Correct answer is (C)

The term $\frac{1}{{\mu }_0{ϵ}_0}$ appears in the formula for the speed of light $c$, which is:

$c=\sqrt{\frac{1}{{\mu }_0{ϵ}_0}}$

where ${\mu }_0$ is the permeability of free space and ${ϵ}_0$ is the permittivity of free space.

The speed of light $c$ has dimensions of length over time ($\mathrm{L}/\mathrm{T}$). Therefore, the term $\frac{1}{{\mu }_0{ϵ}_0}$ has dimensions equal to the square of the speed of light, which is ${\mathrm{L}}^2/{\mathrm{T}}^2$.

Correct answer is (B)

$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}\Rightarrow [\mathrm{L}]=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]\left[{\mathrm{M}}^1{\mathrm{L}}^1{\mathrm{T}}^{-1}\right]$

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

$\begin{array}{rl}\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}\Rightarrow [\tau ]& =\left[{\mathrm{L}}^1\right]\left[{\mathrm{MLT}}^{-2}\right]\\ \\ & =\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]\end{array}$

Stress $\equiv$ Pressure $=\frac{F}{A}\Rightarrow [$ Stress $]=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right]$

Pressure Gradient $=\frac{dP}{dx}\Rightarrow [$ Pressure Gradient $]$ $=\left[{\mathrm{ML}}^{-2}{\mathrm{T}}^{-2}\right]$

Correct answer is (A)

Pressure gradient $=\frac{dp}{dx}=\frac{\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right]}{[\mathrm{L}]}$

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

Energy density $=\frac{\text{ energy }}{\text{ volume }}=\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]}{\left[{\mathrm{L}}^3\right]}$

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

Electric field $=\frac{\text{ Force }}{\text{ ch arge }}=\frac{\left[{\text{ MLT }}^{-2}\right]}{[\text{ A.T }]}$

$=\left[{\mathrm{M}}^1{\mathrm{L}}^1{\mathrm{T}}^{-3}{\mathrm{A}}^{-1}\right]$

Latent heat $=\frac{\text{ heat }}{\text{ mass }}=\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]}{[\mathrm{M}]}$

$=\left[{\mathrm{M}}^0{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]$

Correct answer is (D)

$\begin{array}{rl}& \mathrm{Y}=\frac{\text{ Stress }}{\text{ Strain }}=\frac{\mathrm{F}/\mathrm{A}}{\mathrm{\Delta }\ell /\ell }=\frac{\left[{\mathrm{MLT}}^{-2}\right]}{\left[{\mathrm{L}}^2\right]}=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right]\\ \\ & \mathrm{F}=6\pi \eta \mathrm{rv}\Rightarrow \eta =\frac{\mathrm{F}}{6\pi \mathrm{rv}}\\ \\ & [\eta ]=\frac{\left[{\mathrm{MLT}}^{-2}\right]}{[\mathrm{L}]\left[{\mathrm{LT}}^{-1}\right]}=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-1}\right]\\ \\ & \mathrm{E}=\mathrm{h}v\Rightarrow \mathrm{h}=\frac{\mathrm{E}}{v}=\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]}{\left[{\mathrm{T}}^{-1}\right]}=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}\right]\end{array}$

Work function has same dimension as that of energy, so $[\varphi ]=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]$

Correct answer is (B)

(A) Planck's constant

$\begin{array}{rl}& \mathrm{h}v=\mathrm{E}\\ \\ & \mathrm{h}=\frac{\mathrm{E}}{v}=\frac{{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-2}}{{\mathrm{T}}^{-1}}={\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-1}...(\text{III})\end{array}$

(B) $\mathrm{E}=\mathrm{qV}$

$V=\frac{E}{q}=\frac{M^1{L}^2{T}^{-2}}{A^1{T}^1}=M^1{L}^2{T}^{-3}{A}^{-1}\text{ (IV) }$

(C) $\varphi ($ work function $)=$ energy

$={\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-2}....(\text{I})$

(D) Momentum (p) = F.t

$\begin{array}{rl}& ={\mathrm{M}}^1{\mathrm{L}}^1{\mathrm{T}}^{-2}{\mathrm{T}}^1\\ \\ & ={\mathrm{M}}^1{\mathrm{L}}^1{\mathrm{T}}^{-1}....(\text{II})\end{array}$