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. **Write 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. **Substitute the values into the formula**:
We know that \( R = \sqrt{7}Q \), so substituting the values we have:
\[ \sqrt{7}Q = \sqrt{P^2 + Q^2 + 2PQ \cdot \frac{1}{2}} \]
This simplifies to:
\[ \sqrt{7}Q = \sqrt{P^2 + Q^2 + PQ} \]

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

4. **Rearrange the equation**:
Rearranging the equation, we get:
\[ P^2 + PQ + Q^2 - 7Q^2 = 0 \]
This simplifies to:
\[ P^2 + PQ - 6Q^2 = 0 \]

5. **Factor the quadratic equation**:
We can factor the quadratic equation:
\[ P^2 + PQ - 6Q^2 = 0 \]
This can be factored as:
\[ (P + 3Q)(P - 2Q) = 0 \]

6. **Solve for \( P \)**:
Setting each factor to zero gives us two equations:
- \( P + 3Q = 0 \) → \( P = -3Q \) (not valid since forces cannot be negative)
- \( P - 2Q = 0 \) → \( P = 2Q \)

7. **Find the ratio \( \frac{P}{Q} \)**:
From \( P = 2Q \), we can find the ratio:
\[ \frac{P}{Q} = \frac{2Q}{Q} = 2 \]

### Final Answer:
Thus, the value of \( \frac{P}{Q} \) is \( 2 \).

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