Units and Measurements — NCERT Solutions

(a) Volume of a cube of side 1 cm in m³:

Given: side $= 1\,\text{cm} = 1 \times 10^{-2}\,\text{m}$

$$V = (1\times10^{-2})^3 = 10^{-6}\,\text{m}^3$$

Answer: $10^{-6}\,\text{m}^3$

(b) Surface area of a solid cylinder (radius = 2.0 cm, height = 10.0 cm) in mm²:

Formula: $A = 2\pi r(r + h)$

Convert to mm: $r = 20\,\text{mm}$, $h = 100\,\text{mm}$

$$A = 2\pi \times 20 \times (20 + 100) = 2\pi \times 20 \times 120$$
$$A = 2 \times 3.14159 \times 2400 = 15079.6\,\text{mm}^2$$
$$A \approx 1.5 \times 10^4\,\text{mm}^2$$

Answer: $\approx 1.5 \times 10^4\,\text{mm}^2$

(c) Distance covered in 1 s at 18 km h⁻¹:

$$18\,\text{km h}^{-1} = 18 \times \frac{1000\,\text{m}}{3600\,\text{s}} = 5\,\text{m s}^{-1}$$

Distance in 1 s $= 5 \times 1 = 5\,\text{m}$

Answer: $5\,\text{m}$

(d) Density of lead:

Relative density (specific gravity) $= \dfrac{\text{density of substance}}{\text{density of water}}$

Density of water $= 1\,\text{g cm}^{-3} = 1000\,\text{kg m}^{-3}$

$$\rho_{\text{lead}} = 11.3 \times 1\,\text{g cm}^{-3} = 11.3\,\text{g cm}^{-3}$$
$$\rho_{\text{lead}} = 11.3 \times 1000\,\text{kg m}^{-3} = 1.13 \times 10^4\,\text{kg m}^{-3}$$

Answer: $11.3\,\text{g cm}^{-3}$ or $1.13 \times 10^4\,\text{kg m}^{-3}$

(a) $1\,\text{kg m}^2\,\text{s}^{-2}$ in g cm² s⁻²:

$1\,\text{kg} = 10^3\,\text{g}$, $1\,\text{m} = 10^2\,\text{cm}$

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

Answer: $10^7\,\text{g cm}^2\,\text{s}^{-2}$

(b) 1 m in light years (ly):

Speed of light $c = 3 \times 10^8\,\text{m s}^{-1}$

1 year $= 365.25 \times 24 \times 3600\,\text{s} \approx 3.156 \times 10^7\,\text{s}$

$1\,\text{ly} = 3 \times 10^8 \times 3.156 \times 10^7 = 9.467 \times 10^{15}\,\text{m}$

$$1\,\text{m} = \frac{1}{9.467 \times 10^{15}}\,\text{ly} \approx 1.057 \times 10^{-16}\,\text{ly}$$

Answer: $\approx 1.057 \times 10^{-16}\,\text{ly}$

(c) $3.0\,\text{m s}^{-2}$ in km h⁻²:

$1\,\text{m} = 10^{-3}\,\text{km}$, $1\,\text{s} = \dfrac{1}{3600}\,\text{h}$, so $1\,\text{s}^{-1} = 3600\,\text{h}^{-1}$

$$3.0\,\text{m s}^{-2} = 3.0 \times 10^{-3}\,\text{km} \times (3600)^2\,\text{h}^{-2}$$
$$= 3.0 \times 10^{-3} \times 1.296 \times 10^7\,\text{km h}^{-2}$$
$$= 3.888 \times 10^4\,\text{km h}^{-2} \approx 3.9 \times 10^4\,\text{km h}^{-2}$$

Answer: $3.9 \times 10^4\,\text{km h}^{-2}$

(d) $G = 6.67 \times 10^{-11}\,\text{N m}^2\,\text{kg}^{-2}$ in cm³ s⁻² g⁻¹:

Note: $1\,\text{N} = 1\,\text{kg m s}^{-2}$, so $\text{N m}^2\,\text{kg}^{-2} = \text{kg m s}^{-2} \cdot \text{m}^2 \cdot \text{kg}^{-2} = \text{m}^3\,\text{s}^{-2}\,\text{kg}^{-1}$

Convert: $1\,\text{m}^3 = 10^6\,\text{cm}^3$, $1\,\text{kg}^{-1} = (10^3\,\text{g})^{-1} = 10^{-3}\,\text{g}^{-1}$

$$G = 6.67 \times 10^{-11} \times 10^6\,\text{cm}^3 \times \text{s}^{-2} \times 10^{-3}\,\text{g}^{-1}$$
$$= 6.67 \times 10^{-11} \times 10^3\,\text{cm}^3\,\text{s}^{-2}\,\text{g}^{-1}$$
$$= 6.67 \times 10^{-8}\,\text{cm}^3\,\text{s}^{-2}\,\text{g}^{-1}$$

Answer: $6.67 \times 10^{-8}\,\text{cm}^3\,\text{s}^{-2}\,\text{g}^{-1}$

Given:

• $1\,\text{calorie} = 4.2\,\text{J} = 4.2\,\text{kg m}^2\,\text{s}^{-2}$
• New unit of mass $= \alpha\,\text{kg}$, so $1\,\text{kg} = \dfrac{1}{\alpha}$ (new mass units)
• New unit of length $= \beta\,\text{m}$, so $1\,\text{m} = \dfrac{1}{\beta}$ (new length units)
• New unit of time $= \gamma\,\text{s}$, so $1\,\text{s} = \dfrac{1}{\gamma}$ (new time units)

Dimensional formula of energy: $[M^1 L^2 T^{-2}]$

Express 1 calorie in new units:

$$1\,\text{calorie} = 4.2\,\text{kg m}^2\,\text{s}^{-2}$$

Substituting the conversions:

$$= 4.2 \left(\frac{1}{\alpha}\,\text{new mass unit}\right) \left(\frac{1}{\beta}\,\text{new length unit}\right)^2 \left(\frac{1}{\gamma}\,\text{new time unit}\right)^{-2}$$

$$= 4.2 \times \frac{1}{\alpha} \times \frac{1}{\beta^2} \times \gamma^2 \;\text{(new units of energy)}$$

$$\boxed{1\,\text{calorie} = 4.2\,\alpha^{-1}\beta^{-2}\gamma^{2}\,\text{(in new units)}}$$

This is the required result. $\blacksquare$

Explanation of the statement:

A physical quantity has magnitude only in relation to a chosen standard (unit). Saying something is 'large' or 'small' is meaningful only when we compare it with a reference standard. For example, the size of an atom (~$10^{-10}$ m) is small compared to everyday objects but large compared to a nucleus (~$10^{-15}$ m). Without specifying the comparison standard, such statements are incomplete and meaningless.

(a) Atoms are very small objects.

This statement is incomplete as stated. Reframed:

Atoms are very small objects compared to the ordinary objects around us (e.g., a grain of sand or a human hair).

(b) A jet plane moves with great speed.

This statement is incomplete. Reframed:

A jet plane moves with a speed much greater than that of a car or a train (but much smaller than the speed of light).

(c) The mass of Jupiter is very large.

This statement is incomplete. Reframed:

The mass of Jupiter is very large compared to the mass of the Earth (Jupiter's mass $\approx 318$ times the Earth's mass).

(d) The air inside this room contains a large number of molecules.

This statement is incomplete. Reframed:

The air inside this room contains a very large number of molecules compared to, say, the number of people in the room (or compared to Avogadro's number as a reference).

(e) A proton is much more massive than an electron.

This statement is already meaningful because it specifies a comparison standard (the electron). No reframing needed.

A proton is about 1836 times more massive than an electron.

(f) The speed of sound is much smaller than the speed of light.

This statement is already meaningful because it specifies a comparison standard (the speed of light). No reframing needed.

Speed of sound $\approx 340\,\text{m s}^{-1}$, speed of light $= 3 \times 10^8\,\text{m s}^{-1}$; the speed of sound is much smaller than the speed of light.

Given:

• Speed of light in new units: $c = 1$ (new unit of length per new unit of time)
• Time taken by light to travel from Sun to Earth: $t = 8\,\text{min}\,20\,\text{s}$

Convert time to seconds:
$$t = 8 \times 60 + 20 = 480 + 20 = 500\,\text{s}$$

Using: Distance $=$ Speed $\times$ Time

$$d = c \times t = 1 \times 500 = 500\,\text{new units of length}$$

Answer: The distance between the Sun and the Earth is 500 new units of length.

(This new unit is essentially 1 light-second, and the Sun–Earth distance is 500 light-seconds.)

Concept: Precision of a measuring instrument is determined by its least count (smallest measurement it can make).

(a) Vernier callipers with 20 divisions on sliding scale:

Least count $= \dfrac{1\,\text{MSD}}{\text{number of VSD}} = \dfrac{1\,\text{mm}}{20} = 0.05\,\text{mm} = 5 \times 10^{-5}\,\text{m}$

(b) Screw gauge with pitch 1 mm and 100 divisions:

Least count $= \dfrac{\text{pitch}}{\text{number of divisions}} = \dfrac{1\,\text{mm}}{100} = 0.01\,\text{mm} = 10^{-5}\,\text{m}$

(c) Optical instrument measuring to within a wavelength of light:

Wavelength of visible light $\approx 4000\,\text{Å}$ to $7000\,\text{Å}$, i.e., $\approx 10^{-7}\,\text{m}$

Least count $\approx 10^{-7}\,\text{m}$

Comparison:

• (a): $5 \times 10^{-5}\,\text{m}$
• (b): $10^{-5}\,\text{m}$
• (c): $\sim 10^{-7}\,\text{m}$ ← smallest least count

Answer: The optical instrument (c) is the most precise device, as it has the smallest least count (~$10^{-7}$ m).

Given:

• Magnification of microscope $= 100$
• Average observed width of hair (magnified image) $= 3.5\,\text{mm}$

Formula:
$$\text{Actual thickness} = \frac{\text{Observed width}}{\text{Magnification}}$$

$$\text{Actual thickness} = \frac{3.5\,\text{mm}}{100} = 0.035\,\text{mm}$$

Answer: The estimated thickness of human hair is $\mathbf{0.035\,\text{mm}}$ (i.e., $3.5 \times 10^{-2}\,\text{mm}$).

(a) Estimating the diameter of a thread using a thread and metre scale:

Wind the thread closely and uniformly on a pencil or a cylindrical rod for a known number of turns $n$ (say 20 or more turns), such that the turns are touching each other without gaps or overlaps. Measure the total length $L$ of the winding using the metre scale.

$$\text{Diameter of thread} = \frac{L}{n}$$

This method averages out the error over many turns, giving a more accurate estimate.

(b) Can accuracy of screw gauge be increased arbitrarily by increasing divisions on circular scale?

The least count of a screw gauge is:
$$\text{Least count} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$$

For pitch $= 1.0\,\text{mm}$ and 200 divisions:
$$\text{LC} = \frac{1.0}{200} = 0.005\,\text{mm}$$

No, it is not possible to increase accuracy arbitrarily by simply increasing the number of divisions. Beyond a certain limit:

• The divisions become too small to be read accurately by the human eye.
• Manufacturing imperfections and mechanical backlash introduce errors that cannot be eliminated by finer graduation.
• The instrument's mechanical precision sets a fundamental limit.

Thus, increasing divisions beyond a practical limit does not improve accuracy.

(c) Why does a set of 100 measurements give a more reliable estimate than 5 measurements?

In any measurement, there are random errors that can be positive or negative. When a large number of measurements are taken:

• Random errors tend to cancel out (some are positive, some negative).
• The mean of a large number of observations is closer to the true value.
• The standard error of the mean $= \dfrac{\sigma}{\sqrt{n}}$, where $\sigma$ is the standard deviation and $n$ is the number of measurements.

For $n = 100$: standard error $= \dfrac{\sigma}{10}$

For $n = 5$: standard error $= \dfrac{\sigma}{\sqrt{5}} \approx \dfrac{\sigma}{2.24}$

Since $\dfrac{\sigma}{10} \ll \dfrac{\sigma}{2.24}$, 100 measurements give a far more reliable (precise) estimate of the true diameter.

Given:

• Area of house on slide: $A_1 = 1.75\,\text{cm}^2 = 1.75 \times 10^{-4}\,\text{m}^2$
• Area of house on screen: $A_2 = 1.55\,\text{m}^2$

Concept: If linear magnification is $m$, then areal magnification $= m^2$.

$$m^2 = \frac{A_2}{A_1} = \frac{1.55}{1.75 \times 10^{-4}} = \frac{1.55}{0.000175} = 8857$$

$$m = \sqrt{8857} \approx 94.1$$

Answer: The linear magnification of the projector-screen arrangement is approximately $\mathbf{94.1}$.