If the resultannt of two forces of magnitudes P and Q acting at a point at an angle of `60^(@)` is `sqrt(7)Q`, then P/Q is

Options

A. 1
B. `(3)/(2)`
C. `2`
D. 4

Explanation

To solve the problem, we need to find the ratio \( \frac{P}{Q} \) given that the resultant of two forces \( P \) and \( Q \) acting at an angle of \( 60^\circ \) is \( \sqrt{7}Q \).

### Step-by-Step Solution:

1. **Use the formula for the resultant of two forces**:
The formula for the resultant \( R \) of two forces \( P \) and \( Q \) acting at an angle \( \theta \) is given by:
\[ R = \sqrt{P^2 + Q^2 + 2PQ \cos(\theta)} \]
Here, \( \theta = 60^\circ \), so \( \cos(60^\circ) = \frac{1}{2} \).

2. **Substituting the values into the formula**:
We know that the resultant \( R \) is \( \sqrt{7}Q \). Therefore, we can set up the equation:
\[ \sqrt{7}Q = \sqrt{P^2 + Q^2 + 2PQ \cdot \frac{1}{2}} \]
Simplifying this gives:
\[ \sqrt{7}Q = \sqrt{P^2 + Q^2 + PQ} \]

3. **Square both sides**:
Squaring both sides to eliminate the square root yields:
\[ 7Q^2 = P^2 + Q^2 + PQ \]

4. **Rearranging the equation**:
Rearranging the equation gives:
\[ P^2 + Q^2 + PQ - 7Q^2 = 0 \]
Simplifying further, we have:
\[ P^2 + PQ - 6Q^2 = 0 \]

5. **Dividing by \( Q^2 \)**:
To find \( \frac{P}{Q} \), let \( x = \frac{P}{Q} \). Then \( P = xQ \). Substituting this into the equation gives:
\[ (xQ)^2 + (xQ)Q - 6Q^2 = 0 \]
This simplifies to:
\[ x^2Q^2 + xQ^2 - 6Q^2 = 0 \]
Dividing through by \( Q^2 \) (assuming \( Q \neq 0 \)):
\[ x^2 + x - 6 = 0 \]

6. **Factoring the quadratic equation**:
The quadratic equation \( x^2 + x - 6 = 0 \) can be factored as:
\[ (x - 2)(x + 3) = 0 \]

7. **Finding the solutions**:
Setting each factor to zero gives:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]

8. **Selecting the valid solution**:
Since \( x = \frac{P}{Q} \) represents a ratio of magnitudes, it cannot be negative. Therefore, we discard \( x = -3 \) and accept:
\[ \frac{P}{Q} = 2 \]

### Final Answer:
\[ \frac{P}{Q} = 2 \]