JEE Mains · Physics · STD 11 - Vectors
- JEE Mains 2024
- Hard
- Numerical
Two forces \(\vec{F}_1\) and \(\vec{F}_2\) 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 \(\overrightarrow{\mathrm{F}}_2\) is \(\cos ^{-1}\left(\frac{1}{\mathrm{n}}\right)\). The value of \(|\mathrm{n}|\) is __________.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(6\)
Step-by-step Solution
Detailed explanation
\(\left|\overrightarrow{\mathrm{F}}_1\right|=\mathrm{F}\)
\(\left|\overrightarrow{\mathrm{F}}_{\mathrm{R}}\right|=\left|\overrightarrow{\mathrm{F}}_2\right|=3 \mathrm{~F}\)
\(\mathrm{~F}_{\mathrm{R}}^2=\mathrm{F}_1^2+\mathrm{F}_2^2+2 \mathrm{~F}_1 \mathrm{~F}_2 \cos \theta\)
\(9 \mathrm{~F}^2=\mathrm{F}^2+9 \mathrm{~F}^2+6 \mathrm{~F}^2 \cos \theta\)
\(\cos \theta=-\frac{1}{6}\)
\(\theta=\cos ^{-1}\left(\frac{1}{-6}\right)\)
\(\mathrm{n}=-6\)
\(|\mathrm{n}|=6\)
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