STATEMENT -1 : The escape valocity upon Earth's surface is 11.2 km/s. But velocity required for escaping is little less than 11.2 km/s if launched properly.
and
STATEMENT -2 : Escape velocity is independent of angle of projection. Conservation of mechanical energy uses KE and GPE which are scale quantities.

Options

A. Statement 1- True, Statement -2 is True, Statement -2 is a correct explanation for Statement -11
B. Statement -1 is True, Statement -2 is True, Statement -2 is NOT a correct explanation for Statement -11
C. Statement -1 is True, Statement -2 is False
D. Statement -1 is False, Statement -2 is True

Answer

To analyze the statements provided in the question, we will break down the concepts of escape velocity and its relation to angle of projection and conservation of energy.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
- The escape velocity (\( v_e \)) from the surface of a celestial body, like Earth, is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Escape Velocity Value**:
- For Earth, this value is approximately \( 11.2 \, \text{km/s} \). This means that an object must reach this speed to escape Earth's gravitational pull without any further propulsion.

3. **Statement 1 Analysis**:
- The first statement claims that while the escape velocity is \( 11.2 \, \text{km/s} \), if an object is launched properly (for example, at an optimal angle), it may require slightly less than this speed to escape. This is true because launching at an angle can utilize some of the gravitational potential energy more efficiently, allowing for a lower initial speed to achieve escape.

4. **Statement 2 Analysis**:
- The second statement asserts that escape velocity is independent of the angle of projection. This is correct because escape velocity is a scalar quantity derived from gravitational potential energy and kinetic energy, and does not depend on the direction of the launch. The formula for escape velocity does not include any angle; hence, it remains constant regardless of how the object is launched.

5. **Conclusion**:
- Both statements are true:
- Statement 1 is correct because launching at an optimal angle can indeed allow for slightly less than \( 11.2 \, \text{km/s} \).
- Statement 2 is also correct as escape velocity does not depend on the angle of projection.

6. **Final Assessment**:
- Since both statements are true but do not provide a direct relation to each other, the correct conclusion is that both statements are true independently.