JEE Physics 2025 Question with Solution

Given: Position vector is $\mathbf{r} = (1,1,1)$ and force vector is $\mathbf{F} = (1,-1,1)$.

Find: The magnitude of torque about the origin in the $z$-direction.

Torque is given by

$$\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$$

For the $z$-component,

$$\tau_z = r_x F_y - r_y F_x$$

Substituting the given values,

$$\tau_z = 1 \times (-1) - 1 \times 1 = -2$$

Hence, the magnitude is

$$|\tau_z| = 2$$

Therefore, the correct option is B. The solution states magnitude as $1$, but from the shown working the defensible answer is $2$.

Given: $\mathbf{r} = (1,1,1)$ and $\mathbf{F} = (1,-1,1)$.

Find: The torque component along the $z$-axis.

Using determinant form,

$$\mathbf{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$$

The coefficient of $\hat{k}$ is

$$1 \cdot (-1) - 1 \cdot 1 = -2$$

So the torque in the $z$-direction is $-2$ and its magnitude is $2$.

Therefore, the correct option is B.