A price of length 1 m and mass 1 kg is suspended vertically from one end of a massless string passing on asmooth , fixed pulley . A ball of mass 2 kg is attached to the other end from the position shown . Time taken by the ball to cross the pipe is shown. Time taken by the ball to cross the pipe is

Options

A. `sqrt(3/(2g))`
B. ` sqrt(3/(4g))`
C. `sqrt(3/g)`
D. ` sqrt(3/(8g))`

Answer

To solve the problem, we need to determine the time taken by the ball to cross a pipe of length 1 meter while being suspended by a massless string over a smooth pulley.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Length of the pipe, \( L = 1 \, \text{m} \)
- Mass of the pipe, \( m = 1 \, \text{kg} \)
- Mass of the ball, \( M = 2 \, \text{kg} \)

2. **Calculate the Acceleration of the Ball:**
The acceleration \( a \) of the ball can be calculated using the formula:
\[ a = \frac{M - m}{M + m} \cdot g \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).
Substituting the values:
\[ a = \frac{2 - 1}{2 + 1} \cdot 10 = \frac{1}{3} \cdot 10 = \frac{10}{3} \, \text{m/s}^2 \]

3. **Calculate the Relative Acceleration:**
The relative acceleration \( a_r \) of the ball with respect to the pipe is given by:
\[ a_r = a + a = 2a \]
Substituting the value of \( a \):
\[ a_r = 2 \cdot \frac{10}{3} = \frac{20}{3} \, \text{m/s}^2 \]

4. **Calculate the Time Taken to Cross the Pipe:**
The time \( t \) taken by the ball to cross the pipe can be calculated using the formula:
\[ t = \sqrt{\frac{2L}{a_r}} \]
Substituting the values:
\[ t = \sqrt{\frac{2 \cdot 1}{\frac{20}{3}}} = \sqrt{\frac{2 \cdot 3}{20}} = \sqrt{\frac{6}{20}} = \sqrt{\frac{3}{10}} = \sqrt{\frac{3}{2g}} \quad \text{(since } g \approx 10 \text{)} \]

5. **Final Result:**
The time taken by the ball to cross the pipe is:
\[ t = \sqrt{\frac{3}{2g}} \quad \text{(where } g \text{ is approximately } 10 \text{ m/s}^2\text{)} \]