A block of mass m is placed on a smooth wedge of inclination `theta`. The whole system is accelerated horizontally, so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be

Options

A. mg sin `theta`
B. mg
C. mg/cos `theta`
D. mg cos `theta`

Explanation

To solve the problem of finding the force exerted by the wedge on the block, we can follow these steps:

### Step 1: Understand the Forces Acting on the Block
The block of mass \( m \) is placed on a wedge inclined at an angle \( \theta \). The forces acting on the block are:
- The gravitational force \( mg \) acting downwards.
- The normal force \( N \) exerted by the wedge acting perpendicular to the surface of the wedge.
- A pseudo force \( F_p \) acting horizontally due to the acceleration \( a \) of the wedge.

### Step 2: Resolve the Gravitational Force
The weight of the block can be resolved into two components:
- Perpendicular to the inclined surface: \( mg \cos \theta \)
- Parallel to the inclined surface: \( mg \sin \theta \)

### Step 3: Resolve the Pseudo Force
The pseudo force due to the horizontal acceleration \( a \) can also be resolved:
- The component of the pseudo force acting perpendicular to the inclined surface: \( a \cos \theta \)
- The component acting parallel to the inclined surface: \( a \sin \theta \)

### Step 4: Apply the Condition for No Slipping
Since the block does not slip on the wedge, the forces acting perpendicular to the surface must balance:
\[ N = mg \cos \theta + m a \sin \theta \]

### Step 5: Substitute the Acceleration
From the problem, we know that the wedge is accelerated such that:
\[ a = g \tan \theta \]
This implies:
\[ a = g \frac{\sin \theta}{\cos \theta} \]

### Step 6: Substitute \( a \) into the Normal Force Equation
Substituting \( a \) into the normal force equation gives:
\[ N = mg \cos \theta + m(g \tan \theta) \sin \theta \]
Substituting \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ N = mg \cos \theta + mg \frac{\sin^2 \theta}{\cos \theta} \]

### Step 7: Combine Terms
Now, factor out \( mg \):
\[ N = mg \left( \cos \theta + \frac{\sin^2 \theta}{\cos \theta} \right) \]
Combining the terms gives:
\[ N = mg \left( \frac{\cos^2 \theta + \sin^2 \theta}{\cos \theta} \right) \]
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ N = \frac{mg}{\cos \theta} \]

### Final Result
Thus, the force exerted by the wedge on the block is:
\[ N = \frac{mg}{\cos \theta} \]