If $\overrightarrow{\mathrm{F}}$ is the force acting on a particle having position vector $\vec{r}$ and…

If $\overrightarrow{\mathrm{F}}$ is the force acting on a particle having position vector $\vec{r}$ and $\vec{\tau}$ be the torque of this force about the origin, then
  1. $\overrightarrow{\mathrm{r}} \cdot \vec{\tau} \neq 0$ and $\overrightarrow{\mathrm{F}} \cdot \vec{\tau}=0$
  2. $\overrightarrow{\mathrm{r}} \cdot \vec{\tau} > 0$ and $\overrightarrow{\mathrm{F}} \cdot \vec{\tau} < 0$
  3. $\overrightarrow{\mathrm{r}} \cdot \vec{\tau}=0$ and $\overrightarrow{\mathrm{F}} \cdot \vec{\tau}=0$
  4. $\overrightarrow{\mathrm{r}} \cdot \vec{\tau}=0$ and $\overrightarrow{\mathrm{F}} \cdot \vec{\tau} \neq 0$

Solution

Torque is an axial vector i.e., its direction is always perpendicular to the plane containing vectors $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{F}}$. $\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$ Torque is perpendicular to both $\vec{r}$ and $\vec{F}$ $\therefore \quad \begin{aligned} \vec{\tau} \cdot \overrightarrow{\mathrm{r}} & =0 \\ \overrightarrow{\mathrm{F}} \cdot \vec{\tau} & =0 \end{aligned}$ .

Asked in: NEET 2009 (Screening)

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