Chapter2 Units And Measurements

Chapter Two

UNITS AND MEASUREMENTS

MCQ I

2.1 The number of significant figures in 0.06900 is

(a) 5

(b) 4

(c) 2

(d) 3

Show Answer Answer: (b)

2.2 The sum of the numbers 436.32, 227.2 and 0.301 in appropriate significant figures is

(a) 663.821

(b) 664

(c) 663.8

(d) 663.82

Show Answer Answer: (b)

2.3 The mass and volume of a body are $4.237 \mathrm{~g}$ and $2.5 \mathrm{~cm}^{3}$, respectively. The density of the material of the body in correct significant figures is

(a) $1.6048 \mathrm{~g} \mathrm{~cm}^{-3}$

(b) $1.69 \mathrm{~g} \mathrm{~cm}^{-3}$

(c) $1.7 \mathrm{~g} \mathrm{~cm}^{-3}$

(d) $1.695 \mathrm{~g} \mathrm{~cm}^{-3}$

Show Answer Answer: (c)

2.4 The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give

(a) 2.75 and 2.74

(b) 2.74 and 2.73

(c) 2.75 and 2.73

(d) 2.74 and 2.74

Show Answer Answer: (d)

2.5 The length and breadth of a rectangular sheet are $16.2 \mathrm{~cm}$ and $10.1 \mathrm{~cm}$, respectively. The area of the sheet in appropriate significant figures and error is

(a) $164 \pm 3 \mathrm{~cm}^{2}$

(b) $163.62 \pm 2.6 \mathrm{~cm}^{2}$

(c) $163.6 \pm 2.6 \mathrm{~cm}^{2}$

(d) $163.62 \pm 3 \mathrm{~cm}^{2}$

Show Answer Answer: (a)

2.6 Which of the following pairs of physical quantities does not have same dimensional formula?

(a) Work and torque.

(b) Angular momentum and Planck’s constant.

(c) Tension and surface tension.

(d) Impulse and linear momentum.

Show Answer Answer: (c)

2.7 Measure of two quantities along with the precision of respective measuring instrument is

$\mathrm{A}=2.5 \mathrm{~m} \mathrm{~s}^{-1} \pm 0.5 \mathrm{~m} \mathrm{~s}^{-1}$

$\mathrm{B}=0.10 \mathrm{~s} \pm 0.01 \mathrm{~s}$

The value of $A B$ will be

(a) $(0.25 \pm 0.08) \mathrm{m}$

(b) $(0.25 \pm 0.5) \mathrm{m}$

(c) $(0.25 \pm 0.05) \mathrm{m}$

(d) $(0.25 \pm 0.135) \mathrm{m}$

Show Answer Answer: (a)

2.8 You measure two quantities as $A=1.0 \mathrm{~m} \pm 0.2 \mathrm{~m}, B=2.0 \mathrm{~m} \pm 0.2 \mathrm{~m}$. We should report correct value for $\sqrt{\mathrm{AB}}$ as:

(a) $1.4 \mathrm{~m} \pm 0.4 \mathrm{~m}$

(b) $1.41 \mathrm{~m} \pm 0.15 \mathrm{~m}$

(c) $1.4 \mathrm{~m} \pm 0.3 \mathrm{~m}$

(d) $1.4 \mathrm{~m} \pm 0.2 \mathrm{~m}$

Show Answer Answer: (d)

2.9 Which of the following measurements is most precise?

(a) $5.00 \mathrm{~mm}$

(b) $5.00 \mathrm{~cm}$

(c) $5.00 \mathrm{~m}$

(d) $5.00 \mathrm{~km}$.

Show Answer Answer: (a)

2.10 The mean length of an object is $5 \mathrm{~cm}$. Which of the following measurements is most accurate?

(a) $4.9 \mathrm{~cm}$

(b) $4.805 \mathrm{~cm}$

(c) $5.25 \mathrm{~cm}$

(d) $5.4 \mathrm{~cm}$

Show Answer Answer: (a)

2.11 Young’s modulus of steel is $1.9 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$. When expressed in CGS units of dynes $/ \mathrm{cm}^{2}$, it will be equal to $\left(1 \mathrm{~N}=10^{5}\right.$ dyne, $1 \mathrm{~m}^{2}=10^{4} \mathrm{~cm}^{2}$ )

(a) $1.9 \times 10^{10}$

(b) $1.9 \times 10^{11}$

(c) $1.9 \times 10^{12}$

(d) $1.9 \times 10^{13}$

Show Answer Answer: (c)

2.12 If momentum $(P)$, area $(A)$ and time $(T)$ are taken to be fundamental quantities, then energy has the dimensional formula

(a) $\left(\mathrm{P}^{1} \mathrm{~A}^{-1} \mathrm{~T}^{1}\right)$

(b) $\left(\mathrm{P}^{2} \mathrm{~A}^{1} \mathrm{~T}^{1}\right)$

(c) $\left(\mathrm{P}^{1} \mathrm{~A}^{-1 / 2} \mathrm{~T}^{1}\right)$

(d) $\left(\mathrm{P}^{1} \mathrm{~A}^{1 / 2} \mathrm{~T}^{-1}\right)$

MCQ II

Show Answer Answer: (d)

2.13 On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct:

(a) $y=a \sin 2 \pi t / T$

(b) $y=a \sin v t$.

(c) $y=\frac{a}{T} \sin \left(\frac{t}{a}\right)$

(d) $y=a \sqrt{2}\left(\sin \frac{2 \pi t}{T}-\cos \frac{2 \pi t}{T}\right)$

Show Answer Answer: (b), (c)

2.14 If $P, Q, R$ are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity?

(a) $(P-Q) / R$

(b) $P Q-R$

(c) $P Q / R$

(d) $\left(P R-Q^{2}\right) / R$

(e) $(R+Q) / P$

Show Answer Answer: (a), (e)

2.15 Photon is quantum of radiation with energy $E=h v$ where $v$ is frequency and $h$ is Planck’s constant. The dimensions of $h$ are the same as that of

(a) Linear impulse

(b) Angular impulse

(c) Linear momentum

(d) Angular momentum

Show Answer Answer: (b), (d)

2.16 If Planck’s constant ( $h$ ) and speed of light in vacuum (c) are taken as two fundamental quantities, which one of the following can, in addition, be taken to express length, mass and time in terms of the three chosen fundamental quantities?

(a) Mass of electron $\left(m_{e}\right)$

(b) Universal gravitational constant $(G)$

(c) Charge of electron ( $e$ )

(d) Mass of proton $\left(m_{\mathrm{p}}\right)$

Show Answer Answer: (a), (b), (d)

2.17 Which of the following ratios express pressure?

(a) Force/ Area

(b) Energy/ Volume

(c) Energy/ Area

(d) Force/ Volume

Show Answer Answer: (a), (b)

2.18 Which of the following are not a unit of time?

(a) Second

(b) Parsec

(c) Year

(d) Light year

VSA

Show Answer Answer: (b), (d)

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

Show Answer

Answer: Because, bodies differ in order of magnitude significantly in respect to the same physical quantity. For example, interatomic distances are of the order of angstroms, inter-city distances are of the order of $\mathrm{km}$, and interstellar distances are of the order of light year.

Exemplar Problems-Physics

2.20 The radius of atom is of the order of $1 \AA$ 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?

Show Answer Answer: $\quad 10^{15}$

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

Show Answer Answer: Mass spectrograph

2.22 Express unified atomic mass unit in $\mathrm{kg}$.

Show Answer Answer: 1 $\mathrm{u}=1.67 \times 10^{-27} \mathrm{~kg}$

2.23 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 $q$ to be a dimensionless quantity?

Show Answer Answer: Since $f(\theta)$ is a sum of different powers of $\theta$, it has to be dimensionless

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

SA

Show Answer Answer: Because all other quantities of mechanics can be expressed in terms of length, mass and time through simple relations.

2.25 (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 $(1 / 2)^{\circ}$ 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.

Show Answer

Answer: alt text (a) $\theta=\frac{R_{E}}{60 R_{E}}=\frac{1}{60} \mathrm{rad} 1^{\circ}$

$\therefore$ Diameter of the earth as seen from the moon is about $2^{\circ}$.

(b) At earth-moon distance, moon is seen as $(1 / 2)^{\circ}$ diameter and earth is seen as $2^{\circ}$ diameter. Hence, diameter of earth is 4 times the diameter of moon.

$$ \frac{D_{\text {earth }}}{D_{\text {moon }}}=4 $$

(c) $\frac{r_{\text {sun }}}{r_{\text {moon }}}=400$

(Here $r$ stands for distance, and $D$ for diameter.)

Sun and moon both appear to be of the same angular diameter as seen from the earth.

$\therefore \frac{D_{\text {sun }}}{r_{\text {sun }}}=\frac{D_{\text {moon }}}{r_{\text {moon }}}$

$\therefore \frac{D_{\text {sun }}}{D_{\text {moon }}}=400$

But $\frac{D_{\text {earth }}}{D_{\text {moon }}}=4 \quad \therefore \frac{D_{\text {sun }}}{D_{\text {earth }}}=100$.

2.26 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.

Show Answer Answer: An atomic clock is the most precise time measuring device because atomic oscillations are repeated with a precision of $1 \mathrm{~s}$ in $10^{13} \mathrm{~s}$.

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

Show Answer Answer: 3 $\times 10^{16} \mathrm{~s}$

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

Show Answer Answer: $\quad 0.01 \mathrm{~mm}$

2.29 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.

Show Answer

Answer: $\theta=\left(\pi R_{s}^{2} / R_{es}^{2}\right)\left(\pi R_{m}^{2} / R_{e m}^{2}\right)$

$\Rightarrow \frac{R_{s}}{R_{m}}=\frac{R_{e s}}{R_{e m}}$

2.30 If the unit of force is $100 \mathrm{~N}$, unit of length is $10 \mathrm{~m}$ and unit of time is $100 \mathrm{~s}$, what is the unit of mass in this system of units?

Show Answer Answer: $\quad 10^{5} \mathrm{~kg}$

2.31 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.

Show Answer

Answer: (a) Angle or solid angle

(b) Relative density, etc.

(c) Planck’s constant, universal gravitational constant, etc.

(d) Raynold number

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

Show Answer Answer: $\theta=\frac{l}{r} \Rightarrow l=r \theta \Rightarrow l=31 \times \frac{3.14}{6} \mathrm{~cm}=16.3 \mathrm{~cm}$

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

Show Answer Answer: $4 \times 10^{-2}$ steradian

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

Show Answer

Answer: Dimensional formula of $\omega=\mathrm{T}^{-1}$

Dimensional formula of $k=\mathrm{L}^{-1}$

2.35 Time for 20 oscillations of a pendulum is measured as $t_{1}=39.6 \mathrm{~s}$; $t_{2}=39.9 \mathrm{~s} ; t_{3}=39.5 \mathrm{~s}$. What is the precision in the measurements? What is the accuracy of the measurement?

LA

Show Answer

Answer: (a) Precision is given by the least count of the instrument.

For 20 oscillations, precision $=0.1 \mathrm{~s}$

For 1 oscillation, precision $=0.005 \mathrm{~s}$.

(b) Average time $t=\frac{39.6+39.9+39.5}{3} \mathrm{~s}=39.6 \mathrm{~s}$

Period $=\frac{39.6}{20}=1.98 \mathrm{~s}$

Max. observed error $=(1.995-1.980) \mathrm{s}=0.015 \mathrm{~s}$.

2.36 A new system of units is proposed in which unit of mass is $\alpha \mathrm{kg}$, unit of length $\beta \mathrm{m}$ and unit of time $\gamma \mathrm{s}$. How much will $5 \mathrm{~J}$ measure in this new system?

Show Answer Answer: Since energy has dimensions of $\mathrm{ML}^{2} \mathrm{~T}^{-2}, 1 \mathrm{~J}$ in new units becomes $\gamma^{2} / \alpha \beta^{2}$ J. Hence $5 \mathrm{~J}$ becomes $5 \gamma^{2} / \alpha \beta^{2}$.

2.37 The volume of a liquid flowing out per second of a pipe of length $l$ and radius $r$ is written by a student as

$v=\frac{\pi}{8} \frac{\operatorname{Pr}^{4}}{\eta l}$

where $P$ is the pressure difference between the two ends of the pipe and $\eta$ is coefficent of viscosity of the liquid having dimensional formula $\mathrm{ML}^{-1} \mathrm{~T}^{-1}$.

Check whether the equation is dimensionally correct.

Show Answer Answer: The dimensional part in the expression is $\frac{\rho r^{4}}{\eta l}$. Therefore, the dimensions of the right hand side comes out to be $\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^{4}\right]}{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right][\mathrm{L}]}=\frac{\left[\mathrm{L}^{3}\right]}{[\mathrm{T}]}$, which is volume upon time. Hence, the formula is dimensionally correct.

2.38 A physical quantity $X$ is related to four measurable quantities $a$, $b, c$ and $d$ as follows:

$X=a^{2} b^{3} c^{5 / 2} d^{-2}$.

The percentage error in the measurement of $a, b, c$ and $d$ are $1 %$, $2 %, 3 %$ and $4 %$, respectively. What is the percentage error in quantity $X$ ? If the value of $X$ calculated on the basis of the above relation is 2.763 , to what value should you round off the result.

Show Answer

Answer: The fractional error in $X$ is

$$ \begin{aligned} \frac{\mathrm{d} X}{X} & =\frac{2 \mathrm{~d} a}{a}+\frac{3 \mathrm{~d} b}{b}+\frac{2.5 \mathrm{~d} c}{c}+\frac{2 \mathrm{~d}(d)}{d} \ & =0.235 \quad 0.24 \end{aligned} $$

Since the error is in first decimal, hence the result should be rounded off as 2.8.

2.39 In the expression $P=E l^{2} m^{-5} G^{-2}, E, m, l$ and $G$ denote energy, mass, angular momentum and gravitational constant, respectively. Show that $P$ is a dimensionless quantity.

Show Answer

Answer: Since $\mathrm{E}, l$ and $\mathrm{G}$ have dimensional formulas:

$\mathrm{E} \rightarrow \mathrm{ML}^{2} \mathrm{~T}^{-2}$

$l \rightarrow \mathrm{ML}^{2} \mathrm{~T}^{-1}$

$\mathrm{G} \rightarrow \mathrm{L}^{3} \mathrm{M}^{-1} \mathrm{~T}^{-2}$

Hence, $\mathrm{P}=\mathrm{E} l^{2} m^{-5} G^{-2}$ will have dimensions:

$$ \begin{aligned} {[\mathrm{P}] } & =\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\left[\mathrm{M}^{2} \mathrm{~L}^{4} \mathrm{~T}^{-2}\right]\left[\mathrm{M}^{2} \mathrm{~T}^{4}\right]}{\left[\mathrm{M}^{5}\right]\left[\mathrm{L}^{6}\right]} \ & =\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \end{aligned} $$

Thus, $\mathrm{P}$ is dimensionless.

2.40 If velocity of light $c$, Planck’s constant $h$ and gravitational contant $G$ are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

Show Answer

Answer: $\mathrm{M}, \mathrm{L}, \mathrm{T}$, in terms of new units become

$\mathrm{M} \rightarrow \sqrt{\frac{\mathrm{ch}}{\mathrm{G}}}, \mathrm{L} \rightarrow \sqrt{\frac{\mathrm{hG}}{\mathrm{c}^{3}}}, \mathrm{~T} \rightarrow \sqrt{\frac{\mathrm{hG}}{\mathrm{c}^{5}}}$

2.41 An artificial satellite is revolving around a planet of mass $M$ and radius $R$, in a circular orbit of radius $r$. From Kepler’s Third law about the period of a satellite around a common central body, square of the period of revolution $T$ is proportional to the cube of the radius of the orbit $r$. Show using dimensional analysis, that

$T=\frac{k}{R} \sqrt{\frac{r^{3}}{g}}$,

where $k$ is a dimensionless constant and $g$ is acceleration due to gravity.

Show Answer

Answer: Given $T^{2} \alpha r^{3} \quad T \alpha r^{3 / 2}$. T is also function of $g$ and $R \Rightarrow T \propto g^{x} R^{y}$

$\therefore\left[\mathrm{L}^{\mathrm{o}} \mathrm{M}^{\mathrm{o}} \mathrm{T}^{1}\right]=\left[\mathrm{L}^{3 / 2} \mathrm{M}^{\mathrm{o}} \mathrm{T}^{\mathrm{o}}\right]\left[\mathrm{L}^{1} \mathrm{M}^{\mathrm{o}} \mathrm{T}^{-2}\right]^{\mathrm{x}}\left[\mathrm{L}^{1} \mathrm{M}^{\mathrm{o}} \mathrm{T}^{\mathrm{o}}\right]^{\mathrm{y}}$

For $\mathrm{L}, 0=\frac{3}{2}+x+y$

For T, $1=0-2 x \Rightarrow x=-\frac{1}{2}$

Therefore, $\quad 0=\frac{3}{2}-\frac{1}{2}+y \Rightarrow y=-1$

Thus, $T=k r^{3 / 2} g^{-1 / 2} R^{-1}=\frac{k}{R} \sqrt{\frac{r^{3}}{g}}$

2.42 In an experiment to estimate the size of a molecule of oleic acid $1 \mathrm{~mL}$ of oleic acid is dissolved in $19 \mathrm{~mL}$ of alcohol. Then $1 \mathrm{~mL}$ of this solution is diluted to $20 \mathrm{~mL}$ by adding alcohol. Now 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule.

Read the passage carefully and answer the following questions:

(a) Why do we dissolve oleic acid in alcohol?

(b) What is the role of lycopodium powder?

(c) What would be the volume of oleic acid in each $\mathrm{mL}$ of solution prepared?

(d) How will you calculate the volume of $n$ drops of this solution of oleic acid?

(e) What will be the volume of oleic acid in one drop of this solution?

Show Answer

Answer: (a) Because oleic acid dissolves in alcohol but does not disssolve in water.

(b) When lycopodium powder is spread on water, it spreads on the entire surface. When a drop of the prepared solution is dropped on water, oleic acid does not dissolve in water, it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls. This allows measuring the area where oleic acid spreads.

(c) $\frac{1}{20} \mathrm{~mL} \times \frac{1}{20}=\frac{1}{400} \mathrm{~mL}$

(d) By means of a burette and measuring cylinder and measuring the number of drops.

(e) If $n$ drops of the solution make $1 \mathrm{~mL}$, the volume of oleic acid in one drop will be $(1 / 400) \mathrm{n} \mathrm{mL}$.

2.43 (a) How many astronomical units (A.U.) make 1 parsec?

(b) Consider a sunlike star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be $(1 / 2)^{\circ}$ from the earth. Due to atmospheric fluctuations, eye can’t resolve objects smaller than 1 arc minute.

(c) Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about $1 / 2$ A.U. from the earth. Calculate what size it will appear when seen through the same telescope.

(Comment: This is to illustrate why a telescope can magnify planets but not stars.)

Show Answer

Answer: (a) By definition of parsec

$\therefore 1$ parsec $=\left(\frac{1 \text { A.U. }}{1 \operatorname{arcsec}}\right)$

$1 \mathrm{deg}=3600$ arc sec

$\therefore 1 \operatorname{arcsec}=\frac{\pi}{3600 \times 180}$ radians

$\therefore 1$ parsec $=\frac{3600 \times 180}{\pi}$ A.U. $=206265$ A.U. $\approx 2 \times 10^{5}$ A.U.

(b) At 1 A.U. distance, sun is $\left(1 / 2^{\circ}\right)$ in diameter.

Therefore, at 1 parsec, star is $\frac{1 / 2}{2 \times 10^{5}}$ degree in diameter $=15$ $\times 10^{-5}$ arcmin.

With 100 magnification, it should look $15 \times 10^{-3}$ arcmin. However, due to atmospheric fluctuations, it will still look of about 1 arcmin. It can’t be magnified using telescope.

(c) $\frac{D_{\text {mars }}}{D_{\text {earth }}}=\frac{1}{2}, \frac{D_{\text {earth }}}{D_{\text {sun }}}=\frac{1}{400}$ [from Answer 2.25 (c)]

$\therefore \frac{D_{\text {mars }}}{D_{\text {sun }}}=\frac{1}{800}$.

At 1 A.U. sun is seen as $1 / 2$ degree in diameter, and mars will be seen as $1 / 1600$ degree in diameter.

At $1 / 2$ A.U, mars will be seen as $1 / 800$ degree in diameter. With 100 magnification mars will be seen as $1 / 8$ degree $=\frac{60}{8}=7.5$ arcmin.

This is larger than resolution limit due to atmospheric fluctuations. Hence, it looks magnified.

2.44 Einstein’s mass - energy relation emerging out of his famous theory of relativity relates mass $(m)$ to energy $(E)$ as $E=m c^{2}$, where $c$ is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in $\mathrm{MeV}$, where $1 \mathrm{MeV}=1.6 \times 10^{-13} \mathrm{~J}$; the masses are measured in unified atomic mass unit $(u)$ where $1 \mathrm{u}=1.67 \times 10^{-27} \mathrm{~kg}$.

(a) Show that the energy equivalent of $1 \mathrm{u}$ is $931.5 \mathrm{MeV}$.

(b) A student writes the relation as $1 \mathrm{u}=931.5 \mathrm{MeV}$. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.

Show Answer

Answer: (a) Since $1 \mathrm{u}=1.67 \times 10^{-27} \mathrm{~kg}$, its energy equivalent is $1.67 \times 10^{-27} c^{2}$ in SI units. When converted to $\mathrm{eV}$ and $\mathrm{MeV}$, it turns out to be $1 \mathrm{u} \equiv 931.5 \mathrm{MeV}$.

(b) $1 \mathrm{u} \times c^{2}=931.5 \mathrm{MeV}$.



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