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(A) When the ball falls with a constant terminal velocity,the net force acting on it is zero.The forces acting on the ball are:$1$. Weight of the ball

Vedclass · Accepted Answer

$F=Mg\left(1-\frac{\rho_0}{\rho}\right)$

Vedclass · Answer

(A) When the ball falls with a constant terminal velocity,the net force acting on it is zero.The forces acting on the ball are:$1$. Weight of the ball acting downwards: $W = Mg = \rho Vg$ (where $V$ is the volume of the ball).$2$. Buoyant force acting upwards: $F_B = \rho_0 Vg$.$3$. Viscous force acting upwards: $F_{vis}$.For equilibrium,the downward force must equal the sum of the upward forces:$Mg = F_{vis} + F_B$$F_{vis} = Mg - F_B$$F_{vis} = \rho Vg - \rho_0 Vg$$F_{vis} = \rho Vg \left(1 - \frac{\rho_0}{\rho}\right)$Since $M = \rho V$,we substitute $M$ back into the equation:$F_{vis} = Mg \left(1 - \frac{\rho_0}{\rho}\right)$

$A$ small ball of mass $M$ and density $\rho$ is dropped in a viscous liquid of density $\rho_0$. After some time,the ball falls with a constant velocity. What is the viscous force on the ball?

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