Physics - Unit & Dimension Question with Solution | TestHub

Let,$M=C^a{h}^b{G}^c$

$M{L}^0{T}^0={\left[L{T}^{-1}\right]}^a {\left[M{L}^2{T}^{-1}\right]}^b {\left[M^{-1}{L}^3{T}^{-2}\right]}^c$......(i)

Where,$h=\frac{\mathrm{E}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y}}{\mathrm{F}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{y}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}}$

$=\frac{\left[M{L}^2 T^{-2}\right]}{\left[T^{-1}\right]}=\left[M{L}^2{T}^{-1}\right]$

$C=\frac{\mathrm{M}\mathrm{e}\mathrm{t}\mathrm{r}\mathrm{e}}{\mathrm{S}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}=\left[L{T}^{-1}\right]$

$G=\frac{\mathrm{F}\mathrm{o}\mathrm{r}\mathrm{c}\mathrm{e}\times {\left(\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\right)}^2}{{\left(\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\right)}^2}$

$=\frac{\left[ML{T}^{-2}\right] \left[L^2\right]}{\left[M^2\right]}=\left[M^{-1}{L}^3{T}^{-2}\right]$

Comparing the coefficients M, L, T, of both sides we get

$b-c=1$......(ii)

$a+2b+3c=0$......(iii)

$-\left(a+b+2c\right)=0$.....(iv)

Solve the equations (ii), (iii) and (iv), we get

$a=\frac{1}{2}, b=\frac{1}{2}, c= -\frac{1}{2}$

So,$M=h^{\frac{1}{2}} C^{\frac{1}{2}} G^{-\frac{1}{2}}$