The density of a planet PL1 is thrice that of planet PL2. The acceleration due to gravity at the surface of PL1 is `(1^(th))/(9)` of that at the surface of planet PL2. If the radius of planet PL2 is R, then the radius of planet PL1 will be

Options

A. `R//3`
B. `R//27`
C. 27R
D. 81R

Answer

To solve the problem, we need to relate the densities and gravitational accelerations of the two planets, PL1 and PL2, using the formulas for gravitational acceleration and density.

### Step-by-Step Solution

1. **Understanding the given data:**
- Density of PL1, \( \rho_1 = 3 \rho_2 \) (where \( \rho_2 \) is the density of PL2).
- Acceleration due to gravity at PL1, \( g_1 = \frac{1}{9} g_2 \) (where \( g_2 \) is the acceleration due to gravity at PL2).
- Radius of PL2, \( R_2 = R \).
- We need to find the radius of PL1, \( R_1 \).

2. **Using the formula for gravitational acceleration:**
The formula for gravitational acceleration at the surface of a planet is given by:
\[ g = \frac{G \cdot M}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

3. **Expressing mass in terms of density:**
The mass \( M \) of a planet can be expressed in terms of its density and volume:
\[ M = \rho \cdot V = \rho \cdot \left( \frac{4}{3} \pi R^3 \right) \]
Therefore, for PL1 and PL2:
\[ M_1 = \rho_1 \cdot \left( \frac{4}{3} \pi R_1^3 \right) \quad \text{and} \quad M_2 = \rho_2 \cdot \left( \frac{4}{3} \pi R_2^3 \right) \]

4. **Substituting mass into the gravitational acceleration formula:**
For PL1:
\[ g_1 = \frac{G \cdot M_1}{R_1^2} = \frac{G \cdot \left( 3 \rho_2 \cdot \frac{4}{3} \pi R_1^3 \right)}{R_1^2} = \frac{4 \pi G \rho_2 R_1}{3} \]
For PL2:
\[ g_2 = \frac{G \cdot M_2}{R_2^2} = \frac{G \cdot \left( \rho_2 \cdot \frac{4}{3} \pi R_2^3 \right)}{R_2^2} = \frac{4 \pi G \rho_2 R_2}{3} \]

5. **Setting up the ratio of gravitational accelerations:**
From the problem, we know:
\[ g_1 = \frac{1}{9} g_2 \]
Substituting the expressions for \( g_1 \) and \( g_2 \):
\[ \frac{4 \pi G \rho_2 R_1}{3} = \frac{1}{9} \cdot \frac{4 \pi G \rho_2 R_2}{3} \]

6. **Simplifying the equation:**
The \( \frac{4 \pi G}{3} \) and \( \rho_2 \) terms cancel out:
\[ R_1 = \frac{1}{9} R_2 \]

7. **Substituting \( R_2 \):**
Since \( R_2 = R \):
\[ R_1 = \frac{1}{9} R \]

### Final Answer
The radius of planet PL1 is:
\[ R_1 = \frac{R}{9} \]