A particle is projected vertically upwards with a velocity `sqrt(gR)`, where `R` denotes the radius of the earth and `g` the acceleration due to gravity on the surface of the earth. Then the maximum height ascended by the particle is

Options

A. `R/2`
B. `R`
C. `2R`
D. `(5R)/4`

Explanation

To solve the problem of finding the maximum height ascended by a particle projected vertically upwards with a velocity of \( \sqrt{gR} \), where \( R \) is the radius of the Earth and \( g \) is the acceleration due to gravity, we can use the principle of conservation of energy.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- The initial velocity \( v_0 = \sqrt{gR} \).
- The initial height \( h_0 = 0 \) (at the surface of the Earth).

2. **Kinetic Energy at the Surface:**
- The kinetic energy (KE) at the surface (point 1) is given by:
\[ KE_1 = \frac{1}{2} m v_0^2 = \frac{1}{2} m (\sqrt{gR})^2 = \frac{1}{2} m gR \]

3. **Potential Energy at the Surface:**
- The potential energy (PE) at the surface (point 1) is given by:
\[ PE_1 = -\frac{GMm}{R} \]
- Here, \( G \) is the universal gravitational constant, and \( M \) is the mass of the Earth.

4. **Total Energy at the Surface:**
- The total mechanical energy at point 1 is:
\[ E_1 = KE_1 + PE_1 = \frac{1}{2} m gR - \frac{GMm}{R} \]

5. **At Maximum Height:**
- At the maximum height \( h_{max} \), the velocity becomes zero, so the kinetic energy (KE) at this point (point 2) is:
\[ KE_2 = 0 \]
- The potential energy (PE) at maximum height is:
\[ PE_2 = -\frac{GMm}{R + h_{max}} \]

6. **Total Energy at Maximum Height:**
- The total mechanical energy at point 2 is:
\[ E_2 = KE_2 + PE_2 = 0 - \frac{GMm}{R + h_{max}} = -\frac{GMm}{R + h_{max}} \]

7. **Conservation of Energy:**
- According to the conservation of energy, \( E_1 = E_2 \):
\[ \frac{1}{2} m gR - \frac{GMm}{R} = -\frac{GMm}{R + h_{max}} \]

8. **Simplifying the Equation:**
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} gR - \frac{GM}{R} = -\frac{GM}{R + h_{max}} \]
- Rearranging gives:
\[ \frac{1}{2} gR = \frac{GM}{R} - \frac{GM}{R + h_{max}} \]

9. **Finding a Common Denominator:**
- The right side can be combined:
\[ \frac{GM(R + h_{max}) - GMR}{R(R + h_{max})} = \frac{GMh_{max}}{R(R + h_{max})} \]
- Thus, we have:
\[ \frac{1}{2} gR = \frac{GMh_{max}}{R(R + h_{max})} \]

10. **Substituting \( g = \frac{GM}{R^2} \):**
- Substitute \( g \) into the equation:
\[ \frac{1}{2} \frac{GM}{R^2} R = \frac{GMh_{max}}{R(R + h_{max})} \]
- Simplifying gives:
\[ \frac{1}{2} \frac{GM}{R} = \frac{GMh_{max}}{R(R + h_{max})} \]

11. **Cross Multiplying:**
- Cross multiply to solve for \( h_{max} \):
\[ \frac{1}{2} GM(R + h_{max}) = GMh_{max} \]
- This simplifies to:
\[ \frac{1}{2} GR + \frac{1}{2} Gh_{max} = Gh_{max} \]
- Rearranging gives:
\[ \frac{1}{2} GR = \frac{1}{2} Gh_{max} \]
- Thus, we find:
\[ h_{max} = R \]

### Final Answer:
The maximum height ascended by the particle is \( R \).