If y is force and x is velocity, then dimensions of $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}$$ are A. ML-1T-2 B. M2L-2T-2 C. ML-1T0 D. M2L-2T-4

The correct answer is $\boxed{\text{C}}$, ML-1T0.

The dimensions of force are MLT-2, and the dimensions of velocity are LT-1. Therefore, the dimensions of $\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}$ are $\frac{{M{L^2}T^{-2}}}{{L^2T^{-2}}} = ML^{-1}T^0$.

Here is a brief explanation of each option:

• Option A: MLT-2 / LT-2 = ML-1T-1. This is not the correct answer because the dimensions of the numerator and denominator are not the same.
• Option B: M2L-2T-2. This is not the correct answer because the dimensions of the numerator and denominator are not the same.
• Option C: ML-1T0. This is the correct answer because the dimensions of the numerator and denominator are the same.
• Option D: M2L-2T-4. This is not the correct answer because the dimensions of the numerator and denominator are not the same.