Correct answer is (C)

Moment of inertia, I = Mr²

$\therefore$ [I] = [ML²]

Moment of force, $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

$\therefore$ $[\overrightarrow{\tau }]$ = $[L][ML{T}^{-2}]$ = $[M{L}^2{T}^{-2}]$

Correct answer is (A)

From stokes law, Viscous force F = $6\pi \eta rv$

$\Rightarrow \eta =\frac{F}{6\pi rv}$

$\therefore$ $[\eta ]=\frac{[ML{T}^{-2}]}{[L][L{T}^{-1}]}$

$\Rightarrow [\eta ]=[M{L}^{-1}{T}^{-1}]$

Correct answer is (C)

The velocity of light in vacuum is

c = $\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$;

$\therefore [\frac{1}{{\mu }_0{\epsilon }_0}]$ = [c²] = [L²T^-2]

$\therefore$ Dimension of $\frac{1}{{\mu }_0{\epsilon }_0}$ = [L²T^-2]

Correct answer is (B)

Momentum = mv = [$ML{T}^{-1}$]

Planck's constant, h = $\frac{E}{v}$ = $\frac{[M{L}^2{T}^{-2}]}{[T^{-1}]}$ = $[M{L}^2{T}^{-1}]$

So Momentum and Planck's constant do not have same dimensions.

Correct answer is (A)

Work (W) = $\overrightarrow{F}.\overrightarrow{s}$ = Fs cos$\theta$

= [MLT^-2][L] = [ML²T^-2];

Torque ($\overrightarrow{\tau })=\overrightarrow{r}\times \overrightarrow{F}$ $\Rightarrow \tau =rF\sin \theta$

= [L] [MLT^-2] = [ML²T^-2]

So dimension of torque and work are same.