A metal sphere of mass m and density sigma_{1 | vedclass

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Question

(A) When a sphere falls with terminal velocity,the net force acting on it is zero.The forces acting on the sphere are:$1$. Weight $(W = mg)$ acting do

Vedclass · Accepted Answer

$mg(1 - \frac{\sigma_{2}}{\sigma_{1}})$

Vedclass · Answer

(A) When a sphere falls with terminal velocity,the net force acting on it is zero.The forces acting on the sphere are:$1$. Weight $(W = mg)$ acting downwards.$2$. Buoyant force $(F_{B})$ acting upwards.$3$. Viscous force $(F_{v})$ acting upwards.At terminal velocity: $W = F_{B} + F_{v}$.Therefore,$F_{v} = W - F_{B}$.The weight of the sphere is $W = V \sigma_{1} g$,where $V$ is the volume of the sphere.Since $m = V \sigma_{1}$,we have $V = \frac{m}{\sigma_{1}}$.The buoyant force is $F_{B} = V \sigma_{2} g = (\frac{m}{\sigma_{1}}) \sigma_{2} g = mg(\frac{\sigma_{2}}{\sigma_{1}})$.Substituting these into the equation for viscous force:$F_{v} = mg - mg(\frac{\sigma_{2}}{\sigma_{1}}) = mg(1 - \frac{\sigma_{2}}{\sigma_{1}})$.

$A$ metal sphere of mass $m$ and density $\sigma_{1}$ falls with terminal velocity through a container containing liquid. The density of the liquid is $\sigma_{2}$. The viscous force acting on the sphere is

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