Statement-1: Escape velocity is independent of the angle of projection.
Statement-2: Escape velocity from the surface of earth is `sqrt(2gR)` where `R` is radius of earth.

Options

A. Statement-1 is true, statement-2 is true: Statement-2 is a correct explanation for statement-1
B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1
C. Statement-1 is true, Statement-2 is false.
D. Statement-1 is false, Statement-2 is true.

Answer

To analyze the statements regarding escape velocity, let's break down the concepts step by step:

### Step 1: Understanding Escape Velocity
Escape velocity is defined as the minimum velocity an object must have to break free from the gravitational pull of a celestial body (like Earth) without any further propulsion. This means that the object will reach infinity with zero kinetic energy left.

**Hint:** Remember that escape velocity is about overcoming gravitational potential energy.

### Step 2: Conservation of Mechanical Energy
When an object is projected upwards, its mechanical energy (the sum of kinetic and potential energy) is conserved if we ignore air resistance. The initial mechanical energy when the object is projected is equal to the final mechanical energy when it reaches infinity.

**Hint:** Think about how energy conservation applies to gravitational systems.

### Step 3: Setting Up the Energy Equation
At the surface of the Earth, the initial potential energy (U) of the object is given by:
\[ U = -\frac{GMm}{R} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the object,
- \( R \) is the radius of the Earth.

The initial kinetic energy (K) when the object is projected with escape velocity \( v_e \) is:
\[ K = \frac{1}{2}mv_e^2 \]

At infinity, the potential energy is zero, and the kinetic energy is also zero since we want the object to just reach infinity.

**Hint:** Write down the energy conservation equation clearly.

### Step 4: Applying Energy Conservation
According to the conservation of energy:
\[ K + U = 0 \]
Substituting the expressions for kinetic and potential energy:
\[ \frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0 \]

**Hint:** Focus on how the terms relate to each other.

### Step 5: Solving for Escape Velocity
Rearranging the equation gives:
\[ \frac{1}{2}mv_e^2 = \frac{GMm}{R} \]
Dividing both sides by \( m \) (mass of the object) and multiplying by 2:
\[ v_e^2 = \frac{2GM}{R} \]
Using the relation \( g = \frac{GM}{R^2} \), we can rewrite \( GM \) as \( gR \):
\[ v_e^2 = 2gR \]
Taking the square root:
\[ v_e = \sqrt{2gR} \]

**Hint:** Be careful with the algebra when manipulating the equations.

### Step 6: Conclusion on Statements
- **Statement 1:** "Escape velocity is independent of the angle of projection." - This is true because escape velocity only depends on the gravitational field and radius of the celestial body, not the direction of projection.
- **Statement 2:** "Escape velocity from the surface of Earth is \( \sqrt{2gR} \)." - This is also true as derived above.

Thus, both statements are true, and Statement 2 provides a correct explanation for Statement 1.

**Final Answer:** Both statements are true, and Statement 2 is the correct explanation for Statement 1.