A block of mass'm' is resting on the floor of a lift. The coefficient of friction between the block and the floor is `mu`. When the lift is falling freely, the limiting frictional force between block and surface is

Options

A. `mu` mg
B. `(mu mg)/s`
C. mg
D. zero

Answer

To solve the problem, we need to analyze the forces acting on the block when the lift is falling freely.

### Step-by-Step Solution:

1. **Understanding the Situation**:
- The block of mass \( m \) is resting on the floor of a lift.
- The lift is falling freely, which means it is in free fall under the influence of gravity.

2. **Identifying Forces**:
- When the lift is in free fall, both the lift and the block experience the same acceleration, which is equal to \( g \) (acceleration due to gravity).
- The forces acting on the block are:
- The gravitational force \( mg \) acting downward.
- The normal force \( N \) acting upward from the floor of the lift.

3. **Applying Newton's Second Law**:
- According to Newton's second law, the net force acting on the block is equal to its mass times its acceleration.
- Since the lift is falling freely, the acceleration of the block is also \( g \) downward.

4. **Setting Up the Equation**:
- The net force acting on the block can be expressed as:
\[ mg - N = ma \]
- Here, \( a = g \) (the block is falling with the lift), so we can substitute:
\[ mg - N = mg \]

5. **Solving for Normal Force \( N \)**:
- Rearranging the equation gives:
\[ mg - N = mg \implies N = 0 \]
- This means that the normal force \( N \) acting on the block is zero.

6. **Calculating Limiting Frictional Force**:
- The limiting frictional force \( F_f \) is given by:
\[ F_f = \mu N \]
- Since \( N = 0 \):
\[ F_f = \mu \times 0 = 0 \]

### Final Answer:
The limiting frictional force between the block and the surface when the lift is falling freely is **0**.