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Inverse of a Matrix

A disc is rol...

Question

A disc is rolling (without slipping) on a horizontal surface. $C$ is its centre and $Q$ and $P$ are two points equidistant from $C$ . Let $V_{p}, V_{q}$ and $V_{c}$ be the magnitude of velocities of points $P, Q$ and $C$ respectively, then

V_{Q} > V_{C} > V_{P}

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V_{Q} < V_{C} < V_{P}

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V_{Q} = V_{P}, V_{C} = \frac{1}{2} V_{P}

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V_{Q} < V_{C} > V_{P}

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Solution

The correct option is C $V_{Q} > V_{C} > V_{P}$
From Fig. $(I)$ , we have $O C = r (r a d i u s)$
Therefore, $v = r ω$
Since, $ω = c o n s t a n t$ , therefore $v α r$
Now, form Fig $(I I)$ , it is clear that the distance, $O P < O C < O Q \Rightarrow V_{P} < V_{C} < V_{Q}$ or $V_{Q} > V_{C} > V_{P}$ .

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