A particle is moving in the xy-plane in a cir | vedclass

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(D) The position vector of the particle is $\vec{r} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{j})$.Since the particle is moving in a circular path,its veloc

Vedclass · Accepted Answer

Either $(a)$ or $(b)$

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(D) The position vector of the particle is $\vec{r} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{j})$.Since the particle is moving in a circular path,its velocity vector $\vec{v}$ is always perpendicular to its position vector $\vec{r}$ (i.e.,$\vec{v} \cdot \vec{r} = 0$).Let $\vec{v} = v_x \hat{i} + v_y \hat{j}$.Then,$(v_x \hat{i} + v_y \hat{j}) \cdot \frac{1}{\sqrt{2}}(\hat{i} + \hat{j}) = 0$.This implies $v_x + v_y = 0$,or $v_x = -v_y$.This condition is satisfied by both vectors $\frac{1}{\sqrt{2}}(\hat{i} - \hat{j})$ and $\frac{1}{\sqrt{2}}(\hat{j} - \hat{i})$.Therefore,the velocity can be in either direction depending on whether the particle is moving clockwise or counter-clockwise.Thus,the correct option is $(d)$.

$A$ particle is moving in the $xy$-plane in a circular path with its center at the origin. If at an instant the position of the particle is given by $\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$,then the velocity of the particle is along .......

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