If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's modulus will be:

We begin by recalling the basic definition: Young’s modulus $$Y$$ equals the ratio of normal stress to longitudinal strain. Since strain is dimensionless, the dimension of $$Y$$ is the same as that of stress, that is, force per unit area.

So, in the ordinary $$M\!L\!T$$ system we would write

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

However, in the present problem the fundamental (independent) quantities are not $$M,\;L,\;T$$ but

• speed $$V$$    ($$V=L\,T^{-1}$$),
• acceleration $$A$$    ($$A=L\,T^{-2}$$),
• force $$F$$    ($$F=M\,L\,T^{-2}$$).

We must therefore express the dimension $$M L^{-1} T^{-2}$$ in the form

$$V^{\alpha}\,A^{\beta}\,F^{\gamma}.$$

First we translate each of the three fundamental quantities into $$M,\;L,\;T$$ exponents:

$$V^{\alpha}=L^{\alpha}T^{-\alpha},$$
$$A^{\beta}=L^{\beta}T^{-2\beta},$$
$$F^{\gamma}=M^{\gamma}L^{\gamma}T^{-2\gamma}.$$

Multiplying these three contributions, the overall exponents of $$M,\;L,\;T$$ become

$$M:\ \gamma,$$
$$L:\ \alpha+\beta+\gamma,$$
$$T:\ -\alpha-2\beta-2\gamma.$$

We must equate these to the target exponents in $$M^{1}L^{-1}T^{-2}$$. Hence the system of algebraic equations is

$$\gamma = 1,$$
$$\alpha+\beta+\gamma = -1,$$
$$-\alpha-2\beta-2\gamma = -2.$$

Substituting $$\gamma=1$$ into the second equation gives

$$\alpha+\beta+1=-1\;\;\Longrightarrow\;\;\alpha+\beta=-2.$$

Substituting $$\gamma=1$$ into the third equation yields

$$-\alpha-2\beta-2=-2\;\;\Longrightarrow\;\;-\alpha-2\beta=0\;\;\Longrightarrow\;\;\alpha+2\beta=0.$$

We now solve the simultaneous equations

$$\alpha+\beta=-2,$$
$$\alpha+2\beta=0.$$

Subtracting the first from the second gives

$$(\alpha+2\beta)-(\alpha+\beta)=0-(-2)\;\;\Longrightarrow\;\;\beta=2.$$

Substituting $$\beta=2$$ back into $$\alpha+\beta=-2$$ gives

$$\alpha+2=-2\;\;\Longrightarrow\;\;\alpha=-4.$$

Collecting the exponents, we have

$$\alpha=-4,\quad\beta=2,\quad\gamma=1.$$

Therefore the dimensional formula of Young’s modulus in terms of $$V,\;A,\;F$$ is

$$[Y]=V^{-4}\,A^{2}\,F^{1}=V^{-4}A^{2}F.$$

Among the given alternatives, this corresponds to Option D.

Hence, the correct answer is Option D.