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

Answer

To determine whether the statement "If a planet gets inflated, keeping its density constant, then g will increase" is true or false, we can analyze the relationship between gravitational acceleration (g), density (ρ), and radius (r) of the planet.

### Step-by-Step Solution:

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

2. **Relate mass to density and volume**:
The mass \( M \) of the planet can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho V \]
The volume \( V \) of a sphere (assuming the planet is spherical) is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, we can substitute this into the mass equation:
\[ M = \rho \left(\frac{4}{3} \pi r^3\right) \]

3. **Substitute mass into the gravitational acceleration formula**:
Now substituting \( M \) back into the formula for \( g \):
\[ g = \frac{G \left(\rho \frac{4}{3} \pi r^3\right)}{r^2} \]
Simplifying this gives:
\[ g = \frac{4}{3} G \rho \frac{r^3}{r^2} = \frac{4}{3} G \rho r \]

4. **Analyze the relationship**:
From the derived formula, we see that:
\[ g \propto r \]
This indicates that gravitational acceleration \( g \) is directly proportional to the radius \( r \) of the planet, assuming density \( \rho \) remains constant.

5. **Consider the implications of inflation**:
If the planet gets inflated, it means that the radius \( r \) is increasing while keeping the density \( \rho \) constant. Since \( g \) is directly proportional to \( r \), an increase in \( r \) will lead to an increase in \( g \).

6. **Conclusion**:
Therefore, the statement "If a planet gets inflated, keeping its density constant, then g will increase" is true.

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