If earth has a mass nine times and radius twice to that of a planet `P`.Then `(v_(e))/(3)sqrt(x)ms^(-1)` will be the minimum velocity required by a rocket to pull out of gravitational force of `P`,where `v_e` is escape velocity on earth.The value of `x` is

Options

A. `1`
B. `18`
C. `3`
D. `2`

Answer

To solve the problem, we need to find the minimum escape velocity required for a rocket to pull out of the gravitational force of planet P, given the mass and radius relationships between Earth and planet P.

### Step-by-Step Solution:

1. **Understand Escape Velocity Formula**:
The escape velocity \( v \) from a celestial body is given by the formula:
\[ v = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the body, and \( R \) is its radius.

2. **Define Mass and Radius of Planet P**:
Let the mass of planet P be \( M_P \) and its radius be \( R_P \). According to the problem:
- The mass of Earth \( M_E = 9M_P \)
- The radius of Earth \( R_E = 2R_P \)

3. **Calculate Escape Velocity for Earth**:
The escape velocity for Earth \( v_E \) is:
\[ v_E = \sqrt{\frac{2GM_E}{R_E}} = \sqrt{\frac{2G(9M_P)}{2R_P}} = \sqrt{\frac{9GM_P}{R_P}} \]

4. **Calculate Escape Velocity for Planet P**:
The escape velocity for planet P \( v_P \) is:
\[ v_P = \sqrt{\frac{2GM_P}{R_P}} \]

5. **Relate Escape Velocities**:
Now, we can express \( v_P \) in terms of \( v_E \):
\[ v_P = \sqrt{\frac{2GM_P}{R_P}} = \sqrt{\frac{2G}{R_P}} \cdot M_P \]
Substitute \( M_P \) from \( v_E \):
\[ v_P = \sqrt{\frac{2G}{R_P}} \cdot \frac{R_P}{9} \cdot \frac{v_E}{\sqrt{9}} = \frac{v_E}{3} \]

6. **Set Up the Equation**:
According to the problem, we have:
\[ \frac{v_E}{3\sqrt{x}} = v_P \]
Substituting \( v_P \):
\[ \frac{v_E}{3\sqrt{x}} = \frac{v_E}{3} \]

7. **Solve for x**:
By equating the two expressions:
\[ \sqrt{x} = 1 \]
Squaring both sides gives:
\[ x = 2 \]

### Final Answer:
Thus, the value of \( x \) is **2**.