If J is the angular momentum and E is the kinetic energy, then `(J^2)/E` has the dimensions of

Options

A. Moment of Inertia
B. Power
C. Angular velocity
D. Impulse

Answer

To solve the problem of finding the dimensions of \(\frac{J^2}{E}\), where \(J\) is angular momentum and \(E\) is kinetic energy, we will follow these steps:

### Step 1: Determine the Dimensional Formula for Angular Momentum \(J\)

The formula for angular momentum is given by:

\[ J = mvr \]

where:
- \(m\) is mass,
- \(v\) is velocity,
- \(r\) is the radius (or distance).

The dimensions for each of these quantities are:
- Mass \(m\) has the dimension \([M]\).
- Velocity \(v\) has the dimension \([L][T]^{-1}\) (length per time).
- Radius \(r\) has the dimension \([L]\).

Thus, the dimensional formula for angular momentum \(J\) is:

\[ [J] = [M][L][L][T]^{-1} = [M][L^2][T]^{-1} \]

### Step 2: Find the Dimensional Formula for \(J^2\)

Now, we need to find the dimensions of \(J^2\):

\[ [J^2] = ([M][L^2][T]^{-1})^2 = [M^2][L^4][T^{-2}] \]

### Step 3: Determine the Dimensional Formula for Kinetic Energy \(E\)

The kinetic energy \(E\) is given by the formula:

\[ E = \frac{1}{2}mv^2 \]

The dimensions for kinetic energy are:

\[ [E] = [M][(L[T]^{-1})^2] = [M][L^2][T^{-2}] \]

### Step 4: Calculate the Dimensions of \(\frac{J^2}{E}\)

Now we can find the dimensions of \(\frac{J^2}{E}\):

\[ \frac{J^2}{E} = \frac{[M^2][L^4][T^{-2}]}{[M][L^2][T^{-2}]} \]

When we simplify this expression, we get:

\[ \frac{J^2}{E} = [M^{2-1}][L^{4-2}][T^{-2 - (-2)}] = [M^1][L^2][T^0] = [M][L^2] \]

### Step 5: Identify the Resulting Dimension

The resulting dimension \([M][L^2]\) corresponds to the dimension of moment of inertia, which is defined as:

\[ I = m \cdot r^2 \]

Thus, the dimension of \(\frac{J^2}{E}\) is:

\[ \frac{J^2}{E} \text{ has the dimension of moment of inertia } [M][L^2]. \]

### Final Answer

The correct answer is that \(\frac{J^2}{E}\) has the dimensions of **moment of inertia**.

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