Assertion On the surface of moon, value of g is `(1)/(6)th` the value on the surface of earth. A particle is projected as projectile under similar condition on the surface of moon and on the surface of earth. Then values of T,H and R on the surface of moon will become six times.
Reason T, H and `R prop (1)/(g)`

Options

A. If both Asseration and Reason are correct and Reason is the correct explanation of Assertion
B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion
C. If Assertion is true but Reason is false
D. If Assertion is false but Reason is true

Answer

To solve the problem, we need to analyze the motion of a projectile on the surfaces of the Earth and the Moon, using the given information about the gravitational acceleration (g) on both celestial bodies.

### Step-by-Step Solution:

1. **Understanding the Values of g**:
- The value of gravitational acceleration (g) on the surface of the Earth is approximately \(9.8 \, \text{m/s}^2\).
- On the surface of the Moon, the value of g is \( \frac{1}{6} \) of that on Earth. Therefore,
\[ g_{\text{moon}} = \frac{g_{\text{earth}}}{6} = \frac{9.8}{6} \approx 1.63 \, \text{m/s}^2. \]

2. **Formulas for Projectile Motion**:
- The time of flight (T), maximum height (H), and range (R) of a projectile launched with an initial velocity \(u\) at an angle \(\theta\) are given by:
- Time of flight (T):
\[ T = \frac{2u \sin \theta}{g} \]
- Maximum height (H):
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
- Range (R):
\[ R = \frac{u^2 \sin 2\theta}{g} \]

3. **Calculating Time of Flight, Height, and Range on the Moon**:
- Since \(g_{\text{moon}} = \frac{g_{\text{earth}}}{6}\), we can substitute this into the formulas:
- Time of flight on the Moon (\(T_{\text{moon}}\)):
\[ T_{\text{moon}} = \frac{2u \sin \theta}{g_{\text{moon}}} = \frac{2u \sin \theta}{\frac{g_{\text{earth}}}{6}} = 6 \cdot \frac{2u \sin \theta}{g_{\text{earth}}} = 6T_{\text{earth}} \]
- Maximum height on the Moon (\(H_{\text{moon}}\)):
\[ H_{\text{moon}} = \frac{u^2 \sin^2 \theta}{2g_{\text{moon}}} = \frac{u^2 \sin^2 \theta}{2 \cdot \frac{g_{\text{earth}}}{6}} = 6 \cdot \frac{u^2 \sin^2 \theta}{2g_{\text{earth}}} = 6H_{\text{earth}} \]
- Range on the Moon (\(R_{\text{moon}}\)):
\[ R_{\text{moon}} = \frac{u^2 \sin 2\theta}{g_{\text{moon}}} = \frac{u^2 \sin 2\theta}{\frac{g_{\text{earth}}}{6}} = 6 \cdot \frac{u^2 \sin 2\theta}{g_{\text{earth}}} = 6R_{\text{earth}} \]

4. **Conclusion**:
- From the calculations, we find that the time of flight, maximum height, and range on the Moon are all six times greater than those on the Earth:
\[ T_{\text{moon}} = 6T_{\text{earth}}, \quad H_{\text{moon}} = 6H_{\text{earth}}, \quad R_{\text{moon}} = 6R_{\text{earth}}. \]

### Final Answer:
The assertion is correct, and the reason is also correct as T, H, and R are inversely proportional to g. Hence, the values of T, H, and R on the surface of the Moon will indeed become six times those on the surface of the Earth.