Question

Let $\underset{F}{\to }$ be a force acting on a particle having position vector $\underset{r}{\to }$. Let $\underset{\mathrm{\Gamma }}{\to }$ be the torque of this force about the origin, then
(a) $\overrightarrow{r}.\overrightarrow{\mathrm{\Gamma }}=0\mathrm{and}\overrightarrow{F}.\overrightarrow{\mathrm{\Gamma }}=0$

(b) $\overrightarrow{r}.\overrightarrow{\mathrm{\Gamma }}=0\mathrm{but}\overrightarrow{F}.\overrightarrow{\mathrm{\Gamma }}\ne 0$

(c) $\overrightarrow{r}.\overrightarrow{\mathrm{\Gamma }}\ne 0\mathrm{but}\overrightarrow{F}.\overrightarrow{\mathrm{\Gamma }}=0$

(d) $\overrightarrow{r}.\overrightarrow{\mathrm{\Gamma }}\ne 0\mathrm{and}\overrightarrow{F}.\overrightarrow{\mathrm{\Gamma }}\ne 0$

Answer 1

(a) $\overrightarrow{r}.\overrightarrow{\mathrm{\Gamma }}=0\mathrm{and}\overrightarrow{F}.\overrightarrow{\mathrm{\Gamma }}=0$

We have:
$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$
Thus, $\overrightarrow{\mathrm{\tau }}$ is perpendicular to $\overrightarrow{r}$ and $\overrightarrow{F}$.
Therefore, we have:
$\overrightarrow{r}.\overrightarrow{\mathrm{\tau }}=0\mathrm{and}\overrightarrow{F}.\overrightarrow{\mathrm{\tau }}=0$