We know,

Young's modulus (Y) = $${{{F \over A}} \over {{{\Delta l} \over l}}}$$

$$ \therefore $$ [Y] = $${{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ {{L^2}} \right]}}$$ = [ ML^-1T^-2]

Let [Y] = [V]^x [A]^y [F]^z

$$ \therefore $$ [ ML^-1T^-2] =

[LT^-1]^x [LT^-2]^y [MLT^-2]^z

$$ \Rightarrow $$ [ ML^-1T^-2] =

[ M^z L^{x + y + z} T^{-x -2y - 2z}

For dimensional balance, the dimension on both sides should be same.

So, z = 1

x + y + z = -1

$$ \Rightarrow $$ x + y = -2 ........(1)

and -x -2y - 2z = -2

$$ \Rightarrow $$ x + 2y = 0 ...........(2)

By solving those two equations we get,

x = -4 and y = 2

$$ \therefore $$ [Y] = V^$$-$$4A²F¹