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 $\frac{-v}{\sqrt{2}}\hat{i}+\frac{v}{\sqrt{2}}\hat{j}$