Two spherical planets P and Q have the same uniform density `rho,` masses `M_p and M_Q` and surface areas A and 4A respectively. A spherical planet R also has uniform density `rho` and its mass is `(M_P + M_Q).` The escape velocities from the plantes P,Q and R are `V_P V_Q and V_R` respectively. Then

Options

A. `V_(Q)gtV_(R)gtV_(P)`
B. `V_(R)gtV_(Q)gtV_(P)`
C. `V_(R)//V_(P)=3`
D. `V_(P)//V_(Q)=1/2`

Explanation

To solve the problem, we need to find the relationship between the escape velocities of the three planets P, Q, and R based on their masses and radii.

### Step 1: Understanding Escape Velocity
The escape velocity \( V \) from a planet is given by the formula:
\[ V = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

### Step 2: Relating Surface Area to Radius
Given that the surface area \( A \) of planet P is \( A \) and that of planet Q is \( 4A \), we can relate the radii of the two planets using the formula for surface area of a sphere:
\[ A = 4\pi R^2 \]
For planet P:
\[ A = 4\pi R_P^2 \implies R_P^2 = \frac{A}{4\pi} \implies R_P = \sqrt{\frac{A}{4\pi}} \]
For planet Q:
\[ 4A = 4\pi R_Q^2 \implies R_Q^2 = \frac{4A}{4\pi} \implies R_Q = \sqrt{\frac{4A}{4\pi}} = 2\sqrt{\frac{A}{4\pi}} = 2R_P \]
Thus, we have:
\[ R_Q = 2R_P \]

### Step 3: Relating Mass to Radius
The mass of a planet is given by:
\[ M = \rho \cdot V = \rho \cdot \left(\frac{4}{3}\pi R^3\right) \]
For planet P:
\[ M_P = \rho \cdot \frac{4}{3}\pi R_P^3 \]
For planet Q:
\[ M_Q = \rho \cdot \frac{4}{3}\pi R_Q^3 = \rho \cdot \frac{4}{3}\pi (2R_P)^3 = \rho \cdot \frac{4}{3}\pi \cdot 8R_P^3 = 8M_P \]
Thus, we have:
\[ M_Q = 8M_P \]

### Step 4: Mass of Planet R
The mass of planet R is given as:
\[ M_R = M_P + M_Q = M_P + 8M_P = 9M_P \]

### Step 5: Finding Radius of Planet R
Using the mass of planet R, we can find its radius:
\[ M_R = \rho \cdot \frac{4}{3}\pi R_R^3 \implies 9M_P = \rho \cdot \frac{4}{3}\pi R_R^3 \]
Substituting \( M_P \):
\[ 9\left(\rho \cdot \frac{4}{3}\pi R_P^3\right) = \rho \cdot \frac{4}{3}\pi R_R^3 \]
This simplifies to:
\[ 9R_P^3 = R_R^3 \implies R_R = 9^{1/3} R_P = 3R_P \]

### Step 6: Finding Escape Velocities
Now we can express the escape velocities:
\[ V_P = \sqrt{\frac{2GM_P}{R_P}}, \quad V_Q = \sqrt{\frac{2G(8M_P)}{2R_P}} = \sqrt{\frac{16GM_P}{2R_P}} = \sqrt{\frac{8GM_P}{R_P}} = 2\sqrt{\frac{2GM_P}{R_P}} = 2V_P \]
For planet R:
\[ V_R = \sqrt{\frac{2G(9M_P)}{3R_P}} = \sqrt{\frac{18GM_P}{3R_P}} = \sqrt{\frac{6GM_P}{R_P}} = \sqrt{3}V_P \]

### Step 7: Comparing Escape Velocities
Now we have:
- \( V_P = V_P \)
- \( V_Q = 2V_P \)
- \( V_R = \sqrt{3}V_P \)

Thus, the relationship is:
\[ V_Q > V_R > V_P \]

### Final Answer
The correct relationship is:
\[ V_Q > V_R > V_P \]