An elevator accelerates upwards at a constant rate. A uniform string of length L and mass m supports a small block of mass M that hangs from the ceiling of the elevator. The tension at distance l from the ceiling is T. The acceleration of the elevator is

Options

A. `T/(M+m-(ml)/L)-g`
B. `T/(2M+m-(ml)/L)+g`
C. `T/(M+(ml)/L)-g`
D. `T/(2M+m+(ml)/L)-g`

Answer

To solve the problem, we need to analyze the forces acting on the small block of mass \( M \) and the portion of the string below the point where the tension \( T \) is being measured. The elevator is accelerating upwards, which affects the tension in the string.

### Step-by-Step Solution:

1. **Identify the System**:
- We have a small block of mass \( M \) hanging from a string of length \( L \) and mass \( m \) inside an elevator that is accelerating upwards with acceleration \( a \).

2. **Determine the Mass of the String Below the Point of Interest**:
- The mass density \( \mu \) of the string is given by:
\[ \mu = \frac{m}{L} \]
- The mass of the portion of the string below a distance \( l \) from the ceiling is:
\[ m_1 = \mu \cdot (L - l) = \frac{m}{L} \cdot (L - l) = \frac{m(L - l)}{L} \]

3. **Calculate the Total Mass Below the Point**:
- The total mass hanging below the point where tension \( T \) is measured includes the mass of the block \( M \) and the mass of the string \( m_1 \):
\[ m_{\text{total}} = M + m_1 = M + \frac{m(L - l)}{L} \]

4. **Apply Newton's Second Law**:
- The forces acting on the mass below the point are the tension \( T \) acting upwards and the weight acting downwards, which is the total mass times the effective acceleration due to gravity \( (g + a) \):
\[ T = m_{\text{total}} \cdot (g + a) \]
- Substituting \( m_{\text{total}} \):
\[ T = \left(M + \frac{m(L - l)}{L}\right)(g + a) \]

5. **Rearranging for Acceleration**:
- We can rearrange the equation to solve for \( a \):
\[ a = \frac{T}{M + \frac{m(L - l)}{L}} - g \]

6. **Final Expression**:
- Thus, the acceleration of the elevator \( a \) is given by:
\[ a = \frac{T}{M + \frac{m(L - l)}{L}} - g \]