We know, Young's modulus, $$E = {\sigma \over \varepsilon }$$

$$ \Rightarrow E = {{{F \over A}} \over {{{\Delta L} \over L}}} = {F \over A}\left( {{L \over {\Delta L}}} \right)$$

$$ \Rightarrow [E] = {{[F]} \over {[A]}} = {{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]} \over {\left[ {{L^2}} \right]}}$$

(A) $$ \Rightarrow [E] = [{M^1}{L^{ - 1}}{T^{ - 2}}]$$

(B) we know, Torque = $r\times F$

$$[\tau ] = [r][F]$$

$$ = [L][{M^1}{L^1}{T^{ - 2}}]$$

$$[\tau ] = [{M^1}{L^2}{T^{ - 2}}]$$

(C) We know, $$F = \mu A{u \over y}$$

where, F = force, $\mu$ = coefficient of viscosity

A = area $\frac{u}{y}$ = rate of shear deformation

$$ \Rightarrow \mu = {{Fy} \over {Au}}$$

$$ \Rightarrow [\mu ] = {{[{M^1}{L^1}{T^{ - 2}}][L]} \over {[{L^2}][L{T^{ - 1}}]}}$$

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

(D) We know, $$F = {{G{m_1}{m_2}} \over {{r^2}}}$$ (Newton's law of gravitation)

$$ \Rightarrow G = {{F{r^2}} \over {{m_1}{m_2}}}$$<$$ \Rightarrow G = {{F{r^2}} \over {{m_1}{m_2}}}$$

$$ \Rightarrow [G] = {{[{M^1}{L^1}{T^{ - 2}}][{L^2}]} \over {[{M^2}]}}$$

$$ \Rightarrow [G] = [{M^{ - 1}}{L^3}{T^{ - 2}}]$$

Hence, option 4 is correct.