NCERT Exemplar Solutions Class 11 Physics Units and Measurement — Very Short and Short Answer Questions Q19–35

NCERT Exemplar Problems, Class 11 Physics — Units and Measurement, Very Short Answer Questions 19 to 24 and Short Answer Questions 25 to 35. All 17 questions solved. The book numbers them 2.19 to 2.35.

Three ideas carry this page. A unit is a choice of scale, so the same quantity can be measured in different units, and changing the units of mass, length and time changes every derived unit in a predictable way. Small angles are ratios: an object of size $D$ at distance $r$ subtends $\theta = D/r$ radians, which turns astronomy questions into proportion. Every measurement has an uncertainty, set by the least count of the instrument (precision) and by how far the readings sit from the true value (accuracy).

Key insight. In Questions 30 and 32 to 34, write the dimensions or the defining ratio first and the arithmetic follows in one line. Questions 25 and 29 are the same idea, $\theta = D/r$, used twice.

A note on the syllabus: Questions 20, 21, 25, 26 and 29 draw on the sections on measuring large distances, mass and time (parallax, the mass spectrograph, atomic clocks) that the rationalised NCERT Chapter 1 no longer covers. They are solved here for completeness.

Very Short Answer Questions

Question 19

Why do we have different units for the same physical quantity?

Solution. The sizes of things differ by many orders of magnitude, and a single unit would give awkward numbers at one end of the scale or the other. Interatomic distances are conveniently measured in ångströms ($10^{-10}$ m), distances between cities in kilometres, and distances between stars in light years or parsecs. Choosing a unit close to the size being measured keeps the numbers manageable.

Question 20

The radius of atom is of the order of $1$ Å and radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus?

Solution. $1$ Å $= 10^{-10}$ m and $1$ fermi $= 10^{-15}$ m, so the radii differ by a factor $10^5$. Volume goes as the cube of the radius:

$$\frac{V_{\text{atom}}}{V_{\text{nucleus}}} = \left(\frac{10^{-10}}{10^{-15}}\right)^3 = (10^5)^3 = 10^{15}.$$

Question 21

Name the device used for measuring the mass of atoms and molecules.

Solution. Atoms and molecules are far too light for any balance. In a mass spectrograph they are ionised and sent through electric and magnetic fields; the radius of the path depends on the ratio of mass to charge, so the masses can be read off from where the ions land.

Question 22

Express unified atomic mass unit in kg.

Solution. The unified atomic mass unit is defined as $\tfrac{1}{12}$ of the mass of one carbon-12 atom. One mole of carbon-12 has mass $12$ g and contains $N_A = 6.022\times10^{23}$ atoms, so

$$1\ \text{u} = \frac{1}{12}\times\frac{12\times10^{-3}\ \text{kg}}{6.022\times10^{23}} = 1.66\times10^{-27}\ \text{kg}.$$

Question 23

A function $f(\theta)$ is defined as:

$$f(\theta) = 1 – \theta + \frac{\theta^2}{2!} – \frac{\theta^3}{3!} + \frac{\theta^4}{4!}$$

Why is it necessary for $\theta$ to be a dimensionless quantity?

Solution. Terms can be added only if they have the same dimensions. The first term, $1$, is a pure number, so every other term must be too. If $\theta$ had a dimension, $\theta$, $\theta^2$, $\theta^3$ and $\theta^4$ would each carry a different power of it, and they could not be added to each other or to $1$. The only way for all of them to match is for $\theta$ to have no dimensions.

Question 24

Why length, mass and time are chosen as base quantities in mechanics?

Solution. Every other quantity in mechanics, such as velocity, force, energy, pressure and momentum, can be written in terms of length, mass and time through simple relations. The three are also independent of each other: none can be expressed through the other two. That makes them a complete and minimal set of base quantities.

Short Answer Questions

Question 25

(a) The earth-moon distance is about $60$ earth radius. What will be the diameter of the earth (approximately in degrees) as seen from the moon?

(b) Moon is seen to be of $(\tfrac12)^\circ$ diameter from the earth. What must be the relative size compared to the earth?

(c) From parallax measurement, the sun is found to be at a distance of about $400$ times the earth-moon distance. Estimate the ratio of sun-earth diameters.

Solution. For small angles the angle in radians is size divided by distance.

(a) The earth’s diameter is $2R_E$ and the distance is $60R_E$, so

$$\theta = \frac{2R_E}{60R_E} = \frac{1}{30}\ \text{rad} = \frac{1}{30}\times\frac{180^\circ}{\pi} \approx 1.9^\circ \approx 2^\circ.$$

(b) The moon and the earth are at the same distance from each other, so their angular diameters, as seen from each other, are in the ratio of their diameters:

$$\frac{D_{\text{moon}}}{D_{\text{earth}}} = \frac{(1/2)^\circ}{2^\circ} = \frac14.$$

The earth is about four times the diameter of the moon.

(c) From the earth the sun and the moon look the same size (both about $\tfrac12^\circ$, which is why total eclipses happen). Equal angles mean $D_{\text{sun}}/r_{\text{sun}} = D_{\text{moon}}/r_{\text{moon}}$, so

$$\frac{D_{\text{sun}}}{D_{\text{moon}}} = \frac{r_{\text{sun}}}{r_{\text{moon}}} = 400, \qquad \frac{D_{\text{sun}}}{D_{\text{earth}}} = 400\times\frac14 = 100.$$

The key writes the angle in (a) as $1/60$ rad and calls it about $1^\circ$; that is the angle subtended by one radius, and doubling it gives the same $2^\circ$ for the diameter.

Question 26

Which of the following time measuring devices is most precise?

(a) A wall clock. (b) A stop watch. (c) A digital watch. (d) An atomic clock.

Give reason for your answer.

Solution. A wall clock reads to a second, and a stop watch or digital watch to about a hundredth of a second. An atomic clock counts the oscillations of the radiation from a caesium atom, which repeat with extraordinary regularity, so it keeps time to about one second in $10^{13}$ s. Its least count, and its stability, are far better than any of the others.

Question 27

The distance of a galaxy is of the order of $10^{25}$ m. Calculate the order of magnitude of time taken by light to reach us from the galaxy.

Solution. Light travels at $c = 3\times10^8$ m/s, so

$$t = \frac{10^{25}}{3\times10^8}\ \text{s} = 3.3\times10^{16}\ \text{s}.$$

The order of magnitude is $10^{16}$ s, roughly a billion years.

Question 28

The vernier scale of a travelling microscope has $50$ divisions which coincide with $49$ main scale divisions. If each main scale division is $0.5$ mm, calculate the minimum inaccuracy in the measurement of distance.

Solution. The smallest distance the instrument can resolve is its least count, the difference between one main scale division and one vernier division. Since $50$ VSD $= 49$ MSD, one VSD $= \tfrac{49}{50}$ MSD, so

$$\text{LC} = 1\ \text{MSD} – 1\ \text{VSD} = \frac{1}{50}\ \text{MSD} = \frac{0.5}{50}\ \text{mm} = 0.01\ \text{mm}.$$

Question 29

During a total solar eclipse the moon almost entirely covers the sphere of the sun. Write the relation between the distances and sizes of the sun and moon.

Solution. Seen from the earth, the moon just covers the sun, so both subtend the same angle at the observer. With $R_s$ and $R_m$ the radii of the sun and moon and $R_{es}$ and $R_{em}$ their distances from the earth,

$$\theta = \frac{R_s}{R_{es}} = \frac{R_m}{R_{em}} \quad\Longrightarrow\quad \frac{R_s}{R_m} = \frac{R_{es}}{R_{em}}.$$

The ratio of the sizes equals the ratio of the distances. (The key states the same thing as equal solid angles, $\pi R_s^2/R_{es}^2 = \pi R_m^2/R_{em}^2$.)

Question 30

If the unit of force is $100$ N, unit of length is $10$ m and unit of time is $100$ s, what is the unit of mass in this system of units?

Solution. From $F = ma$, mass has the form $M = \dfrac{F}{a} = \dfrac{F\,T^2}{L}$. The new unit of mass is therefore built from the new units of force, length and time in the same way:

$$M = \frac{100\ \text{N}\times(100\ \text{s})^2}{10\ \text{m}} = \frac{100\times10^4}{10}\ \text{kg} = 10^5\ \text{kg}.$$

Question 31

Give an example of

(a) a physical quantity which has a unit but no dimensions.

(b) a physical quantity which has neither unit nor dimensions.

(c) a constant which has a unit.

(d) a constant which has no unit.

Solution.

• (a) Plane angle (radian) or solid angle (steradian). Each is a ratio of lengths (arc over radius, area over radius squared), so it is dimensionless, but it is still given a unit.
• (b) Strain or relative density (also refractive index). Each is a ratio of two like quantities, so both the unit and the dimensions cancel.
• (c) The gravitational constant $G$ ($\text{N m}^2\,\text{kg}^{-2}$) or Planck’s constant $h$ (J s).
• (d) $\pi$, $e$, or the Reynolds number (a pure number).

Question 32

Calculate the length of the arc of a circle of radius $31.0$ cm which subtends an angle of $\dfrac{\pi}{6}$ at the centre.

Solution. By the definition of the radian, $\theta = l/r$, so

$$l = r\theta = 31.0\times\frac{\pi}{6}\ \text{cm} = 16.23\ \text{cm} \approx 16.2\ \text{cm},$$

rounded to the three significant figures of the radius.

Printed key slip. The key works $31\times\dfrac{3.14}{6}$ and writes $16.3$ cm. That product is $16.22$, and with the exact $\pi$ it is $16.23$; either way the answer is $16.2$ cm.

Question 33

Calculate the solid angle subtended by the periphery of an area of $1\ \text{cm}^2$ at a point situated symmetrically at a distance of $5$ cm from the area.

Solution. The area is small compared with the square of the distance, so it can be treated as a patch of a sphere of radius $r = 5$ cm. By the definition of the steradian,

$$\Omega = \frac{A}{r^2} = \frac{1\ \text{cm}^2}{(5\ \text{cm})^2} = \frac{1}{25} = 4\times10^{-2}\ \text{sr}.$$

Question 34

The displacement of a progressive wave is represented by $y = A\sin(\omega t – kx)$, where $x$ is distance and $t$ is time. Write the dimensional formula of (i) $\omega$ and (ii) $k$.

Solution. The argument of a sine must be dimensionless, so each term inside it is a pure number. $\omega t$ dimensionless means $[\omega] = [t]^{-1}$, and $kx$ dimensionless means $[k] = [x]^{-1}$.

Question 35

Time for $20$ oscillations of a pendulum is measured as $t_1 = 39.6$ s; $t_2 = 39.9$ s; $t_3 = 39.5$ s. What is the precision in the measurements? What is the accuracy of the measurement?

Solution. Precision. The readings are recorded to $0.1$ s, which is the least count of the clock, so each measurement of $20$ oscillations is precise to $0.1$ s. For one oscillation this is $0.1/20 = 0.005$ s.

Accuracy. The mean time is

$$\bar t = \frac{39.6 + 39.9 + 39.5}{3} = 39.67\ \text{s} \approx 39.7\ \text{s},$$

so the period is $T = 39.67/20 = 1.98$ s. The reading furthest from the mean is $39.9$ s, which gives $39.9/20 = 1.995$ s, so the largest deviation in the period is $1.995 – 1.983 = 0.012 \approx 0.01$ s. The accuracy is therefore about $\pm0.01$ s, and the period is $T = 1.98 \pm 0.01$ s.

Printed key slip. The key writes the mean as $39.6$ s (truncating $39.67$) and so takes the period as $1.980$ s, giving a maximum error of $1.995 – 1.980 = 0.015$ s. With the mean rounded correctly the deviation is $0.012$ s. Either way the period is good to about a hundredth of a second.

Common mistakes

• Using the radius instead of the diameter in Question 25(a). One radius subtends $1/60$ rad; the full disc subtends $1/30$ rad, about $2^\circ$.
• Inverting the conversion in Question 30. Mass is $FT^2/L$, so a bigger unit of time makes the unit of mass much bigger.
• Taking the main scale division as the least count in Question 28. The vernier resolves $\tfrac{1}{50}$ of it.
• Truncating instead of rounding the mean in Question 35. $39.67$ s rounds to $39.7$ s, not $39.6$ s.
• Saying an angle has no unit in Question 31. It is dimensionless, but the radian is still its unit.

Practise next

• The vernier scale of a calliper has $20$ divisions that coincide with $19$ main scale divisions, and each main scale division is $1$ mm. Find the least count.

Show answer

$0.05$ mm. One VSD is $\tfrac{19}{20}$ MSD, so the least count is $1 – \tfrac{19}{20} = \tfrac{1}{20}$ of $1$ mm, that is $0.05$ mm.

• In a system of units the unit of force is $10$ N, the unit of length is $2$ m and the unit of time is $5$ s. What is the unit of mass?

Show answer

$125$ kg. Mass is $FT^2/L$, so the unit is $10\times5^2/2 = 125$ kg.

• Exemplar Units and Measurement: Long Answer Questions — the next questions, on new systems of units, error propagation, Planck units and astronomical estimates.
• Exemplar Units and Measurement: Multiple Choice Questions — significant figures, errors and dimensions in objective form.
• NCERT textbook Chapter 1 — Units and Measurement for the theory of SI units, significant figures, errors and dimensions.