Least count $$ = \frac{{0.5}}{{50}} = 0.01\,mm$$

Let dimensions of length is related as,
$$\eqalign{ & L = {\left[ c \right]^x}{\left[ G \right]^y}{\left[ {\frac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right]^z} \Rightarrow \frac{{{e^2}}}{{4\pi {\varepsilon _0}}} = M{L^3}{T^{ - 2}} \cr & L = {\left[ {L{T^{ - 1}}} \right]^x}{\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]^y}{\left[ {M{L^3}{T^{ - 2}}} \right]^z} \cr & [L] = \left[ {{L^{x + 3y + 3z}}\,{M^{ - y + z}}\,{T^{ - x - 2y - 2z}}} \right] \cr} $$
Comparing both sides
$$z = y = \frac{1}{2},x = - 2\,\,{\text{Hence,}}\,L = {c^{ - 2}}{\left[ {G \cdot \frac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right]^{\frac{1}{2}}}$$

As we know, time period of a simple pendulum
$$T = 2\pi \sqrt {\frac{L}{g}} \Rightarrow g = \frac{{4{\pi ^2}L}}{{{T^2}}}$$
The maximum percentage error in $$g$$
$$\eqalign{ & \frac{{\Delta g}}{g} \times 100 = \frac{{\Delta L}}{L} \times 100 + 2\left( {\frac{{\Delta T}}{T} \times 100} \right) \cr & = 2\% + 2\left( {3\% } \right) = 8\% \cr} $$

$$\eqalign{ & {\text{As}}\,f = C{m^x}{k^y} \cr & \therefore \left( {{\text{Dimension of}}\,f} \right) = C{\left( {{\text{dimension of }}m} \right)^x} \times {\left( {{\text{dimensions of }}k} \right)^y} \cr & \left[ {{T^{ - 1}}} \right] = C{\left[ M \right]^x}{\left[ {M{T^{ - 2}}} \right]^y}\,......\left( {\text{i}} \right)\,\left( {{\text{where,}}\,k = \frac{{{\text{force}}}}{{{\text{length}}}}} \right) \cr} $$
Applying the principle of homogeneity of dimensions, we get
$$\eqalign{ & x + y = 0, - 2y = - 1\,{\text{or}}\,y = \frac{1}{2} \cr & \therefore x = - \frac{1}{2} \cr} $$

Surface tension
$$S = \frac{{{\text{Force}}}}{{{\text{Length}}}} = \frac{N}{m} = N{m^{ - 1}}$$

Torque $$\tau = r \times F$$
Dimensions of $$\tau = {\text{dimension of}}\,r \times {\text{dimension of }}F$$
$$\eqalign{ & = \left[ L \right]\left[ {ML{T^{ - 2}}} \right] \cr & = \left[ {M{L^2}{T^{ - 2}}} \right] \cr} $$

$${\text{Force}}\,F = \left[ {ML{T^{ - 2}}} \right] = \left( {10g} \right)\left( {10\,cm} \right){\left( {0.01s} \right)^{ - 2}}$$
Changing these units into $$MKS$$  system
$$\eqalign{ & F = \left( {{{10}^{ - 2}}\,kg} \right)\left( {{{10}^{ - 1}}m} \right){\left( {{{10}^{ - 1}}s} \right)^{ - 2}} \cr & = {10^{ - 1}}N \cr & = 0.1\,N \cr} $$

$$\eqalign{ & V = \frac{4}{3}\pi {r^3}; \cr & \frac{{\Delta V}}{V} \times 100 = 3\left( {\frac{{\Delta r}}{r}} \right) \times 100 = 3 \times 1\% = 3\% \cr} $$

$$\eqalign{ & {\text{Energy carried by photon is given by }}E = hv \cr & \Rightarrow h = {\text{Planc's constant}} = \frac{E}{v} \cr & \therefore \left[ h \right] = \frac{{\left[ {M{L^2}\;{T^{ - 2}}} \right]}}{{\left[ {{T^{ - 1}}} \right]}} = \left[ {M{L^2}\;{T^{ - 1}}} \right] \cr & {\text{and}}\,I = {\text{moment of inertia }} = M{R^2} \cr & \Rightarrow \left[ I \right] = \left[ {M{L^2}} \right] \cr & {\text{Hence,}}\,\frac{{\left[ h \right]}}{{\left[ I \right]}} = \frac{{\left[ {M{L^2}\;{T^{ - 1}}} \right]}}{{\left[ {M{L^2}} \right]}} = \left[ {{T^{ - 1}}} \right] \cr & = \frac{1}{{[\;T]}} = {\text{dimension of frequency}} \cr} $$
Alternative
$$\eqalign{ & \frac{h}{I} = \frac{{\frac{E}{v}}}{I} \cr & = \frac{{E \times T}}{I} = \frac{{\left( {\frac{{kg - {m^2}}}{{{s^2}}}} \right) \times s}}{{\left( {kg - {m^2}} \right)}} \cr & = \frac{1}{{\;s}} = \frac{1}{{{\text{ time }}}} = {\text{frequency}} \cr} $$
Thus, dimensions of $$\frac{h}{I}$$ is same as that of frequency.

Here $$A,B$$  and $$C$$ must have same unit as of electric field.
$$\eqalign{ & {\text{So,}}\,\frac{A}{{{x^3}}} = \frac{N}{C} \cr & \Rightarrow \frac{A}{{{m^3}}} \equiv \frac{N}{C} \cr & \Rightarrow A \equiv \frac{{N - {m^3}}}{C} \cr & {\text{and}}\,By = \frac{V}{m} \cr & \Rightarrow Bm \equiv \frac{V}{{{m^2}}} \cr & \Rightarrow B \equiv \frac{V}{{{m^2}}} \cr & {\text{and}}\,C{z^2} \equiv \frac{N}{C} \cr & \Rightarrow C{m^2} \equiv \frac{N}{C} \cr & \Rightarrow C \equiv \frac{N}{{C - {m^2}}} \cr} $$