1. Write the dimensional formulas in MLT system

   Young’s modulus has the same dimension as stress: $Y = \frac{\text{force}}{\text{area}}$ so $[Y] = \frac{[F]}{[A]} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}$

   The chosen fundamental quantities are:

   • Speed: $[V] = LT^{-1}$
   • Area: $[A] = L^2$
   • Force: $[F] = MLT^{-2}$

2. Assume Young’s modulus in terms of the new fundamental units

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

   Substituting dimensions: $ML^{-1}T^{-2} = (MLT^{-2})^x (L^2)^y (LT^{-1})^z$

3. Expand powers

   Right-hand side becomes: $M^x L^{x+2y+z} T^{-2x-z}$

   Equating powers of $M, L, T$:

   • For $M$: $x=1$
   • For $T$: $-2x-z=-2$
   • For $L$: $x+2y+z=-1$

4. Solve for $x,y,z$

   From $x=1$

   Using time equation: $-2(1)-z=-2 \Rightarrow -2-z=-2 \Rightarrow z=0$

   Using length equation: $1+2y+0=-1$ $2y=-2 \Rightarrow y=-1$

5. Final dimension in new system

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

   So the dimension of Young’s modulus is: $\boxed{FA^{-1}V^0}$

6. Option check

   • A: $FA^{-1}V^0$ ✅
   • B: $FA^2V^{-1}$ ❌
   • C: $FA^2V^{-2}$ ❌
   • D: $FA^2V^{-3}$ ❌

Therefore, the correct option is A.