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.

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We know that, dimensions of $(h)=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right]$ (From $\left.\mathrm{E}=h f\right]$ Dimensions of $(c)=\left[\mathrm{L}^{1} \mathrm{~T}^{-1}\right] \quad(c$ is velocity $)$

Dimensions of gravitational constant

$(\mathrm{G})=\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right] \quad\left(\text { From } \mathrm{F}=\frac{\mathrm{G} m_{1} m_{2}}{r^{2}}\right)$

$(i)$ Let $\mathrm{m} \propto c^{a} h^{b} \mathrm{G}^{c}$

$\Rightarrow \mathrm{m}=k c^{a} h^{b} \mathrm{G}^{c}$

where, $k$ is a dimensionless constant of proportionality. Substituting dimensions of each term in Eq.$ (i)$, we get

$\left[\mathrm{ML}^{0} \mathrm{~T}^{0}\right] =\left[\mathrm{LT}^{-1}\right]^{x} \times\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]^{y}\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]^{z}$

$=\left[\mathrm{M}^{b-c} \mathrm{~L}^{a+2 b+3 c} \mathrm{~T}^{-a-b-2 c}\right]$

Comparing powers of same terms on both sides, we get

$b-c=1\ldots \text { (ii) }$

$a+2 b+3 c=0\ldots\text { (iii) }$

$-a-b-2 c=0\ldots\text { (iv) }$

$\ldots \text { (ii) }$

$\ldots \text { (iii) }$

Adding Eqs. $(ii)$, $(iii)$ and $(iv)$, we get

$2 b=1 \Rightarrow b=\frac{1}{2}$

Substituting value of $\mathrm{b}$ in eq. $(ii)$, we get

$c=-\frac{1}{2}$

From eq. $(iv)$,

$a=-b-2 c$

Substituting values of $b$ and $c$, we get

$a=-\frac{1}{2}-2\left(-\frac{1}{2}\right)=\frac{1}{2}$

Similar Questions

Let us consider an equation

$\frac{1}{2} m v^{2}=m g h$

where $m$ is the mass of the body. velocity, $g$ is the acceleration do gravity and $h$ is the height. whether this equation is dimensionally correct. 

Given below are two statements: One is labelled as Assertion $(A)$ and other is labelled as Reason $(R)$.
Assertion $(A)$ : Time period of oscillation of a liquid drop depends on surface tension $(S)$, if density of the liquid is $p$ and radius of the drop is $r$, then $T = k \sqrt{ pr ^{3} / s ^{3 / 2}}$ is dimensionally correct, where $K$ is dimensionless.
Reason $(R)$: Using dimensional analysis we get $R.H.S.$ having different dimension than that of time period.
In the light of above statements, choose the correct answer from the options given below.

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