Physics - Unit & Dimension Question with Solution | TestHub

Given here: $\left[B\right]={\left[e\right]}^{\alpha }{\left[m_e\right]}^{\beta }{\left[h\right]}^{\gamma }{\left[k\right]}^{\delta }$

The dimensions of $\left[e\right]=\left[\mathrm{Current}\times \mathrm{time}\right]=\left[IT\right]$ and $\left[m_e\right]=\left[M\right]$, Planck's constant $\left[h\right]=\frac{\mathrm{Energy}}{\mathrm{Frequency}}=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[T^{-1}\right]}=\left[M{L}^2{T}^{-1}\right]$ and $\left[k\right]=\frac{1}{\left[{\epsilon }_0\right]}={\left[M^{-1}{L}^{-3}{T}^4{I}^2\right]}^{-1}=\left[M{L}^3{T}^{-4}{I}^{-2}\right]$

Using Principle of homogeneity,

$\left[M^1{T}^{-2}{I}^{-1}\right]={\left[IT\right]}^{\alpha }{\left[M\right]}^{\beta }{\left[M{L}^2{T}^{-1}\right]}^{\gamma }{\left[M{L}^3{T}^{-4}{I}^{-2}\right]}^{\delta }$

Comparing both sides, we get

 $\beta +\gamma +\delta =1 ...\left(\mathrm{i}\right)$

$2\gamma +3\delta =0 ...\left(\mathrm{ii}\right)$

$\alpha -\gamma -4\delta =-2 ...\left(\mathrm{iii}\right)$

$\alpha -2\delta =-1 ...\left(\mathrm{iv}\right)$

On solving above four equations, we get

$\alpha =3, \beta =-4, \gamma =3$ and $\delta =2$

So, $\alpha +\beta +\gamma +\delta =4$