If the angle $\alpha$ between 2 forces of equal magnitude is reduced to $\alpha -\frac{\mathrm{\pi }}{3}$, then the magnitude of their resultant becomes $\sqrt{3}$ times the earlier one. The angle $\alpha$ is

The correct option is B

$\frac{2\mathrm{\pi }}{3}$

Step 1: Given data

The magnitude of the force is $F$

The angle between the forces is $\alpha$

Resultant force when the angle is $\alpha$ is $R$

Resultant force when the angle is $\alpha -\frac{\mathrm{\pi }}{3}$ is $R\text{'}$

Step 2: Formula Used

The magnitude of resultant of two forces is given by $R=\sqrt{F^2+F^2+2F^2\cos \alpha }$

Step 3: Solution

The magnitude of resultant when the angle is $\alpha$ is $R=\sqrt{F^2+F^2+2F^2\cos \alpha }$

Magnitude of resultant when angle is $\alpha -\frac{\mathrm{\pi }}{3}$ . Replace $R\text{'}$ by $\sqrt{3}R$ since the magnitude of their resultant becomes $\sqrt{3}$ times the earlier one.

$$\begin{gathered} R\text{'}=\sqrt{F^2+F^2+2F^2\cos \left(\alpha -\frac{\mathrm{\pi }}{3}\right)} \\ \Rightarrow \sqrt{3}R=\sqrt{F^2+F^2+2F^2\cos \left(\alpha -\frac{\mathrm{\pi }}{3}\right)} \end{gathered}$$

Substitute $R=\sqrt{F^2+F^2+2F^2\cos \alpha }$ in the above expression.

$$\begin{gathered} \sqrt{3}\left(\sqrt{F^2+F^2+2F^2\cos \alpha }\right)=\sqrt{F^2+F^2+2F^2\cos \left(\alpha -\frac{\mathrm{\pi }}{3}\right)} \\ \Rightarrow {\left[\sqrt{3}\left(\sqrt{2F^2\left(1+\cos \alpha \right)}\right)\right]}^2={\left[\sqrt{2F^2\left(1+\cos \left(\alpha -\frac{\mathrm{\pi }}{3}\right)\right)}\right]}^2 \\ \Rightarrow 2F^2\left[3\left(1+\cos \alpha \right)\right]=2F^2\left(1+\cos \left(\alpha -\frac{\mathrm{\pi }}{3}\right)\right) \\ \Rightarrow 3+3\cos \alpha =1+\cos \left(\alpha -\frac{\mathrm{\pi }}{3}\right) \end{gathered}$$

On substituting the value of $\alpha =\frac{2\mathrm{\pi }}{3}$ from option B in the above equation, we get left hand side of the equation equal to the right hand side as shown below:

$$\begin{gathered} \Rightarrow 3+3\cos \left(\frac{2\mathrm{\pi }}{3}\right)=1+\cos \left(\frac{2\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{3}\right) \\ \Rightarrow 3+3\cos \left(\frac{2\mathrm{\pi }}{3}\right)=1+\cos \left(\frac{2\mathrm{\pi }-\mathrm{\pi }}{3}\right) \\ \Rightarrow 3+3\cos \left(\frac{2\mathrm{\pi }}{3}\right)=1+\cos \left(\frac{\mathrm{\pi }}{3}\right) \\ \Rightarrow 3\cos \left(\frac{2\mathrm{\pi }}{3}\right)-\cos \left(\frac{\mathrm{\pi }}{3}\right)=1-3 \\ \Rightarrow 3\times \left(\frac{-1}{2}\right)-\left(\frac{1}{2}\right)=-2\left[\because \cos \left(\frac{2\mathrm{\pi }}{3}\right)=\left(\frac{-1}{2}\right);\cos \left(\frac{\mathrm{\pi }}{3}\right)=\left(\frac{1}{2}\right)\right] \\ \Rightarrow \frac{-3-1}{2}=-2 \\ \Rightarrow \frac{-4}{2}=-2 \\ \Rightarrow -2=-2 \\ \Rightarrow L.H.S=R.H.S \end{gathered}$$

Hence, option B is correct i.e., angle $\alpha$ is $\frac{2\mathrm{\pi }}{3}$ .