Physics - Unit & Dimension Question with Solution | TestHub

The dimensions of the required quantities can be written as

$$\begin{gathered} E=h\nu \\ \Rightarrow \left[h\right]=\frac{\left[E\right]}{\left[\nu \right]} \\ \Rightarrow \left[h\right]=\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]}{\left[{\mathrm{T}}^{-1}\right]} \\ =\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}\right] \end{gathered}$$

$\begin{array}{rcl}\left[L\right]& =& \left[mvr\right]\\ & =& \left[\mathrm{M}\right]\left[{\mathrm{LT}}^{-1}\right]\left[\mathrm{L}\right]\\ & =& \left[{\mathrm{ML}}^2 {\mathrm{T}}^{-1}\right]\end{array}$

$\begin{array}{rcl}\left[p\right]& =& \left[mv\right]\\ & =& \left[\mathrm{M}\right]\left[{\mathrm{LT}}^{-1}\right]\\ & =& \left[{\mathrm{ML}}^1 {\mathrm{T}}^{-1}\right]\end{array}$

$\begin{array}{rcl}\left[\tau \right]& =& \left[Fr\right]\\ & =& \left[{\mathrm{MLT}}^{-2}\right]\left[\mathrm{L}\right]\\ & =& \left[{\mathrm{ML}}^2 {\mathrm{T}}^{-2}\right]\end{array}$

where, $h$  is Planck's constant, $L$ is angular momentum, $p$ is linear momentum and $\tau$ is moment of force.

From the above analysis, it can be concluded that Statement I is true, but statement II is false.