Question

The ratio of dimensions of Planck’s constant and that of inertia has the dimension of?
(A). Frequency
(B). Velocity
(C). Angular momentum
(D). Time

Answer 1

- Hint: Dimensional formula of a physical quantity can easily be calculated if the student knows the correct formula. The formula for Planck’s constant is \[h=\dfrac{E}{\nu }\] ,and for moment of inertia is \[I=M{{R}^{2}}\]

Complete step-by-step solution -

Firstly, we will calculate the dimensional formula of each physical quantity asked in the question.
Formula for Planck’s constant \[h\] is:
\[h=\dfrac{E}{\nu }\]
On writing the dimensional formula we get,
\[h=\dfrac{\left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]}{\left[ {{M}^{0}}{{L}^{0}}{{T}^{-1}} \right]}\]
\[h=\left[ {{M}^{1}}{{L}^{2}}{{T}^{-1}} \right]\]
Now, formula for Moment of inertia \[I\] is:
\[I=M{{R}^{2}}\]
On writing the dimensional formula we get,
\[I=\left[ {{M}^{1}}{{L}^{2}}{{T}^{0}} \right]\]
Now, we take the ratio of dimensional formulas of \[h\] and \[I\]
We get,
\[\begin{align}
  & \dfrac{h}{I}=\dfrac{\left[ {{M}^{1}}{{L}^{2}}{{T}^{-1}} \right]}{\left[ {{M}^{1}}{{L}^{2}}{{T}^{0}} \right]} \\
 & \dfrac{h}{I}=\left[ {{M}^{0}}{{L}^{0}}{{T}^{-1}} \right] \\
\end{align}\]
Here \[\left[ {{M}^{0}}{{L}^{0}}{{T}^{-1}} \right]\] is dimensionally equal to the formula of frequency
So, the correct answer is A. Frequency

Additional Information:
Principle of homogeneity of dimensions comes in very handy while solving dimensional formula questions. According to the principle the powers of M,L,T on either side of the equation are always equal.
The SI units and Dimensional formula of some important physical quantities to remember are:
Work, Energy of all kinds = \[J,[{{M}^{1}}{{L}^{2}}{{T}^{-2}}]\]
Power =\[W,[{{M}^{1}}{{L}^{2}}{{T}^{-3}}]\]
Planck’s Constant (h) = \[Js,[{{M}^{1}}{{L}^{2}}{{T}^{-1}}]\]
Angular displacement (\[\theta\])=$rad,[{{M}^{0}}{{L}^{0}}{{T}^{0}}]$.
Angular velocity (\[\omega\])=\[rad{{s}^{-1}}[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\]
Force constant ($\dfrac{\text{force}}{\text{displacement}}$) = $N{{m}^{-1}},\left[ {{M}^{1}}{{L}^{0}}{{T}^{-2}} \right]$
Coefficient of elasticity ($\dfrac{\text{stress}}{\text{strain}}$) = $N{{m}^{-2}},\left[ {{M}^{1}}{{L}^{-1}}{{T}^{-2}} \right]$
Angular frequency \[(\omega )=,rad{{s}^{-1}}[{{M}^{0}}{{L}^{0}}{{T}^{-1}}]\]
Angular momentum \[I\omega =kg{{m}^{2}}{{s}^{-1}}[{{M}^{1}}{{L}^{2}}{{T}^{-1}}]\]

Note: While solving dimensional formula questions students must note that every physical quantity must be expressed in its absolute units only. The MKS (meter- kilogram-second) system of units should be used instead of CGS (centimetre-gram-second) to get answers in the appropriate SI units.