$$\eqalign{ & 1\,{\sec ^ * } = 5\,\sec ,1k{g^ * } = 20\,kg,1\,{m^ * } = 10\,m \cr & 1\,watt = 1\,kg\,{m^2}{\sec ^{ - 3}} \cr & = \frac{1}{{20}}k{g^ * }{\left( {\frac{1}{{10}}{m^ * }} \right)^2}{\left( {\frac{1}{5}{{\sec }^ * }} \right)^{ - 3}} \cr & = \frac{1}{{20}} \times \frac{1}{{100}} \times 125\,k{g^ * }{m^{ * 2}}{\sec ^{ * - 3}} = \frac{1}{{16}}wat{t^ * } \cr} $$

Electric flux $${\phi _E} = \overrightarrow E \,.\,\overrightarrow S $$
$$\therefore $$ Dimensionally $${\phi _E} \ne E$$

Use principle of homogeneity.

Except Gravitational constant other are ratios of same dimensional quantity

$$\eqalign{ & \frac{{{\text{Planck's constant}}}}{{{\text{Moment of inertia}}}} = \frac{{\frac{{2\pi I\omega }}{n}}}{I}\,\,\left[ {{\text{As}}\,\,\frac{{nh}}{{2\pi }} = I\omega } \right] \cr & = \frac{{2\pi I\left( {2\pi f} \right)}}{{nI}} \cr & = \left( {\frac{{4{\pi ^2}}}{n}.f} \right) \cr & = \left[ {{T^{ - 1}}} \right] \cr} $$

Dimensionally $$v = \frac{a}{t} \Rightarrow a = L{T^{ - 1}} \times T = L$$
also dimensionally $$v = b{t^3} \Rightarrow b = \frac{{L{T^{ - 1}}}}{{{T^3}}} = L{T^{ - 4}}$$

As we know that emf induced in the inductors is given by $$e = L\frac{{di}}{{dt}}$$
$$L = \frac{{edt}}{{di}} = \frac{W}{q} \cdot \frac{{dt}}{{di}} = \frac{{\left[ {M{L^2}{T^{ - 2}}} \right]\left[ T \right]}}{{\left[ {AT} \right]\left[ A \right]}} = \left[ {M{L^2}{T^{ - 2}}{A^{ - 2}}} \right]$$

$$\eqalign{ & Y = \frac{{4mgL}}{{\pi {D^2}\ell }} \cr & = \frac{{4 \times 1 \times 9.8 \times 2}}{{\pi {{\left( {0.4 \times {{10}^{ - 3}}} \right)}^2} \times \left( {0.8 \times {{10}^{ - 3}}} \right)}} \cr & = 2.0 \times {10^{11}}\,N/{m^2} \cr & {\text{Now, }}\frac{{\Delta Y}}{Y} = \frac{{2\Delta D}}{D} + \frac{{\Delta \ell }}{\ell } \cr & \left[ {\because {\text{ the value of }}m,{\text{ }}g{\text{ and }}L{\text{ are exact}}} \right] \cr & = 2 \times \frac{{0.01}}{{0.4}} + \frac{{0.05}}{{0.8}} \cr & = 2 \times 0.025 + 0.0625 \cr & = 0.05 + 0.0625 \cr & = 0.1125 \cr & \Rightarrow \Delta Y = 2 \times {10^{11}} \times 0.1125 \cr & = 0.225 \times {10^{11}} \cr & = 0.2 \times {10^{11\,}}N/{m^2} \cr} $$
Note: We can also take value of $$y$$ from options given without calculating it as it is same in all options.
$$\therefore Y = \left( {2 \pm 0.2} \right) \times {10^{11}}N/{m^2}$$

$$\eqalign{ & {\text{Momentum}} = m\nu = \left[ {ML{T^{ - 1}}} \right] \cr & {\text{Planck's constant}},{\mkern 1mu} \,h = \frac{E}{\nu } = \frac{{M{L^2}{T^{ - 2}}}}{{{T^{ - 1}}}} = M{L^2}{T^{ - 1}} \cr} $$

30 Divisions of Vernier scale coincide with 29 divisions of main scales.
Therefore 1 V.S.D. $$ = \frac{{29}}{{30}}$$   M.S.D.
Least count = 1 M.S.D. $$-$$ 1 V.S.D.
= 1 M.S.D. $$ - \frac{{29}}{{30}}$$   M.S.D.
= $$\frac{1}{{30}}$$ M.S.D. = $$\frac{1}{{30}} \times {0.5^ \circ }$$
= 1 minute