NEET 2020 Physics Question 149 Solution

The torque (moment of force) $$\vec{\tau}$$ acting on a particle having position vector $$\vec{r}$$ with respect to the origin is defined as the cross product: $$\vec{\tau} = \vec{r} \times \vec{F}$$ Note that the cross product is anti-commutative ($$\vec{r} \times \vec{F} = -(\vec{F} \times \vec{r})$$), so the order of vectors is critical.

Given: $$\vec{r} = 2\hat{k} \text{ m}$$ $$\vec{F} = 3\hat{j} \text{ N}$$

Recall the right-handed Cartesian cyclic cross product relations: - $$\hat{i} \times \hat{j} = \hat{k}$$ - $$\hat{j} \times \hat{k} = \hat{i}$$ - $$\hat{k} \times \hat{i} = \hat{j}$$ In reverse (counter-cyclic) order: - $$\hat{k} \times \hat{j} = -\hat{i}$$ Therefore: $$\vec{\tau} = (2\hat{k}) \times (3\hat{j}) = (2 \times 3)(\hat{k} \times \hat{j}) = 6(-\hat{i}) = -6\hat{i} \text{ N m}$$

$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 2 \\ 0 & 3 & 0 \end{vmatrix} = \hat{i}(0 - 6) - \hat{j}(0 - 0) + \hat{k}(0 - 0) = -6\hat{i}\text{ N m}$$ This confirms Option (3).