Question

What is the dimension of the ratio of the dimensions of Planck’s constant and that of moment of inertia?

Answer 1

Hint: Find the dimension formula of Planck’s constant by using the equation of energy of a wave i.e. $E=h\nu$, where $h$ is Planck’s constant, $\nu$ is frequency of the wave and $E$ is the energy of the wave. Then find the dimensional formula of moment of inertia. Divide them to find the ratio.

Complete step by step answer:

Planck’s constant is a universal constant used in the equation to show the relationship between the energy of a wave and its frequency i.e. $E=h\nu$ where, $h$ is Planck’s constant, $\nu$ is frequency of the wave and $E$ is the energy of the wave. So, $h={\displaystyle \frac{E}{\nu }}$. Now find the ratio of the dimensions of energy and frequency to find the dimension of Planck’s constant.
Dimension of energy is the same as the dimension of work done which is a product of force and displacement. Therefore, $\left[\text{E}\right]\text{=}\left[\text{Force}\right]\left[\text{displacement}\right]$ (dimensions are denoted by square brackets). Therefore, $\left[E\right]=\left[ML{T}^{-2}\right]\left[L\right]$, where M, L, T are the dimensions of mass, length and time respectively. The dimension of frequency is $\left[T^{-1}\right]$. Now, divide dimension of energy and dimension of frequency to get the dimension of Planck’s constant.
Therefore, $$\left[h\right]={\displaystyle \frac{\left[E\right]}{\left[\nu \right]}}={\displaystyle \frac{\left[ML{T}^{-2}\right]\left[L\right]}{\left[T^{-1}\right]}}=\left[M{L}^2{T}^{-1}\right]$$ …………. (1)
Now, find the dimension of moment of inertia. Moment of inertia of a body or a particle is the tendency of that body or particle to resist the motion while rotating. For a point size particle, it is equal to the product of its mass and the perpendicular distance of the that particle from the axis of rotation i.e. $I=M{d}^2$, where is $I$ the moment of inertia, $M$ is the mass of the particle and $d$ is the perpendicular distance as mentioned before.
Therefore, $\left[I\right]=\left[M\right]\left[d^2\right]=\left[M{L}^2\right]$ …………. (2)
Now, ratio of the dimensions of Planck’s constant and that of moment of inertia will be, ${\displaystyle \frac{\left[h\right]}{\left[I\right]}}={\displaystyle \frac{\left[M{L}^2{T}^{-1}\right]}{\left[M{L}^2\right]}}=\left[T^{-1}\right]$
As you can see, the dimension of the ratio of the dimensions of Planck’s constant and that of moment of inertia is reciprocal of time. That means the ratio has the dimension of frequency.

Note: Other than solving the given question by using dimensions, we can also solve it by using units of the physical quantities.

The unit of energy is $kg{m}^2{s}^{-2}$ .
The unit of frequency is $s^{-1}$.

Therefore, the unit of Planck’s constant will be the ratio of units of energy and frequency i.e. $kg{m}^2{s}^{-1}$.

The unit of mass is kg.
The unit of distance (length) is m.
Therefore, the unit of moment of inertia is the product of the units of mass and square of the distance i.e. $kg{m}^2$.
Now, divide the units of Planck’s constant and moment of inertia to get the ratio.
$\text{ratio = }{\displaystyle \frac{\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{-1}}}{\text{kg}{\text{m}}^{\text{2}}}}\text{ = }{\text{s}}^{\text{-1}}$
As you can see the ratio of the units of the Planck’s constant and that of moment of inertia is the unit of frequency.
Therefore, sometimes you can use units in place of dimensions for solving certain questions.