$$\eqalign{ & {\text{Moment of Inertia,}}\,\,I = M{r^2} \cr & \left[ I \right] = \left[ {M{L^2}} \right] \cr & {\text{Moment of force,}}\,\,\,\overrightarrow \tau = \overrightarrow r \times \overrightarrow F \cr & \left[ {\overrightarrow \tau } \right] = \left[ L \right]\left[ {ML{T^{ - 2}}} \right] = \left[ {M{L^2}{T^{ - 2}}} \right] \cr} $$

Viscous force on a sphere of radius $$r$$ is
$$\eqalign{ & F = 6\pi \eta rv \Rightarrow \eta = \frac{F}{{6\pi rv}} \cr & \left[ \eta \right] = \frac{{\left[ F \right]}}{{\left[ r \right]\left[ v \right]}} = \frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ L \right]\left[ {L{T^{ - 1}}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 1}}} \right] \cr} $$
NOTE
The above expression is obtained from Stokes’ law.

We have, $$c = \frac{1}{{\sqrt {{\mu _0}{ \in _0}} }} \Rightarrow {\mu _0} = \frac{1}{{{ \in _0}{c^2}}}$$
∴ units of $${\mu _0} = \frac{1}{{{\text{units}}\,{\text{of}}\,{ \in _0}{c^2}}}$$
$$\eqalign{ & = \frac{1}{{{N^{ - 1}}{C^2}{m^{ - 2}}{{\left( {m{s^{ - 1}}} \right)}^2}}} = N{s^2}{C^{ - 2}} \cr & = \left( {kgm{s^{ - 2}}} \right){s^2}{C^{ - 2}} \cr & = kgm{C^{ - 2}} \cr} $$

Reading $$=$$ M.S.R. $$+$$ Number of division of V.S. matching the main scale division (1 M.S.D. $$-$$ 1 V.S.D.)
$$\eqalign{ & = 5.10 + 24\left( {0.05 - \frac{{2.45}}{{50}}} \right) \cr & = 5.124\,cm \cr} $$
Option (B) is correct

$$\because $$ Reading of Vernier = Main scale reading + Vernier scale reading × least count
Main scale reading = 58.5
Vernier scale reading = 09 division
least count of Vernier = $$\frac{{{{0.5}^ \circ }}}{{30}}$$
Thus
$$\eqalign{ & R = {58.5^ \circ } + 9 \times \frac{{{{0.5}^ \circ }}}{{30}} \cr & R = {58.65^ \circ } \cr} $$

20 divisions on the Vernier scale = 16 divisions of main scale
$$\therefore $$ 1 division on the Vernier scale $$ = \frac{{16}}{{20}}$$   divisions of main scale $$ = \frac{{16}}{{20}} \times 1\,mm = 0.8\,mm$$
We know that least count $$= 1\, MSD - 1\,VSD = 1\, mm - 0.8\, mm = 0.2\ mm$$

From Biot-Savart law
$$dB = \frac{{{\mu _0}}}{{4\pi }}\frac{{Idl\sin \theta }}{{{r^2}}}$$
$$Idl = $$  current element
$$r =$$  displacement vector
$${\mu _0} = \frac{{4\pi {r^2}\left( {dB} \right)}}{{Idl\sin \theta }} = \frac{{\left[ {{L^2}} \right]\left[ {M{T^{ - 2}}{A^{ - 1}}} \right]}}{{\left[ A \right]\left[ L \right]}} = \left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]$$

$$\eqalign{ & {\text{We know that }}F = qvB \cr & \therefore B = \frac{F}{{qv}} = \frac{{ML{T^2}}}{{C \times L{T^{ - 1}}}} = M{T^{ - 1}}{C^{ - 1}} \cr} $$

$${\text{Unit of}}\,\,{\text{impulse}} \times {\text{time}}\,{\text{ = }}\left( {kg\,m{s^{ - 2}} \times s} \right) \times s = kg\,m$$

$$\eqalign{ & V = \pi {r^2}\ell \cr & \frac{{\Delta V}}{V}\% = 2\frac{{\Delta r}}{r}\% + 2\frac{{\Delta \ell }}{\ell }\% \cr & = 2 \times \frac{{0.01}}{2} \times 100\% + \frac{{0.1}}{5} \times 100\% \cr & = 3\% \cr} $$