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 the velocities gained by a particle falling to the Earth's surface from two different heights, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Define the scenario
We have two cases:
1. A particle falling from infinity.
2. A particle falling from a height of 10 times the radius of the Earth (10R).

### Step 2: Calculate the velocity for the first case (falling from infinity)
- The gravitational potential energy (U) at infinity is considered to be zero.
- The gravitational potential energy at the surface of the Earth (U) 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 particle, and \( R \) is the radius of the Earth.

- The work done (W) in bringing the particle from infinity to the Earth's surface is:
\[ W = U_{\text{surface}} - U_{\text{infinity}} = -\frac{GMm}{R} - 0 = -\frac{GMm}{R} \]

- This work done is converted into kinetic energy (KE) when the particle reaches the surface:
\[ KE = \frac{1}{2} mv_1^2 \]

- Setting the work done equal to the kinetic energy:
\[ -\frac{GMm}{R} = \frac{1}{2} mv_1^2 \]

- Canceling \( m \) from both sides, we find:
\[ -\frac{GM}{R} = \frac{1}{2} v_1^2 \]

- Rearranging gives:
\[ v_1^2 = \frac{2GM}{R} \]
\[ v_1 = \sqrt{\frac{2GM}{R}} \]

### Step 3: Calculate the velocity for the second case (falling from height 10R)
- The gravitational potential energy at a height of 10R is:
\[ U = -\frac{GMm}{R + 10R} = -\frac{GMm}{11R} \]

- The work done in bringing the particle from height 10R to the surface is:
\[ W = U_{\text{surface}} - U_{\text{height}} = -\frac{GMm}{R} - \left(-\frac{GMm}{11R}\right) \]
\[ W = -\frac{GMm}{R} + \frac{GMm}{11R} = \frac{GMm}{11R} - \frac{11GMm}{11R} = \frac{10GMm}{11R} \]

- This work done is converted into kinetic energy:
\[ \frac{10GMm}{11R} = \frac{1}{2} mv_2^2 \]

- Canceling \( m \) gives:
\[ \frac{10GM}{11R} = \frac{1}{2} v_2^2 \]

- Rearranging gives:
\[ v_2^2 = \frac{20GM}{11R} \]
\[ v_2 = \sqrt{\frac{20GM}{11R}} \]

### Step 4: Calculate the ratio of the velocities
- Now, we find the ratio \( \frac{v_1}{v_2} \):
\[ \frac{v_1}{v_2} = \frac{\sqrt{\frac{2GM}{R}}}{\sqrt{\frac{20GM}{11R}}} \]

- Simplifying gives:
\[ \frac{v_1}{v_2} = \frac{\sqrt{2}}{\sqrt{20/11}} = \frac{\sqrt{2} \cdot \sqrt{11}}{\sqrt{20}} = \frac{\sqrt{22}}{\sqrt{20}} = \frac{\sqrt{11}}{\sqrt{10}} \]

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