Three forces are acting on a particle of mass m initially in equilibrium if the first two forces `(vec(R)_(1)and vec(R)_(2))` are perpendicular to each other and suddenly the third force `(vec(R)_(3))` is removed, then the magnitude of acceleration of the particle is:

Options

A. `1/m|vec(R)_(2)|`
B. `1/m|vec(R)_(1)+vec(R)_(2)|`
C. `1/m|vec(R)_(1)-vec(R)_(2)|`
D. `1/m|vec(R)_(1)|`

Explanation

To solve the problem, we need to analyze the forces acting on the particle and apply Newton's laws of motion. Here’s a step-by-step solution:

### Step 1: Understand the initial conditions
The particle of mass \( m \) is initially in equilibrium, meaning the net force acting on it is zero. This implies that the sum of all forces acting on the particle is balanced.

### Step 2: Identify the forces acting on the particle
We have three forces acting on the particle:
- \( \vec{R}_1 \)
- \( \vec{R}_2 \)
- \( \vec{R}_3 \)

Given that \( \vec{R}_1 \) and \( \vec{R}_2 \) are perpendicular to each other, we can represent them as vectors in a coordinate system.

### Step 3: Apply the equilibrium condition
Since the particle is in equilibrium, we have:
\[ \vec{R}_1 + \vec{R}_2 + \vec{R}_3 = 0 \]
This means:
\[ \vec{R}_3 = -(\vec{R}_1 + \vec{R}_2) \]

### Step 4: Remove the third force
Now, if the third force \( \vec{R}_3 \) is suddenly removed, the new net force acting on the particle becomes:
\[ \vec{F}_{net} = \vec{R}_1 + \vec{R}_2 \]

### Step 5: Calculate the magnitude of the net force
Since \( \vec{R}_1 \) and \( \vec{R}_2 \) are perpendicular, we can use the Pythagorean theorem to find the magnitude of the net force:
\[ |\vec{F}_{net}| = \sqrt{|\vec{R}_1|^2 + |\vec{R}_2|^2} \]

### Step 6: Apply Newton's second law
According to Newton's second law:
\[ \vec{F}_{net} = m \vec{a} \]
where \( \vec{a} \) is the acceleration of the particle. Therefore, we can express the acceleration as:
\[ \vec{a} = \frac{\vec{F}_{net}}{m} \]

### Step 7: Substitute the net force into the equation
Substituting the expression for the net force, we get:
\[ \vec{a} = \frac{\sqrt{|\vec{R}_1|^2 + |\vec{R}_2|^2}}{m} \]

### Step 8: Determine the magnitude of acceleration
The magnitude of the acceleration is then given by:
\[ |\vec{a}| = \frac{|\vec{R}_1| + |\vec{R}_2|}{m} \]

### Conclusion
Thus, the magnitude of acceleration of the particle after the removal of the third force is:
\[ |\vec{a}| = \frac{|\vec{R}_1| + |\vec{R}_2|}{m} \]

### Final Answer
The correct option is:
\[ \frac{1}{m} \times |\vec{R}_1 + \vec{R}_2| \]

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