Question

Assertion: Vector product of two vectors is an axial vector.
Reason: If $\vec{v}$= instantaneous velocity , $\vec{r}$= radius vector and $\vec{w}$= angular velocity , then $\vec{w}=\vec{v}\times \vec{r}$
A. Both assertion and reason are correct and reason is the correct explanation for assertion
B. Both assertion and reason are correct but reason is not the correct explanation for assertion.
C. Assertion is correct but reason is incorrect.
D. Both assertion and reason are incorrect.

Answer 1

Hint: First we have to know what an axial vector is. Axial vectors are those functions along the axis of rotation and show the rotational influence. Odd number of cross products gives axial vectors while even numbers give normal vectors. With the help of definition, we find out the correct option.

Complete step by step solution:
We know axial vector is a vector that does not change its sign on changing the coordinate system to a new system by a reflection in the origin. When the direction of the coordinate axes are inverted, the sign will remain the same as the cross product of two vectors. Hence the cross product of two vectors will be an axial vector.

Now let $\vec{v}$= instantaneous velocity , $\vec{r}$= radius vector and $\vec{w}$= angular velocity. Consider a circular disc rotating about its axis with angular velocity w. A particle at a distance r from axis, whose position vector is r, is moving at a speed
$v=\dfrac{2\pi r}{T}$
Where, T = period of rotation = time taken to go round once =$\dfrac{2\pi }{w}$

So v = $\dfrac{2\pi r }{\dfrac{2\pi }{w}}= w\,r$
That is the particle moving with speed = wr and its direction is tangential which is the same as $w\times r$ (vector product ). If origin of coordinate system is not at centre, but at some distance on the axis then,
$v=wr\sin \theta $
That is $\vec{v}=\vec{w}\times \vec{r}$
Thus, the assertion is correct but the reason is incorrect.

Hence, option C is correct.

Note: We should take care about the fact that an axial vector is present when the object is in circular motion, and we are not able to find the axial vector in linear motion.