If the resultant of two forces of magnitudes `p` and `2p` is perpendicular to `p`, then the angle between the forces is

Options

A. `(2pi)/3`
B. `(3pi)/4`
C. `(4pi)/5`
D. `(5pi)/6`

Explanation

To solve the problem, we need to find the angle between two forces of magnitudes \( p \) and \( 2p \) when their resultant is perpendicular to the force \( p \).

### Step-by-Step Solution:

1. **Understanding the Forces**:
We have two forces:
- \( \vec{F_1} = p \)
- \( \vec{F_2} = 2p \)

2. **Resultant Force**:
The resultant force \( \vec{R} \) of these two forces can be expressed using the law of cosines. The angle between the two forces is denoted as \( \alpha \).

3. **Condition of Perpendicularity**:
According to the problem, the resultant \( \vec{R} \) is perpendicular to \( \vec{F_1} \). This means that the angle between \( \vec{R} \) and \( \vec{F_1} \) is \( 90^\circ \).

4. **Using the Formula for Resultant**:
The magnitude of the resultant \( R \) can be calculated as:
\[ R = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos \alpha} \]
Substituting the values of \( F_1 \) and \( F_2 \):
\[ R = \sqrt{p^2 + (2p)^2 + 2 \cdot p \cdot 2p \cdot \cos \alpha} \]
\[ R = \sqrt{p^2 + 4p^2 + 4p^2 \cos \alpha} \]
\[ R = \sqrt{5p^2 + 4p^2 \cos \alpha} \]

5. **Condition for Perpendicularity**:
Since \( R \) is perpendicular to \( p \), we can use the relationship:
\[ R \sin(90^\circ) = F_1 \cos \theta + F_2 \cos \alpha \]
Here, \( \theta \) is the angle between \( \vec{R} \) and \( \vec{F_2} \), which leads us to:
\[ 0 = p + 2p \cos \alpha \]
Rearranging gives:
\[ 2p \cos \alpha = -p \]
\[ \cos \alpha = -\frac{1}{2} \]

6. **Finding the Angle**:
The angle \( \alpha \) for which \( \cos \alpha = -\frac{1}{2} \) is:
\[ \alpha = 120^\circ \quad \text{(or } \frac{2\pi}{3} \text{ radians)} \]

### Final Answer:
The angle between the forces is \( 120^\circ \).

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