The dimensions of Planck's constant equals to that of

The correct option is angular momentum.

To find which quantity has the same dimensions as Planck's constant, let's derive the dimensional formulas for both Planck's constant and the given options.

1. Dimensions of Planck's Constant ($h$):

According to Planck's quantum theory, the energy ($E$) of a photon is directly proportional to its frequency ($\nu$):

$E=h\nu$

Rearranging the equation to solve for Planck's constant ($h$):

$h=\frac{E}{\nu }$

The dimensional formula of energy ($E$) is:

$\left[E\right]=\left[{\mathrm{ML}}^2{T}^{-2}\right]$

The dimensional formula of frequency ($\nu$, which is reciprocal of time period) is:

$\left[\nu \right]=\left[T^{-1}\right]$

Substituting these dimensions into the equation for $h$:

$\left[h\right]=\frac{\left[{\mathrm{ML}}^2{T}^{-2}\right]}{\left[T^{-1}\right]}=\left[{\mathrm{ML}}^2{T}^{-1}\right]$

2. Dimensions of Angular Momentum ($L$):

Angular momentum is defined as the product of the position vector ($r$) and linear momentum ($p$):

$L=r\times p=r\times \left(mv\right)$

Where:
- $m$ is mass, with dimension $\left[M\right]$
- $v$ is velocity, with dimension $\left[{\mathrm{LT}}^{-1}\right]$
- $r$ is distance, with dimension $\left[L\right]$

Substituting these dimensions into the formula for angular momentum:

$\left[L\right]=\left[L\right]\times \left[M\right]\times \left[{\mathrm{LT}}^{-1}\right]=\left[{\mathrm{ML}}^2{T}^{-1}\right]$

Comparing the two results, we find that the dimensions of Planck's constant and angular momentum are identical, both being equal to $\left[{\mathrm{ML}}^2{T}^{-1}\right]$.

3. Dimensions of the Other Options:

- Energy: $\left[{\mathrm{ML}}^2{T}^{-2}\right]$
- Momentum: $\left[{\mathrm{MLT}}^{-1}\right]$
- Power: $\left[{\mathrm{ML}}^2{T}^{-3}\right]$