Question

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

• A. $\frac{1}{\sqrt{2}}(\hat{i}-\hat{j})$
• B. $\frac{1}{\sqrt{2}}(\hat{j}-\hat{i})$
• C. $\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$
• D. Either (1) or (2)

Answer 1

Answer: (D)

$\vec{r}=\frac{1}{2} i+\frac{1}{\sqrt{2}} j$
$\vec{v} \cdot \vec{r}=0$ as velocity is always tangential to the path.
$\left(v_{x} \hat{i}+v_{y} \hat{j}\right) \cdot \frac{1}{2}(\vec{i}+\hat{j})=0$
$v_{x}+v_{y}=0$
$\Rightarrow v_{x}=-v_{y}$
or $v_{y}=-v_{x}$
$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$
$\Rightarrow \sqrt{2} v_{x}=v$
$\Rightarrow v_{x}=\frac{v}{\sqrt{2}}$
$v_{y}=-\frac{v}{\sqrt{2}}$
or $v_{x}=-\frac{v}{\sqrt{2}}, v_{y}=\frac{v}{\sqrt{2}}$
So, possible value of $v \Rightarrow v_{x} \hat{i}+v_{y} \hat{j}$
$\Rightarrow \frac{v}{\sqrt{2}} \hat{i}-\frac{v}{\sqrt{2}} \hat{j}$
or $\frac{-v}{\sqrt{2}} \hat{i}+\frac{v}{\sqrt{2}} \hat{j}$