Question

Which of the following combinations has the dimension of electrical resistance (${ϵ}_0$ is the permittivity of vacuum, and ${\mu }_0$ is the permeability of vacuum)?

• A. $\frac{{\mu }_0}{{ϵ}_0}$
• B. $\frac{{ϵ}_0}{{\mu }_0}$
• C. $\sqrt{\frac{{ϵ}_0}{{\mu }_0}}$
• D. $\sqrt{\frac{{\mu }_0}{{ϵ}_0}}$

Answer 1

The correct option is D $\sqrt{\frac{{\mu }_0}{{ϵ}_0}}$

As we know,
Dimension of $\left[{ϵ}_0\right]=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$
Dimension of $\left[{\mu }_0\right]=\left[ML{T}^{-2}{A}^{-2}\right]$
Dimension of $\left[R\right]=\left[M{L}^2{T}^{-3}{A}^{-2}\right]$
$\left[R\right]={\left[{\epsilon }_0\right]}^a{\left[{\mu }_0\right]}^b$ ......(i)

By putting all the dimensions in equation (𝑖) we get,
$\left[M{L}^2{T}^{-3}{A}^{-2}\right]=\left[M^{-a}{L}^{-3a}{T}^{4a}{A}^{2a}\right]\left[M^b{L}^b{T}^{-2b}{A}^{-2b}\right]$

Comparing the equation on both sides
$a-b=-\mathrm{1...}\left(ii\right)$
$3a-b=-\mathrm{2...}\left(iii\right)$
$4a-2b=-\mathrm{3...}\left(iv\right)$
$2a-2b=-\mathrm{2...}\left(v\right)$

From equation 𝑖𝑣 & 𝑣 we get,
$2a=-1$
$a=\frac{-1}{2}$

From equation (iii)& (iv) we get,
$b=\frac{1}{2}$

By putting the value of $a$ & $b$ we get,
$\left[R\right]=\sqrt{\frac{{\mu }_0}{{ϵ}_0}}$

Final Answer: (c)