Rotational Motion

JEE Main level, numerical answer, medium

120 of 280

Question 120 of 280

A force F⃗=(i^+2j^+3k^)N\vec{F}=(\hat{i}+2\hat{j}+3\hat{k})N acts at a point (4i^+3j^−k^)m(4\hat{i}+3\hat{j}-\hat{k})m. Then the magnitude of torque about the point (i^+2j^+k^)m(\hat{i}+2\hat{j}+\hat{k})m will be xN−m.\sqrt{x}N-m.The value of xx is..........

Show answer and solution

Answer: 195

r⃗=(4−1)i^+(3−2)j^+(−1−1)k^\vec{r}=\left(4-1\right)\hat{i}+\left(3-2\right)\hat{j}+\left(-1-1\right)\hat{k}
=3i^+j^−2k^=3\hat{i}+\hat{j}-2\hat{k}
τ=r⃗×F⃗=∣i^jk^31−2123∣\tau =\vec{r}\times \vec{F}=\left|\begin{matrix}\hat{i} & j & \hat{k} \\ 3 & 1 & -2 \\ 1 & 2 & 3\end{matrix}\right|
=i^(7)−j^(11)+k^(5)=7i^−11j^+5k^=\hat{i}\left(7\right)-\hat{j}\left(11\right)+\hat{k}\left(5\right)=7\hat{i}-11\hat{j}+5\hat{k}
=49+121+25=195=\sqrt{49+121+25}=\sqrt{195}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Rotational dynamics and rolling simulationDrag the sliders and watch the numbers in this question change.Revise Rotational Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Rotational Motion (Torque), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.