Physics · Rotational Dynamics

JEE Main 2025 — 29 January, Morning Shift — Question 40

The coordinates of a particle with respect to origin in a given reference frame is (1,1,1)(1,1,1) meters. If a force of F→=i^−j^+k^\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} acts on the particle, then the magnitude of torque (with respect to origin) in zz-direction is \qquad .

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

τ⃗=r→×F→=∣i^j^k^1111−11∣\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{k}\\ 1 & 1 & 1 \\1 & -1 & 1\end{array}\right|

τ⃗=k^(−1−1)=−2k^\vec{\tau}=\hat{\mathrm{k}}(-1-1)=-2 \hat{\mathrm{k}} ∣τ⃗∣=2Nm|\vec{\tau}|=2 \mathrm{Nm}

Answer key and solution verified before publishing.

Practise Rotational Dynamics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2025
Subject
Physics
Chapter
Rotational Dynamics
Topic
Torque, Equation of Motion and Toppling