A is a point on the circle with radius 8 and centre at $O$. A particle P is moving on the circumference of the circle starting from A. M. is the foot of the perpendicular from $P$ on $O A$ and $\angle \mathrm{POM}=\theta$. When $O M=4$ and $\frac{d \theta}{d t}=6$ radians $/ \mathrm{sec}$, then the rate of change of PM is (in units/sec)

• A

  $24 \sqrt{3}$
• Correct answer
  B

  24
• C

  $15 \sqrt{3}$
• D

  $48 \sqrt{3}$

Solution

When $O M=4 \Rightarrow \cos \theta=\frac{O M}{O P}=\frac{1}{2}$
$\begin{aligned} & \text { Now, } P M=O P \sin \theta \Rightarrow P M=8 \sin \theta \\ & \Rightarrow \frac{d(P M)}{d t}=8 \cos \theta \frac{d \theta}{d t} \\ & \Rightarrow 8 \times \frac{1}{2} \times 6=24\end{aligned}$

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