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(B) The particle moves with a constant speed $v$ in a circular path. Let the initial velocity be $\vec{v}_i = v\hat{i}$.After moving to the diametrica

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The momentum changes by $2mv$

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(B) The particle moves with a constant speed $v$ in a circular path. Let the initial velocity be $\vec{v}_i = v\hat{i}$.After moving to the diametrically opposite point,the velocity becomes $\vec{v}_f = -v\hat{i}$.The change in momentum is $\Delta \vec{p} = m\vec{v}_f - m\vec{v}_i = m(-v\hat{i}) - m(v\hat{i}) = -2mv\hat{i}$.The magnitude of the change in momentum is $|\Delta \vec{p}| = 2mv$.Since the speed $v$ is uniform,the kinetic energy $K = (1/2)mv^2$ remains constant throughout the motion.Therefore,the change in kinetic energy is zero.Thus,the correct option is $B$.

List-$I$ (Defects)	List-$II$ (Shown by)
$a$. Frenkel defect	$i$. Non-ionic solids and density of the solid decreases
$b$. Schottky defect	$ii$. Non-ionic solids and density of the solid increases
$c$. Vacancy defect	$iii$. Ionic solids and density of the solid decreases
$d$. Interstitial defect	$iv$. Ionic solids and density of the solid remains constant

$A$ particle is moving in a circle with uniform speed $v$. In moving from a point to another diametrically opposite point:

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