Question

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

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

Answer 1

The correct option is C $\overrightarrow{r}.\overrightarrow{\tau }=0$ and $\overrightarrow{F}.\overrightarrow{\tau }=0$

We know torque is defined as,
$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=rF\sin \theta \hat{n}$
$\hat{n}$ is the unit vector which shows the direction of torque. From right hand thumb rule, its direction is perpendicular to both $\overrightarrow{F}$ and $\overrightarrow{r}$.
$\Rightarrow \overrightarrow{F}\perp \overrightarrow{\tau }$ and $\overrightarrow{r}\perp \overrightarrow{\tau }...\left(1\right)$
Also, we know dot product of two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is
$\overrightarrow{A}.\overrightarrow{B}=AB\cos \theta$
For $\overrightarrow{r}\perp \overrightarrow{\tau }$, dot product is,
$\overrightarrow{r}.\overrightarrow{\tau }=0$
Similarly, $\overrightarrow{F}$ and $\overrightarrow{\tau }$ are perpendicular $(\theta ={90}^{\circ })$
$\Rightarrow \overrightarrow{F}.\overrightarrow{\tau }=0$
$\therefore$Option $\left(c\right)$ is correct.