The resultant of two forces $3P$ and $2P$ is $R$, if the first force is doubled, the resultant is also doubled. The angle between the forces is

The correct option is B

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

Step 1: Give data

First force$\left(F_1\right)=3P$

Second force$\left(F_2\right)=2P$

Resultant force$\left(F_{combined}\right)=R$

Step 2: Formula used

The force is a vector quantity. The angle $\theta$ between two vectors $F_1,F_2$ can be obtained using the following formula:

$F_{combined}=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \theta }$

Step 3: Compute resultant force when $F_1=3P\&F_2=2P$

Substitute the known values in the formula $F_{combined}=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \theta }$,

$$\begin{gathered} R=\sqrt{{\left(3P\right)}^2+{\left(2P\right)}^2+2\times 3P\times 2P\cos \theta } \\ R=\sqrt{13P^2+12P^2\cos \theta } \\ R=P\sqrt{13+12\cos \theta }·······\left(1\right) \end{gathered}$$

Step 4: When the first force is doubled

When the first force is doubled i.e $F_1=6P$ then the resultant force also becomes double i.e $F_{combined}=2R$ . So, the resultant force is,

$$\begin{gathered} 2R=\sqrt{{\left(6P\right)}^2+{\left(2P\right)}^2+2\times 6P\times 2P\cos \theta } \\ 2R=\sqrt{40P^2+24P^2\cos \theta } \\ 2R=2P\sqrt{10+6\cos \theta } \\ R=P\sqrt{10+6\cos \theta }·······\left(2\right) \end{gathered}$$

Step 5: Compute the angle

Equate the equation $\left(1\right)$ and equation $\left(2\right)$,

$\sqrt{13+12\cos \theta }=\sqrt{10+6\cos \theta }$

Take the square of both sides,

$$\begin{gathered} 13+12\cos \theta =10+6\cos \theta \\ 6\cos \theta =-3 \\ \cos \theta =-\frac{1}{2} \\ \theta ={120}^{°} \end{gathered}$$

Thus, the angle between the force is ${120}^{\theta }or\frac{2\mathrm{\pi }}{3}.$

Hence, option B is the correct answer.