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. `(4E_(0))/3`
C. `4E_(0)`
D. `8E_(0)`

Explanation

To solve the problem of finding the energy of the explosion when a stationary body explodes into four identical fragments, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Mass of Each Fragment:**
Let the mass of each fragment be denoted as \( M \).

2. **Kinetic Energy of the Fragments:**
Given that three of the fragments have the same kinetic energy \( E_0 \), we can denote the kinetic energies of these fragments as:
\[ E_1 = E_2 = E_3 = E_0 \]

3. **Momentum of Each Fragment:**
The momentum \( P \) of a fragment can be expressed in terms of its kinetic energy:
\[ P_i = \sqrt{2mE_i} \]
For the first three fragments, we have:
\[ P_1 = P_2 = P_3 = \sqrt{2M E_0} \]

4. **Direction of the Fragments:**
Since the three fragments are flying mutually perpendicular to each other, we can treat their momenta as components along the x, y, and z axes.

5. **Resultant Momentum of the First Three Fragments:**
The resultant momentum \( P_4 \) of the fourth fragment, which flies in the direction opposite to the resultant of \( P_1, P_2, \) and \( P_3 \), can be calculated using the Pythagorean theorem:
\[ P_4 = \sqrt{P_1^2 + P_2^2 + P_3^2} \]
Substituting the values:
\[ P_4 = \sqrt{(2M E_0) + (2M E_0) + (2M E_0)} = \sqrt{6M E_0} \]

6. **Kinetic Energy of the Fourth Fragment:**
The kinetic energy \( E_4 \) of the fourth fragment can be calculated using its momentum:
\[ E_4 = \frac{P_4^2}{2M} = \frac{(P_4)^2}{2M} = \frac{6M E_0}{2M} = 3E_0 \]

7. **Total Energy of the Explosion:**
The total energy of the explosion \( E \) is the sum of the kinetic energies of all four fragments:
\[ E = E_1 + E_2 + E_3 + E_4 = E_0 + E_0 + E_0 + 3E_0 = 6E_0 \]

### Final Answer:
The energy of the explosion is:
\[ \boxed{6E_0} \]