Question
If $$F$$ is the force acting on a particle having position vector $$r$$ and $$\tau $$ be the torque of this force about the origin, then
A.
$$r \cdot \tau \ne 0\,{\text{and}}\,F \cdot \tau = 0$$
B.
$$r \cdot \tau > 0\,{\text{and}}\,F \cdot \tau < 0$$
C.
$$r \cdot \tau = 0\,{\text{and}}\,F \cdot \tau = 0$$
D.
$$r \cdot \tau = 0\,{\text{and}}\,F \cdot \tau \ne 0$$
Answer :
$$r \cdot \tau = 0\,{\text{and}}\,F \cdot \tau = 0$$
Solution :
$$\tau = r \times F,$$ where $$r =$$ position vector
$$F = {\text{force}} \Rightarrow \tau = \left| r \right| \cdot \left| F \right|\sin \theta $$
Torque is perpendicular to both $$r$$ and $$F.$$ So, dot product of two vectors will be zero.
$$\therefore \tau \cdot r = 0 \Rightarrow F \cdot \tau = 0$$