Two planets of radii in the ratio `2 : 3` are made from the materials of density in the ratio `3 : 2`. Then the ratio of acceleration due to gravity `g_(1)//g_(2)` at the surface of two planets will be

Options

A. (a)1
B. (b)2.25
C. (c)`0.50`
D. (d)`0.12`

Answer

To solve the problem, we need to find the ratio of the acceleration due to gravity \( g_1/g_2 \) at the surfaces of two planets, given their radii and densities.

### Step-by-Step Solution:

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

2. **Expressing Mass in Terms of Density**:
The mass \( M \) of a planet can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho \cdot V \]
The volume \( V \) of a sphere (planet) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the mass can be rewritten as:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

3. **Substituting Mass in the Gravity Formula**:
Substituting the expression for mass into the formula for \( g \):
\[ g = \frac{G \cdot \left(\rho \cdot \frac{4}{3} \pi R^3\right)}{R^2} \]
Simplifying this gives:
\[ g = \frac{G \cdot \rho \cdot \frac{4}{3} \pi R^3}{R^2} = G \cdot \rho \cdot \frac{4}{3} \pi R \]

4. **Calculating \( g_1 \) and \( g_2 \)**:
For planet 1:
\[ g_1 = G \cdot \rho_1 \cdot \frac{4}{3} \pi R_1 \]
For planet 2:
\[ g_2 = G \cdot \rho_2 \cdot \frac{4}{3} \pi R_2 \]

5. **Finding the Ratio \( g_1/g_2 \)**:
To find the ratio of the gravitational accelerations:
\[ \frac{g_1}{g_2} = \frac{G \cdot \rho_1 \cdot \frac{4}{3} \pi R_1}{G \cdot \rho_2 \cdot \frac{4}{3} \pi R_2} \]
The constants \( G \), \( \frac{4}{3} \), and \( \pi \) cancel out:
\[ \frac{g_1}{g_2} = \frac{\rho_1 \cdot R_1}{\rho_2 \cdot R_2} \]

6. **Substituting the Ratios**:
Given that the radii are in the ratio \( R_1:R_2 = 2:3 \) and the densities are in the ratio \( \rho_1:\rho_2 = 3:2 \):
\[ \frac{g_1}{g_2} = \frac{3/2 \cdot 2/3} \]

7. **Simplifying the Ratio**:
Simplifying the above expression:
\[ \frac{g_1}{g_2} = \frac{3 \cdot 2}{2 \cdot 3} = 1 \]

### Final Answer:
Thus, the ratio of the acceleration due to gravity at the surfaces of the two planets is:
\[ \frac{g_1}{g_2} = 1 \]