A point $P$ is moving in uniform circular motion with radius $3 \mathrm{~m}$. Let at some instant the acceleration of the point is $\mathbf{a}=(6 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^2$, the position vector is $\mathbf{r}$ and velocity vector is $\mathbf{v}$. Choose the correct statement.

• A

  $v \cdot a=0$ and $r \times a \neq 0$
• B

  $v \cdot a \neq 0$ and $r \times a \neq 0$
• Correct answer
  C

  $v \cdot a=0$ and $r \times a=0$
• D

  $v \cdot a \neq 0$ and $r \times a=0$

Solution

The given situation is shown in the following figure
Acceleration, $\mathbf{a}=6 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}$ Clearly, acceleration $\mathbf{a}$ is perpendicular to velocity $\mathbf{v}$, hence $\mathbf{v} \cdot \mathbf{a}=0$ Again, position vector $\mathbf{r}$ is parallel to acceleration a, hence $\mathbf{r} \times \mathbf{a}=0$

Hence, the correct answer is option: 'C'.