To determine the correct option, we compare the dimensional formula of Planck's constant with the given physical quantities:

1. Planck's Constant ($h$): From the relation $E = h\nu$, where $E$ is energy and $\nu$ is frequency: $[h] = \frac{[E]}{[\nu]} = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$ Alternatively, the unit is Joule-second ($J\cdot s$).

2. Analyze Options:

• Energy/Work: Dimensions are $[ML^2T^{-2}]$.
• Linear Momentum ($p = mv$): Dimensions are $[MLT^{-1}]$.
• Angular Momentum ($L = mvr$): Dimensions are $[M][LT^{-1}][L] = [ML^2T^{-1}]$.

3. Conclusion: The dimensions of Planck's constant $[ML^2T^{-1}]$ are identical to those of Angular momentum .