Question

If $\frac{8\pi {\epsilon }_{0}Kx}{Q^2}$ is a dimensionless quantity, ${\epsilon }_0$ permitivity of free space, $K$ - energy, $Q$ - charge. Then the dimensions of $x$ are :

• A. $ML{T}^2$
• B. $ML{T}^{-1}$
• C. $M^{0}L{T}^0$
• D. $M{L}^{-1}{T}^{-1}$

Answer 1

The correct option is C $M^{0}L{T}^0$

$8\pi \frac{E_{0}kx}{Q^2}=M^0{L}^0{T}^0$

Let $x=M^{\alpha }{L}^{\beta }{T}^{\gamma }$
$Q=I^1{T}^1$
$K=M^1{L}^2{T}^{-2}$
$E_0=M^{-1}{L}^{-3}{I}^2{T}^4$
$\Rightarrow \frac{\left(M^{-1}{L}^{-3}{I}^2{T}^4\right)\left(M^{\alpha }{L}^{\beta }{T}^{\gamma }\right)\left(M^1{L}^2{T}^{-2}\right)}{I^2{T}^2}=M^0{L}^0{T}^0$
$\alpha =0\beta =1,\gamma =0$
Hence, $x=M^{0}L{T}^0$