Two particles of masses m and `M(Mgtm)` are connected by a cord that passes over a massless, frictionless pulley. The tension T in the string and the acceleration a of the particles is

Options

A. `(sqrt2Mmg)/(M+m)`
B. `(Mmg)/(M+m)`
C. `(2Mmg)/(M+m)`
D. `(sqrt3Mmg)/(M+m)`

Answer

To solve the problem of two particles of masses \( m \) and \( M \) (where \( M > m \)) connected by a cord over a massless, frictionless pulley, we need to find the tension \( T \) in the string and the acceleration \( a \) of the particles.

### Step-by-Step Solution:

1. **Identify Forces Acting on Each Mass:**
- For mass \( m \) (the lighter mass), the forces acting on it are:
- Tension \( T \) acting upward.
- Weight \( mg \) acting downward.
- For mass \( M \) (the heavier mass), the forces acting on it are:
- Weight \( Mg \) acting downward.
- Tension \( T \) acting upward.

2. **Write the Equations of Motion:**
- For mass \( m \):
\[ T - mg = ma \quad \text{(Equation 1)} \]
- For mass \( M \):
\[ Mg - T = Ma \quad \text{(Equation 2)} \]

3. **Add the Two Equations:**
- Adding Equation 1 and Equation 2:
\[ (T - mg) + (Mg - T) = ma + Ma \]
- This simplifies to:
\[ Mg - mg = (m + M)a \]

4. **Solve for Acceleration \( a \):**
- Rearranging the equation gives:
\[ (M - m)g = (m + M)a \]
- Thus, the acceleration \( a \) is:
\[ a = \frac{(M - m)g}{m + M} \]

5. **Substitute \( a \) Back to Find Tension \( T \):**
- Substitute \( a \) into Equation 1:
\[ T - mg = m\left(\frac{(M - m)g}{m + M}\right) \]
- Rearranging gives:
\[ T = mg + m\left(\frac{(M - m)g}{m + M}\right) \]

6. **Simplify the Expression for Tension \( T \):**
- Factor out \( g \):
\[ T = g\left(m + \frac{m(M - m)}{m + M}\right) \]
- Combine the terms:
\[ T = g\left(\frac{m(m + M) + m(M - m)}{m + M}\right) \]
- This simplifies to:
\[ T = g\left(\frac{mM + mg}{m + M}\right) \]

7. **Final Expression for Tension \( T \):**
- The final expression for tension \( T \) is:
\[ T = \frac{2mMg}{m + M} \]

### Summary of Results:
- The acceleration \( a \) of the particles is:
\[ a = \frac{(M - m)g}{m + M} \]
- The tension \( T \) in the string is:
\[ T = \frac{2mMg}{m + M} \]