If E,M,J and G respectively denote energy , mass angular momentum and gravitational constant then `(EJ^(2))/(M^(2)G^(2))` has the dimensions of .

Options

A. length
B. angle
C. mass
D. time

Answer

To find the dimensions of the expression \(\frac{E J^2}{M^2 G^2}\), we need to determine the dimensions of each variable involved: energy (E), mass (M), angular momentum (J), and gravitational constant (G).

### Step-by-Step Solution:

1. **Identify the dimensions of each variable:**
- **Energy (E)**: The dimension of energy is given by the formula for work, which is force times distance. The dimension of force is \(M L T^{-2}\) (mass times acceleration), thus:
\[ [E] = [F] \cdot [L] = (M L T^{-2}) \cdot (L) = M L^2 T^{-2} \]

- **Mass (M)**: The dimension of mass is simply:
\[ [M] = M^1 L^0 T^0 \]

- **Angular Momentum (J)**: The dimension of angular momentum is given by the product of moment of inertia and angular velocity. The dimension of moment of inertia is \(M L^2\) and angular velocity is \(T^{-1}\):
\[ [J] = [I] \cdot [\omega] = (M L^2) \cdot (T^{-1}) = M L^2 T^{-1} \]

- **Gravitational Constant (G)**: The gravitational constant can be derived from the formula for gravitational force:
\[ F = \frac{G m_1 m_2}{r^2} \implies G = \frac{F r^2}{m_1 m_2} \]
The dimensions of force are \(M L T^{-2}\), and distance squared is \(L^2\):
\[ [G] = \frac{[F] \cdot [L^2]}{[M]^2} = \frac{(M L T^{-2}) \cdot (L^2)}{M^2} = M^{-1} L^3 T^{-2} \]

2. **Substituting the dimensions into the expression:**
Now we substitute the dimensions into the expression \(\frac{E J^2}{M^2 G^2}\):
\[ \frac{E J^2}{M^2 G^2} = \frac{(M L^2 T^{-2}) (M L^2 T^{-1})^2}{(M^1)^2 (M^{-1} L^3 T^{-2})^2} \]

3. **Calculating \(J^2\) and \(G^2\)**:
- For \(J^2\):
\[ [J^2] = (M L^2 T^{-1})^2 = M^2 L^4 T^{-2} \]
- For \(G^2\):
\[ [G^2] = (M^{-1} L^3 T^{-2})^2 = M^{-2} L^6 T^{-4} \]

4. **Putting it all together**:
Now substituting these into the expression:
\[ \frac{(M L^2 T^{-2}) (M^2 L^4 T^{-2})}{M^2 (M^{-2} L^6 T^{-4})} \]
Simplifying the numerator:
\[ = \frac{M^3 L^6 T^{-4}}{M^2 M^{-2} L^6 T^{-4}} = \frac{M^3 L^6 T^{-4}}{L^6 T^{-4}} = M^3 \]

5. **Final dimensions**:
The final dimensions simplify to:
\[ [\frac{E J^2}{M^2 G^2}] = M^0 L^0 T^0 \]
This indicates that the expression is dimensionless.

### Conclusion:
The dimensions of \(\frac{E J^2}{M^2 G^2}\) are \(M^0 L^0 T^0\), which corresponds to a dimensionless quantity or an angle.