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Units and Measurements — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsUnits and Measurements1 Mark
Question
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.
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Answer

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

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