text{m}\\\\ &= 2.5 \times 10^{-11}\ \text{m} \end{aligned}\] Volume of one hydrogen atom \[\begin{aligned} V_{\text{atom}} &= \frac{4}{3}\pi r^3\\\\ &= \frac{4}{3}\pi (2.5 \times 10^{-11})^3\ \text{m}^3\\\\ &= \frac{4}{3}\pi \times 15.625 \times 10^{-33}\ \text{m}^3\\\\ & =4.19 \times 15.625 \times 10^{-33}\ \text{m}^3\\\\ & =65.46875 \times 10^{-33}\ \text{m}^3\\\\ &\approx 6.55 \times 10^{-32}\ \text{m}^3\\\\ \end{aligned}\]

Total atomic volume for 1 mole

Avogadro number: \[ N_A \approx 6.02 \times 10^{23}\ \text{mol}^{-1}. \] Total volume: \[\begin{aligned} V_{\text{total}} & = N_A V_{\text{atom}}\\\\ & = (6.02 \times 10^{23})(6.55 \times 10^{-32})\ \text{m}^3 \\\\ & =6.02 \times 6.55 \approx 39.5,\quad\\\\ & =10^{23} \times 10^{-32} = 10^{-9}\\\\ & = 39.5 \times 10^{-9}\ \text{m}^3\\\\ & \approx 3.95 \times 10^{-8}\ \text{m}^3 \end{aligned}\] Answer \[ \boxed{V_{\text{total}} \approx 4 \times 10^{-8}\ \text{m}^3} \]

Q15. 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 ?

Solution

Given:
Molar volume at STP: \(V_m = 22.4\ \text{L} = 22.4 \times 10^{-3}\ \text{m}^3\).
Size (diameter) of a hydrogen molecule: \(d \approx 1\ \text{Å} = 10^{-10}\ \text{m}\).
Assume molecule is spherical.

Volume of one hydrogen molecule

Radius: \[\begin{aligned} r &= \frac{d}{2} = \frac{1 \times 10^{-10}}{2}\\\\ &= 0.5 \times 10^{-10}\ \text{m}\\\\ &= 5 \times 10^{-11}\ \text{m} \end{aligned}\] Volume of one molecule: \[\begin{aligned} V_{\text{mol}} &= \frac{4}{3}\pi r^3\\\\ &= \frac{4}{3}\pi (5 \times 10^{-11})^3\ \text{m}^3\\\\ &= \frac{4}{3}\pi \times 125 \times 10^{-33}\\\\ &= 4.19 \times 125 \times 10^{-33}\\\\ &= 523.75 \times 10^{-33}\ \text{m}^3\\\\ &\approx 5.24 \times 10^{-31}\ \text{m}^3 \end{aligned}\]

Atomic volume of 1 mole of hydrogen

Avogadro number: \[ N_A \approx 6.02 \times 10^{23}. \] Total “atomic” (molecular) volume for 1 mole: \[\begin{aligned} V_{\text{atomic}} &= N_A V_{\text{mol}}\\ &\approx (6.02 \times 10^{23})(5.24 \times 10^{-31})\ \text{m}^3\\ &\approx 31.6 \times 10^{-8}\ \text{m}^3\\ &= 3.16 \times 10^{-7}\ \text{m}^3 \end{aligned}\]

Ratio of molar volume to atomic volume

\[\begin{aligned} \frac{V_m}{V_{\text{atomic}}} &= \frac{2.24 \times 10^{-2}}{3.16 \times 10^{-7}}\\\\ &= \frac{2.24}{3.16} \times 10^{5}\\\\ &\approx 0.71 \times 10^{5}\\\\ &\approx 7.1 \times 10^{4} \end{aligned}\] So, \[ \boxed{\dfrac{V_m}{V_{\text{atomic}}} \sim 10^{4}} \]

(If one compares \(V_m\) with one molecule’s volume, the ratio is \(\dfrac{2.24\times 10^{-2}}{5.24\times 10^{-31}} \sim 10^{28}\))

Why is this ratio so large?

• The molecules are extremely small compared to the average separation between them in a gas at STP, so each molecule occupies only a tiny fraction of the container volume.
• In kinetic theory, an ideal gas is treated as point particles with negligible own volume; most of the molar volume is effectively **empty space**, hence the very large ratio.

Q16. 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).

Solution

Nearby objects and distant objects subtend very different angles at your eye when you move, so their apparent motion on your retina is very different. This effect is called motion parallax. ​

• As the train moves, your eye position changes continuously relative to the outside scene.


• For nearby objects (trees, poles, houses close to the track), a small sideways shift of your eye causes a large change in the direction from your eye to the object. The image of the object sweeps quickly across your retina, so it appears to move rapidly in the opposite direction to the train’s motion.
​
• For distant objects (far hills, the Moon, stars), the same sideways shift produces only a tiny change in direction, so their images shift negligibly on the retina. They therefore appear almost stationary, or to share your motion, because their parallax (apparent angular shift) is extremely small due to the large distance.

​ Thus, the closer an object is, the faster it appears to move across your field of view for the same motion of the observer; very distant objects appear practically fixed in direction even though you are moving rapidly.

Q17. The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding \(\mathrm{10^7\ K}\), 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 = \(\mathrm{2.0 ×10^30\ kg}\), radius of the Sun = \(\mathrm{7.0 × 10^8\ m}\).

Solution

Given:
Mass of the Sun: \(M = 2.0 \times 10^{30}\ \text{kg}\)
Radius of the Sun: \(R = 7.0 \times 10^{8}\ \text{m}\)

We want its average density:

\[ \rho = \frac{M}{V} \]

where \(V\) is the volume of a sphere:

\[\begin{aligned} V &= \frac{4}{3}\pi R^3\\\\ &=\frac{4}{3}\pi (7.0 \times 10^{8})^3\\\\ &=\frac{4}{3}\pi \times 343 \times 10^{24}\\\\ &= \frac{4}{3} \times 3.14 \times 3.43 \times 10^{26}\\\\ &= 4.19 \times 3.43 \times 10^{26}\\\\ &\approx 1.44 \times 10^{27}\ \text{m}^3 \end{aligned}\]

Density

\[\begin{aligned} \rho &= \frac{M}{V}\\\\ &= \frac{2.0 \times 10^{30}}{1.44 \times 10^{27}}\ \text{kg m}^{-3}\\\\ &\approx 1.39 \times 10^{3}\ \text{kg m}^{-3} \end{aligned}\] So the average density of the Sun is \[ \boxed{\rho \approx 1.4 \times 10^{3}\ \text{kg m}^{-3}} \]

Range comparison

Typical densities:

• Solids/liquids: \(\sim 10^{3}\ \text{kg m}^{-3}\) (e.g. water \(\approx 10^{3}\ \text{kg m}^{-3}\))
• Gases at STP: \(\sim 1\ \text{kg m}^{-3}\) or less

Therefore, the Sun’s average density (\(\sim 1.4\times 10^{3}\ \text{kg m}^{-3}\)) lies in the range of solids and liquids, not in the range of ordinary gases, even though its matter is in the plasma (high‑temperature gas) state.