The resultant of two forces 3P and 2P is R. If the first force is doubled then resultant is also doubled.The angle between the two forces is

Options

A. `60^@`
B. `120^@`
C. `30^@`
D. `135^@`

Explanation

To solve the problem, we need to find the angle between the two forces, 3P and 2P, given that when the first force is doubled, the resultant also doubles.

### Step-by-Step Solution:

1. **Understanding the Resultant of Two Forces**:
The resultant \( R \) of two forces \( F_1 \) and \( F_2 \) acting at an angle \( \theta \) is given by the formula:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} \]
In our case, \( F_1 = 3P \) and \( F_2 = 2P \). Therefore, the resultant can be expressed as:
\[ R = \sqrt{(3P)^2 + (2P)^2 + 2(3P)(2P) \cos \theta} \]
Simplifying this gives:
\[ R = \sqrt{9P^2 + 4P^2 + 12P^2 \cos \theta} \]
\[ R = \sqrt{13P^2 + 12P^2 \cos \theta} \]

2. **Doubling the First Force**:
When the first force is doubled, the new force becomes \( 6P \). The new resultant \( R' \) is given by:
\[ R' = \sqrt{(6P)^2 + (2P)^2 + 2(6P)(2P) \cos \theta} \]
Simplifying this gives:
\[ R' = \sqrt{36P^2 + 4P^2 + 24P^2 \cos \theta} \]
\[ R' = \sqrt{40P^2 + 24P^2 \cos \theta} \]

3. **Setting Up the Equation**:
According to the problem, the new resultant \( R' \) is double the original resultant \( R \):
\[ R' = 2R \]
Substituting the expressions for \( R \) and \( R' \):
\[ \sqrt{40P^2 + 24P^2 \cos \theta} = 2 \sqrt{13P^2 + 12P^2 \cos \theta} \]

4. **Squaring Both Sides**:
Squaring both sides to eliminate the square roots:
\[ 40P^2 + 24P^2 \cos \theta = 4(13P^2 + 12P^2 \cos \theta) \]
Expanding the right side:
\[ 40P^2 + 24P^2 \cos \theta = 52P^2 + 48P^2 \cos \theta \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ 40P^2 - 52P^2 + 24P^2 \cos \theta - 48P^2 \cos \theta = 0 \]
\[ -12P^2 - 24P^2 \cos \theta = 0 \]
Dividing through by \( -12P^2 \):
\[ 1 + 2 \cos \theta = 0 \]
This simplifies to:
\[ 2 \cos \theta = -1 \]
\[ \cos \theta = -\frac{1}{2} \]

6. **Finding the Angle**:
The angle \( \theta \) for which \( \cos \theta = -\frac{1}{2} \) is:
\[ \theta = 120^\circ \]

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