Ravi can throw a ball at a speed on the earth which can cross a river of width `10 m`. Ravi reaches on an imaginary planet whose mean density is twice that of the earth. Find out the maximum possible radius of the planet so that if Ravi throws the ball at the same speed it may escape from the planet. Given radius of the earth `= 6.4 xx 10^(6) m`.

Explanation

To solve the problem, we need to find the maximum possible radius of an imaginary planet where Ravi can throw a ball at the same speed he used on Earth to cross a river of width 10 meters. The mean density of this imaginary planet is twice that of Earth.

### Step-by-Step Solution:

1. **Understanding the Range Formula**:
The range \( R \) of a projectile launched at an angle \( \theta \) with an initial speed \( V \) is given by the formula:
\[ R = \frac{V^2 \sin(2\theta)}{g} \]
For maximum range, \( \theta \) should be \( 45^\circ \), thus \( \sin(2\theta) = 1 \). Therefore, the maximum range simplifies to:
\[ R = \frac{V^2}{g} \]

2. **Setting Up the Equation for Earth**:
We know that Ravi can throw the ball to cross a river of width \( 10 \, m \). Therefore, we can set up the equation:
\[ 10 = \frac{V^2}{g_e} \]
where \( g_e \) is the acceleration due to gravity on Earth.

3. **Expressing \( V^2 \)**:
Rearranging the equation gives:
\[ V^2 = 10 g_e \]

4. **Relating Gravity to Density**:
The acceleration due to gravity \( g \) can be expressed in terms of the mass and radius of the planet:
\[ g = \frac{G M}{R^2} \]
where \( M \) is the mass of the planet and \( G \) is the gravitational constant. The mass \( M \) can also be expressed in terms of density \( \rho \) and volume \( V \):
\[ M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi R^3 \]
Thus, we can write:
\[ g = \frac{G \rho \cdot \frac{4}{3} \pi R^3}{R^2} = \frac{4 \pi G \rho R}{3} \]

5. **Substituting for \( g \)**:
Now substituting this expression for \( g \) back into the equation for \( V^2 \):
\[ V^2 = 10 \cdot \frac{4 \pi G \rho_e R_e}{3} \]
where \( \rho_e \) is the density of Earth and \( R_e \) is the radius of Earth.

6. **Setting Up the Equation for the Imaginary Planet**:
For the imaginary planet, we know that the density \( \rho_p \) is twice that of Earth:
\[ \rho_p = 2 \rho_e \]
The escape velocity condition gives:
\[ \frac{1}{2} V^2 = g_p \cdot R_p \]
where \( g_p = \frac{4 \pi G \rho_p R_p}{3} \).

7. **Equating the Two Expressions**:
Equating the two expressions for \( V^2 \):
\[ 10 \cdot \frac{4 \pi G \rho_e R_e}{3} = 2 \cdot \frac{4 \pi G (2 \rho_e) R_p}{3} \]

8. **Simplifying the Equation**:
Canceling common terms gives:
\[ 10 R_e = 4 R_p \]
Rearranging gives:
\[ R_p = \frac{10}{4} R_e = \frac{5}{2} R_e \]

9. **Substituting the Radius of Earth**:
Given \( R_e = 6.4 \times 10^6 \, m \):
\[ R_p = \frac{5}{2} \times 6.4 \times 10^6 = 16 \times 10^6 \, m \]

### Final Answer:
The maximum possible radius of the imaginary planet is:
\[ R_p = 16 \times 10^6 \, m \]