A container of mass M containing liquid of ma | vedclass

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(D) Let the total mass of the system be $M_{total} = M + m_0 + m$. According to the center of mass motion, $F_{ext} = M_{total} a_{cm}$. The external

Vedclass · Accepted Answer

Reading of the weighing machine  <  $(M + m + m_0)g$ if the density of the ball $
eq$ the density of the liquid.

Vedclass · Answer

(D) Let the total mass of the system be $M_{total} = M + m_0 + m$. According to the center of mass motion, $F_{ext} = M_{total} a_{cm}$. The external forces acting on the system are the normal force $N$ from the weighing machine and the total weight $(M + m_0 + m)g$. Thus, $(M + m_0 + m)g - N = M_{total} a_{cm}$, where $a_{cm}$ is the downward acceleration of the center of mass of the system. Since the ball is released from rest, it will accelerate (either downward if $\rho_{ball} > \rho_{liquid}$ or upward if $\rho_{ball} In either case, the center of mass of the system will accelerate downward. Therefore, $a_{cm} > 0$, which implies $N = (M + m_0 + m)g - M_{total} a_{cm}$. Thus, the reading of the weighing machine $N$ must be less than the total weight $(M + m + m_0)g$ whenever the ball is accelerating relative to the container.

$A$ container of mass $M$ containing liquid of mass $m_0$ is kept on a weighing machine. $A$ ball of mass $m$ is released from rest from the given position. Choose the correct option just after the ball is released.

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