Physics · Vectors and Scalars

JEE Main 2024 — 4 April, Shift 1 — Question 57

Two forces $\overrightarrow{\mathrm{F}}_{1}$ and $\overrightarrow{\mathrm{F}}_{2}$ are acting on a body. One force has magnitude thrice that of the other force and the resultant of the two forces is equal to the force of larger magnitude. The angle between $\vec{F}_{1}$ and $\vec{F}_{2}$ is $\cos ^{-1}\left(\frac{1}{n}\right)$ . The value of $|n|$ is $\qquad$ .

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

$\left|\overrightarrow{\mathrm{F}}_{1}\right|=\mathrm{F}$ $\left|\overrightarrow{\mathrm{F}}_{\mathrm{R}}\right|=\left|\overrightarrow{\mathrm{F}}_{2}\right|=3 \mathrm{~F}$

$\mathrm{F}_{\mathrm{R}}^{2}=\mathrm{F}_{1}^{2}+\mathrm{F}_{2}^{2}+2 \mathrm{~F}_{1} \mathrm{~F}_{2} \cos \theta$

$9 \mathrm{~F}^{2}=\mathrm{F}^{2}+9 \mathrm{~F}^{2}+6 \mathrm{~F}^{2} \cos \theta$

$\cos \theta=-\frac{1}{6}$

$\theta=\cos ^{-1}\left(\frac{1}{-6}\right)$

$\mathrm{n}=-6$

$|\mathrm{n}|=6$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Physics
Chapter: Vectors and Scalars
Topic: Properties and addition/Subtraction of Vectors