Least count $$ = \frac{{0.5}}{{50}} = 0.01\,mm$$
Zero error = 5 $$× L.C. = 5 × 0.01\,mm = 0.05\,mm$$
Diameter of ball = [Reading on main scale] + [Reading on circular scale $$× L.C.$$  ] $$-$$ Zero error
= 0.5 × 2 + 25 × 0.01 $$-$$ 0.05
= 1.20 $$mm$$

Given, $$R = \frac{{\rho \ell }}{{\pi {r^2}}},$$   then
$$\eqalign{ & \frac{{\Delta R}}{R} \times 100 \cr & = \frac{{\Delta \rho }}{\rho } \times 100 + \frac{{\Delta \ell }}{\ell } \times 100 + 2\frac{{\Delta r}}{r} \times 100 \cr & = 1\% + 2\% + 2 \times 3\% \cr & = 9\% \cr} $$

$$\eqalign{ & {\text{From Stokes law}}\,F = 6\pi \eta rv \Rightarrow \eta = \frac{F}{{6\pi rv}} \cr & \therefore \eta = \frac{{ML{T^{ - 2}}}}{{\left[ L \right]\left[ {L{T^{ - 1}}} \right]}} \Rightarrow \eta = \left[ {M{L^{ - 1}}{T^{ - 1}}} \right] \cr} $$

The mean value of refractive index,
$$\eqalign{ & \mu = \frac{{1.34 + 1.38 + 1.32 + 1.36}}{4} = 1.35\,{\text{and}} \cr & \Delta \mu = \frac{{\left| {\left( {1.35 - 1.34} \right)} \right| + \left| {\left( {1.35 - 1.38} \right)} \right| + \left| {\left( {1.35 - 1.32} \right)} \right| + \left| {\left( {1.35 - 1.36} \right)} \right|}}{4} = 0.02 \cr & {\text{Thus}}\,\,\frac{{\Delta \mu }}{\mu } \times 100 = \frac{{0.02}}{{1.35}} \times 100 = 1.48 \cr} $$

Least count of screw gauge $$ = \frac{{0.5}}{{50}}\,mm = 0.01\,mm$$
$$\therefore $$ Reading = [Main scale reading + circular scale reading × L.C. ] $$-$$ (zero error)
= [3 + 35 × 0.01] $$-$$ ($$-$$ 0.03) = 3.38 $$mm$$

Given, $$f = \frac{p}{{2\ell }}\sqrt {\frac{T}{\mu }} $$    Taking log of both sides
$$\log f = \log \left( {\frac{p}{2}} \right) - \log \ell + \frac{1}{2}\log T - \frac{1}{2}\log \mu $$
Differentiating partially on both sides,
$$\eqalign{ & \frac{{df}}{f} = 0 - \frac{{d\ell }}{\ell } + \frac{1}{2}\frac{{dT}}{T} - \frac{1}{2}\frac{{d\mu }}{\mu } \cr & {\text{or}}\,\,\frac{{df}}{f} \times 100 = \left( { - \frac{{d\ell }}{\ell } \times 100} \right) + \left( {\frac{1}{2}\frac{{dT}}{T} \times 100} \right) - \left( {\frac{1}{2}\frac{{d\mu }}{\mu } \times 100} \right) \cr & = \left( { - 1} \right) + \frac{1}{2}\left( { - 2} \right) - \frac{1}{2}\left( 4 \right) \cr & = - 1 - 1 - 2 \cr & = - 4\% \cr} $$

$$\eqalign{ & T = {\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]^a}{\left[ {{L^{ - 3}}M} \right]^b}{\left[ {M{T^{ - 2}}} \right]^c} \cr & \Rightarrow b + a + c = 0, - a - 3b = 0, - 2a - 2c = 1 \cr & \Rightarrow a = - \frac{3}{2},b = \frac{1}{2},c = 1 \cr} $$

$$\eqalign{ & \left( {\text{A}} \right){\text{Force}} = {\text{Mass}} \times {\text{acceleration}} \cr & {\text{or}}\,F = ma \cr & = \left[ M \right]\left[ {{\text{L}}{{\text{T}}^{ - 2}}} \right] = \left[ {ML{T^{ - 2}}} \right] \cr & {\text{Torque}} = {\text{Moment}}\,{\text{of}}\,{\text{inertia}} \times {\text{angular}}\,{\text{acceleration}} \cr & {\text{or}}\,\tau = I \times \alpha = \left[ {M{L^2}} \right]\left[ {{T^{ - 2}}} \right] = \left[ {M{L^2}\;{T^{ - 2}}} \right] \cr & \left( {\text{B}} \right){\text{Work}} = {\text{Force}} \times {\text{displacement}} \cr & {\text{or }}W = F \times d = \left[ {ML{T^{ - 2}}} \right]\left[ L \right] = \left[ {M{L^2}\;{T^{ - 2}}} \right] \cr & {\text{Energy }} = \frac{1}{2} \times {\text{ mass }} \times {\text{ }}{\left( {{\text{velocity}}} \right)^2} \cr & {\text{or}}\,K = \frac{1}{2}m{v^2} = \left[ M \right]{\left[ {L{T^{ - 1}}} \right]^2} = \left[ {M{L^2}{T^{ - 2}}} \right] \cr & \left( {\text{C}} \right)\,{\text{Force as discussed above}} \cr & \left[ F \right] = \left[ {ML{T^{ - 2}}} \right] \cr & {\text{Impulse}} = {\text{ Force}} \times {\text{time - interval}} \cr & \therefore \left[ I \right] = \left[ {ML{T^{ - 2}}} \right]\left[ T \right] = \left[ {ML{T^{ - 1}}} \right] \cr & \left( {\text{D}} \right)\,{\text{Linear}}\,{\text{momentum}} = {\text{Mass}} \times {\text{velocity}} \cr & {\text{or}}\,p = mv \cr & \therefore \left[ p \right] = \left[ M \right]\left[ {L{T^{ - 1}}} \right] = \left[ {ML{T^{ - 1}}} \right] \cr & {\text{Angular momentum}} = {\text{Moment of inertia}} \times {\text{angular velocity}} \cr & {\text{or,}}\,\left[ L \right] = \left[ I \right] \times \left[ \omega \right] \cr & \therefore \left[ L \right] = \left[ {M{L^2}} \right]\left[ {{T^{ - 1}}} \right] = \left[ {M{L^2}{T^{ - 1}}} \right] \cr} $$
Hence, we observe that choice (B) is correct.
NOTE
In this problem, the momentum of inertia and impulse are denoted by same symbol $$I.$$

As given, $$P = \frac{{{a^3}{b^2}}}{{cd}}$$
$$\eqalign{ & \therefore \frac{{\Delta P}}{P} \times 100 = \left( {\frac{{3\Delta a}}{a} + \frac{{2\Delta b}}{b} + \frac{{\Delta c}}{c} + \frac{{\Delta d}}{d}} \right) \times 100 \cr & = 3\frac{{\Delta a}}{a} \times 100 + 2\frac{{\Delta b}}{b} \times 100 + \frac{{\Delta c}}{c} \times 100 + \frac{{\Delta d}}{d} \times 100 \cr & = 3 \times 1 + 2 \times 2 + 3 + 4 \cr & = 3 + 4 + 3 + 4 = 14\% \cr} $$

Unitless quantity.