The square of the resultant of two forces `4 N` and `3N` 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^(@)`

Explanation

To solve the problem step by step, we need to analyze the forces given and apply the relevant physics concepts.

### Step 1: Understand the Problem
We have two forces, \( F_1 = 4 \, \text{N} \) and \( F_2 = 3 \, \text{N} \). The problem states that the square of the resultant of these two forces when they are at an angle \( \theta \) exceeds the square of the resultant when they are mutually perpendicular by 12.

### Step 2: Calculate the Resultant When Forces are Perpendicular
When the forces are perpendicular, we can use the Pythagorean theorem to find the resultant \( R \):
\[ R_{\perpendicular} = \sqrt{F_1^2 + F_2^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{N} \]
Thus, the square of the resultant is:
\[ R_{\perpendicular}^2 = 5^2 = 25 \]

### Step 3: Set Up the Equation for the General Case
For the case where the angle between the forces is \( \theta \), the resultant \( R \) can be calculated using the formula:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} \]
Substituting the values of \( F_1 \) and \( F_2 \):
\[ R = \sqrt{4^2 + 3^2 + 2 \cdot 4 \cdot 3 \cos \theta} = \sqrt{16 + 9 + 24 \cos \theta} = \sqrt{25 + 24 \cos \theta} \]
Thus, the square of the resultant is:
\[ R^2 = 25 + 24 \cos \theta \]

### Step 4: Set Up the Equation Based on the Problem Statement
According to the problem, the square of the resultant when at angle \( \theta \) exceeds the square of the resultant when they are perpendicular by 12:
\[ R^2 - R_{\perpendicular}^2 = 12 \]
Substituting the values we have:
\[ (25 + 24 \cos \theta) - 25 = 12 \]
This simplifies to:
\[ 24 \cos \theta = 12 \]

### Step 5: Solve for \( \cos \theta \)
Dividing both sides by 24 gives:
\[ \cos \theta = \frac{12}{24} = \frac{1}{2} \]

### Step 6: Determine 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 \( 60^\circ \).

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