R is the resultant of forces P and Q acting on a particle at O. If P is reversed, Q remaining the same, the resultant becomes R'. If R and R' are perpendicular to each other, then

Options

A. P=Q
B. P=2Q
C. 3P=Q
D. none

Answer

To solve the problem, we need to analyze the forces acting on a particle at point O and how the resultant changes when one of the forces is reversed.

### Step-by-Step Solution:

1. **Understanding the Forces**:
Let \( \vec{P} \) and \( \vec{Q} \) be the two forces acting on the particle at point O. The resultant force when both forces act in their original directions is denoted as \( \vec{R} \).

2. **Resultant Force Calculation**:
The resultant force \( \vec{R} \) can be expressed using the law of cosines:
\[ R = \sqrt{P^2 + Q^2 + 2PQ \cos \alpha} \]
where \( \alpha \) is the angle between the forces \( \vec{P} \) and \( \vec{Q} \).

3. **Reversing Force P**:
When force \( \vec{P} \) is reversed, it becomes \( -\vec{P} \), and the resultant force is now denoted as \( \vec{R'} \). The new resultant can be calculated as:
\[ R' = \sqrt{(-P)^2 + Q^2 + 2(-P)Q \cos \alpha} \]
Simplifying this gives:
\[ R' = \sqrt{P^2 + Q^2 - 2PQ \cos \alpha} \]

4. **Condition of Perpendicular Resultants**:
According to the problem, \( \vec{R} \) and \( \vec{R'} \) are perpendicular to each other. Therefore, we can use the property of perpendicular vectors:
\[ R^2 + R'^2 = (P + Q)^2 \]
This means:
\[ R^2 + R'^2 = R^2 + R'^2 = P^2 + Q^2 + 2PQ \cos \alpha + P^2 + Q^2 - 2PQ \cos \alpha \]
Simplifying this gives:
\[ R^2 + R'^2 = 2(P^2 + Q^2) \]

5. **Final Relation**:
Since \( R \) and \( R' \) are perpendicular, we can equate:
\[ R^2 + R'^2 = P^2 + Q^2 + 2PQ \cos \alpha + P^2 + Q^2 - 2PQ \cos \alpha = 2(P^2 + Q^2) \]
Thus, we derive the relationship between \( P \) and \( Q \):
\[ P^2 = Q^2 \sin^2 \alpha \]

### Conclusion:
The relationship derived from the conditions given in the problem is:
\[ P^2 = Q^2 \sin^2 \alpha \]