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(C) The velocity vector of a particle moving in a circle is always tangential to the path.Let the initial velocity at point $P$ be $\vec{v}_i$ and the

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$2v \sin 20^o$

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(C) The velocity vector of a particle moving in a circle is always tangential to the path.Let the initial velocity at point $P$ be $\vec{v}_i$ and the final velocity at point $Q$ be $\vec{v}_f$.The magnitude of both velocities is $v$, so $|\vec{v}_i| = |\vec{v}_f| = v$.The angle between the two velocity vectors is equal to the angle subtended by the arc $PQ$ at the center, which is $	heta = 40^o$.The magnitude of the change in velocity is given by $|\Delta \vec{v}| = |\vec{v}_f - \vec{v}_i|$.Using the vector subtraction formula, $|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos 	heta}$.Substituting the values: $|\Delta \vec{v}| = \sqrt{v^2 + v^2 - 2v^2 \cos 	heta} = \sqrt{2v^2(1 - \cos 	heta)}$.Using the trigonometric identity $1 - \cos 	heta = 2 \sin^2(	heta/2)$:$|\Delta \vec{v}| = \sqrt{2v^2(2 \sin^2(	heta/2))} = \sqrt{4v^2 \sin^2(	heta/2)} = 2v \sin(	heta/2)$.Given $	heta = 40^o$, the magnitude of change in velocity is $2v \sin(40^o/2) = 2v \sin 20^o$.

$A$ particle is moving on a circular path of radius $r$ with uniform speed $v$. The magnitude of change in velocity when the particle moves from $P$ to $Q$ is $(\angle POQ = 40^o)$.

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