JEE Main 2025 — Rotational Motion Question with Solution

First, torque ($\vec{\tau }$) is found by taking the cross product of the position vector ($\vec{r}$) and the force vector ($\vec{F}$):

$\vec{\tau }=\vec{r}\times \vec{F}$

Here, $\vec{r}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{F}=2\hat{i}+\hat{j}+2\hat{k}$.

We use a determinant to find the cross product: $$\vec{\tau }=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2& 1& 2\end{array}\right|$$

Expanding the determinant:

$\vec{\tau }=\hat{i}[(1)(2)-(1)(1)]-\hat{j}[(1)(2)-(1)(2)]+\hat{k}[(1)(1)-(1)(2)]$

This gives: $\vec{\tau }=\hat{i}(2-1)-\hat{j}(2-2)+\hat{k}(1-2)$

So, $\vec{\tau }=\hat{i}-0\hat{j}-\hat{k}$