JEE Main 2024 — Laws Of Motion Question with Solution

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, $$|n|=6$$.