Rotational Motion

JEE Main level, numerical answer, medium

119 of 280

Question 119 of 280

A force F⃗=4i^+3j^+4k^\vec{F}=4\hat{i}+3\hat{j}+4\hat{k} is applied on an intersection point of x=2x=2 plane and xx-axis. The magnitude of torque of this force about a point (2,3,4)\left(2,3,4\right) is ________.
(Round off to the Nearest Integer)

Show answer and solution

Answer: 20

τ⃗=r⃗×F⃗\vec{\tau }=\vec{r}\times \vec{F}
r⃗=(2i^)−(2i^+3j^+4k^)=−3j^−4k^\vec{r}=\left(2\hat{i}\right)-\left(2\hat{i}+3\hat{j}+4\hat{k}\right)=-3\hat{j}-4\hat{k}
and F⃗=4i^+3j^+4k^\vec{F}=4\hat{i}+3\hat{j}+4\hat{k}
τ⃗=r⃗×F⃗=∣i^j^k^0−3−4434∣\vec{\tau }=\vec{r}\times \vec{F}=\left|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -4 \\ 4 & 3 & 4\end{matrix}\right|
=i^(−12+12)−j^(0+16)+k^(0+12)=\hat{i}\left(-12+12\right)-\hat{j}\left(0+16\right)+\hat{k}\left(0+12\right)
=−16i^+12k^=-16\hat{i}+12\hat{k}
∴∣τ⃗∣=162+122=20∴|\vec{\tau }|=\sqrt{{16}^{2}+{12}^{2}}=20

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.