A mass m rests on another mass M. The co-efficient of static friction between the surfaces of m and M is `mu`. M rests on smooth frictionless horizontal plane. The maximum force applied horizontally on M for which m will move along with M without slipping is

Options

A. mg + `mu` mg
B. `(M + m) mu g `
C. `(mM)/(M + m) "" mu g `
D. `mu mg`

Explanation

To solve the problem, we need to find the maximum force \( F \) that can be applied horizontally on mass \( M \) such that mass \( m \) moves along with \( M \) without slipping.

### Step-by-step Solution:

1. **Identify Forces Acting on Mass \( m \)**:
- The gravitational force acting on mass \( m \) is \( mg \) (downward).
- The normal force \( N \) exerted by mass \( M \) on mass \( m \) is equal to the gravitational force, hence \( N = mg \).
- The static friction force \( f_s \) acts to the right (opposing the relative motion) and is given by \( f_s = \mu N \).

2. **Set Up the Equations**:
- The maximum static friction force that can act on mass \( m \) is:
\[ f_s = \mu N = \mu mg \]
- For mass \( m \) to move along with mass \( M \) without slipping, the friction force must equal the net force required to accelerate mass \( m \) at the same acceleration \( a \) as mass \( M \):
\[ f_s = ma \]
- Therefore, we have:
\[ \mu mg = ma \]

3. **Solve for Acceleration \( a \)**:
- From the equation \( \mu mg = ma \), we can cancel \( m \) (assuming \( m \neq 0 \)):
\[ a = \mu g \]

4. **Determine the Force \( F \) on Mass \( M \)**:
- The total force \( F \) applied to mass \( M \) must be equal to the total mass being accelerated (both \( M \) and \( m \)):
\[ F = (M + m)a \]
- Substituting \( a = \mu g \):
\[ F = (M + m)(\mu g) \]

5. **Final Expression for Maximum Force**:
- Thus, the maximum force \( F \) that can be applied to mass \( M \) without causing mass \( m \) to slip is:
\[ F = \mu g (M + m) \]

### Conclusion:
The maximum force applied horizontally on \( M \) for which \( m \) will move along with \( M \) without slipping is:
\[ F = \mu g (M + m) \]