NCERT Solutions for Class 11 Physics Chapter 1: Units and Measurements

Question 1.1

Fill in the blanks:
(a) The volume of a cube of side $1\text{ cm}$ is equal to $\dots\dots \text{m}^3$
(b) The surface area of a solid cylinder of radius $2.0\text{ cm}$ and height $10.0\text{ cm}$ is equal to $\dots\dots (\text{mm})^2$
(c) A vehicle moving with a speed of $18\text{ km h}^{-1}$ covers $\dots\dots \text{m}$ in $1\text{ s}$
(d) The relative density of lead is $11.3$. Its density is $\dots\dots \text{g cm}^{-3}$ or $\dots\dots \text{kg m}^{-3}$.

Solution:

Question 1.2

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

Solution:

Question 1.3

A calorie is a unit of heat (energy in transit) and it equals about $4.2\text{ J}$ where $1\text{ J} = 1\text{ kg m}^2\text{ s}^{-2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha\text{ kg}$, the unit of length equals $\beta\text{ m}$, the unit of time is $\gamma\text{ s}$. Show that a calorie has a magnitude $4.2\alpha^{-1}\beta^{-2}\gamma^2$ in terms of the new units.

Solution:

$$1\text{ cal} = 4.2\text{ kg m}^2\text{ s}^{-2}$$
Since $1\text{ kg} = \alpha^{-1}\text{ new mass unit}$, $1\text{ m} = \beta^{-1}\text{ new length unit}$, and $1\text{ s} = \gamma^{-1}\text{ new time unit}$:
$$1\text{ cal} = 4.2(\alpha^{-1})(\beta^{-1})^2(\gamma^{-1})^{-2} = 4.2\alpha^{-1}\beta^{-2}\gamma^2\text{ new energy units}.$$

Question 1.4

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.

Solution:

A dimensional quantity becomes meaningfully “large” or “small” only when it is compared with another quantity of the same dimensions. Suitable statements are:
– An atom is very small compared with a football or another macroscopic object.
– A jet plane moves much faster than a bicycle.
– The mass of Jupiter is very large compared with the mass of the Earth or a laboratory object.
– The room contains many more air molecules than a small closed box.
– A proton is about $1836$ times as massive as an electron.
– The speed of sound in air is much smaller than the speed of light in vacuum.

Question 1.5

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 $8\text{ min}$ and $20\text{ s}$ to cover this distance?

Solution:

$$t = 8\text{ min } 20\text{ s} = 8 \times 60 + 20 = 500\text{ s}$$
$$\text{Distance} = \text{speed} \times \text{time} = 1 \times 500 = 500\text{ new length units}.$$

Question 1.6

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 $1\text{ mm}$ and $100$ divisions on the circular scale
(c) an optical instrument that can measure length to within a wavelength of light?

Solution:

The optical instrument has the smallest least count and is therefore the most precise.

Question 1.7

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 $3.5\text{ mm}$. What is the estimate on the thickness of hair?

Solution:

$$\text{Actual thickness} = \frac{\text{apparent width}}{\text{magnification}} = \frac{3.5}{100} = 0.035\text{ mm} = 3.5 \times 10^{-5}\text{ m}.$$

Question 1.8

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 $1.0\text{ mm}$ 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) Wind the thread tightly in $N$ adjacent turns around a smooth cylindrical rod. Measure the total length $L$ occupied by the turns. The mean thread diameter is $d = L/N$. Repeating the measurement at several places improves reliability.

(b) No. Increasing the number of divisions reduces the nominal least count only up to the limit set by the screw quality, backlash, zero error, surface irregularities and the observer’s ability to read the scale.

(c) Random errors fluctuate in sign and magnitude. Taking the mean of a larger number of observations reduces their effect, so $100$ observations generally give a more reliable estimate than $5$ observations.

Question 1.9

The photograph of a house occupies an area of $1.75\text{ cm}^2$ on a $35\text{ mm}$ slide. The slide is projected on to a screen, and the area of the house on the screen is $1.55\text{ m}^2$. What is the linear magnification of the projector-screen arrangement.

Solution:

$$\text{Area on slide} = 1.75\text{ cm}^2 = 1.75 \times 10^{-4}\text{ m}^2$$
$$\text{Areal magnification} = \frac{1.55}{1.75 \times 10^{-4}} = 8.857 \times 10^3$$
$$\text{Linear magnification} = \sqrt{\text{areal magnification}} = \sqrt{8.857 \times 10^3} \approx 94.1.$$

Question 1.10

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

Solution:

Question 1.11

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

Solution:

Length $l = 4.234\text{ m}$ ($4\text{ sig figs}$), Breadth $b = 1.005\text{ m}$ ($4\text{ sig figs}$), Thickness $t = 0.0201\text{ m}$ ($3\text{ sig figs}$).
$$\text{Area} = lb = 4.234 \times 1.005 = 4.25517\text{ m}^2 \approx 4.255\text{ m}^2\text{ ($4$ significant figures)}.$$
$$\text{Volume} = lbt = 4.234 \times 1.005 \times 0.0201 = 0.0855289\text{ m}^3 \approx 0.0855\text{ m}^3\text{ ($3$ significant figures)}.$$

Question 1.12

The mass of a box measured by a grocer’s balance is $2.30\text{ kg}$. Two gold pieces of masses $20.15\text{ g}$ and $20.17\text{ g}$ 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:

$$\text{Total mass} = 2.300\text{ kg} + 0.02015\text{ kg} + 0.02017\text{ kg} = 2.34032\text{ kg} \approx 2.340\text{ kg}.$$
$$\text{Difference} = 20.17\text{ g} – 20.15\text{ g} = 0.02\text{ g}.$$
For addition or subtraction, the result is rounded to the least number of decimal places among the measured quantities expressed in the same unit.

Question 1.13

A famous relation in physics relates ‘moving mass’ $m$ to the ‘rest mass’ $m_0$ 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: $m = \frac{m_0}{\sqrt{1 – v^2}}$. Guess where to put the missing $c$.

Solution:

The quantity subtracted from $1$ must be dimensionless. Since $v$ and $c$ both have dimensions of speed ($\text{LT}^{-1}$), $v^2$ must be divided by $c^2$ to make the ratio dimensionless.
Thus, the correct relation is:
$$m = \frac{m_0}{\sqrt{1 – \frac{v^2}{c^2}}}.$$

Question 1.14

The unit of length convenient on the atomic scale is known as an angstrom and is denoted by $\text{\AA}$: $1\text{ \AA} = 10^{-10}\text{ m}$. The size of a hydrogen atom is about $0.5\text{ \AA}$. What is the total atomic volume in $\text{m}^3$ of a mole of hydrogen atoms?

Solution:

$$r = 0.5\text{ \AA} = 0.5 \times 10^{-10}\text{ m}$$
$$\text{Volume of one atom} = \frac{4}{3}\pi r^3 = 5.24 \times 10^{-31}\text{ m}^3$$
$$\text{Volume of one mole} = N_A \times \text{volume of one atom} = 6.022 \times 10^{23} \times 5.24 \times 10^{-31} \approx 3.16 \times 10^{-7}\text{ m}^3.$$

Question 1.15

One mole of an ideal gas at standard temperature and pressure occupies $22.4\text{ L}$ (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 $1\text{ \AA}$). Why is this ratio so large?

Solution:

$$V_m = 22.4\text{ L} = 2.24 \times 10^{-2}\text{ m}^3$$
$$\text{Approximate material volume of one mole} \approx 3.16 \times 10^{-7}\text{ m}^3$$
$$\frac{V_m}{V_a} \approx \frac{2.24 \times 10^{-2}}{3.16 \times 10^{-7}} \approx 7.1 \times 10^4.$$
The ratio is very large because gas molecules are separated by distances much greater than their own size; most of a gas volume is empty space.

Question 1.16

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).

Solution:

The apparent motion depends on the rate of change of the line of sight. For nearby objects, a small displacement of the observer produces a large angular change, so they appear to sweep backward rapidly. For very distant objects, the same displacement causes an extremely small angular change, so they appear almost stationary.

Question 1.17

The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding $10^7\text{ K}$, and its outer surface at a temperature of about $6000\text{ K}$. 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 $= 2.0 \times 10^{30}\text{ kg}$, radius of the Sun $= 7.0 \times 10^8\text{ m}$.

Solution:

$$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (7.0 \times 10^8)^3 \approx 1.44 \times 10^{27}\text{ m}^3$$
$$\rho = \frac{M}{V} = \frac{2.0 \times 10^{30}}{1.44 \times 10^{27}} \approx 1.4 \times 10^3\text{ kg m}^{-3}.$$
The Sun is plasma, yet its mean density is comparable to that of liquids and some solids because its enormous self-gravity compresses the material strongly.

Additional Question 1

A parsec is the distance at which one astronomical unit subtends an angle of $1\text{ arcsecond}$. Express one parsec in metres.

Solution:

$$1\text{ arcsecond} = \frac{\pi / 180}{3600}\text{ rad} = 4.848 \times 10^{-6}\text{ rad}$$
$$1\text{ pc} = \frac{1\text{ AU}}{1\text{ arcsecond in radians}} = \frac{1.496 \times 10^{11}\text{ m}}{4.848 \times 10^{-6}\text{ rad}} \approx 3.09 \times 10^{16}\text{ m}.$$

Additional Question 2

Alpha Centauri is $4.29\text{ light-years}$ away. Express this distance in parsecs and find the total angular shift when observed from Earth positions six months apart.

Solution:

Since $1\text{ pc} = 3.26\text{ ly}$:
$$D = \frac{4.29}{3.26} \approx 1.32\text{ pc}$$
The annual parallax angle is $p = \frac{1}{D} \approx 0.758\text{ arcsecond}$. Observations six months apart use opposite ends of Earth’s orbital diameter, so the total angular shift is approximately $2p \approx 1.52\text{ arcseconds}$.

Additional Question 3

Give examples from modern science where very precise measurements of length, time or mass are required.

Solution:

– Ultrashort laser pulses measure processes occurring over intervals near $10^{-15}\text{ s}$.
– Atomic clocks measure time with fractional uncertainties far below one part in $10^{12}$.
– X-ray diffraction determines interatomic and interplanar distances of order $10^{-10}\text{ m}$.
– Mass spectrometers distinguish atomic and molecular masses with very high precision.
– Radar and satellite navigation require precise position and time measurements to determine speed and distance.

Additional Question 4

Suggest methods for making rough estimates of the following quantities:
(a) Mass of monsoon rain over India
(b) Mass of an elephant
(c) Wind speed during a storm
(d) Number of hairs on a head
(e) Air molecules in a room

Solution:

Additional Question 5

A book gives four formulas for the displacement $y$ in periodic motion. Rule out the dimensionally wrong formulas:
(a) $y = a \sin\left(\frac{2\pi t}{T}\right)$
(b) $y = a \sin(vt)$
(c) $y = \frac{a}{T} \sin\left(\frac{t}{a}\right)$
(d) $y = \frac{a}{\sqrt{2}} \left[\sin\left(\frac{2\pi t}{T}\right) + \cos\left(\frac{2\pi t}{T}\right)\right]$.

Solution:

Additional Question 6

A student remembers the relativistic mass formula as $m = \frac{m_0}{\sqrt{1 – v^2}}$ but has omitted $c$. Restore $c$ using dimensional analysis.

Solution:

The quantity subtracted from $1$ must be dimensionless. Since $v$ and $c$ both have dimensions $LT^{-1}$, $\frac{v^2}{c^2}$ is dimensionless.
$$m = \frac{m_0}{\sqrt{1 – \frac{v^2}{c^2}}}.$$

Additional Question 7

Jupiter is $824.7\text{ million kilometres}$ from Earth and has angular diameter $35.72\text{ arcseconds}$. Calculate its diameter.

Solution:

$$D = 824.7 \times 10^6\text{ km}; \quad \theta = 35.72 \times 4.848 \times 10^{-6}\text{ rad} = 1.732 \times 10^{-4}\text{ rad}$$
$$\text{Diameter } d = \theta D \approx 1.732 \times 10^{-4} \times 824.7 \times 10^6\text{ km} \approx 1.43 \times 10^5\text{ km}.$$

Additional Question 8

A person walking in vertical rain with speed $v$ slants an umbrella by angle $\theta$ from the vertical. Can $\tan\theta = v$ be correct? Suggest a correct relation.

Solution:

No. $\tan\theta$ is dimensionless but $v$ has dimensions of speed. If $u$ is the vertical speed of rain relative to the ground, the horizontal relative speed is $v$ and the vertical relative speed is $u$. Therefore,
$$\tan\theta = \frac{v}{u}.$$

Additional Question 9

Two caesium clocks may differ by only $0.02\text{ s}$ after $100\text{ years}$. What accuracy does this imply for measuring a $1\text{ s}$ interval?

Solution:

$$100\text{ years} \approx 100 \times 365.25 \times 24 \times 3600 = 3.156 \times 10^9\text{ s}$$
$$\text{Fractional error} = \frac{0.02}{3.156 \times 10^9} \approx 6.3 \times 10^{-12}.$$
Thus, a $1\text{ s}$ interval is measured with an uncertainty of order $10^{-11}\text{ to } 10^{-12}\text{ s}$.

Additional Question 10

Estimate the average mass density of a sodium atom, assuming its diameter is $2.5\text{ \AA}$. Compare it with crystalline sodium, whose density is $970\text{ kg m}^{-3}$.

Solution:

$$r = 1.25 \times 10^{-10}\text{ m}; \quad V = \frac{4}{3}\pi r^3 \approx 8.18 \times 10^{-30}\text{ m}^3$$
$$\text{Mass of one Na atom} = \frac{23 \times 10^{-3}}{6.022 \times 10^{23}} \approx 3.82 \times 10^{-26}\text{ kg}$$
$$\rho_{\text{atom}} \approx \frac{3.82 \times 10^{-26}}{8.18 \times 10^{-30}} \approx 4.7 \times 10^3\text{ kg m}^{-3}.$$
This is the same order of magnitude as $970\text{ kg m}^{-3}$. The estimate treats the atom as a hard sphere; real crystals contain packing gaps and the quoted atomic “size” is only approximate.

Additional Question 11

Nuclear radius follows $r = r_0 A^{1/3}$ with $r_0 \approx 1.2\text{ fm}$. Show that nuclear density is nearly constant and estimate it for sodium.

Solution:

$$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi r_0^3 A; \quad \text{mass} \approx A \times 1.66 \times 10^{-27}\text{ kg}$$
$$\rho = \frac{A(1.66 \times 10^{-27})}{\frac{4}{3}\pi r_0^3 A} = \frac{3(1.66 \times 10^{-27})}{4\pi r_0^3}.$$
The factor $A$ cancels, so the density is approximately independent of the nucleus.
For $r_0 = 1.2 \times 10^{-15}\text{ m}$, $\rho \approx 2.3 \times 10^{17}\text{ kg m}^{-3}$.
This is about $10^{14}$ times the estimated average density of a sodium atom, showing that almost all atomic mass is concentrated in the tiny nucleus.

Additional Question 12

A laser pulse sent to the Moon returns in $2.56\text{ s}$. Find the radius of the lunar orbit.

Solution:

$$\text{One-way time} = \frac{2.56}{2} = 1.28\text{ s}$$
$$\text{Earth-Moon distance} = c \times 1.28 = 3.0 \times 10^8 \times 1.28 = 3.84 \times 10^8\text{ m} = 3.84 \times 10^5\text{ km}.$$

Additional Question 13

A SONAR echo returns after $77.0\text{ s}$. If the speed of sound in water is $1450\text{ m s}^{-1}$, find the distance of the reflecting submarine.

Solution:

$$\text{Distance} = \frac{vt}{2} = \frac{1450 \times 77.0}{2} = 5.5825 \times 10^4\text{ m} \approx 55.8\text{ km}.$$

Additional Question 14

How far away is a quasar whose light takes $3.0\text{ billion years}$ to reach Earth? Express the answer in kilometres.

Solution:

$$t = 3.0 \times 10^9 \times 365.25 \times 24 \times 3600 \approx 9.47 \times 10^{16}\text{ s}$$
$$D = ct = 3.0 \times 10^8 \times 9.47 \times 10^{16} \approx 2.84 \times 10^{25}\text{ m} = 2.84 \times 10^{22}\text{ km}.$$

Additional Question 15

During a total solar eclipse, the Moon and Sun have nearly equal angular diameters. Using Earth-Moon distance $3.84 \times 10^8\text{ m}$, Earth-Sun distance $1.496 \times 10^{11}\text{ m}$ and Sun diameter $1.39 \times 10^9\text{ m}$, estimate the Moon’s diameter.

Solution:

$$\frac{D_{\text{moon}}}{D_{\text{sun}}} \approx \frac{R_{\text{moon}}}{R_{\text{sun}}}$$
$$D_{\text{moon}} = 1.39 \times 10^9 \times \frac{3.84 \times 10^8}{1.496 \times 10^{11}} \approx 3.57 \times 10^6\text{ m} \approx 3570\text{ km}.$$

Additional Question 16

Construct a combination of fundamental constants having the dimensions of time, of very large magnitude, as discussed by P. A. M. Dirac. What would a significant coincidence with the age of the universe imply?

Solution:

One possible time-like combination is:
$$t = \frac{(ke^2)^2}{m_p m_e^2 c^3 G}$$
Dimensional analysis gives $[t] = T$. Substitution gives a very large timescale, of order $10^{16}\text{ s}$ (roughly $10^9\text{ years}$). The question is intended as an exploration of large-number coincidences rather than a precise determination of the universe’s age.
If such a coincidence were fundamental rather than accidental, it could suggest that at least one supposedly constant dimensionless combination of physical constants changes over cosmological time. Any meaningful test must ultimately be framed in terms of dimensionless ratios.