Mathematics in Physics

JEE Main level, numerical answer, medium

9 of 121

Question 9 of 121

Two forces Fˉ1\bar{F}_1 and Fˉ2\bar{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 F⃗1\vec{F}_1 and F⃗2\vec{F}_2 is cos⁡−1(1n)\cos ^{-1}\left(\frac{1}{n}\right). The value of ∣n∣|n| is _____.

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Answer: 6

∣F⃗1∣=F∣F→R∣=∣F→2∣=3 F FR2=F12+F22+2 F1 F2cos⁡θ9 F2=F2+9 F2+6 F2cos⁡θcos⁡θ=−16θ=cos⁡−1(1−6)n=−6∣n∣=6\begin{aligned} & \left|\vec{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\end{aligned}

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