A bullet is fired vertically upwards with a velocity `upsilon` from the surface of a spherical planet when it reaches its maximum height, its acceleration due to the planet's gravity is `(1)/(4)th` of its value at the surface of the planet. If the escape velocity from the planet is `V_("escape") = upsilon sqrt(N)`, then the value of `N` is : (ignore energy loss due to atmosphere).

Options

A. 2
B. 3
C. 4
D. 5

Explanation

To solve the problem, we need to find the value of \( N \) in the escape velocity formula \( V_{\text{escape}} = v \sqrt{N} \) given that the acceleration due to gravity at the maximum height of a bullet fired upwards is \( \frac{1}{4} \) of its value at the surface of the planet.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let \( g \) be the acceleration due to gravity at the surface of the planet.
- At maximum height \( h \), the acceleration due to gravity is \( \frac{g}{4} \).

2. **Using the Formula for Gravity**:
- The formula for gravitational acceleration at a distance \( r \) from the center of the planet is:
\[ g' = \frac{GM}{r^2} \]
- At the surface of the planet (radius \( R \)):
\[ g = \frac{GM}{R^2} \]
- At height \( h \) (total distance from the center is \( R + h \)):
\[ g' = \frac{GM}{(R + h)^2} \]
- Given that \( g' = \frac{g}{4} \):
\[ \frac{GM}{(R + h)^2} = \frac{1}{4} \cdot \frac{GM}{R^2} \]

3. **Setting Up the Equation**:
- Cancel \( GM \) from both sides (assuming \( G \) and \( M \) are non-zero):
\[ \frac{1}{(R + h)^2} = \frac{1}{4R^2} \]
- Cross-multiplying gives:
\[ 4R^2 = (R + h)^2 \]

4. **Expanding and Solving for \( h \)**:
- Expanding the right side:
\[ 4R^2 = R^2 + 2Rh + h^2 \]
- Rearranging gives:
\[ 3R^2 = 2Rh + h^2 \]
- This is a quadratic equation in \( h \):
\[ h^2 + 2Rh - 3R^2 = 0 \]

5. **Using the Quadratic Formula**:
- The quadratic formula is \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = 2R \), and \( c = -3R^2 \).
- Discriminant:
\[ b^2 - 4ac = (2R)^2 - 4(1)(-3R^2) = 4R^2 + 12R^2 = 16R^2 \]
- Therefore:
\[ h = \frac{-2R \pm 4R}{2} = R \quad \text{(taking the positive root)} \]

6. **Finding Escape Velocity**:
- The escape velocity \( V_{\text{escape}} \) from the planet is given by:
\[ V_{\text{escape}} = \sqrt{2gR} \]
- Substituting \( g = \frac{GM}{R^2} \):
\[ V_{\text{escape}} = \sqrt{2 \cdot \frac{GM}{R^2} \cdot R} = \sqrt{\frac{2GM}{R}} \]

7. **Relating to Given Velocity**:
- We know \( v = \sqrt{gR} \) (the initial velocity).
- Thus:
\[ V_{\text{escape}} = \sqrt{2} \cdot v \]
- Therefore, we can equate:
\[ V_{\text{escape}} = v \sqrt{2} \]

8. **Conclusion**:
- From the equation \( V_{\text{escape}} = v \sqrt{N} \), we find:
\[ N = 2 \]

### Final Answer:
The value of \( N \) is \( 2 \).