If 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.
We know that, dimensions of
$$ \text { (h) }=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right] $$
Dimensions of
(c) $=\left[\mathrm{LT}^{-1}\right]$
Dimensions of gravitational constant $(G)=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]$
(i) Let $$ m \propto c^x h^y G^z $$
$$ \Rightarrow m=k c^x h^y G^z $$ ....(i)
where, $k$ is a dimensionless constant of proportionality.
Substituting dimensions of each term in Eq. (i), we get
$$ \begin{aligned} {\left[\mathrm{ML}^0 \mathrm{~T}^0\right] } & =\left[\mathrm{LT}^{-1}\right]^x \times\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^y\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^z \\\\ & =\left[\mathrm{M}^{y-z} \mathrm{~L}^{x+2 y+3 z} \mathrm{~T}^{-x-y-2 z}\right] \end{aligned} $$
Comparing powers of same terms on both sides, we get
$$ \begin{array}{r} y-z=1 \,\,\,\, ....(ii) \\ x+2 y+3 z=0 \,\,\,\, ....(iii) \\ -x-y-2 z=0 \,\,\,\, ....(iv) \end{array} $$
Adding Eqs. (ii), (iii) and (iv), we get
$$ 2 y=1 \Rightarrow y=\frac{1}{2} $$
Substituting value of $y$ in Eq. (ii), we get
$$ z=-\frac{1}{2} $$
From Eq. (iv)
$$ x=-y-2 z $$
Substituting values of $y$ and $z$, we get
$$ x=-\frac{1}{2}-2\left(-\frac{1}{2}\right)=\frac{1}{2} $$
Putting values of $x, y$ and $z$ in Eq. (i), we get
$$ \begin{aligned} & m=k c^{1 / 2} h^{1 / 2} G^{-1 / 2} \\\\ & \Rightarrow \quad m=k \sqrt{\frac{c h}{G}} \\ & \end{aligned} $$
$$ \begin{aligned} & \text{Let} \quad L \propto C^x h^y G^z \\\\ & \Rightarrow L=k c^x h^y G^z \,\,\,\, ....(v) \end{aligned} $$
where, $k$ is a dimensionless constant.
Substituting dimensions of each term in Eq. (v), we get
$$ \begin{aligned} {\left[\mathrm{M}^0 \mathrm{LT}^0\right] } & =\left[\mathrm{LT}^{-1}\right]^x \times\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^y \times\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^z \\\\ & =\left[\mathrm{M}^{y-z} \mathrm{~L}^{x+2 y+3 z} \mathrm{~T}^{-x-y-2 z}\right] \end{aligned} $$
On comparing powers of same terms, we get
$$ \begin{array}{r} y-z=0 \,\,\,\,....(vi)\\ x+2 y+3 z=1 \,\,\,\,....(vii)\\ -x-y-2 z=0 \,\,\,\,....(viii) \end{array} $$
Adding Eqs. (vi), (vii) and (viii), we get
$$ \begin{array}{r} 2 y=1 \\\Rightarrow y=\frac{1}{2} \end{array} $$
Substituting value of $y$ in Eq. (vi), we get
$$ z=\frac{1}{2} $$
From Eq. (viii),
$$ x=-y-2 z $$
Substituting values of $y$ and $z$, we get
$$ x=-\frac{1}{2}-2\left(\frac{1}{2}\right)=-\frac{3}{2} $$
Putting values of $x, y$ and $z$ in Eq. (v), we get
$$ \begin{aligned} & L=k c^{-3 / 2} h^{1 / 2} G^{1 / 2} \\\\ & L=k \sqrt{\frac{h G}{c^3}} \end{aligned} $$
(iii) Let $T \propto C^x h^y G^z$
$$ \Rightarrow \quad T=k c^x h^y G^z $$ ....(ix)
where, $k$ is a dimensionless constant.
Substituting dimensions of each term in Eq. (ix), we get
$$ \begin{aligned} {\left[M^0 L^0 T\right.} & =\left[L T^{-1}\right]^x \times\left[M^2 T^{-1}\right]^y \times\left[M^{-1} L^3 T^{-2}\right]^z \\\\ & =\left[M^{y-z} L^{x+2 y+3 z} T^{-x-y-2 z}\right] \end{aligned} $$
On comparing powers of same terms, we get
$$ \begin{array}{r} y-z=0 \,\,\,\,....(x) \\ x+2 y+3 z=0 \,\,\,\,....(xi) \\ -x-y-2 z=1 \,\,\,\,....(xii) \end{array} $$
Adding Eqs. (x), (xi) and (xii), we get
$$ \begin{aligned} 2 y & =1 \\ \Rightarrow y & =\frac{1}{2} \end{aligned} $$
Substituting value of $y$ in Eq. (x), we get
$$ z=y=\frac{1}{2} $$
From Eq. (xii),
$$ x=-y-2 z-1 $$
Substituting values of $y$ and $z$, we get
$$ x=-\frac{1}{2}-2\left(\frac{1}{2}\right)-1=-\frac{5}{2} $$
Putting values of $x, y$ and $z$ in Eq. (ix), we get
$$ \begin{aligned} T & =k c^{-5 / 2} h^{1 / 2} G^{1 / 2} \\\\ T & =k \sqrt{\frac{h G}{c^5}} \end{aligned} $$
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.
By Kepler’s third law, $T^2 \propto r^3 \Rightarrow T \propto r^{3/2}$
We know that $T$ is a function of $R$ and $g$.
Let $T \propto r^{3/2}R^ag^b \Rightarrow T = kr^{3/2}R^ag^b$ .....(i)
where, $k$ is a dimensionless constant of proportionality.
Substituting the dimensions of each term in Eq. (i), we get
$[M^0L^0 T] = k[L^3]^{1/2}[L]^{a}[LT^{-2}]^b = k[L^{a + b + 3/2}T^{-2b}]$
On comparing the powers of same terms, we get
$a + b + 3/2 = 0$ ...(ii)
$-2b = 1 \Rightarrow b = -1/2$ ...(iii)
From Eq. (ii), we get $a-1 / 2+3 / 2=0 \Rightarrow a=-1$
Substituting the values of $a$ and $b$ in Eq. (i), we get
$T = kr^{3/2}R^{-1}g^{-1/2}$
$ T = \frac{k}{R}\sqrt{\frac{r^3}{g}} $
Note: When we are applying formulae, we should be careful about $r$ (radius of orbit) and $R$ (radius of planet).
In an experiment to estimate the size of a molecule of oleic acid 1 mL of oleic acid is dissolved in 19 mL of alcohol. Then 1 mL of this solution is diluted to 20 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 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?
(a) Oleic acid does not dissolve in water hence, it is dissolved in alcohol.
(b) Lycopodium powder spreads over the entire surface of water when it is sprinkled evenly. When a drop of prepared solution 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 therefore, measure the area over which oleic acid spreads.
(c) In each mL of solution prepared volume of oleic acid $=\frac{1}{20} \mathrm{~mL} \times \frac{1}{20}=\frac{1}{400} \mathrm{~mL}$
(d) Volume of $n$ drops of this solution of oleic acid can be calculated by means of a burette and measuring cylinder and measuring the number of drops.
(e) If $n$ drops of the solution make 1 mL , the volume of oleic acid in one drop will be $\frac{1}{(400) n} \mathrm{~mL}$.
(a) How many astronomical units (AU) 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)° from the earth. Due to atmospheric fluctuations, eye cannot 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 AU from the earth. Calculate what size it will appear when seen through the same telescope.
(a) By definition,
1 parsec $=$ Distance at which 1 AU long arc subtends an angle of 1 s .
$\therefore \quad 1 \text { parsec }=\left(\frac{1 \mathrm{AU}}{1 \operatorname{arc~sec}}\right)$
$1 \mathrm{deg}=3600 \text { arc sec }$
$\therefore \quad 1 \text { parsec }=\frac{\pi}{3600 \times 180} \mathrm{rad}$
$\therefore \quad 1$ parsec $=\frac{3600 \times 180}{\pi} \mathrm{AU}=206265 \mathrm{AU} \approx 2 \times 10^5 \mathrm{AU}$
(b) Sun's diameter is $\left(\frac{1}{2}\right)^{\circ}$ at 1 AU .
Therefore, at 1 parsec, star is $\frac{1 / 2}{2 \times 10^5}$ degree in diameter $=15 \times 10^{-5} \mathrm{arc} \,\mathrm{min}$.
With 100 magnification, it should look $15 \times 10^{-3} \mathrm{arc\,min}$. However, due to atmospheric fluctuations, it will still look of about 1 arc min. It cannot be magnified using telescope.
(c) Given that $$ \frac{D_{\text {mars }}}{D_{\text {earth }}}=\frac{1}{2} $$ ...(i)
where $D$ represents diameter.
From answer 25(e)
we know that,
$$ \begin{aligned} & \frac{D_{\text {earth }}}{D_{\text {sun }}}=\frac{1}{100} \\\\ & \frac{D_{\text {mars }}}{D_{\text {sun }}}=\frac{1}{2} \times \frac{1}{100}\,\,\,\, \text{[from Eq. (i)]}\end{aligned} $$
$\begin{aligned} & \text { At } 1 \mathrm{AU} \text { sun's diameter }=\left(\frac{1}{2}\right)^{\circ} \\\\ & \therefore \quad \text { mar's diameter }=\frac{1}{2} \times \frac{1}{200}=\frac{1}{400} \end{aligned}$
$\text { At } \frac{1}{2} \mathrm{AU}, \text { mar's diameter }=\frac{1}{400} \times 2^{\circ}=\left(\frac{1}{200}\right)^{\circ}$
With 100 magnification, Mar's diameter $=\frac{1}{200} \times 100^{\circ}=\left(\frac{1}{2}\right)^{\circ}=30^{\prime}$
This is larger than resolution limit due to atmospheric fluctuations. Hence, it looks magnified.
Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass ($m$) to energy ($E$) as $ E = mc^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 MeV, where $1$ MeV $= 1.6 \times 10^{-13}$ J; the masses are measured in unified atomic mass unit ($u$) where, $1u = 1.67 \times 10^{-27}$ kg.
(a) Show that the energy equivalent of $1u$ is $931.5$ MeV.
(b) A student writes the relation as $1u = 931.5$ MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.
In this problem, we have to apply Einstein’s mass-energy relation. $E = mc^2$, to calculate the energy equivalent of the given mass.
(a) We know that
$$ 1 \text{ amu} = 1u = 1.67 \times 10^{-27} \text{kg} $$
Applying $ E = mc^2 $
Energy $E$ = $( 1.67 \times 10^{-27})( 3 \times 10^8)^2$ J
= $ 1.67 \times 9 \times 10^{-11} $ J
$$ E = \dfrac{ 1.67 \times 9 \times 10^{-11}}{ 1.6 \times 10^{-13}} \text{MeV} \approx 939.4 \text{MeV} \approx 931.5 \text{MeV} $$
(b) The dimensionally correct relation is
$$ 1 \text{ amu} \times c^2 = 1u \times c^2 = 931.5 \text{MeV} $$