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(D) The frictional resistance $f_{r}$ is given as: $f_{r} \propto A$ and $f_{r} \propto v^{2}$ $f_{r} = k A v^{2}$ (where $k$ is a constant). The down

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$r^{1/2}$

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(D) The frictional resistance $f_{r}$ is given as: $f_{r} \propto A$ and $f_{r} \propto v^{2}$ $f_{r} = k A v^{2}$ (where $k$ is a constant). The downward force (weight) is given as: $W = mg = \frac{4}{3} \pi r^{3} \rho g$. At terminal velocity,the downward force is balanced by the frictional resistance (ignoring buoyancy for simplicity as it is usually negligible in air): $mg = f_{r}$ $\frac{4}{3} \pi r^{3} \rho g = k (\pi r^{2}) v^{2}$ Simplifying the equation: $\frac{4}{3} r \rho g = k v^{2}$ Since $\rho$,$g$,and $k$ are constants,we have: $r \propto v^{2}$ $v \propto r^{1/2}$

$A$ solid metallic sphere of radius $r$ is allowed to fall freely through air. If the frictional resistance due to air is proportional to the cross-sectional area and to the square of the velocity,then the terminal velocity of the sphere is proportional to which of the following?

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