Question

The ratio of the dimensions of Plancks constant and that of the moment of inertia is the dimension of

• A. frequency
• B. velocity
• C. angular momentum
• D. time

Answer 1

Answer: (A)

$E=hv$
$\Rightarrow h=$ Plancks constant $=\frac{E}{V}$
$\therefore [h]=\frac{[E]}{[v]}=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[T^{-1}\right]}$
$=\left[M{L}^2{T}^{-1}\right]$
and $I=$ moment of inertia $=M{R}^2$
$\Rightarrow [I]=[M]\left[L^2\right]=\left[M{L}^2\right]$
Hence, $\frac{[h]}{[I]}=\frac{\left[M{L}^2{T}^{-1}\right]}{\left[M{L}^2\right]}=\left[T^{-1}\right]$
$=\frac{1}{[T]}=$ dimensions of frequency
Alternative: $\frac{h}{I}=\frac{E/v}{I}$
$=\frac{E\times T}{I}=\frac{\left(kg-m^2/s^2\right)\times s}{\left(kg-m^2\right)}$
$=\frac{1}{s}=\frac{1}{\text{ time }}=$ frequency
Thus, dimensions of $\frac{h}{I}$ is same as of frequency.