What is the velocity v of a metallic ball of | vedclass

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(A) The forces acting on the metallic ball are: $1$. Weight $W = V\rho g = \frac{4}{3}\pi r^3 \rho g$ (downwards)$2$. Buoyant force $F_B = V\sigma g =

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$\frac{r^2 g}{9\eta}(\rho - 2\sigma)$

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(A) The forces acting on the metallic ball are: $1$. Weight $W = V\rho g = \frac{4}{3}\pi r^3 \rho g$ (downwards)$2$. Buoyant force $F_B = V\sigma g = \frac{4}{3}\pi r^3 \sigma g$ (upwards)$3$. Viscous drag force $F_v = 6\pi \eta r v$ (upwards)According to Newton's second law,$F_{net} = ma = V\rho a$.$V\rho g - V\sigma g - 6\pi \eta r v = V\rho a$.Given that the acceleration $a = g/2$,we substitute this into the equation:$V\rho g - V\sigma g - 6\pi \eta r v = V\rho (g/2)$.$V\rho g - V\sigma g - V\rho g/2 = 6\pi \eta r v$.$V\rho g/2 - V\sigma g = 6\pi \eta r v$.Substituting $V = \frac{4}{3}\pi r^3$:$\frac{4}{3}\pi r^3 g (\frac{\rho}{2} - \sigma) = 6\pi \eta r v$.$\frac{4}{3}\pi r^3 g (\frac{\rho - 2\sigma}{2}) = 6\pi \eta r v$.$\frac{2}{3}\pi r^3 g (\rho - 2\sigma) = 6\pi \eta r v$.$v = \frac{2\pi r^3 g (\rho - 2\sigma)}{3 \cdot 6\pi \eta r} = \frac{r^2 g}{9\eta}(\rho - 2\sigma)$.Thus,the correct option is $A$.

What is the velocity $v$ of a metallic ball of radius $r$ falling in a tank of liquid at the instant when its acceleration is one-half that of a freely falling body? (The densities of metal and of liquid are $\rho$ and $\sigma$ respectively,and the viscosity of the liquid is $\eta$).

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