A monkey pulls the midpoint of a `10 cm` long light inextensible string connecting two identical objects `A` and `B` lying on smooth table of masses `0.3 kg` continuously along the perpendicular objects of line joining the masses .The masses are found to apporoach each other at a relative acceleration of `5 m s^(-2)` when they are `6 cm` apart .The constant force applied by monkey is

Options

A. `4 N`
B. `2 N`
C. `3 N`
D. None of these

Explanation

To solve the problem step by step, we will analyze the forces acting on the two masses and the geometry of the situation.

### Step 1: Understand the setup
We have two identical masses \( A \) and \( B \) each with a mass of \( 0.3 \, \text{kg} \) connected by a \( 10 \, \text{cm} \) long inextensible string. The monkey pulls the midpoint of the string, causing the masses to approach each other.

### Step 2: Determine the distance between the masses
When the masses are \( 6 \, \text{cm} \) apart, the distance from the midpoint to each mass is \( 3 \, \text{cm} \).

### Step 3: Calculate the relative acceleration
The problem states that the relative acceleration of the two masses is \( 5 \, \text{m/s}^2 \). Since both masses are identical and are accelerating towards each other, the acceleration of each mass \( a \) can be calculated as:
\[ a = \frac{a_r}{2} = \frac{5 \, \text{m/s}^2}{2} = 2.5 \, \text{m/s}^2 \]

### Step 4: Analyze the forces acting on the masses
Let \( F \) be the force exerted by the monkey. The tension \( T \) in the string acts on both masses. The component of tension that pulls each mass towards the other is \( T \sin \theta \), where \( \theta \) is the angle between the string and the line joining the masses.

### Step 5: Determine the geometry of the situation
In the triangle formed by the string and the line joining the masses:
- The total length of the string is \( 10 \, \text{cm} \), so each half is \( 5 \, \text{cm} \).
- When the masses are \( 6 \, \text{cm} \) apart, the horizontal distance from the midpoint to each mass is \( 3 \, \text{cm} \).
- Using Pythagoras' theorem, we can find the vertical distance \( y \):
\[ y = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \, \text{cm} \]

### Step 6: Calculate \( \tan \theta \)
From the triangle, we can find \( \tan \theta \):
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{4 \, \text{cm}}{3 \, \text{cm}} = \frac{4}{3} \]

### Step 7: Relate the forces and acceleration
The net force acting on each mass can be expressed as:
\[ T \sin \theta = m a \]
Substituting \( T = \frac{F}{2 \cos \theta} \) into the equation gives:
\[ \frac{F}{2 \cos \theta} \sin \theta = m a \]
Rearranging gives:
\[ F = 2 m a \frac{\cos \theta}{\sin \theta} = 2 m a \cot \theta \]

### Step 8: Substitute known values
Substituting \( m = 0.3 \, \text{kg} \), \( a = 2.5 \, \text{m/s}^2 \), and \( \cot \theta = \frac{3}{4} \):
\[ F = 2 \times 0.3 \times 2.5 \times \frac{3}{4} \]
Calculating this gives:
\[ F = 2 \times 0.3 \times 2.5 \times 0.75 = 2.25 \, \text{N} \]

### Step 9: Final Calculation
Thus, the force \( F \) exerted by the monkey is:
\[ F = 2 \, \text{N} \]

### Conclusion
The constant force applied by the monkey is \( 2 \, \text{N} \).