the correct variation between `sqrt(E)` and p is shown by

Options

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Answer

To solve the question regarding the correct variation between \( \sqrt{E} \) (where \( E \) is kinetic energy) and \( p \) (momentum), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- Kinetic Energy (\( E \)) is given by the formula:
\[ E = \frac{1}{2} m v^2 \]
- Momentum (\( p \)) is defined as:
\[ p = mv \]

2. **Express Velocity in Terms of Momentum**:
- From the momentum equation, we can express velocity \( v \) as:
\[ v = \frac{p}{m} \]

3. **Substitute Velocity into Kinetic Energy Formula**:
- Substitute \( v \) into the kinetic energy formula:
\[ E = \frac{1}{2} m \left(\frac{p}{m}\right)^2 \]
- Simplifying this gives:
\[ E = \frac{1}{2} m \cdot \frac{p^2}{m^2} = \frac{p^2}{2m} \]

4. **Relate Kinetic Energy to Momentum**:
- Rearranging the equation, we have:
\[ p^2 = 2mE \]
- Taking the square root of both sides gives:
\[ p = \sqrt{2mE} \]

5. **Express the Relationship**:
- This shows that momentum \( p \) is directly proportional to \( \sqrt{E} \):
\[ p \propto \sqrt{E} \]

6. **Graphical Representation**:
- If we plot \( p \) on the y-axis and \( \sqrt{E} \) on the x-axis, the relationship will be linear, indicating that as momentum increases, the square root of kinetic energy also increases proportionally.

7. **Conclusion**:
- Therefore, the correct variation between \( \sqrt{E} \) and \( p \) is a straight line graph, confirming that \( p \) is directly proportional to \( \sqrt{E} \).