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'=9g`
C. `g'=g//7`
D. `g'=27g`

Explanation

To solve the problem, we need to find the relationship between the acceleration due to gravity on the surface of Earth (g) and the acceleration due to gravity on the surface of a new planet (g'). 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 are given that the new planet has the same density as Earth but is 3 times larger in size. This means the radius of the new planet (R') is 3 times the radius of Earth (R).

\[ R' = 3R \]

2. **Formula for Acceleration Due to Gravity**:
The formula for the acceleration due to gravity at the surface of a planet is given by:

\[ g = \frac{GM}{R^2} \]

where G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet.

3. **Expressing Mass in Terms of Density**:
The mass (M) of the planet can be expressed in terms of its density (ρ) and volume (V). The volume of a sphere is given by:

\[ V = \frac{4}{3} \pi R^3 \]

Therefore, the mass can be expressed as:

\[ M = \rho V = \rho \left(\frac{4}{3} \pi R^3\right) \]

4. **Substituting Mass in the Gravity Formula**:
Substituting the expression for mass into the formula for gravity, we get:

\[ g = \frac{G \left(\rho \frac{4}{3} \pi R^3\right)}{R^2} = \frac{4}{3} \pi G \rho R \]

5. **Calculating Gravity for Earth (g)**:
For Earth, we denote the density as ρ and the radius as R. Thus, the acceleration due to gravity on Earth (g) is:

\[ g = \frac{4}{3} \pi G \rho R \]

6. **Calculating Gravity for the New Planet (g')**:
For the new planet, since the density is the same (ρ) but the radius is R' = 3R, we can write:

\[ g' = \frac{4}{3} \pi G \rho R' = \frac{4}{3} \pi G \rho (3R) = 3 \left(\frac{4}{3} \pi G \rho R\right) = 3g \]

7. **Final Relation**:
Thus, we find that the acceleration due to gravity on the new planet (g') is:

\[ g' = 3g \]

### Conclusion:
The acceleration due to gravity on the new planet is three times that of Earth. Therefore, the answer is:

\[ g' = 3g \]