Body A is placed on frictionless wedge making an angle `theta` with the horizon. The horizontal acceleration towards left to be imparted to the wedge for the body A to freely fall vertically, is-

Options

A. (a)`g sin theta`
B. (b)` g cos theta`
C. (c)` g tan theta`
D. (d)` g cot theta`

Answer

To solve the problem of finding the horizontal acceleration that needs to be imparted to the wedge so that body A falls vertically, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a wedge inclined at an angle \( \theta \) with the horizontal.
- A body A is placed on this frictionless wedge.
- We need to find the horizontal acceleration \( a \) of the wedge towards the left such that body A falls vertically.

2. **Forces Acting on Body A**:
- The weight of body A acting downwards is \( mg \), where \( m \) is the mass of body A and \( g \) is the acceleration due to gravity.
- The normal force \( N \) exerted by the wedge on body A acts perpendicular to the surface of the wedge.

3. **Components of Forces**:
- The normal force can be resolved into two components:
- Horizontal component: \( N \sin \theta \)
- Vertical component: \( N \cos \theta \)

4. **Condition for Vertical Fall**:
- For body A to fall vertically, the horizontal acceleration of the wedge \( a \) must equal the horizontal component of the force acting on body A due to the normal force.
- Therefore, we can write the equation:
\[ N \sin \theta = ma \]

5. **Vertical Forces on Body A**:
- The vertical forces acting on body A must also balance out. The weight \( mg \) must equal the vertical component of the normal force:
\[ mg = N \cos \theta \]

6. **Solving for Normal Force**:
- From the second equation, we can express \( N \):
\[ N = \frac{mg}{\cos \theta} \]

7. **Substituting Normal Force**:
- Substitute the expression for \( N \) into the first equation:
\[ \frac{mg}{\cos \theta} \sin \theta = ma \]

8. **Simplifying the Equation**:
- This simplifies to:
\[ \frac{mg \sin \theta}{\cos \theta} = ma \]

9. **Canceling Mass**:
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{g \sin \theta}{\cos \theta} = a \]

10. **Final Expression**:
- Therefore, the required horizontal acceleration \( a \) of the wedge is:
\[ a = g \tan \theta \]

### Final Answer:
The horizontal acceleration towards the left that needs to be imparted to the wedge for body A to fall vertically is \( a = g \tan \theta \).