QJEEM24Ph455c: JEE Main 2024 Physics - Class 11 JEEM24Ph455c Solution: Two forces and are acting on a body. One force has magnitude…

Let's denote the magnitude of the smaller force as $F$ , hence the magnitude of the larger force is $3F$ . The resultant force $\vec{R}$ is equal in magnitude to the larger force, which means $|\vec{R}| = 3F$ . When two forces $\vec{F}_1$ and $\vec{F}_2$ act on a body, the magnitude of their resultant $\vec{R}$ can be found using the law of vector addition:

$|\vec{R}| = \sqrt{|\vec{F}_1|^2 + |\vec{F}_2|^2 + 2|\vec{F}_1||\vec{F}_2|\cos\theta}$ ,

where $\theta$ is the angle between $\vec{F}_1$ and $\vec{F}_2$ . Given that in our case $|\vec{R}| = 3F$ , $|\vec{F}_1| = F$ and $|\vec{F}_2| = 3F$ , by substituting these values into the equation, we get:

$3F = \sqrt{F^2 + (3F)^2 + 2(F)(3F)\cos\theta}$

$9F^2 = F^2 + 9F^2 + 6F^2\cos\theta$

Simplifying this equation by subtracting $10F^2$ from both sides gives:

$-F^2 = 6F^2\cos\theta$

Dividing both sides by $-F^2$ gives:

$-1 = -6\cos\theta$

Therefore, $\cos\theta = \frac{1}{6}$ .

It is given that the angle between $\vec{F}_1$ and $\vec{F}_2$ is $\cos^{-1}\left(\frac{1}{n}\right)$ , hence comparing this with the above result, we find that $n = 6$ . Therefore, $|n| = 6$ .