JEE Main 2025 — Rotational Motion Question with Solution

The torque $\vec{\tau }$ acting on the particle with respect to the origin can be calculated using the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:

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

Given the position vector $\vec{r}=(1,1,1)\text{m}$ and the force vector $\vec{F}=\hat{i}-\hat{j}+\hat{k}$, we need to calculate the cross product:

$$\vec{\tau }=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 1& -1& 1\end{array}\right|$$

Calculating the determinant, we have:

$$\vec{\tau }=\hat{i}\left(1\cdot 1-1\cdot (-1)\right)-\hat{j}\left(1\cdot 1-1\cdot 1\right)+\hat{k}\left(1\cdot (-1)-1\cdot 1\right)$$

This simplifies to:

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

$$\vec{\tau }=2\hat{i}-0\hat{j}-2\hat{k}$$

The torque vector is $\vec{\tau }=2\hat{i}-2\hat{k}$.

To find the magnitude of the torque in the z-direction, we look at the $\hat{k}$ component:

$${\tau }_z=-2$$

The magnitude of torque in the z-direction is:

$$|{\tau }_z|=2\text{Nm}$$

Thus, the magnitude of the torque in the z-direction is 2 Newton-meters.