The square of the resultant of two forces 4N and 3 N exceeds the square of the resultant of the two forces by 12 when they are mutually perpendicular. The angle between the vectors is

Options

A. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `120^(@)`

Answer

To solve the problem, we need to find the angle between two forces, given that the square of the resultant of two forces (4N and 3N) exceeds the square of the resultant of the same forces when they are mutually perpendicular by 12.

### Step-by-Step Solution:

1. **Understand 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:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} \]

2. **Calculate the Resultant When Forces are Perpendicular**:
When the forces are mutually perpendicular, \( \theta = 90^\circ \) and \( \cos 90^\circ = 0 \). Thus, the resultant becomes:
\[ R_{\perpendicular} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{N} \]
The square of this resultant is:
\[ R_{\perpendicular}^2 = 25 \]

3. **Set Up the Equation for the Given Condition**:
According to the problem, the square of the resultant when the angle is \( \theta \) exceeds the square of the resultant when they are perpendicular by 12:
\[ R^2 - R_{\perpendicular}^2 = 12 \]
Substituting \( R_{\perpendicular}^2 = 25 \):
\[ R^2 - 25 = 12 \]
Therefore:
\[ R^2 = 37 \]

4. **Express the Resultant in Terms of the Angle \( \theta \)**:
Now, we can express \( R^2 \) using the formula for the resultant:
\[ R^2 = F_1^2 + F_2^2 + 2F_1F_2 \cos \theta \]
Substituting \( F_1 = 4 \, \text{N} \) and \( F_2 = 3 \, \text{N} \):
\[ R^2 = 4^2 + 3^2 + 2 \cdot 4 \cdot 3 \cos \theta \]
This simplifies to:
\[ R^2 = 16 + 9 + 24 \cos \theta = 25 + 24 \cos \theta \]

5. **Set the Two Expressions for \( R^2 \) Equal**:
Now we can set the two expressions for \( R^2 \) equal:
\[ 25 + 24 \cos \theta = 37 \]

6. **Solve for \( \cos \theta \)**:
Rearranging gives:
\[ 24 \cos \theta = 37 - 25 \]
\[ 24 \cos \theta = 12 \]
\[ \cos \theta = \frac{12}{24} = \frac{1}{2} \]

7. **Find the Angle \( \theta \)**:
The angle \( \theta \) for which \( \cos \theta = \frac{1}{2} \) is:
\[ \theta = 60^\circ \]

### Final Answer:
The angle between the vectors is \( 60^\circ \).