The ratio of the dimensions of plank's constant and that of the moment of inertia is the dimension of

Options

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

Explanation

To solve the problem of finding the ratio of the dimensions of Planck's constant (h) and the moment of inertia (I), we will follow these steps:

### Step 1: Determine the dimensions of Planck's constant (h)
Planck's constant relates energy (E) and frequency (ν) through the equation:
\[ E = h \nu \]
From this equation, we can express Planck's constant as:
\[ h = \frac{E}{\nu} \]

The dimensions of energy (E) are given by:
\[ [E] = [M][L^2][T^{-2}] \]
where:
- [M] = mass
- [L] = length
- [T] = time

The dimensions of frequency (ν) are:
\[ [\nu] = [T^{-1}] \]

Now substituting these into the expression for Planck's constant:
\[ [h] = \frac{[E]}{[\nu]} = \frac{[M][L^2][T^{-2}]}{[T^{-1}]} = [M][L^2][T^{-2}] \cdot [T] = [M][L^2][T^{-1}] \]

### Step 2: Determine the dimensions of moment of inertia (I)
The moment of inertia (I) is defined as:
\[ I = m r^2 \]
where:
- m = mass
- r = radius (length)

Thus, the dimensions of moment of inertia are:
\[ [I] = [M][L^2] \]

### Step 3: Calculate the ratio of the dimensions of Planck's constant to the moment of inertia
Now we can find the ratio of the dimensions of Planck's constant to the dimensions of moment of inertia:
\[ \text{Ratio} = \frac{[h]}{[I]} = \frac{[M][L^2][T^{-1}]}{[M][L^2]} \]

### Step 4: Simplify the ratio
When we simplify this ratio, we see that the mass ([M]) and length squared ([L^2]) cancel out:
\[ \text{Ratio} = [T^{-1}] \]

### Step 5: Interpret the result
The dimension [T^{-1}] corresponds to frequency, which is measured in Hertz (Hz). Therefore, the ratio of the dimensions of Planck's constant to the moment of inertia gives us the dimension of frequency.

### Final Answer
The ratio of the dimensions of Planck's constant and that of the moment of inertia is the dimension of frequency.

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