Imagine a new planet having the same density as that of earth but `3` times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is `g` and that on the new plane is `g` , then :

Options

A. `g' = 3g`
B. `g' = (g)/(g)`
C. `g' = 9g`
D. `g' = 27g`

Answer

To solve the problem, we need to find the relationship between the acceleration due to gravity on the surface of the new planet (g') and that on Earth (g), given that the new planet has the same density as Earth but is three times larger in size.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a new planet with a radius that is 3 times that of Earth (let's denote the radius of Earth as \( r \), so the radius of the new planet is \( R = 3r \)).
- The density of the new planet is the same as that of Earth, denoted as \( \rho \).

2. **Formula for Acceleration due to Gravity**:
- The acceleration due to gravity 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.

3. **Mass of the New Planet**:
- The mass \( M \) of a planet can be expressed in terms of its density and volume:
\[ M = \rho V \]
- The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- Therefore, the mass of the new planet can be expressed as:
\[ M' = \rho \left(\frac{4}{3} \pi (3r)^3\right) = \rho \left(\frac{4}{3} \pi 27r^3\right) = 27 \left(\rho \frac{4}{3} \pi r^3\right) = 27M \]
where \( M \) is the mass of Earth.

4. **Acceleration due to Gravity on the New Planet**:
- Now, we can calculate the acceleration due to gravity on the new planet \( g' \):
\[ g' = \frac{G M'}{(3r)^2} = \frac{G (27M)}{9r^2} = \frac{27GM}{9r^2} = 3 \frac{GM}{r^2} = 3g \]

5. **Conclusion**:
- Therefore, the acceleration due to gravity on the new planet is:
\[ g' = 3g \]

### Final Answer:
The acceleration due to gravity on the new planet is \( g' = 3g \).