Question

Match List-I with List-II

List-I	List-II
$\left(a\right)$ Capacitance, $C$	$\left(i\right)M^1{L}^1{T}^{-3}{A}^{-1}$
$\left(b\right)$ Permitivity of free space, ${ϵ}_0$	$\left(ii\right)M^{-1}{L}^{-3}{T}^4{A}^2$
$\left(c\right)$ permeabilityof free space, ${\mu }_0$	$\left(iii\right)M^{-1}{L}^{-2}{T}^4{A}^2$
$\left(d\right)$ electric field, $E$	$\left(iv\right)M^1{L}^1{T}^{-2}{A}^{-2}$

Choose the correct answer from the options given below.

• A. $\left(a\right)\to \left(iii\right),\left(b\right)\to \left(ii\right),\left(c\right)\to \left(iv\right),\left(d\right)\to \left(i\right)$
• B. $\left(a\right)\to \left(iii\right),\left(b\right)\to \left(iv\right),\left(c\right)\to \left(ii\right),\left(d\right)\to \left(i\right)$
• C. $\left(a\right)\to \left(iv\right),\left(b\right)\to \left(ii\right),\left(c\right)\to \left(iii\right),\left(d\right)\to \left(i\right)$
• D. $\left(a\right)\to \left(iv\right),\left(b\right)\to \left(iii\right),\left(c\right)\to \left(ii\right),\left(d\right)\to \left(i\right)$

Answer 1

The correct option is A $\left(a\right)\to \left(iii\right),\left(b\right)\to \left(ii\right),\left(c\right)\to \left(iv\right),\left(d\right)\to \left(i\right)$

We know that, $Q=CV$
$\left[C\right]=\left[\frac{Q}{V}\right]=\frac{[A\times T]}{\left[M{L}^2{T}^{-3}{A}^{-1}\right]}$
$=\left[M^{-1}{L}^{-2}{T}^4{A}^2\right]$
Similarly,
$\left[E\right]=\left[\frac{F}{q}\right]=\frac{\left[ML{T}^{-2}\right]}{\left[AT\right]}$
$=\left[ML{T}^{-3}{A}^{-1}\right]$

From the Coloumb's law, we have
$F=\frac{q_1{q}_2}{4\pi {ϵ}_0{r}^2}$
$\left[{ϵ}_0\right]=\frac{\left[q_1{q}_2\right]}{\left[F{r}^2\right]}=\frac{[AT{]}^2}{\left[ML{T}^{-2}\right]\left[L^2\right]}=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$

We know that, speed of light $c=\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}$
${\mu }_0=\frac{1}{{ϵ}_0{c}^2}$
$\left[{\mu }_0\right]=\frac{1}{\left[M^{-1}{L}^{-1}{T}^4{A}^2\right]{\left[L{T}^{-1}\right]}^2}$

$=\left[M^1{L}^1{T}^{-2}{A}^{-2}\right]$
Hence, option (a) is corrcet.