A stationary body explodes in to four identical fragments such that three of them fly mutually perpendicular to each other, each with same `KE(E_(0))`. The energy of explosion will be

Options

A. `6E_(0)`
B. `3E_(0)`
C. `4E_(0)`
D. `2E_(0)`

Explanation

To solve the problem, we need to analyze the explosion of a stationary body into four identical fragments, three of which move mutually perpendicular to each other, each with the same kinetic energy \(E_0\). We will find the total energy of the explosion.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- The body is stationary before the explosion, which means its initial momentum is zero.
- After the explosion, it breaks into four identical fragments.

2. **Momentum Conservation:**
- Since the initial momentum is zero, the total momentum after the explosion must also be zero.
- Let the momentum of the three fragments moving along the x, y, and z axes be represented as \( \vec{p_1}, \vec{p_2}, \) and \( \vec{p_3} \).
- The fourth fragment's momentum, \( \vec{p_4} \), can be expressed as:
\[ \vec{p_4} = -(\vec{p_1} + \vec{p_2} + \vec{p_3}) \]

3. **Kinetic Energy of the Fragments:**
- Each of the three fragments has a kinetic energy \(E_0\). The relationship between kinetic energy and momentum is given by:
\[ E = \frac{p^2}{2m} \]
- Therefore, for each of the three fragments:
\[ E_0 = \frac{p_1^2}{2m} = \frac{p_2^2}{2m} = \frac{p_3^2}{2m} \]
- This implies that the magnitudes of their momenta are equal:
\[ p_1 = p_2 = p_3 = p_0 \]

4. **Calculating the Fourth Fragment's Momentum:**
- The total momentum of the three fragments can be calculated as:
\[ \vec{p_1} + \vec{p_2} + \vec{p_3} = p_0 \hat{i} + p_0 \hat{j} + p_0 \hat{k} \]
- The magnitude of this total momentum is:
\[ |\vec{p_1} + \vec{p_2} + \vec{p_3}| = \sqrt{p_0^2 + p_0^2 + p_0^2} = \sqrt{3}p_0 \]
- Therefore, the momentum of the fourth fragment is:
\[ p_4 = \sqrt{3}p_0 \]

5. **Kinetic Energy of the Fourth Fragment:**
- The kinetic energy of the fourth fragment can be calculated as:
\[ E_4 = \frac{p_4^2}{2m} = \frac{(\sqrt{3}p_0)^2}{2m} = \frac{3p_0^2}{2m} \]
- Since \(E_0 = \frac{p_0^2}{2m}\), we can express \(E_4\) in terms of \(E_0\):
\[ E_4 = 3E_0 \]

6. **Total Energy of the Explosion:**
- The total energy of the explosion is the sum of the kinetic energies of all four fragments:
\[ E_{\text{total}} = E_0 + E_0 + E_0 + E_4 = 3E_0 + 3E_0 = 6E_0 \]

### Final Answer:
The energy of the explosion will be \(6E_0\).