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(A) When a ball reaches terminal velocity $V$,the net force acting on it is zero.The forces acting on the ball are its weight $(mg)$ acting downwards,

Vedclass · Accepted Answer

$\frac{m - m'}{r}$

Vedclass · Answer

(A) When a ball reaches terminal velocity $V$,the net force acting on it is zero.The forces acting on the ball are its weight $(mg)$ acting downwards,the buoyant force $(m'g)$ acting upwards,and the viscous drag force $(F_v = 6 \pi \eta r V)$ acting upwards.Equating the forces: $mg = m'g + 6 \pi \eta r V$.Rearranging for the terminal velocity $V$: $6 \pi \eta r V = (m - m')g$.Thus,$V = \frac{(m - m')g}{6 \pi \eta r}$.Since $g$,$\pi$,and $\eta$ are constants for a given system,the terminal velocity $V$ is proportional to $\frac{m - m'}{r}$.

$A$ ball of mass $m$ and radius $r$ is gently released in a viscous liquid. The mass of the liquid displaced by it is $m'$ such that $m > m'$. The terminal velocity is proportional to

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