Question

Let \[\overrightarrow F\] be a force acting on a particle having position vector \[\overrightarrow r.\] Let \[\overrightarrow\Gamma\] be the torque of this force about the origin, then __________ .

Options

• \[\overrightarrow{r}  .  \overrightarrow{\Gamma}  = 0\text{ and }\overrightarrow{F}  .  \overrightarrow{\Gamma}  = 0\]

• \[\overrightarrow{r}  .  \overrightarrow{\Gamma}  = 0\text{ but }\overrightarrow{F}  .  \overrightarrow{\Gamma}  \ne 0\]

• \[\overrightarrow{r}  .  \overrightarrow{\Gamma}  \ne 0\text{ but }\overrightarrow{F}  .  \overrightarrow{\Gamma}  = 0\]

• \[\overrightarrow{r}  .  \overrightarrow{\Gamma}  \ne 0\text{ and }\overrightarrow{F}  .  \overrightarrow{\Gamma}  \ne 0\]

Answer 1

\[\overrightarrow{r}  .  \overrightarrow{\Gamma}  = 0\text{ and }\overrightarrow{F}  .  \overrightarrow{\Gamma}  = 0\]

 

We have

\[\overrightarrow{\Gamma}  =  \overrightarrow{r}  \times  \overrightarrow{F}\]

Thus,

\[\overrightarrow{\Gamma}\] is perpendicular to \[\overrightarrow{r}\] and \[\overrightarrow{F}.\]

Therefore, we have

\[\overrightarrow{r}  .  \overrightarrow{\Gamma}  = 0\text{ and }\overrightarrow{F}  .  \overrightarrow{\Gamma}  = 0\]