If speed V, area A and force F are chosen as fundamental units, then the dimension of Young's modulus will be:

We know that Young’s modulus $$Y$$ is defined as the ratio of stress to strain. Because strain is a pure number (dimensionless), the dimension of $$Y$$ is the same as the dimension of stress.

First, let us recall that stress is force per unit area, so

$$\text{Stress}=\dfrac{\text{Force}}{\text{Area}}.$$

In the ordinary $$M\!L\!T$$ (mass-length-time) system the dimensional formula of force is $$M\,L\,T^{-2},$$ while the dimensional formula of area is $$L^{2}.$$ Therefore, the dimensional formula of Young’s modulus in the $$M\!L\!T$$ system becomes

$$[Y]=\dfrac{M\,L\,T^{-2}}{L^{2}}=M\,L^{-1}\,T^{-2}.$$

Now the question asks us to re-express this same quantity using a new set of fundamental units: speed $$V,$$ area $$A$$ and force $$F.$$ We write the unknown dimensional formula as

$$[Y]=F^{\,a}\,A^{\,b}\,V^{\,c},$$

where $$a,\,b,\,c$$ are the exponents we must determine.

Next we translate each new fundamental unit back into the familiar $$M\!L\!T$$ language:

$$[F]=M\,L\,T^{-2},\qquad [A]=L^{2},\qquad [V]=L\,T^{-1}.$$

Substituting these into $$F^{\,a}\,A^{\,b}\,V^{\,c}$$ gives

$$F^{\,a}\,A^{\,b}\,V^{\,c}=\left(M\,L\,T^{-2}\right)^{a}\left(L^{2}\right)^{b}\left(L\,T^{-1}\right)^{c}=M^{a}\,L^{\,a+2b+c}\,T^{-2a-c}.$$

This result must match the already known $$M\!L\!T$$ exponents of Young’s modulus, namely $$M^{1}\,L^{-1}\,T^{-2}.$$ Hence we equate exponents term by term:

Mass exponent: $$a=1.$$(1)

Length exponent: $$a+2b+c=-1.$$(2)

Time exponent: $$-2a-c=-2.$$(3)

Using equation (1) $$a=1$$ inside equation (3) gives

$$-2(1)-c=-2\;\;\Longrightarrow\;\;-2-c=-2\;\;\Longrightarrow\;\;c=0.$$

With $$a=1$$ and $$c=0,$$ substitute into equation (2):

$$1+2b+0=-1\;\;\Longrightarrow\;\;2b=-2\;\;\Longrightarrow\;\;b=-1.$$

Thus we have determined

$$a=1,\qquad b=-1,\qquad c=0.$$

Putting these values back, the dimensional formula of Young’s modulus in the $$F\!A\!V$$ system is

$$[Y]=F^{1}\,A^{-1}\,V^{0}=F\,A^{-1}.$$

This matches option D, which is written as $$FA^{-1}V^{0}.$$

Hence, the correct answer is Option D.