Units & Measurements — NCERT Exercises

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 kmh^-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) The volume of a cube of side 1 cm is equal to 10^-6 m³.

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to 1.51 x 10⁴ mm².

(c) A vehicle moving with a speed of 18 kmh^-1 covers 5 m in 1 s.

(d) The relative density of lead is 11.3. Its density is 11.3 g cm^-3 or 11.3 x 10³ kg m^-3.

Reason —

(a) Given,

• Side of the cube, a = 1 cm

Since 1 cm = 10^-2 m, the volume of the cube is

$\text V = \text a^3 = (1\ \text{cm})^3 \\[1em] = (10^{-2}\ \text m)^3 \\[1em] = 10^{-6}\ \text m^3$

(b) Given,

• Radius of the cylinder, r = 2.0 cm = 20 mm
• Height of the cylinder, h = 10.0 cm = 100 mm

The total surface area of a solid cylinder of radius r and height h is

$\text A = 2\pi \text{rh} + 2\pi \text r^2 = 2\pi \text r(\text h + \text r)$

Substituting the values,

$\text A = 2 \times 3.14 \times 20 \times (100 + 20) \\[1em] = 125.6 \times 120 \\[1em] = 15072\ \text{mm}^2 \\[1em] = 1.51 \times 10^4\ \text{mm}^2$

(c) Given,

• Speed of the vehicle, v = 18 km h^-1

Since 1 km = 1000 m and 1 h = 3600 s, converting km h^-1 into m s^-1,

$\text v = 18 \times \dfrac{1\ \text{km}}{1\ \text h} \\[1em] = 18 \times \dfrac{1000\ \text m}{3600\ \text s} \\[1em] = 18 \times \dfrac{5\ \text m}{18\ \text s} \\[1em] = 5\ \text{m s}^{-1}$

Hence, the vehicle covers 5 m in 1 s.

(d) Given,

• Relative density of lead = 11.3

The relative density of a substance is the ratio of its density to the density of water, so

$\text{Relative density of lead} = \dfrac{\text{Density of lead}}{\text{Density of water}}$

Taking the density of water as 1 g cm^-3,

$\text{Density of lead} = 11.3 \times 1\ \text{g cm}^{-3} = 11.3\ \text{g cm}^{-3}$

Since 1 g = 10^-3 kg and 1 cm = 10^-2 m,

$\text{Density of lead} = 11.3 \times \dfrac{10^{-3}\ \text{kg}}{(10^{-2}\ \text m)^3} \\[1em] = 11.3 \times 10^{-3} \times 10^{6}\ \text{kg m}^{-3} \\[1em] = 11.3 \times 10^{3}\ \text{kg m}^{-3}$

Fill in the blanks by suitable conversion of units:

(a) 1 kg m² s^-2 = ............... g cm² s^-2.

(b) 1 m = ............... light year (ly).

(c) 3 m s^-2 = ............... km h^-2.

(d) G = 6.67 x 10^-11 N m² kg^-2 = ............... cm³ s^-2 g^-1.

Answer

(a) 1 kg m² s^-2 = 10⁷ g cm² s^-2.

(b) 1 m = 1.06 x 10^-16 light year (ly).

(c) 3 m s^-2 = 3.888 x 10⁴ km h^-2.

(d) G = 6.67 x 10^-11 N m² kg^-2 = 6.67 x 10^-8 cm³ s^-2 g^-1.

Reason —

(a) Since 1 kg = 10³ g and 1 m = 10² cm,

$1\ \text{kg m}^2\text{ s}^{-2} = (10^3\ \text g)(10^2\ \text{cm})^2\ \text s^{-2} \\[1em] = 10^3 \times 10^4\ \text{g cm}^2\text{ s}^{-2} \\[1em] = 10^{7}\ \text{g cm}^2\text{ s}^{-2}$

(b) One light year is the distance travelled by light in vacuum in one year, and

$1\ \text{ly} = 9.46 \times 10^{15}\ \text m$

Therefore,

$1\ \text m = \dfrac{1}{9.46 \times 10^{15}}\ \text{ly} \\[1em] = 1.06 \times 10^{-16}\ \text{ly}$

(c) Since 1 m = 10^-3 km and 1 s = $\dfrac{1}{3600}$ h,

$3\ \text{m s}^{-2} = 3 \times (10^{-3}\ \text{km}) \times \left(\dfrac{1}{3600}\ \text h\right)^{-2} \\[1em] = 3 \times 10^{-3} \times (3600)^2\ \text{km h}^{-2} \\[1em] = 3 \times 10^{-3} \times 1.296 \times 10^{7}\ \text{km h}^{-2} \\[1em] = 3.888 \times 10^{4}\ \text{km h}^{-2}$

(d) Writing the newton in terms of the fundamental units, 1 N = 1 kg m s^-2. Therefore,

$\text G = 6.67 \times 10^{-11}\ \text{N m}^2\text{ kg}^{-2} \\[1em] = 6.67 \times 10^{-11}\ (\text{kg m s}^{-2})\ \text m^2\ \text{kg}^{-2} \\[1em] = 6.67 \times 10^{-11}\ \text{m}^3\text{ s}^{-2}\text{ kg}^{-1}$

Since 1 m = 10² cm and 1 kg = 10³ g,

$\text G = 6.67 \times 10^{-11} (10^2\ \text{cm})^3\ \text s^{-2}\ (10^3\ \text g)^{-1} \\[1em] = 6.67 \times 10^{-11} \times 10^{6} \times 10^{-3}\ \text{cm}^3\text{ s}^{-2}\text{ g}^{-1} \\[1em] = 6.67 \times 10^{-8}\ \text{cm}^3\text{ s}^{-2}\text{ g}^{-1}$

A calorie is a unit of heat (or energy) and it equals about 4.2 J, where 1 J = 1 kg m² s^-2. Suppose we use a system of units in which the unit of mass is α kg, the unit of length equals β m and the unit of time is γ s. Show that the magnitude of a calorie in terms of the new units is 4.2 α^-1 β^-2 γ².

Answer

Given,

• Magnitude of a calorie in SI units, n₁ = 4.2 (since 1 calorie = 4.2 J)
• First system of units (SI) : unit of mass M₁ = 1 kg, unit of length L₁ = 1 m, unit of time T₁ = 1 s
• New system of units : unit of mass M₂ = α kg, unit of length L₂ = β m, unit of time T₂ = γ s

The dimensional formula of heat (energy) is

$[\text{energy}] = [\text{M}^1\text{L}^2\text{T}^{-2}]$

so the dimensions in mass, length and time are a = 1, b = 2 and c = -2.

The magnitude of a physical quantity remains the same in every system of units, that is,

$\text n_1\text u_1 = \text n_2\text u_2$

Hence the numerical value in the new system is

$\text n_2 = \text n_1 \left[\dfrac{\text M_1}{\text M_2}\right]^\text a \left[\dfrac{\text L_1}{\text L_2}\right]^\text b \left[\dfrac{\text T_1}{\text T_2}\right]^\text c$

Substituting the values,

$\text n_2 = 4.2 \left[\dfrac{1\ \text{kg}}{\alpha\ \text{kg}}\right]^1 \left[\dfrac{1\ \text m}{\beta\ \text m}\right]^2 \left[\dfrac{1\ \text s}{\gamma\ \text s}\right]^{-2} \\[1em] = 4.2 \left[\dfrac{1}{\alpha}\right] \left[\dfrac{1}{\beta}\right]^2 \left[\dfrac{1}{\gamma}\right]^{-2} \\[1em] = 4.2\ \alpha^{-1}\ \beta^{-2}\ \gamma^{2}$

Hence, the magnitude of a calorie in terms of the new units is 4.2 α^-1 β^-2 γ².

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 has 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 dimensional quantity has a definite unit, and its measured value depends on the unit chosen. Calling such a quantity large or small has no meaning by itself, because the same quantity may appear large when compared with one standard and small when compared with another. A quantity can be called large or small only when it is compared with a standard (or reference) quantity of the same kind. Hence the statements are reframed by including a suitable standard of comparison.

(a) An atom is very small compared to the objects we see around us in daily life. So the statement should be reframed as : Atoms are very small objects compared to a pinhead (or any object of everyday size).

(b) The speed of a jet plane is great compared to the speed of an ordinary vehicle such as a car or a train. So the statement should be reframed as : A jet plane moves with a speed greater than that of a superfast train (or an ordinary car).

(c) The mass of Jupiter is very large compared to the mass of other planets. So the statement should be reframed as : The mass of Jupiter is very large compared to the mass of the Earth (or any other planet).

(d) The number of air molecules in this room is large compared to the number of molecules present in a small region, such as the air contained in a test tube. So the statement should be reframed as : The air inside this room has a large number of molecules compared to the number contained in a test tube.

(e) This statement is already correct, since the mass of a proton is compared with the mass of an electron, which serves as the standard of comparison. Both are of the same kind (mass), so no reframing is necessary.

A proton is much more massive than an electron.

(f) This statement is also correct, since the speed of sound is compared with the speed of light, which serves as the standard of comparison. Both are of the same kind (speed), so no reframing is necessary.

The speed of sound is much smaller than the speed of light.

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?

Answer

Given,

• Time taken by light to travel from the sun to the earth, t = 8 min 20 s = (8 × 60) + 20 = 500 s
• Speed of light in vacuum, c = 3 × 10⁸ m s^-1

The new unit of length is so chosen that the speed of light in vacuum is unity, that is, light travels 1 new unit of length in 1 s. Hence

$1\ \text{new unit of length} = 3 \times 10^8\ \text m$

The distance between the sun and the earth in SI units is

$\text s = \text c \times \text t \\[1em] = (3 \times 10^8\ \text{m s}^{-1}) \times (500\ \text s) \\[1em] = 1.5 \times 10^{11}\ \text m$

The magnitude of a physical quantity remains the same in every system of units, that is, n₁u₁ = n₂u₂. Here

• n₁ = 1.5 × 10¹¹, u₁ = 1 m
• u₂ = 3 × 10⁸ m

Therefore,

$\text n_2 = \dfrac{\text n_1 \text u_1}{\text u_2} = \dfrac{1.5 \times 10^{11}\ \text m}{3 \times 10^{8}\ \text m} \\[1em] = 0.5 \times 10^{3} \\[1em] = 500$

Hence, the distance between the sun and the earth is 500 new units of length.

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

Answer

The precision of a measuring instrument is determined by its least count. The smaller the least count, the greater is the precision. Hence the most precise device is the one having the smallest least count.

(a) For the vernier callipers,

$\text{Least count} = \dfrac{\text{Value of one main scale division}}{\text{Number of divisions on the sliding scale}} \\[1em] = \dfrac{1\ \text{mm}}{20} \\[1em] = 5 \times 10^{-2}\ \text{mm} = 5 \times 10^{-5}\ \text m$

(b) For the screw gauge,

$\text{Least count} = \dfrac{\text{Pitch}}{\text{Number of divisions on the circular scale}} \\[1em] = \dfrac{1\ \text{mm}}{100} \\[1em] = 10^{-2}\ \text{mm} = 10^{-5}\ \text m$

(c) For the optical instrument, the least count is of the order of the wavelength of light, which for visible light is about 5000 Å,

$\text{Least count} = 5000 \times 10^{-10}\ \text m = 5 \times 10^{-7}\ \text m$

Comparing the three least counts, the optical instrument has the smallest least count.

Hence, the optical instrument is the most precise device for measuring length.

A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes observations and finds the mean width of the hair in the field of view of the microscope to be 3.5 mm. What is the estimated thickness of hair?

Answer

Given,

• Magnification of the microscope, m = 100
• Observed (apparent) width of the hair in the field of view = 3.5 mm

The magnification of a microscope is the ratio of the apparent width to the actual width, that is,

$\text{Magnification} = \dfrac{\text{Apparent width of the hair}}{\text{Actual width of the hair}}$

Therefore,

$\text{Actual width of the hair} = \dfrac{\text{Apparent width of the hair}}{\text{Magnification}} \\[1em] = \dfrac{3.5\ \text{mm}}{100} \\[1em] = 0.035\ \text{mm}$

Hence, the estimated thickness of the hair is 0.035 mm.

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. Is it possible to increase the accuracy of the 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

(a) Wind the thread closely and uniformly on a cylindrical body such as a pencil, so that the successive turns touch one another without any gap or overlap. Let the number of turns be n. Measure with the metre scale the length L of the portion of the pencil covered by these n turns. Then the diameter of the thread is

$\text d = \dfrac{\text L}{\text n}$

The estimate can be improved by increasing the number of turns n.

(b) The least count of a screw gauge is

$\text{Least count} = \dfrac{\text{Pitch}}{\text{Number of divisions on the circular scale}}$

so increasing the number of divisions on the circular scale reduces the least count and thus increases the precision. However, this cannot be done indefinitely, because on increasing the number of divisions the divisions come so close together that they can no longer be resolved by the eye, and the reading itself becomes uncertain.

Hence, the accuracy of the screw gauge cannot be increased arbitrarily by increasing the number of divisions on the circular scale.

(c) The readings of the diameter differ from one another because of random errors, which occur irregularly and are equally likely to be positive or negative. Random errors are minimised by taking a large number of observations of the same quantity and then taking their arithmetic mean, since the positive and negative errors tend to cancel one another.

Hence, the mean of 100 measurements is a more reliable estimate of the diameter than the mean of only 5 measurements.

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

Given,

• Area of the house on the slide (area of the object), A_o = 1.75 cm²
• Area of the house on the screen (area of the image), A_i = 1.55 m² = 1.55 × 10⁴ cm²

The areal magnification is the ratio of the area of the image to the area of the object,

$\text{Areal magnification} = \dfrac{\text A_\text i}{\text A_\text o} = \dfrac{1.55 \times 10^4\ \text{cm}^2}{1.75\ \text{cm}^2} \\[1em] = 8857.14$

Since area is proportional to the square of the length, the areal magnification is the square of the linear magnification. Therefore,

$\text{Linear magnification} = \sqrt{\text{Areal magnification}} \\[1em] = \sqrt{8857.14} \\[1em] = 94.1$

Hence, the linear magnification of the projector-screen arrangement is 94.1.

Note: The value 8857.14 is the areal magnification, not the linear magnification. The linear magnification is its square root, 94.1.

State the number of significant figures in the following:

(i) 0.007 m²,

(ii) 2.64 x 10²⁴ kg,

(iii) 0.2370 g cm^-3,

(iv) 6.320 J,

(v) 6.032 N m²,

(vi) 0.0006032 m²

Answer

(i) One.

Reason — All the initial zeros on the right of the decimal point but on the left of the first non-zero digit are not significant, since they only fix the position of the decimal point. Hence in 0.007 only the digit 7 is significant.

(ii) Three.

Reason — All the non-zero digits are significant. In 2.64 × 10²⁴ the significant figures are 2, 6 and 4; the power of ten only fixes the position of the decimal point and is not counted.

(iii) Four.

Reason — All the zeros to the right of the decimal point which come after a non-zero digit are significant. Hence in 0.2370 the significant figures are 2, 3, 7 and 0.

(iv) Four.

Reason — All the non-zero digits are significant, and the trailing zero to the right of the decimal point is also significant. Hence in 6.320 the significant figures are 6, 3, 2 and 0.

(v) Four.

Reason — All the zeros lying between two non-zero digits are significant. Hence in 6.032 the significant figures are 6, 0, 3 and 2.

(vi) Four.

Reason — The initial zeros on the right of the decimal point but on the left of the first non-zero digit are not significant, whereas the zero lying between the digits 6 and 3 is significant. Hence in 0.0006032 the significant figures are 6, 0, 3 and 2.

The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m and 2.01 cm respectively. Give area and the volume of the sheet to correct significant figures.

Answer

Given,

• Length, l = 4.234 m (4 significant figures)
• Breadth, b = 1.005 m (4 significant figures)
• Thickness, t = 2.01 cm = 0.0201 m (3 significant figures)

Area of the sheet.

The total surface area of the rectangular sheet is

$\text A = 2(\text{lb} + \text{bt} + \text{tl})$

Substituting the values,

$\text A = 2[(4.234 \times 1.005) + (1.005 \times 0.0201) + (0.0201 \times 4.234)] \\[1em] = 2[4.25517 + 0.0202005 + 0.0851034] \\[1em] = 2 \times 4.3604739 \\[1em] = 8.7209478 \text { m}^2$

In multiplication, the result is rounded off to the least number of significant figures in the given data, which is 3. Hence,

$\text A = 8.72 \text { m}^2$

Volume of the sheet.

The volume of the rectangular sheet is

$\text V = \text l \times \text b \times \text t$

Substituting the values,

$\text V = 4.234 \times 1.005 \times 0.0201 \\[1em] = 0.0855289... \\[1em] = 0.0855 \text { m}^3$

Here also the result is rounded off to 3 significant figures, the least number of significant figures in the given data.

Hence, the area of the sheet is 8.72 m² and the volume of the sheet is 0.0855 m³.

The mass of a box measured by a grocer's balance is 2.3 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 mass of the pieces to correct significant figures?

Answer

Given,

• Mass of the box, m = 2.3 kg
• Mass of the first gold piece, m₁ = 20.15 g = 0.02015 kg
• Mass of the second gold piece, m₂ = 20.17 g = 0.02017 kg

(a) The total mass of the box is

$\text M = \text m + \text m_1 + \text m_2 \\[1em] = 2.3 + 0.02015 + 0.02017 \\[1em] = 2.34032 \text { kg}$

In addition, the result is rounded off to the same number of decimal places as the quantity having the least number of decimal places. Here the mass of the box, 2.3 kg, has only one decimal place. Hence

$\text M = 2.3 \text { kg}$

Hence, the total mass of the box is 2.3 kg.

(b) The difference in the masses of the two gold pieces is

$\text m_2 - \text m_1 = 20.17 - 20.15 \\[1em] = 0.02 \text { g}$

Both the masses are given up to two decimal places, so the difference is also written up to two decimal places.

Hence, the difference in the mass of the pieces is 0.02 g.

A famous relation in physics relates 'moving mass' m to the 'rest mass' m_o of a particle in terms of its speed v and speed of light c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost but forgets where to put the constant c. He writes $\text m =\dfrac{\text m_\text o}{(1 - \text v^2)^{1/2}}$. Guess, where to put the missing c?

Answer

The given relation is

$\text m = \dfrac{\text m_\text o}{(1 - \text v^2)^{1/2}}$

According to the principle of homogeneity of dimensions, only quantities of the same dimensions can be added to or subtracted from one another. In the bracket, the number 1 is dimensionless, so the quantity subtracted from it, that is v², must also be dimensionless.

But the speed v is not dimensionless; its dimensional formula is

$[\text v] = [\text{L}\text{T}^{-1}]$

Hence v² by itself cannot be subtracted from 1. To make this term dimensionless, v² must be divided by a quantity having the same dimensions as v². The speed of light c has the same dimensions as v,

$[\text c] = [\text{L}\text{T}^{-1}]$

so the correct dimensionless term is $\dfrac{\text v^2}{\text c^2}$, since

$\left[\dfrac{\text v^2}{\text c^2}\right] = \dfrac{[\text{L}^2\text{T}^{-2}]}{[\text{L}^2\text{T}^{-2}]} = [\text{M}^0\text{L}^0\text{T}^0]$

which is dimensionless.

Therefore, the constant c must be placed with v so that v² is replaced by $\dfrac{\text v^2}{\text c^2}$, and the correct relation is

$\text m = \dfrac{\text m_\text o}{\left(1 - \dfrac{\text v^2}{\text c^2}\right)^{1/2}}$

Hence, the missing constant c should be placed as $\dfrac{\text v^2}{\text c^2}$ in place of v².

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

Given,

• Size (radius) of a hydrogen atom, r = 0.5 Å = 0.5 × 10^-10 m
• Avogadro's number, N_A = 6.023 × 10²³ mol^-1

Assuming the hydrogen atom to be a sphere of radius r, the volume of one atom is

$\text V_{\text{atom}} = \dfrac{4}{3}\pi \text r^3$

Substituting the values,

$\text V_{\text{atom}} = \dfrac{4}{3} \times 3.14 \times (0.5 \times 10^{-10}\ \text m)^3 \\[1em] = \dfrac{4}{3} \times 3.14 \times 1.25 \times 10^{-31} \\[1em] = 5.24 \times 10^{-31}\ \text m^3$

One mole of hydrogen contains N_A atoms, so the total atomic volume of one mole of hydrogen atoms is

$\text V_{\text{total}} = \text N_\text A \times \text V_{\text{atom}} \\[1em] = (6.023 \times 10^{23}) \times (5.24 \times 10^{-31}) \\[1em] = 3.16 \times 10^{-7}\ \text m^3$

Hence, the total atomic volume of a mole of hydrogen atoms is 3.16 × 10^-7 m³.

One mole of an ideal gas at STP occupies 22.4 L (molar volume). What is the ratio of molar volume to atomic volume of a mole of hydrogen? Why is this ratio so large? Take the size (diameter) of hydrogen molecule to be about 1 Å.

Answer

Given,

• Molar volume of hydrogen at STP, V_molar = 22.4 L = 22.4 × 10^-3 m³
• Size (diameter) of a hydrogen molecule = 1 Å, so radius r = 0.5 Å = 0.5 × 10^-10 m
• Avogadro's number, N_A = 6.023 × 10²³ mol^-1

Treating the hydrogen molecule as a sphere of radius r, the atomic volume of one mole of hydrogen is

$\text V_{\text{atomic}} = \dfrac{4}{3}\pi \text r^3 \times \text N_\text A$

Substituting the values,

$\text V_{\text{atomic}} = \dfrac{4}{3} \times 3.14 \times (0.5 \times 10^{-10}\ \text m)^3 \times (6.023 \times 10^{23}) \\[1em] = 3.152 \times 10^{-7}\ \text m^3$

Therefore, the required ratio is

$\dfrac{\text V_{\text{molar}}}{\text V_{\text{atomic}}} = \dfrac{22.4 \times 10^{-3}\ \text m^3}{3.152 \times 10^{-7}\ \text m^3} \\[1em] = 7.1 \times 10^{4}$

Hence, the ratio of the molar volume to the atomic volume of a mole of hydrogen is 7.1 × 10⁴.

This ratio is so large because in a gas the intermolecular separation is very much greater than the size of a molecule. The molecules of a gas occupy only a very small part of the total volume of the gas, the rest being empty space.

Explain the common observation clearly: If you look out of the window of a fast-moving train, the nearby trees, electric poles, etc., appear to move rapidly opposite to the train's motion ; but the distant objects (hill tops, moon, stars; etc.) appear to be stationary. Explain this observation.

Answer

This observation is explained by the change in the line of sight, that is, the line joining the object to the eye of the observer.

For a nearby object such as a tree or an electric pole, the distance from the eye is small. As the train moves, the line of sight of the tree turns through a large angle in a short time, so its direction changes rapidly. Hence the tree appears to move rapidly in a direction opposite to the motion of the train.

For a distant object such as a hill top, the moon or a star, the distance from the eye is extremely large. As the train moves, the line of sight of such an object turns through an extremely small angle, so its direction remains practically unchanged. Hence the distant object appears to be stationary, that is, it appears to move along with the observer.

Hence, the nearby objects appear to move rapidly in the backward direction because the direction of the line of sight changes rapidly, while the distant objects appear stationary because the direction of the line of sight remains practically unchanged.

The sun is a hot plasma (ionised matter) with its inner core at a temperature exceeding 10⁷ K, and its outer surface at about 6000 K. At such high temperatures, no substance remains in solid or liquid phase. In what range do you expect the mass density of sun, in the range of densities of solids, liquids or gases? Check your guess using the data, mass of sun, M = 2.0 x 10³⁰ kg and radius of sun, R = 7.0 x 10⁸ m.

Answer

Given,

• Mass of the sun, M = 2.0 × 10³⁰ kg
• Radius of the sun, R = 7.0 × 10⁸ m

Treating the sun as a sphere of radius R, its volume is

$\text V = \dfrac{4}{3}\pi \text R^3$

The mass density of the sun is

$\rho = \dfrac{\text{Mass of the sun}}{\text{Volume of the sun}} = \dfrac{\text M}{\dfrac{4}{3}\pi \text R^3}$

Substituting the values,

$\rho = \dfrac{2.0 \times 10^{30}\ \text{kg}}{\dfrac{4}{3} \times 3.14 \times (7.0 \times 10^8\ \text m)^3} \\[1em] = \dfrac{2.0 \times 10^{30}}{1.437 \times 10^{27}} \\[1em] = 1.39 \times 10^{3}\ \text{kg m}^{-3}$

This value lies in the range of the densities of solids and liquids (about 10³ kg m^-3), and is very much greater than the densities of ordinary gases (about 1 kg m^-3). Although the sun is a hot plasma, its enormous gravitational attraction compresses the matter to such a high density.

Hence, the mass density of the sun is 1.39 × 10³ kg m^-3, which lies in the range of the densities of solids and liquids.