Question

If $\overrightarrow{F}$ is a force acting on a particle having the position vector $\overrightarrow{r}$ and $\overrightarrow{\tau }$ is the torque due to this force about the origin, then-

• A. $\overrightarrow{r}\cdot \overrightarrow{\tau }=0$
• B. $\overrightarrow{r}\cdot \overrightarrow{\tau }\ne 0$
• C. $\overrightarrow{F}\cdot \overrightarrow{\tau }=0$
• D. $\overrightarrow{F}\cdot \overrightarrow{\tau }\ne 0$

Answer 1

Answer: (A), (C)

$\overrightarrow{F}\cdot \overrightarrow{\tau }=0$

As we know that cross product of the two vectors is always perpendicular to both individual vectors.

Torque on the particle about origin is,

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

So,
$\overrightarrow{\tau }\perp \left(\overrightarrow{r}\text{and}\overrightarrow{F}\right)$

$\overrightarrow{r}\perp (\overrightarrow{r}\times \overrightarrow{F});\overrightarrow{F}\perp (\overrightarrow{r}\times \overrightarrow{F})$

$\therefore \overrightarrow{r}\cdot \overrightarrow{\tau }=\overrightarrow{r}\cdot (\overrightarrow{r}\times \overrightarrow{F})=0$

$\overrightarrow{F}\cdot \overrightarrow{\tau }=\overrightarrow{F}\cdot (\overrightarrow{r}\times \overrightarrow{F})=0$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, options $\left(\text{A}\right)$ and $\left(\text{C}\right)$ are the correct answer.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Why this question ? This question involves the idea of torque on a particle. Torque is always perpendicular to the plane formed by force vector and position vector.