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. `90^(@)`
D. `180^(@)`

Explanation

To solve the problem step by step, we will analyze the forces and their resultant using vector addition principles.

### Step 1: Understand the Given Forces
We have two forces:
- \( F_1 = 3P \)
- \( F_2 = 2P \)

The resultant of these two forces is given as \( R \).

### Step 2: Write the Expression for Resultant
The magnitude of the resultant \( R \) of two forces can be expressed using the formula:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} \]
Substituting the values of \( F_1 \) and \( F_2 \):
\[ R = \sqrt{(3P)^2 + (2P)^2 + 2(3P)(2P) \cos \theta} \]
\[ R = \sqrt{9P^2 + 4P^2 + 12P^2 \cos \theta} \]
\[ R = \sqrt{13P^2 + 12P^2 \cos \theta} \]

### Step 3: Resultant with First Force Doubled
When the first force is doubled, \( F_1 \) becomes \( 6P \). The new resultant \( R' \) is given as:
\[ R' = \sqrt{(6P)^2 + (2P)^2 + 2(6P)(2P) \cos \theta} \]
\[ R' = \sqrt{36P^2 + 4P^2 + 24P^2 \cos \theta} \]
\[ R' = \sqrt{40P^2 + 24P^2 \cos \theta} \]

### Step 4: Relate the Resultants
According to the problem, when the first force is doubled, the resultant is also doubled:
\[ R' = 2R \]
Substituting the expressions for \( R \) and \( R' \):
\[ \sqrt{40P^2 + 24P^2 \cos \theta} = 2\sqrt{13P^2 + 12P^2 \cos \theta} \]

### Step 5: Square Both Sides
Squaring both sides to eliminate the square roots:
\[ 40P^2 + 24P^2 \cos \theta = 4(13P^2 + 12P^2 \cos \theta) \]
\[ 40P^2 + 24P^2 \cos \theta = 52P^2 + 48P^2 \cos \theta \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ 40P^2 - 52P^2 = 48P^2 \cos \theta - 24P^2 \cos \theta \]
\[ -12P^2 = 24P^2 \cos \theta \]
Dividing both sides by \( 12P^2 \):
\[ -1 = 2 \cos \theta \]
\[ \cos \theta = -\frac{1}{2} \]

### Step 7: Finding the Angle
The angle \( \theta \) for which \( \cos \theta = -\frac{1}{2} \) is:
\[ \theta = 120^\circ \]

### Conclusion
Thus, the angle between the two forces is \( 120^\circ \).