A rocket of mass `m_0` has attained a speed equal to its exhaust speed and that time the mass of the rocket is `m`. Then the ratio `m_0/m` is (neglect gravity)

Options

A. (a) `2.718`
B. (b) `7.8`
C. (c) `3.14`
D. (d) `4`

Explanation

To solve the problem, we will use the concept of the rocket equation, also known as Tsiolkovsky's rocket equation. The equation relates the change in velocity of a rocket to the effective exhaust velocity and the mass of the rocket.

### Step-by-step Solution:

1. **Understanding the Given Information**:
- The initial mass of the rocket is \( m_0 \).
- The mass of the rocket at a certain time is \( m \).
- The speed of the rocket at this time is equal to its exhaust speed \( V_r \).

2. **Using the Rocket Equation**:
The rocket equation is given by:
\[ \Delta V = V_r \ln\left(\frac{m_0}{m}\right) \]
where \( \Delta V \) is the change in velocity, \( V_r \) is the effective exhaust velocity, \( m_0 \) is the initial mass, and \( m \) is the final mass.

3. **Setting the Change in Velocity**:
Since it is given that the speed of the rocket is equal to its exhaust speed, we can set:
\[ \Delta V = V_r \]
Thus, we can rewrite the equation as:
\[ V_r = V_r \ln\left(\frac{m_0}{m}\right) \]

4. **Canceling \( V_r \)**:
Since \( V_r \) is not zero, we can divide both sides by \( V_r \):
\[ 1 = \ln\left(\frac{m_0}{m}\right) \]

5. **Exponentiating Both Sides**:
To eliminate the natural logarithm, we exponentiate both sides:
\[ e^1 = \frac{m_0}{m} \]
This simplifies to:
\[ \frac{m_0}{m} = e \]

6. **Final Result**:
Therefore, the ratio \( \frac{m_0}{m} \) is equal to \( e \), which is approximately 2.718.

### Conclusion:
The ratio \( \frac{m_0}{m} \) is \( e \).