Question

If the resultant of two forces of magnitudes $P$ and $Q$ acting at a point at an angle of $60°$ is $\sqrt{7}Q$, then $\frac{P}{Q}$ is

• A. $1$
• B. $\frac{3}{2}$
• C. $2$
• D. $4$

Answer 1

The correct option is C

$2$

Step 1: Given data:

The resultant of two forces of magnitudes $P$ and $Q$ acting at an angle of ${60}^{°}$is $\sqrt{7}Q$.

Step 2: To find:

The value of $\frac{P}{Q}$

Step 3: Calculating $\frac{P}{Q}$

The resultant of two vectors $P$ and $Q$ is given by

$R=\sqrt{P^2+Q^2+2PQ\cos \theta }$

So, $\sqrt{7}Q=\sqrt{P^2+Q^2+2PQ\times \frac{1}{2}}$

Since, $\cos {60}^{°}=\frac{1}{2}$

$\sqrt{7}Q=\sqrt{P^2+Q^2+2PQ\times \frac{1}{2}}$

Squaring on both sides,

$\Rightarrow 7Q^2=P^2+Q^2+PQ$

Dividing by $Q^2$ on both sides,

$$\begin{gathered} \Rightarrow 7=\frac{P^2}{Q^2}+1+\frac{P}{Q} \\ \Rightarrow \frac{P^2}{Q^2}+\frac{P}{Q}-6=0 \end{gathered}$$

Let $\frac{P}{Q}=x$,

$$\begin{gathered} \Rightarrow x^2+x-6=0 \\ \Rightarrow x^2+3x-2x-6=0 \\ \Rightarrow x(x+3)-2(x-3)=0 \\ \Rightarrow (x-2)(x+3)=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow x=2orx=-3 \\ \Rightarrow \frac{P}{Q}=2or\frac{P}{Q}=-3 \end{gathered}$$

$\frac{P}{Q}$ cannot be negative as the magnitudes are positive.

Therefore $\frac{P}{Q}=2$

Hence, option C is the correct answer.