Mass of moon is `7.34xx10^(22)kg`. If the acceleration due gravity on the moon is `1.4m//s^(2)`, the radius of the moon `(G=6.667xx10^(-11)Nm^(2)//kg^(2))`

Options

A. `0.56xx10^(4)m`
B. `1.87xx10^(6)m`
C. `1.92xx10^(6)m`
D. `1.01xx10^(8)m`

Explanation

To find the radius of the moon given its mass, the acceleration due to gravity on its surface, and the gravitational constant, we can use the formula for acceleration due to gravity:

\[ g = \frac{G \cdot M}{R^2} \]

Where:
- \( g \) is the acceleration due to gravity (1.4 m/s²),
- \( G \) is the gravitational constant (6.67 × 10⁻¹¹ N m²/kg²),
- \( M \) is the mass of the moon (7.34 × 10²² kg),
- \( R \) is the radius of the moon.

### Step-by-Step Solution:

1. **Rearranging the Formula**:
We need to isolate \( R \) in the formula. Rearranging gives us:
\[ R^2 = \frac{G \cdot M}{g} \]
Therefore,
\[ R = \sqrt{\frac{G \cdot M}{g}} \]

2. **Substituting the Values**:
Now, we substitute the known values into the equation:
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( M = 7.34 \times 10^{22} \, \text{kg} \)
- \( g = 1.4 \, \text{m/s}^2 \)

So we have:
\[ R = \sqrt{\frac{(6.67 \times 10^{-11}) \cdot (7.34 \times 10^{22})}{1.4}} \]

3. **Calculating the Numerator**:
First, we calculate the numerator:
\[ (6.67 \times 10^{-11}) \cdot (7.34 \times 10^{22}) = 4.89558 \times 10^{12} \]

4. **Dividing by g**:
Now we divide this result by \( g \):
\[ \frac{4.89558 \times 10^{12}}{1.4} = 3.49399 \times 10^{12} \]

5. **Taking the Square Root**:
Finally, we take the square root to find \( R \):
\[ R = \sqrt{3.49399 \times 10^{12}} \approx 1.87 \times 10^6 \, \text{m} \]

6. **Final Result**:
Thus, the radius of the moon is approximately:
\[ R \approx 1.87 \times 10^6 \, \text{m} \]

### Conclusion:
The radius of the moon is approximately \( 1.87 \times 10^6 \, \text{m} \).