2.11 Exercise Problems

Q1: Why do we have different units for the same physical quantity?

Answer: The magnitude of any given physical quantity may vary over a wide range, therefore, different units of the same physical quantity are required. For example:
1.Mass ranges from \(10^{-30} \mathrm{~kg}\) (for an electron) to \(10^{53} \mathrm{~kg}\) (for the known universe). We need different units to measure them like milligram, gram, kilogram, etc.
2. The length of a pen can be easily measured in \(\mathrm{cm}\), the height of a tree can be measured in metres, the distance between two cities can be measured in kilometers and the distance between two heavenly bodies can be measured in light year.

Q2: The radius of atom is of the order of 1 Å and radius of the nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of the nucleus?

Answer: According to the question,
Radius of atom 1 Å \(=10^{-10} \mathrm{~m} \quad\) Radius of nucleus \(\approx 1\) fermi \(=10^{-15} \mathrm{~m}\)
Volume of atom \(V_{\text {Atom }}=\frac{4}{3} \pi R_A^3 \quad\) Volume of nucleus \(V_{\text {nucleus }}=\frac{4}{3} \pi R_N^3\)
\(
\frac{V_{\text {Atom }}}{V_{\text {Nucleus }}}=\frac{\frac{4}{3} \pi R_A^3}{\frac{4}{3} \pi R_N^3}=\left(\frac{R_A}{R_N}\right)^3=\left(\frac{10^{-10}}{10^{-15}}\right)^3=10^{15}
\)
Mass of one mole of \({ }_6 \mathrm{C}^{12}\) atom \(=12 \mathrm{~g}\)
Number of atoms in one mole \(=\) Avogadro’s number \(=6.023 \times 10^{23}\)
\(\therefore \quad\) Mass of one \({ }_6 \mathrm{C}^{12}\) atom \(=\frac{12}{6.023 \times 10^{23}} \mathrm{~g}\)
\(
\begin{aligned}
& 1 \mathrm{amu}=\frac{1}{12} \times \text { mass of one }{ }_6 C^{12} \text { atom } \\
& \therefore \quad 1 \mathrm{amu}=\left(\frac{1}{12} \times \frac{12}{6.023 \times 10^{23}}\right) \mathrm{g}=1.67 \times 10^{-24} \mathrm{~g} \\
& =1.67 \times 10^{-27} \mathrm{~kg} \quad\left(\because 1 \mathrm{~g}=10^{-3} \mathrm{~kg}\right) \\
&
\end{aligned}
\)

Q3: Name the device used for measuring the mass of atoms and molecules.

Answer:  A mass spectrograph is a device which is used for measuring the mass of atoms and molecules.

Q4: Express unified atomic mass unit in kg.

Answer: The unified atomic mass unit is the standard unit that is used for indicating mass on an atomic or molecular scale (atomic mass). One unfiled atomic mass unit is approximately the mass of one nucleon (either a single proton or neutron) and is numerically equivalent to \(1 \mathrm{~g} / \mathrm{mol}\). It is defined as one-twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state.

Q5: 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?

Answer: We know, the function \(f(\theta)\) is a sum of different powers of \(\theta\). And it is a dimensionless quantity. This is the limitation of dimensional analysis that we cannot add the powers of a dimensional quantity. By the principle of homogeneity as RHS is dimensionless, hence LHS should also be dimensionless.

Important point: To avoid confusion we can assume the similar expression to understand which is a function of \(x\) instead of \(\theta\).
\(
f(x)=1-x+\frac{x^2}{2 !}-\frac{x^3}{3 !}+\frac{x^4}{4 !}+\ldots
\)
In fact, in this case \(x\) has dimensions.

Q6: Why length, mass and time are chosen as base quantities in mechanics?

Answer: Normally each physical quantity requires a unit or standard for its specification, so it appears that there must be as many units as there are physical quantities. However, it is not so. It has been found that if in mechanics we choose arbitrarily units of any three physical quantities we can express the units of all other physical quantities in mechanics in terms of these. So, length, mass and time are chosen as base quantities in mechanics because
(i) Length, mass and time cannot be derived from one another, that is these quantities are independent.
(ii) All other quantities in mechanics can be expressed in terms of length, mass and time.

Q7: (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.

Answer: (a) As the distance between moon and the earth is greater than the radius of the earth, then the radius of the earth can be treated as an arc. According to the problem,
\(
R_E=\text { length of } \operatorname{arc}
\)
Distance between moon and earth \(=60 R_E\)
So, angle subtended at distance \(r\) due to an arc of length \(l\) is
\(
\begin{aligned}
\theta_E & =\frac{l}{r}=\frac{2 R_E}{60 R_E}=\frac{1}{30} \mathrm{rad} \\
& =\frac{1}{30} \times \frac{180^{\circ}}{\pi} \text { degree }=\frac{6^{\circ}}{3.14} \text { degree }=1.9^{\circ} \approx 2^{\circ}
\end{aligned}
\)
Hence, the angle is subtended by the diameter of the earth \(2 \theta=2^{\circ}\).

(b) According to the problem, moon is seen as \(\left(\frac{1}{2}\right)^{\circ}\) diameter from earth and earth is seen as \(2^{\circ}\) diameter from moon.
As \(\theta\) is proportional to diameter,
Hence, \(\frac{\text { Diameter of earth }}{\text { Diameter of moon }}=\frac{2}{\left(\frac{1}{2}\right)}=4\)
(c) From parallax measurement given that Sun is at a distance of about 400 times the earth-moon distance, hence, \(\frac{r_{\text {sun }}}{r_{\text {moon }}}=400\)
(Suppose, 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.
\(
\begin{aligned}
& \therefore \frac{D_{\text {sun }}}{r_{\text {sun }}}=\frac{D_{\text {moon }}}{r_{\text {moon }}} \Rightarrow \frac{D_{\text {sun }}}{D_{\text {moon }}}=400 \\
& \text { But } \frac{D_{\text {earth }}}{D_{\text {moon }}}=4 \Rightarrow \frac{D_{\text {sun }}}{D_{\text {earth }}}=100
\end{aligned}
\)

Q8: Which of the following time-measuring devices is most precise?
(a) A wall clock.
(b) A stopwatch.
(c) A digital watch.
(d) An atomic clock.
Give reason for your answer.

Answer: Option (d) is correct because a clock can measure time correctly up to one second. A stopwatch can measure time correctly up to a fraction of a second. A digital watch can measure time up to a fraction of second whereas an atomic clock is the most accurate timekeeper and is based on characteristic frequencies of radiation emitted by certain atoms having the precision of about 1 second in 300,000 years. So an atomic clock can measure time most precisely as the precision of this clock is about \(1 \mathrm{~s}\) in \(10^{13} \mathrm{~s}\).

Q9: 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.

Answer: According to the problem, the distance of the galaxy \(=10^{25} \mathrm{~m}\).
Speed of light \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
Hence, the time taken by light to reach us from the galaxy is
\(
\begin{aligned}
t=\frac{\text { Distance }}{\text { Speed }} & =\frac{10^{25} \mathrm{~m}}{3 \times 10^8 \mathrm{~m} / \mathrm{s}} \approx \frac{1}{3} \times 10^{17} \\
& =\frac{10}{3} \times 10^{16}=3.33 \times 10^{16} \mathrm{~s}
\end{aligned}
\)

Q10: The vernier scale of a traveling 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.

Answer: The diagram of Vernier caliper is shown below.

According to the problem, 50 divisions of Vernier scale which coincide with 49 main scale divisions.
\(
50 \mathrm{VSD}=49 M S D
\)
\(\Rightarrow 1 M S D=\frac{50}{49} V S D \quad\) or \(\quad 1 V S D=\frac{49}{50} M S D\)
where \(M S D=\) Main scale division and \(V S D=\) Vernier scale division.
We know that
Minimum inaccuracy \(=\) Vernier constant
\(
=1 M S D-1 V S D
\)
\(
=1 M S D-\frac{49}{50} M S D=\frac{1}{50} M S D
\)
It is given in the problem that \(1 M S D=0.5 \mathrm{~mm}\)
Hence, minimum inaccuracy \(=\frac{1}{50} \times 0.5 \mathrm{~mm}=\frac{1}{100}=0.01 \mathrm{~mm}\)

Q11: 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. 

Answer: Key point: In geometry, a solid angle (symbol: \(\Omega\) or \(w\) ) is the two-dimensional angle in three-dimensional space that an object subtends at a point. It is a measure of how large the object appears to an observer looking from that point. In the International System of Units (SI), a solid angle is expressed in a dimensionless unit called a steradian (symbol: sr).

A small object nearby may subtend the same solid angle as a larger object farther away. For example, although the Moon is much smaller than the Sun, it is also much closer to Earth. The diagram given below shows that the moon almost entirely covers the sphere of the sun.
\(R_{\mathrm{me}}=\) Distance of moon from earth
\(R_{\mathrm{se}}=\) Distance of sun from earth
Let the solid angle made by the sun and moon is \(d \Omega\), we can write

\(
d \Omega=\frac{A_{\mathrm{sun}}}{R_{\mathrm{se}}^2}=\frac{A_{\mathrm{moon}}}{R_{\mathrm{me}}^2}
\)

Here, \(A_{\text {sun }}=\) Area of the sun
\(A_{\text {moon }}=\) Area of the moon
\(
\begin{aligned}
& \Rightarrow \theta=\frac{\pi R_{\mathrm{s}}^2}{R_{\mathrm{se}}^2}=\frac{\pi R_{\mathrm{m}}^2}{R_{\mathrm{me}}^2} \\
& \Rightarrow\left(\frac{R_{\mathrm{s}}}{R_{\mathrm{se}}}\right)^2=\left(\frac{R_{\mathrm{m}}}{R_{\mathrm{me}}}\right)^2 \\
& \Rightarrow \frac{R_{\mathrm{s}}}{R_{\mathrm{se}}}=\frac{R_{\mathrm{m}}}{R_{\mathrm{me}}} \text { or } \quad \frac{R_{\mathrm{s}}}{R_{\mathrm{m}}}=\frac{R_{\mathrm{se}}}{R_{\mathrm{me}}}
\end{aligned}
\)
(Here, the radius of the sun and moon represent their sizes respectively)

Q12: 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?

Answer: First write the dimensions of each quantity and then relate them.
Force \([F]=\left[M L T^{-2}\right]=100 \mathrm{~N}\)
Length \([L]=[L]=10 \mathrm{~m}\)
Time \([t]=[T]=100 \mathrm{~s}\)
Substituting values of \(L\) and \(T\) from Eqs. (ii) and (iii) in Eq. (i), we get
\(
\begin{aligned}
& M \times 10 \times(100)^{-2}=100 \\
\Rightarrow \quad & \frac{M \times 10}{100 \times 100}=100 \\
\Rightarrow & M=100 \times 1000 \mathrm{~kg}=10^5 \mathrm{~kg}
\end{aligned}
\)

Q13: 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.

Answer: (a) Solid angle \(\Omega=\frac{A}{r^2}\) steradian and a Plane angle \(\theta=\frac{L}{r}\) radian. Both are dimensionless but have units.
(b) Specific density \(=\frac{\text { density of medium }}{\text { density of water at } 4^{\circ} \mathrm{C}}\)
It is a ratio of two same quantities. So, it is a unitless and dimensionless constant.
(c) Gravitational constant \((G)=6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)
(d) Reynold’s number is a constant which has no unit.

Q14: 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.

Answer: Key concept:
Plane Angle \(\alpha=\frac{\text { Arc length }}{\text { Radius }}=\frac{s}{r}\)

According to the problem, \(\theta=\frac{\pi}{6}=\frac{s}{31} \mathrm{~cm}\)
Hence, length of \(\operatorname{arc}=s=31 \times \frac{\pi}{6} \mathrm{~cm}=\frac{31 \times 3.14}{6} \mathrm{~cm}=16.22 \mathrm{~cm}\)
Rounding off to three significant figures it would be \(16.2 \mathrm{~cm}\).

Q15: 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.

Answer: Solid angle, \(\Omega=\frac{\text { Area }}{{\text { (Radial distance })^2}^2}\)
\(
\begin{gathered}
=\frac{1 \mathrm{~cm}^2}{(5 \mathrm{~cm})^2}=\frac{1}{25}=4 \times 10^{-2} \text { steradian } \\
\left(\because \text { Area }=1 \mathrm{~cm}^2, \text { distance }=5 \mathrm{~cm}\right)
\end{gathered}
\)
Important point: Please keep in mind that solid angle is for 3-D figure like sphere, cone etc, and plane angle is for plane objects or 2-D figures like circle, arc etc.

Q16: 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 (1) \(\omega\) and (ii) \(k\).

Answer: We have to apply the principle of homogeneity to solve this problem. The principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be the same, i.e., dimensions of LHS and RHS should be equal.
According to the problem
\(
y=A \sin (\omega t-k x)
\)
Here \(y=[L]\) hence
\(
A \sin (\omega t-k x)=[L]
\)
Here \(A=[L]\), which peak value of \(y\)
So, \(\omega t-k x\) should be dimensionless,
(i) \([\omega t]=\) constant
\(
\Rightarrow \quad[\omega]=\left[T^{-1}\right]
\)
(ii) \([k x]=\) constant
\(
\Rightarrow \quad[k]=\left[L^{-1}\right]
\)

Q17: 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?

Answer: According to the problem, time for 20 oscillations of a pendulum, \(\mathrm{t}_1=39.6 \mathrm{~s}, \mathrm{t}_2=39.9 \mathrm{~s}\) and \(\mathrm{t}_3=39.5 \mathrm{~s}\)
It is quite obvious from these observations that the least count of the watch is \(0.1 \mathrm{~s}\). As measurements have only one decimal place. Precision in the measurement = Least count of the measuring instrument \(=0.1\) s Precision in 20 oscillations \(=0.1\)
Therefore, Precision in 1 oscillation \(=\frac{0.1}{20}=0.005\)
(ii) Mean value of time for 20 oscillations is given by
\(
\begin{aligned}
t & =\frac{t_1+t_2+t_3}{3} \\
& =\frac{39.6+39.9+39.5}{3}=39.66 \mathrm{~s}
\end{aligned}
\)
Mean time period of the second pendulum \(=\frac{39.66}{20} \approx 1.98 \mathrm{~s}\)
Rounding off the time period of second pendulum \(=2 \mathrm{~s}\)
Measured time period of the second pendulum \(=2-0.005=1.995 \mathrm{~s}\)
Accuracy of measurement is the máximum observed error and is given by \(=1.995-1.980=0.015 \mathrm{~s}\)

Q18: 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?

Answer: For solving this problem, dimensions of physical quantity will remain the same whatever be the system of units of its measurement.
Let the physical quantity be \(Q=n_1 u_1=n_2 u_2\)
Let \(M_1, L_1, T_1\), and \(M_2, L_2, T_2\) are units of mass, length and time in given the systems.
So, \(n_2=n_1\left[\frac{M_1}{M_2}\right]^a \times\left[\frac{L_1}{L_2}\right]^b \times\left[\frac{T_1}{T_2}\right]^c\)
We know that dimension of energy \([U]=\left[M L^2 T^{-2}\right]\)
According to the problem, \(M_1=1 \mathrm{~kg}, L_1=1 \mathrm{~m}, T_1=1 \mathrm{~s}\)
\(
M_2=\alpha \mathrm{kg}, L_2=\beta \mathrm{m}, T_2=\gamma \mathrm{s}
\)
Substituting the values, we get CBSELabs.com
\(
\begin{aligned}
n_2 & =5\left[\frac{M_1}{M_2}\right] \times\left[\frac{L_1}{L_2}\right]^2 \times\left[\frac{T_1}{T_2}\right]^{-2} \\
& =5\left[\frac{1}{\alpha} \mathrm{kg}\right] \times\left[\frac{1}{\beta} \mathrm{m}\right]^2 \times\left[\frac{1}{\gamma} \mathrm{s}\right]^{-2} \\
& =5 \times \frac{1}{\alpha} \times \frac{1}{\beta^2} \times \frac{1}{\gamma^{-2}}=\frac{5 \gamma^2}{\alpha \beta^2} \mathrm{~J}
\end{aligned}
\)
This is the required value of energy in the new system of units.

Q19: 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 the coefficient of viscosity of the liquid having the dimensional formula \(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\).
Check whether the equation is dimensionally correct.

Answer: If the dimensions of LHS of an equation is equal to the dimensions of RHS, then the equation is said to be dimensionally correct.
According to the problem, the volume of a liquid flowing out per second of a pipe is given by \(V=\frac{\pi}{8} \frac{p r^4}{\eta l}\)
(where, \(V=\) rate of volume of liquid per unit time)
Dimension of given physical quantities,
\(
\begin{aligned}
& {[V]=\frac{\text { Dimension of volume }}{\text { Dimension of time }}=\frac{\left[L^3\right]}{[T]}=\left[L^3 T^{-1}\right],[p]=\left[M L^{-1} T^{-2}\right],} \\
& {[\eta]=\left[M L^{-1} T^{-1}\right],[l]=[L],[r]=[L]} \\
& \text { LHS }=[V]=\frac{\left[L^3\right]}{[T]}=\left[L^3 T^{-1}\right] \\
& \text { RHS }=\frac{\left[M L^{-1} T^{-2}\right] \times\left[L^4\right]}{\left[M L^{-1} T^{-1}\right] \times[L]}=\left[L^3 T^{-1}\right]
\end{aligned}
\)
Dimensionally, L.H.S. = R.H.S.
Therefore, the equation is correct dimensionally.

Q20: 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?

Answer:
Percentage error in quantity \(X\) is given by, \(\frac{\Delta x}{x} \times 100\)
According to the problem, the physical quantity is \(X=a^2 b^3 c^{5 / 2} d^{-2}\)
percentage error in \(a=\left(\frac{\Delta a}{a} \times 100\right)=1 \%\)
percentage error in \(b=\left(\frac{\Delta b}{b} \times 100\right)=2 \%\)
percentage error in \(c=\left(\frac{\Delta c}{c} \times 100\right)=3 \%\)
percentage error in \(d=\left(\frac{\Delta d}{d} \times 100\right)=4 \%\)
The maximum percentage error in \(X\) is
\(
\begin{aligned}
\frac{\Delta X}{X} \times 100 & = \pm\left[2\left(\frac{\Delta a}{a} \times 100\right)+3\left(\frac{\Delta b}{b} \times 100\right)+\frac{5}{2}\left(\frac{\Delta c}{c} \times 100\right)+2\left(\frac{\Delta d}{d} \times 100\right)\right] \\
& = \pm\left[2(1)+3(2)+\frac{5}{2}(3)+2(4)\right] \% \\
& = \pm\left[2+6+\frac{15}{2}+8\right]= \pm 23.5 \%
\end{aligned}
\)
\(\therefore \quad\) Percentage error in quantity \(X= \pm 23.5 \%\)
Mean absolute error in \(X= \pm 0.235= \pm 0.24\) (rounding-off upto two significant digits)
On the basis of these values, the value of \(X\) should have two significant digits only.
\(
\therefore \quad X=2.8
\)

Q21: In the expression \(P=E l^2 m \tau^5 G^{-2}, E, m, l\) and \(G\) denote energy, mass, angular momentum and the gravitational constant, respectively. Show that \(P\) is a dimensionless quantity.

Answer: According to the problem, the expression is \(P=E l^2 \mathrm{~m}^{-5} G^{-2}\) where \(E\) is energy \([E]=\left[M L^2 T^{-2}\right], m\) is mass \([m]=[M]\), \(L\) is angular momentum \([L]=\left[M L^2 T^{-1}\right], G\) is gravitational constant \([G]=\left[M^{-1} L^2 T^{-2}\right]\)
Substituting dimensions of each physical quantity in the given expression,
\(
\begin{aligned}
{[P] } & =\left[M L^2 T^{-2}\right] \times\left[M L^2 T^{-1}\right]^2 \times[M]^{-5} \times\left[M^{-1} L^3 T^{-2}\right]^{-2} \\
& =\left[M^{1+2-5+2} L^{2+4-6} T^{-2-2+4}\right] \\
& =\left[M^0 L^0 T^0\right]
\end{aligned}
\)
This shows that \(P\) is a dimensionless quantity.

Q22: If the velocity of light \(c\), Planck’s constant \(h\), and gravitational constant \(G\) are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

Answer: We have to apply the principle of homogeneity to solve this problem. The principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be the same, i.e., dimensions of LHS and RHS should be equal, We know that, the dimensions of
\(
[h]=\left[M L^2 T^{-1}\right],[c]=\left[L T^{-1}\right],[G]=\left[M^{-1} L^3 T^{-2}\right]
\)
(i) Let \(\mathrm{m} \propto c^x h^y G^z\)
\(\Rightarrow m=k c^a h^b G^c \dots(i)\)
where, \(k\) is a dimensionless constant of proportionality.
Substituting the dimensions of each term in Eq. (i), we get
\(
\left[M L^0 T^0\right]=\left[L T^{-1}\right]^a \times\left[M L^2 T^{-1}\right]^b\left[M^{-1} L^3 T^{-2}\right]^{\mathrm{c}}
\)
Comparing powers of the same terms on both sides, we get
\(
\begin{aligned}
& b-c=1 \dots(ii) \\
& a+2 b+3 c=0 \dots(iii) \\
& -a-b-2 c=0 \dots(iv)
\end{aligned}
\)
Adding Eqs. (ii), (iii) and (iv), we get
\(
2 b=1 \Rightarrow b=\frac{1}{2}
\)
Substituting value of \(b\) in Eq. (ii), we get
\(
c=-\frac{1}{2} 
\)
From Eq. (iv)
\(
a=-b-2 c
\)
Substituting values of \(b\) and \(c\), we get
\(
a=-\frac{1}{2}-2\left(-\frac{1}{2}\right)=\frac{1}{2}
\)
Putting values of \(a, b\) and \(c\) in Eq. (i), we get
\(
m=k c^{1 / 2} h^{1 / 2} G^{-1 / 2}=k \sqrt{\frac{c h}{G}}
\)
(ii) Let \(L \propto c^a h^b G^c\)
\(
\Rightarrow \quad L=k c^a h^b G^c \dots(v)
\)
where \(k\) is a dimensionless constant.
Substituting the dimensions of each term in Eq. (v), we get
\(
\begin{aligned}
{\left[M^0 L T^0\right] } & =\left[L T^{-1}\right]^a \times\left[M L^2 T^{-1}\right]^b \times\left[M^{-1} L^3 T^{-2}\right]^c \\
& =\left[M^{b-c} L^{a+2 b+3 c} T^{-a-b-2 c}\right]
\end{aligned}
\)
On comparing powers of same terms, we get
\(
\begin{aligned}
& b-c=0 \dots(vi) \\
& a+2 b+3 c=1 \dots(vii) \\
& -a-b-2 c=0 \dots(viii)
\end{aligned}
\)
Adding Eqs. (vi), (vii) and (viii), we get
\(
2 b=1 \quad \Rightarrow \quad b=\frac{1}{2}
\)
Substituting value of \(b\) in Eq. (vi), we get
\(
c=\frac{1}{2}
\)
From Eq. (viii), \(a=-b-2 c\)
Substituting values of \(b\) and \(c\), we get
\(
a=-\frac{1}{2}-2\left(\frac{1}{2}\right)=-\frac{3}{2}
\)
Putting values of \(a, b\) and \(c\) in Eq. (v), we get
\(
L=k c^{-3 / 2} h^{1 / 2} G^{1 / 2}=k \sqrt{\frac{h G}{c^3}}
\)
(iii) Let \(T \propto c^a h^b G^c\)
\(
\Rightarrow T=c^a h^b G^c \dots(ix)
\)
where, \(k\) is a dimensionless constant.
Substituting the dimensions of each term in Eq. (ix), we get
\(
\begin{aligned}
{\left[M^0 L^0 T^1\right] } & =\left[L T^{-1}\right]^a \times\left[M L^2 T^{-1}\right]^b \times\left[M^{-1} L^3 T^{-2}\right]^c \\
& =\left[M^{b-c} L^{a+2 b+3 c} T^{a-b-2 c}\right]
\end{aligned}
\)
On comparing powers of same terms, we get
\(
\begin{aligned}
& b-c=0 \dots(x) \\
& a+2 b+3 c=1 \dots(xi) \\
& -a-b-2 c=1 \dots(xii)
\end{aligned}
\)
Adding Eqs. (x), (xi) and (xii), we get
\(
2 b=1 \quad \Rightarrow \quad b=\frac{1}{2}
\)
Substituting the value of \(b\) in Eq. (x), we get
\(
c=b=\frac{1}{2}
\)
From Eq. (xii),
\(
a=-b-2 c-1
\)
Substituting values of \(b\) and \(c\), we get
\(
a=-\frac{1}{2}-2\left(\frac{1}{2}\right)-1=-\frac{5}{2}
\)
Putting values of \(a, b\) and \(c\) in Eq. (ix), we get
\(
T=k c^{-5 / 2} h^{1 / 2} G^{1 / 2}=k \sqrt{\frac{h G}{c^5}}
\)

Q23: 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
\(
T=\frac{k}{R} \sqrt{\frac{r^3}{g}} \text {, }
\)
where \(k\) is a dimensionless constant and \(g\) is acceleration due to gravity.

Answer: According to Kepler’s third law, \(T^2 \propto a^3\) i.e., square of time period \(\left(T^2\right)\) of a satellite revolving around a planet, is proportional to the cube of the radius of the orbit \(\left(a^3\right)\).
We have to apply Kepler’s third law,
\(
T^2 \propto r^3 \Rightarrow T \propto r^{3 / 2}
\)
Also, \(T\) depends on \(R\) and \(g\).
Let \(T \propto r^{3 / 2} g^a R^b\)
\(
\Rightarrow T=k r^{3 / 2} R^a g^b \dots(i)
\)
where, \(k\) is a dimensionless constant of proportionality.
Writing the dimensions of various quantities on both the sides, we get
\(
\begin{aligned}
{\left[M^0 L^0 T\right] } & =[L]^{3 / 2}\left[L T^{-2}\right]^a[L]^b \\
& =\left[M^0 L^{a+b+3 / 2} T^{-2 a}\right]
\end{aligned}
\)
On comparing the dimensions of both sides, we get
\(
\begin{aligned}
& a+b+\frac{3}{2}=0 \dots(ii) \\
& -2 a=1 \quad \Rightarrow a=-1 / 2 \dots(iii)
\end{aligned}
\)
From Eq. (ii), we get
\(
b-\frac{1}{2}+\frac{3}{2}=0 \Rightarrow b=-1
\)
Substituting the values of \(a\) and \(b\) in Eq. (i), we get
\(
T=k r^{3 / 2} R^{-1} g^{-1 / 2} \Rightarrow T=\frac{k}{R} \sqrt{\frac{r^3}{g}}
\)

Q24: 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 the 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 the 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?

Answer: (a) Since Oleic acid does not dissolve in water, hence it is dissolved in alcohol.
(b) Lycopodium powder spreads on the entire surface of the water when it is sprinkled evenly. When a drop of prepared solution of oleic acid and alcohol is dropped on water, oleic acid does not dissolve in water. Instead, it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls. We can thus be able to measure the area over which oleic acid spreads.
(c) Since 20 mL (1 mL oleic acid + 19 mL alcohol) contains 1 mL of oleic acid, oleic acid in each mL of the solution \(=1 / 20 \mathrm{~mL}\). Further, as this \(1 \mathrm{~mL}\) is diluted to \(20 \mathrm{~mL}\) by adding alcohol. In each \(\mathrm{mL}\) of solution prepared, volume of oleic acid \(=1 / 20 \mathrm{~mL} \times 1 / 20=1 / 400 \mathrm{~mL}\)
(d) Volume of \(n\) drops of this solution of oleic acid can be calculated by means of a burette (used to make solution in the form of countable drops) and measuring cylinder and measuring the number of drops.
(e) As \(1 \mathrm{~mL}\) of solution contains \(n\) number of drops, then the volume of oleic acid in one drop will be = \(1 /(400) \mathrm{n} \mathrm{mL}\)

Q25: (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, the 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.

Answer: (a)1 A.U. long arc subtends the angle of \(1 \mathrm{~s}\) or 1 arc sec at a distance of 1 parsec.
Thus, angle \(1 \mathrm{sec}=\frac{1 A \cdot U \text {. }}{1}\) parsec
\(
\text { Thus, } 1 \text { parsec }=\frac{1 A . U}{1} \operatorname{arcsec}
\)
\(
\text { Thus, } 1 \text { parsec }=\frac{630(3600)}{11}
\)
\(
\begin{aligned}
& =206182.8^{\circ} \\
& =2 \times 10^5 A . U
\end{aligned}
\)
(b)Angle of the sun’s diameter \(\left(\frac{1}{2}\right)^o\) is subtended by 1 A.U. since the distance from the sun increases angle subtended in the same ratio.
Now,
\(2 \times 10^5 A . U\). will form an angle of \(\theta=\left(\frac{1}{4 \times 10^5}\right)^o\), since the diameter is the same angle subtended on earth by 1 parsec will be same.
If the sunlike star is at 2 parsec the angle becomes half \(=\left(1.25 \times 10^{-6}\right)^{\circ}\)
Thus, angle \(=75 \times 10^{-6} \mathrm{~min} \quad\)
When it is seen with a telescope that has a magnification of 100, the angle formed will be \(7.5 \times 10^{-3}\) min, viz., less than a minute.
Hence, it can’t be observed by a telescope.
(c) Given that \(\frac{D_{\text {mars }}}{D_{\text {earth }}}=\frac{1}{2}\)
where \(D\) represents diameter.
We know that, \(\frac{D_{\text {earth }}}{D_{\text {sun }}}=\frac{1}{100}\)
\(\therefore \quad \frac{D_{\text {mars }}}{D_{\text {sun }}}=\frac{1}{2} \times \frac{1}{100}[\) from Eq. (i)]
At 1 AU Sun’s diameter \(=\left(\frac{1}{2}\right)^{\circ}\)
\(\therefore\) Diameter of Mars \(=\frac{1}{2} \times \frac{1}{200}=\left(\frac{1}{400}\right)^{\circ}\)
At \(\frac{1}{2} \mathrm{AU}\), Mars’ diameter
\(
=\frac{1}{400} \times 2^{\circ}=\left(\frac{1}{200}\right)^{\circ}
\)
With 100 magnification, Mars’ diameter
\(
=\frac{1}{200} \times 100^{\circ}=\left(\frac{1}{2}\right)^{\circ}=30^{\prime}
\)
This is larger than the resolution limit due to atmospheric fluctuations. Hence, it looks magnified.

Q26: 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 the speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at the nuclear level is usually measured in \(\mathrm{MeV}\), where \(1 \mathrm{MeV}=1.6 \times 10^{-13} \mathrm{~J}\); the masses are measured in the unified atomic mass unit \((\mathrm{u})\) where \(\mathrm{lu}=1.67 \times 10^{-27} \mathrm{~kg}\).
(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.

Answer: (a) We can apply Einstein’s mass-energy relation in this problem, \(E=m c^2\), to calculate the energy equivalent of the given mass.
Here
\(
\begin{aligned}
1 \mathrm{amu} & =1 u=1.67 \times 10^{-27} \mathrm{~kg} \\
\text { Applying } \quad & =m c^2 \\
\text { Energy } E & =\left(1.67 \times 10^{-27}\right)\left(3 \times 10^8\right)^2 \mathrm{~J}=1.67 \times 9 \times 10^{-11} \mathrm{~J} \\
E & =\frac{1.67 \times 9 \times 10^{-11}}{1.6 \times 10^{-13}} \mathrm{MeV} \\
& =939.4 \mathrm{MeV} \approx 931.5 \mathrm{MeV}
\end{aligned}
\)
(b) As \(E=m c^2 \Rightarrow m=\frac{E}{c^2}\)
According to this, \(1 u=\frac{931.5 \mathrm{MeV}}{c^2}\)
Hence the dimensionally correct relation \(1 \mathrm{amu} \times c^2=1 u \times c^2\)
\(
=931.5 \mathrm{MeV}
\)

Q27: Find the dimensional formulae of the following quantities :
(a) the universal constant of gravitation \(G\),
(b) the surface tension \(S\),
(c) the thermal conductivity \(k\) and
(d) the coefficient of viscosity \(\eta\).
Some equations involving these quantities are
\(
\begin{aligned}
& F=\frac{G m_1 m_2}{r^2}, \quad S=\frac{\rho g r h}{2}, \\
& Q=k \frac{A\left(\theta_2-\theta_1\right) t}{d} \text { and } F=-\eta A \frac{v_2-v_1}{x_2-x_1} \\
&
\end{aligned}
\)
where the symbols have their usual meanings.

Answer: (a) \(F=G \frac{m_1 m_2}{r^2}\) or, \(\quad G=\frac{F r^2}{m_1 m_2}\) or, \(\quad[G]=\frac{[F] \mathrm{L}^2}{\mathrm{M}^2}=\frac{\mathrm{MLT}^{-2} \cdot \mathrm{L}^2}{\mathrm{M}^2}=\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\).
(b) \(\quad S=\frac{\rho g r h}{2}\)
or, \(\quad[S]=[\rho][g] \mathrm{L}^2=\frac{\mathrm{M}}{\mathrm{L}^3} \cdot \frac{\mathrm{L}}{\mathrm{T}^2} \cdot \mathrm{L}^2=\mathrm{MT}^{-2}\).
\(\begin{array}{ll}\text { (c) } & Q=k \frac{A\left(\theta_2-\theta_1\right) t}{d} \\ \text { or, } & k=\frac{Q d}{A\left(\theta_2-\theta_1\right) t}\end{array}\)
Here, \(Q\) is the heat energy having dimension \(\mathrm{ML}^2 \mathrm{~T}^{-2}, \theta_2-\theta_1\) is temperature, \(A\) is area, \(d\) is thickness and \(t\) is time. Thus,
\(
[k]=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~L}}{\mathrm{~L}^2 \mathrm{KT}}=\mathrm{MLT}^{-3} \mathrm{~K}^{-1}
\)
\(
\begin{aligned}
& \text { (d) } \quad F=-\eta A \frac{v_2-v_1}{x_2-x_1} \\
& \text { or, } \mathrm{MLT}^{-2}=[\eta] \mathrm{L}^2 \frac{\mathrm{L} / \mathrm{T}}{\mathrm{L}}=[\eta] \frac{\mathrm{L}^2}{\mathrm{~T}} \\
& \text { or, } \quad[\eta]=\mathrm{ML}^{-1} \mathrm{~T}^{-1} \text {. } \\
&
\end{aligned}
\)

Q28: Find the dimensional formulae of
(a) the charge \(Q\),
(b) the potential \(V\),
(c) the capacitance \(C\), and
(d) the resistance \(R\).
Some of the equations containing these quantities are
\(
Q=I t, U=V I t, Q=C V \text { and } V=R I
\)
where \(I\) denotes the electric current, \(t\) is time and \(U\) is energy.

Answer: (a) \(Q=\) It. \(\quad\) Hence, \([Q]=\) IT.
(b) \(U=V I t\)
or, \(\mathrm{ML}^2 \mathrm{~T}^{-2}=[V] \mathrm{IT}\) or, \([V]=\mathrm{ML}^2 \mathrm{I}^{-1} \mathrm{~T}^{-3}\).
(c) \(Q=C V\)
or, \(\mathrm{IT}=[C] \mathrm{ML}^2 \mathrm{I}^{-1} \mathrm{~T}^{-3} \quad\) or, \([C]=\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{I}^2 \mathrm{~T}^4\).
(d) \(V=R I\)
or, \(R=\frac{V}{I} \quad\) or, \([R]=\frac{\mathrm{ML}^2 \mathrm{I}^{-1} \mathrm{~T}^{-3}}{\mathrm{I}}=\mathrm{ML}^2 \mathrm{I}^{-2} \mathrm{~T}^{-3}\).

Q29: The SI and CGS units of energy are joule and erg respectively. How many ergs are equal to one joule?

Answer: Dimensionally, Energy \(=\) mass \(\times(\text { velocity })^2\)
\(
\begin{aligned}
& =\operatorname{mass} \times\left(\frac{\text { length }}{\text { time }}\right)^2=\mathrm{ML}^2 \mathrm{~T}^{-2} . \\
\text { Thus, } 1 \text { joule } & =(1 \mathrm{~kg})(1 \mathrm{~m})^2(1 \mathrm{~s})^{-2} \\
\text { and } 1 \mathrm{erg} & =(1 \mathrm{~g})(1 \mathrm{~cm})^2(1 \mathrm{~s})^{-2} \\
\frac{1 \text { joule }}{1 \mathrm{erg}} & =\left(\frac{1 \mathrm{~kg}}{1 \mathrm{~g}}\right)\left(\frac{1 \mathrm{~m}}{1 \mathrm{~cm}}\right)^2\left(\frac{1 \mathrm{~s}}{1 \mathrm{~s}}\right)^{-2} \\
& =\left(\frac{1000 \mathrm{~g}}{1 \mathrm{~g}}\right)\left(\frac{100 \mathrm{~cm}}{1 \mathrm{~cm}}\right)^2=1000 \times 10000=10^7 .
\end{aligned}
\)
So, \(\quad 1\) joule \(=10^7\) erg.

Q30: Young’s modulus of steel is \(19 \times 10^{10} \mathrm{~N} / \mathrm{m}^2\). Express it in dyne \(/ \mathrm{cm}^2\). Here dyne is the CGS unit of force.

Answer: The unit of Young’s modulus is \(\mathrm{N} / \mathrm{m}^2\).
This suggests that it has dimensions of \(\frac{\text { Force }}{\text { (distance }^2}\).
Thus,
\(
[Y]=\frac{[F]}{\mathrm{L}^2}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}
\)
\(\mathrm{N} / \mathrm{m}^2\) is in SI units.
\(\mathrm{So}\)
\(
1 \mathrm{~N} / \mathrm{m}^2=(1 \mathrm{~kg})(1 \mathrm{~m})^{-1}(1 \mathrm{~s})^{-2}
\)
and
so, \(\frac{1 \mathrm{~N} / \mathrm{m}^2}{1 \mathrm{dyne} / \mathrm{cm}^2}=\left(\frac{1 \mathrm{~kg}}{1 \mathrm{~g}}\right)\left(\frac{1 \mathrm{~m}}{1 \mathrm{~cm}}\right)^{-1}\left(\frac{1 \mathrm{~s}}{1 \mathrm{~s}}\right)^{-2}\)
\(
=1000 \times \frac{1}{100} \times 1=10
\)
or, \(\quad 1 \mathrm{~N} / \mathrm{m}^2=10 \mathrm{dyne} / \mathrm{cm}^2\)
or, \(19 \times 10^{10} \mathrm{~N} / \mathrm{m}^2=19 \times 10^{11} \mathrm{dyne} / \mathrm{cm}^2\).

Q31: If velocity, time and force were chosen as basic quantities, find the dimensions of mass.

Answer:

\(
\begin{aligned}
& \text { Dimensionally, Force }=\text { mass } \times \text { acceleration } \\
& =\text { mass } \times \frac{\text { velocity }}{\text { time }} \\
& \text { or, } \\
& \text { mass }=\frac{\text { force } \times \text { time }}{\text { velocity }} \\
& \text { or, } \\
& {[\text { mass }]=F T V^{-1} \text {. }} \\
&
\end{aligned}
\)

Q32: Test dimensionally if the equation \(v^2=u^2+2 a x\) may be correct.

Answer: There are three terms in this equation \(v^2, u^2\) and \(2 a x\). The equation may be correct if the dimensions of these three terms are equal.
\(
\begin{aligned}
& {\left[v^2\right]=\left(\frac{\mathrm{L}}{\mathrm{T}}\right)^2=\mathrm{L}^2 \mathrm{~T}^{-2} ;} \\
& {\left[u^2\right]=\left(\frac{\mathrm{L}}{\mathrm{T}}\right)^2=\mathrm{L}^2 \mathrm{~T}^{-2} ;} \\
& {[2 a x]=[a][x]=\left(\frac{\mathrm{L}}{\mathrm{T}^2}\right) \mathrm{L}=\mathrm{L}^2 \mathrm{~T}^{-2} .}
\end{aligned}
\)
and
Thus, the equation may be correct.

Q33: The distance covered by a particle in time \(t\) is given by \(x=a+b t+c t^2+d t^3\); find the dimensions of \(a, b, c\) and \(d\).

Answer: The equation contains five terms. All of them should have the same dimensions. Since \([x]=\) length, each of the remaining four must have the dimension of length.
Thus,
\(
\text { Thus, } \begin{aligned}
{[a] } & =\text { length }=\mathrm{L} & & \\
{[b t] } & =\mathrm{L}, & & \text { or, }[b]=\mathrm{LT}^{-1} \\
{\left[c t^2\right] } & =\mathrm{L}, & & \text { or, }[c]=\mathrm{LT}^{-2} \\
\text { and }\left[d t^3\right] & =\mathrm{L}, & & \text { or, }[d]=\mathrm{LT}^{-3} .
\end{aligned}
\)

Q34: If the centripetal force is of the form \(m^a v^b r^c\), find the values of \(a, b\) and \(c\).

Answer: Dimensionally,
\(
\begin{aligned}
& \text { Force }=(\text { Mass })^a \times(\text { velocity })^b \times(\text { length })^c \\
& \text { or, } \mathrm{MLT}^{-2}=\mathrm{M}^a\left(\mathrm{~L}^b \mathrm{~T}^{-b}\right) \mathrm{L}^c=\mathrm{M}^a \mathrm{~L}^{b+c} \mathrm{~T}^{-b} \\
& \text { Equating the exponents of similar quantities, } \\
& a=1, b+c=1,-b=-2 \\
& \text { or, } a=1, b=2, c=-1 \quad \text { or, } F=\frac{m v^2}{r} .
\end{aligned}
\)

Q35: When a solid sphere moves through a liquid, the liquid opposes the motion with a force \(F\). The magnitude of \(F\) depends on the coefficient of viscosity \(\eta\) of the liquid, the radius \(r\) of the sphere and the speed \(v\) of the sphere. Assuming that \(F\) is proportional to different powers of these quantities, guess a formula for \(F\) using the method of dimensions.

Answer: Suppose the formula is \(F=k \eta^a r^b v^c\).
Then,
\(
\begin{aligned}
\mathrm{MLT}^{-2} & =\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]^a \mathrm{~L}^b\left(\frac{\mathrm{L}}{\mathrm{T}}\right)^c \\
& =\mathrm{M}^a \mathrm{~L}^{-a+b+c} \mathrm{~T}^{-a-c}
\end{aligned}
\)
Equating the exponents of M, L and T from both sides,
\(
\begin{aligned}
a & =1 \\
-a+b+c & =1 \\
-a-c & =-2
\end{aligned}
\)
Solving these, \(a=1, b=1\), and \(c=1\).
Thus, the formula for \(F\) is \(F=k \eta r v\).

Q36: The heat produced in a wire carrying an electric current depends on the current, the resistance and the time. Assuming that the dependence is of the product of powers type, guess an equation between these quantities using dimensional analysis. The dimensional formula of resistance is \(\mathrm{ML}^2 \mathrm{I}^{-2} \mathrm{~T}^{-3}\) and heat is a form of energy.

Answer: Let the heat produced be \(H\), the current through the wire be \(I\), the resistance be \(R\) and the time be \(t\). Since heat is a form of energy, its dimensional formula is \(\mathrm{ML}^2 \mathrm{~T}^{-2}\).
Let us assume that the required equation is
\(
H=k I^a R^b t^c
\)
where \(k\) is a dimensionless constant.
Writing dimensions of both sides,
\(
\begin{aligned}
\mathrm{ML}^2 \mathrm{~T}^{-2} & =\mathrm{I}^a\left(\mathrm{ML}^2 \mathrm{I}^{-2} \mathrm{~T}^{-a}\right)^b \mathrm{~T}^c \\
& =\mathrm{M}^b \mathrm{~L}^{2 b} \mathrm{~T}^{-a b+c} \mathrm{I}^{a-2 b}
\end{aligned}
\)
Equating the exponents,
\(
\begin{aligned}
b & =1 \\
2 b & =2 \\
-3 b+c & =-2 \\
a-2 b & =0
\end{aligned}
\)
Solving these, we get, \(a=2, b=1\) and \(c=1\).
Thus, the required equation is \(H=k I^2 R t\).

Q37: The metre is defined as the distance travelled by light in \(\frac{1}{299,792,458}\) second. Why didn’t people choose some easier number such as \(\frac{1}{300,000,000}\) second? Why not 1 second?

Answer: The speed of light in vacuum is \(299,792,458 \mathrm{~m} / \mathrm{s}\).
Then time taken by light to cover a distance of 1 metre in vacuum \(=\frac{1}{299,792,458} \mathrm{~S}\)
Hence, the metre is defined as the distance travelled by light in \(\frac{1}{299,792,458} \mathrm{~S}\)
As \(300,000,000 \mathrm{~m} / \mathrm{s}\) is an approximate speed of light in vacuum, it cannot be used to define the metre.
The distance travelled by light in one second is \(299,792,458 \mathrm{~m}\). This is a large quantity and cannot be used as a base unit. So, the metre is not defined in terms of second.

Q38: What are the dimensions of :
(a) volume of a cube of edge \(a\),
(b) volume of a sphere of radius \(a\),
(c) the ratio of the volume of a cube of edge \(a\) to the volume of a sphere of radius \(a\)?

Answer: (a) Volume of a cube of edge \(a, \mathrm{~V}=\mathrm{a} \times \mathrm{a} \times \mathrm{a}\) i.e \([\mathrm{V}]=\mathrm{L} \times \mathrm{L} \times \mathrm{L}=\mathrm{L}^3\)
(b) Volume of a sphere of radius \(a, V=\frac{4}{3} \pi(a)^3\) i.e \([V]=L \times L \times L=L^3\)
(c) The ratio of the volume of the cube to the volume of the sphere is a dimensionless quantity.

Q39: Suppose you are told that the linear size of everything in the universe has been doubled overnight. Can you test this statement by measuring sizes with a metre stick? Can you test it by using the fact that the speed of light is a universal constant and has not changed? What will happen if all the clocks in the universe also start running at half the speed?

Answer: The validity of this statement cannot be tested by measuring sizes with a metre stick, because the size of the metre stick has also got doubled overnight.
Yes, it can be verified by using the fact that speed of light is a universal constant and has not changed.
If the linear size of everything in the universe is doubled and all the clocks in the universe starts running at half the speed, then we cannot test the validity of this statement by any method.

Q40: If all the terms in an equation have same units, is it necessary that they have same dimensions? If all the terms in an equation have same dimensions, is it necessary that they have same units?

Answer: Yes, if all the terms in an equation have the same units, it is necessary that they have the same dimension.
No, if all the terms in an equation have the same dimensions, it is not necessary that they have the same unit. It is because two quantities with different units can have the same dimension, but two quantities with different dimensions cannot have the same unit. For example angular frequency and frequency, both have the dimensions \(\left[T^{-1}\right]\) but units of angular frequency is rad/s and frequency is Hertz. Another example is energy per unit volume and pressure. Both have the dimensions of \(\left[M L^{-1} T^{-2}\right]\) but units of pressure is \(\mathrm{N} / \mathrm{m}^2\) and that of energy per unit volume is \(\mathrm{J} / \mathrm{m}^3\)

Q41: If two quantities have same dimensions, do they represent same physical content?

Answer: No, even if two quantities have the same dimensions, they may represent different physical contents. Example: Torque and energy have the same dimension, but they represent different physical contents.

Q42: It is desirable that the standards of units be easily available, invariable, indestructible and easily reproducible. If we use foot of a person as a standard unit of length, which of the above features are present and which are not?

Answer: If we use the foot of a person as a standard unit of length, features that will not be present are variability, destructibility and reproducible nature and the feature that will be present is the availability of the foot of a person to measure any length.

Q43: Suggest a way to measure :
(a) the thickness of a sheet of paper,
(b) the distance between the sun and the moon.

Answer: (a) The thickness of a sheet of paper can roughly be determined by measuring the height of a stack of paper.
Example: Let us consider a stack of 100 sheets of paper. We will use a ruler to measure its height. In order to determine the thickness of a sheet of paper, we will divide the height of the stack with the number of sheets (i.e., 100).
(b) The distance between the Sun and the Moon can be measured by using Pythagoras theorem when the Earth makes an angle of \(90^{\circ}\) with the Sun and the Moon. We already know the distances from the Sun to the Earth and from the Earth to the Moon. However, these distances keep on changing due to the revolution of the Moon around the Earth and the revolution of the Earth around the Sun.

Q44: 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.

Answer: An object is said to be large or small when it is compared by the some other standard object of the same type. For example, the size of a ship is very large as compared to the size of a bicycle. The comparison of following statement in their reframe is:
(a). Atoms are very small in comparison to a tennis ball.
(b) A jet plane moves with great speed in comparison to the bicycle.
(c) The mass of Jupiter is very large in comparison to the mass of a bob of a simple pendulum.
(d) The air inside the room contains a larger number of molecules in comparison to the number of molecules in a small box.
(e) a proton is much more massive than an electron in comparison to the difference in the mass of a basketball and a baseball.
(f) the speed of sound is much smaller than the speed of light in comparison to the difference between the speeds of a bicycle and the speed of a car.

Q45: 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 \(1.0 \mathrm{~mm}\) 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?

Answer: (a) Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a metre scale. The diameter of the thread is given by the relation,
\(
\text { Diameter }=\frac{\text { Length of thread }}{\text { Number of turns }}
\)
(b) We know that least count = Pitch/number of divisions on circular scale When number of divisions on circular scale is increased, the least count is decreased. Hence the accuracy is increased. However, this is only a theoretical idea. Practically speaking, increasing the number of ‘turns would create many difficulties.

As an example, the low resolution of the human eye would make observations difficult. The nearest divisions would not clearly be distinguished as separate. Moreover, it would be technically difficult to maintain uniformity of the pitch of the screw throughout its length.
(c) Due to random errors, a large number of observations will give a more reliable result than the smaller number of observations. This is due to the fact that the probability (chance) of making a positive random error of a given magnitude is equal to that of making a negative random error of the same magnitude. Thus in a large number of observations, positive and negative errors are likely to cancel each other. Hence more reliable results can be obtained.

Q46: One mole of an ideal gas at standard temperature and pressure occupies \(22.4 \mathrm{~L}\) (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of the hydrogen molecule to be about 1 Å ). Why is this ratio so large?

Answer: Radius of hydrogen atom, \(r=0.5 Å =0.5 \times 10^{-10} \mathrm{~m}\)
Volume of hydrogen atom \(=(4 / 3) \pi r^3\)
\(
\begin{aligned}
& =(4 / 3) \times(22 / 7) \times\left(0.5 \times 10^{-10}\right)^3 \\
& =0.524 \times 10^{-30} \mathrm{~m}^3
\end{aligned}
\)
Now, 1 mole of hydrogen contains \(6.023 \times 10^{23}\) hydrogen atoms.
\(\therefore\) Volume of 1 mole of hydrogen atoms, \(\mathrm{V}_{\mathrm{a}}=6.023 \times 10^{23} \times 0.524 \times 10^{-30}\)
\(
=3.16 \times 10^{-7} \mathrm{~m}^3
\)
Molar volume of 1 mole of hydrogen atoms at STP,
\(
\begin{aligned}
& \mathrm{V}_{\mathrm{m}}=22.4 \mathrm{~L}=22.4 \times 10^{-3} \mathrm{~m}^3 \\
& \therefore \frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{V}_{\mathrm{a}}}=\frac{22.4 \times 10^{-3}}{3.6 \times 10^{-7}}=7.08 \times 10^4
\end{aligned}
\)
Hence, the molar volume is \(7.08 \times 10^4\) times higher than the atomic volume. The ratio is so large because inter-atomic separation in hydrogen gas is large.

Q46: 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).

Answer: An imaginary line joining between the human eyes to a particular object is known as line of sight. When the person moves with fast speed the line of sight of close object changes rapidly. But the line of sight remains same for the distant object.
Similarly, when the passenger is looking outside through the window of a fast-moving train, the passenger observes that the nearby trees, houses, etc seems to move rapidly because the line of sight between the passenger and the nearby things changes rapidly as the train moves very fast.
But when the passenger looks for distant objects like hilltops and mountains, they seem to be stationary due to the line of sight remains the same between the passenger and the distant object.

Near objects make greater angle than distant (far off) objects at the eye of the observer. When you are moving, the angular change is less for distant objects than nearer objects. So, these distant objects seem to move along with you, but the nearer objects in opposite direction.

Q47: The Principle of ‘parallax’ is used in the determination of distances of very distant stars. The baseline \(A B\) is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit \(\approx 3 \times 10^{11} \mathrm{~m}\). However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1 ” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1″ (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?

Answer: Distance of the earth from the Sun =
\(
\begin{aligned}
& =\mathrm{b}=1 / 2 \times \text { diameter of earth’s orbit } \\
& \Rightarrow \mathrm{b}=0.5 \times 3 \times 10^{11} \mathrm{~m} \\
& \mathrm{~b}=1.5 \times 10^{11} \mathrm{~m} \\
& 1 \text { parsec }=\mathrm{D}=\mathrm{b} / \theta \\
& =\frac{1.5 \times 10^{11} \mathrm{~m}}{1^{11}}=\frac{1.5 \times 10^{11}}{\left(\frac{1}{3600}\right)} \\
& =\frac{1.5 \times 10^{11} \mathrm{~m}}{\frac{1}{3600} \times \frac{\pi}{180}}=3 \times 10^{16} \mathrm{~m}
\end{aligned}
\)

\(
\cong 3 \times 10^{16} \mathrm{~m} \text {; as a unit of length } 1 \text { parsec is defined to be equal to } 3.084 \times 10^{16} \mathrm{~m} \text {. }
\)

Q48: The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?

Answer: As we know, 1 light year \(=9.46 \times 10^{15} \mathrm{~m}\)
\(\therefore 4.29\) light years \(=4.29 \times 9.46 \times 10^{15}=4.058 \times 10^{16} \mathrm{~m}\)
Also, 1 parsec \(=3.08 \times 10^{16} \mathrm{~m}\)
\(\therefore 4.29\) light-years \(=4.508 \times 10^{16} / 3.80 \times 10^{16}=1.318\) parsec \(=1.32\) parsec
As a parsec distance subtends a parallax angle of 1 ” for a basis of radius of Earth’s orbit around the sun (r). In present problem base is the distance between two location of the Earth six months apart in its orbit around the Sun = dimeter of Earth’s orbit ( \(b=2 r\) ).
\(\therefore\) Parallax angle subtended by 1 parsec distance at this basis \(=2\) second (by definition of parsec).
\(\therefore\) Parallax angle subtended by the star Alpha Centauri at the given basis \(\theta=1.32 \times 2=2.64\).

Q49: Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass, etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

Answer: Precise measurement of a physical quantity is truly needed.
For example,
1) Ultra-shot laser pulses (time interval \(1015 \mathrm{~s}\) ) are used to measure time intervals in several physical and chemical processes.
2) \(\mathrm{X}\)-ray spectroscopy is used to determine the inter-atomic separation or inter-planer spacing.
3) The development of mass spectrometer helps to measure mass of an atom precisely.

Q50: Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity) :
(a) the total mass of rain-bearing clouds over India during the Monsoon
(b) the mass of an elephant
(c) the wind speed during a storm
(d) the number of strands of hair on your head
(e) the number of air molecules in your classroom.

Answer: (a) Rainfall is measured extensively covering the entire nation along with an approximate demarcation of the areas involved. This will give the data for computing the total volume of rainfall in the country. The projection of the trend for future can be forecast in general based on the statistical data collected over years. An average rainfall of about \(100 \mathrm{~cm}\) during monsoon in India is already recorded by the meteorologist.
(b) Consider a ship of known base area floating in the sea. Measure its depth \(d_1\) in the sea. The volume of water displaced by the ship:
\(\mathrm{V}_1=\) Ad. Now, move an elephant on the ship and measure the depth of the ship, \(d_2\) in this case. The volume of water displaced by the ship with the elephant on board: \(\mathrm{V}_2=\mathrm{Ad}_2\). The volume of water displaced by the elephant= \(\mathrm{Ad}_2-\mathrm{Ad}_1\). The density of water= \(\mathrm{D}\). Mass of elephant \(=\mathrm{D} \times\left(\mathrm{Ad}_2-\mathrm{Ad}_1\right)\)
(c) Wind speed during a storm can be measured by an anemometer. As the wind blows, it rotates. The rotation made by the anemometer in one second gives the value of wind speed.
or Wind speed can be estimated by floating a gas-filled balloon at a known height h. At the onset of wind, the angle drift of balloon in one second can be estimated. Using simple trigonometry, we can estimate the wind speed.
(d) Area of the head surface carrying hair \(=\mathrm{A}\)
With the help of a screw gauge, the diameter and hence, the radius of a hair can be determined. Let it be \(r\).
Area of one hair \(=r^2\)
Number of strands of hair \(=\) Total surface area \(/\) Area of one hair \(=\frac{\mathrm{A}}{\mathrm{r}^2}\)
(e) Let the volume of the room be \(V\)
One mole of air at NTP occupies 22.41 i.e., \(22.4 \times 10^{-3} \mathrm{~m}^3\) volume.
Number of molecules in one mole \(=6.023 \times 10^{23}\)
Number of molecules in the room of volume \(\mathrm{V}\)
\(=6.023 \times 10^{23} \mathrm{~V} / 22.4 \times 10^{-3}=134.915 \times 10^{26} \mathrm{~V}=1.35 \times 10^{28} \mathrm{~V}\)

Q51: The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding \(10^7 \mathrm{~K}\), and its outer surface at a temperature of about \(6000 \mathrm{~K}\). 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 \(=2.0 \times 10^{30} \mathrm{~kg}\), radius of the Sun \(=7.0 \times 10^8 \mathrm{~m}\)

Answer:

\(
\begin{aligned}
& \text { Actual Density of the Sun }=\frac{\text { Mass }}{\text { V olume }} \\
& =\frac{2 \times 10^{30} \mathrm{~kg}}{\frac{4}{3} \pi \times\left(7 \times 10^8\right)^3 \mathrm{~m}^3} \\
& =1.4 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}
\end{aligned}
\)

\(1.4 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\); the mass density of the Sun is in the range of densities of liquids/solids and not gases. This high density arises due to inward gravitational attraction on outer layers due to the inner layers of the Sun.

Q52: When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72 ” of arc. Calculate the diameter of Jupiter.

Answer: Given, the distance of the planet Jupiter from the Earth is 824.7 million kilometers and the angular diameter is 35.72 ” of arc.
Let \(D\) be the distance of Jupiter from the Earth and \(d\) be the diameter of the planet Jupiter.
The angular diameter of the Jupiter is given as,
\(
\begin{aligned}
& \theta=\frac{d}{D} \\
& d=\theta D
\end{aligned}
\)
Substitute the required values in the above equation.
\(
d=\theta D=35.72 ” \times 824.7 \times 10^6 \mathrm{~km}=35.72 ” \times 4.874 \times 10^{-6} \times 824.7 \times 10^9 \mathrm{~m}=1.429 \times 10^5
\)
Thus, the diameter of the planet Jupiter is \(1.429 \times 10^5 \mathrm{~km}\).

Q53: A man walking briskly in the rain with speed \(v\) must slant his umbrella forward making an angle \(\theta\) with the vertical. A student derives the following relation between \(\theta\) and \(v: \tan \theta=v\) and checks that the relation has a correct limit: as \(v \rightarrow 0, \theta \rightarrow 0\), as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct? If not, guess the correct relation.

Answer: According to the principle of homogeneity of dimensional equations,
Dimensions of L.H.S. = Dimensions of R.H.S.
Here, \(v=\tan \theta\)
i. e., \(\left[L^1 T^{-1}\right]=\) dimensionless, which is incorrect.
Correcting the L.H.S., we. get
The correct formula is \(\tan \theta=v / v^{\prime}\) where \(v^{\prime}\) is the speed of rainfall.

Q54: Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase : \(970 \mathrm{~kg} \mathrm{~m}^{-3}\). Are the two densities of the same order of magnitude? If so, why?

Answer: It is given that radius of the sodium atom, \(R=2.5 Å=2.5 \times 10^{-10} \mathrm{~m}\)
\(
\text { No. of atoms in } 1 \text { mole } N=6.023 \times 10^{23}
\)
\(
\text { Volume of sodium atom } V .=\frac{4}{3} \pi R^3
\)
\(
\text { volume of } 1 \text { mole of sodium }=N \times V
\)
Mass of one-mole atom of sodium, \(M=23 \mathrm{~g}=23 \times 10^{-3} \mathrm{~kg}\)
The average density of 1-mole atoms of sodium \(\rho=\frac{M}{V}=0.7 \times 10^{3} \mathrm{kgm}^{-3}\)
The density of sodium in its crystalline phase \(=970 \mathrm{kgm}^{-3}\) \(=0.97 \times 10^3 \mathrm{kgm}^{-3}\)
Yes, both densities are in the same no. of power. Because atoms are bound strongly in solid form. Therefore, atomic density is close to the density of a solid.

In the solid phase atoms are tightly packed, so the atomic mass density is close to the mass density of the solid.

Q55: The unit of length convenient on the nuclear scale is a fermi : \(1 \mathrm{f}=10^{-15} \mathrm{~m}\). Nuclear sizes obey roughly the following empirical relation :
\(
r=r_0 A^{1 / 3}
\)
where \(r\) is the radius of the nucleus, \(A\) its mass number, and \(r_0\) is a constant equal to about, \(1.2 \mathrm{f}\). Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of the sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

Answer: Radius of nucleus \(r\) is given by the relation,
\(
r=r_0 A^{\frac{1}{3}} \ldots \text {… (i) }
\)
\(
r_0=1.2 f=1.2 \times 10^{-15} \mathrm{~m}
\)
Volume of nucleus \(\mathrm{V}=\frac{4}{3} \pi r^3\)
\(
=\frac{4}{3} \pi\left(r_0 A^{\frac{1}{3}}\right)^3=\frac{4}{3} \pi r_0^3 A \ldots \text {..(ii) }
\)
Now, the mass of a nuclei \(M\) is equal to its mass number i.e.,
\(
M=A \mathrm{~amu}=A \times 1.66 \times 10^{-27} \mathrm{~kg}
\)
Density of nucleus
\(
\begin{aligned}
& \rho=(\text { “Mass of nucleus”)/”Volume of nucleus” } \\
& =\frac{A \times 1.66 \times 10^{-27}}{\frac{4}{3} \pi r^{\wedge} 3 A}=\frac{3 \times 1.66 \times 10^{-27}}{4 \pi r_0^3} \mathrm{~kg} / \mathrm{m}^3
\end{aligned}
\)
This relation shows that nuclear mass depends only on constant \(r_0\). Hence, the nuclear mass densities of all nuclei are nearly the same.
The density of sodium nucleus is given by,
\(
\begin{aligned}
& \rho_{\text {sodium }}=\frac{3 \times 1.66 \times 10^{-27}}{4 \times 3.14 \times\left(1.2 \times 10^{-15}\right)^3} \\
& =\frac{4.98}{21.71} \times 10^{18}\cong 0.3 \times 10^{18} \mathrm{~kg} \mathrm{~m}^{-3} \text { – Nuclear density is typically } 10^{15} \text { times atomic density of matter. }
\end{aligned}
\)

Q56: It is a well-known fact that during a total solar eclipse, the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

Answer: From examples 2.3 and 2.4, we get \(\theta=1920\) and \(S=3.8452 \times 10^8 \mathrm{~m}\). During the total solar eclipse, the disc of the moon completely covers the disc of the sun, so the angular diameter of both the sun and the moon must be equal. Angular diameter of the moon, \(\theta=\) angular diameter of the sun
\(
=1920^{\prime}=1920 \times 4.85 \times 10^{-6} \mathrm{rad}\left[1^{\prime}=4.85 \times 10^{-6} \mathrm{rad}\right]
\)
the earth-moon distance, \(S=3.8452 \times 10^8 \mathrm{~m}\). the diameter of the moon,
\(
\begin{aligned}
& D=\theta \times S \\
& =1920 \times 4.85 \times 10^{-6} \times 3.8452 \times 10^8 \mathrm{~m}=35806.5024 \times 10^2 \mathrm{~m}=3581 \\
& \times 10^3 \mathrm{~m} = 3581 \mathrm{~km}
\end{aligned}
\)

Q57: A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant \(G\), he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe ( \(\sim 15\) billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?

Answer: Trying out with basic constants of atomic physics (speed of light c, the charge on electron e, the mass of electron \(m_e\) mass of proton \(m_p\) ) and universal gravitational constant \(\mathrm{G}\), we can arrive at a quantity which has the dimensions of time. Once such quantity is \(t\) = \(
e^4 /\left(16 \pi^2 \epsilon_0^2 m_{\mathrm{p}} \mathrm{m}_{\mathrm{e}}^2 \mathrm{c}^3 G\right)
\)
Putting
\(
\begin{aligned}
& e=1.6 \times 10^{-19} \mathrm{C}, \frac{1}{4 \pi \mathrm{in}_0}=9 \times 10^9, c=3 \times 10^8 \mathrm{~m} / \mathrm{s} \text { and } G=6.67 \\
& \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2} \\
& m_p =1.67 \times 10^{-27} \mathrm{~kg}, m_e=9 \times 10^{-31} \mathrm{~kg} \\
& t=\left(1.6 \times 10^{-19}\right)^4 \times\left(9 \times 10^9\right)^2 \times x \frac{1}{1.67 \times 10^{-27}\left(9 \times 10^{-31}\right)^2\left(3 \times 10^8\right)^3 \times 6.67 \times 10^{-11}} \\
& t=2.18 \times 10^{16} \mathrm{sec} \text {. This time is of the order of the age of the universe. }
\end{aligned}
\)