If a planet gets inflated, keeping its density constant, then g will increase.

Answer

To solve the problem, we need to analyze how the acceleration due to gravity (g) changes when a planet is inflated while keeping its density constant. We will use the formula for gravitational acceleration and the relationship between mass, density, and volume.

### Step-by-Step Solution:

1. **Understanding the Formula for Gravitational Acceleration**:
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,
- \( R \) is the radius of the planet.

2. **Relating Mass and Density**:
The mass \( M \) of the planet can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho \cdot V \]
For a spherical planet, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, we can substitute this into the mass equation:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

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

4. **Analyzing the Effect of Increasing Radius**:
From the derived formula, we see that \( g \) is directly proportional to the radius \( R \) of the planet when density \( \rho \) is constant:
\[ g \propto R \]
Therefore, if the radius \( R \) increases (as the planet gets inflated), the acceleration due to gravity \( g \) will also increase.

5. **Conclusion**:
Since we have established that increasing the radius of the planet while keeping the density constant results in an increase in \( g \), the statement "If a planet gets inflated, keeping its density constant, then g will increase" is **true**.

### Final Answer:
The statement is **true**.

---