A particle falls on earth :
(i) from infinity. (ii) from a height 10 times the radius of earth. The ratio of the velocities gained on reaching at the earth's surface is :

Options

A. `sqrt(11) : sqrt(10)`
B. `sqrt(10) : sqrt(11)`
C. `10 : 11`
D. `11 : 10`

Answer

To solve the problem of finding the ratio of velocities gained by a particle falling to the Earth's surface from two different heights, we can use the principle of conservation of energy.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two scenarios:
(i) A particle falling from infinity.
(ii) A particle falling from a height of 10 times the radius of the Earth (10R).

2. **Using Conservation of Energy**:
- The total mechanical energy (kinetic + potential) at any point must remain constant if we ignore air resistance.
- The potential energy (U) at a distance \( r \) from the center of the Earth is given by:
\[ U = -\frac{GMm}{r} \]
- The kinetic energy (K) is given by:
\[ K = \frac{1}{2} mv^2 \]

3. **Case 1: Falling from Infinity**:
- At infinity, the total energy (E) is zero:
\[ E = K + U = 0 + 0 = 0 \]
- As the particle falls to the Earth's surface (at distance \( R \)):
\[ E = \frac{1}{2} mv_1^2 - \frac{GMm}{R} \]
- Setting the total energy at the surface equal to zero:
\[ 0 = \frac{1}{2} mv_1^2 - \frac{GMm}{R} \]
- Rearranging gives:
\[ \frac{1}{2} mv_1^2 = \frac{GMm}{R} \]
- Canceling \( m \) and solving for \( v_1 \):
\[ v_1^2 = \frac{2GM}{R} \implies v_1 = \sqrt{\frac{2GM}{R}} \]

4. **Case 2: Falling from a Height of 10R**:
- The total energy at a height of 10R:
\[ E = K + U = 0 - \frac{GMm}{11R} \]
- As the particle falls to the Earth's surface:
\[ E = \frac{1}{2} mv_2^2 - \frac{GMm}{R} \]
- Setting the total energy equal to the potential energy at height:
\[ -\frac{GMm}{11R} = \frac{1}{2} mv_2^2 - \frac{GMm}{R} \]
- Rearranging gives:
\[ \frac{GMm}{R} - \frac{GMm}{11R} = \frac{1}{2} mv_2^2 \]
- Simplifying the left side:
\[ \frac{GMm}{R} \left(1 - \frac{1}{11}\right) = \frac{1}{2} mv_2^2 \]
\[ \frac{GMm}{R} \cdot \frac{10}{11} = \frac{1}{2} mv_2^2 \]
- Canceling \( m \) and solving for \( v_2 \):
\[ v_2^2 = \frac{20GM}{11R} \implies v_2 = \sqrt{\frac{20GM}{11R}} \]

5. **Finding the Ratio of Velocities**:
- The ratio \( \frac{v_1}{v_2} \) is:
\[ \frac{v_1}{v_2} = \frac{\sqrt{\frac{2GM}{R}}}{\sqrt{\frac{20GM}{11R}}} = \sqrt{\frac{2}{\frac{20}{11}}} = \sqrt{\frac{2 \cdot 11}{20}} = \sqrt{\frac{22}{20}} = \sqrt{\frac{11}{10}} \]

### Final Answer:
The ratio of the velocities gained on reaching the Earth's surface from the two heights is:
\[ \frac{v_1}{v_2} = \sqrt{\frac{11}{10}} \]