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Question

(C) The terminal velocity $v_t$ of a sphere of radius $r$ and density $\rho$ falling through a liquid of density $\sigma$ and viscosity $\eta$ is give

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) The terminal velocity $v_t$ of a sphere of radius $r$ and density $ho$ falling through a liquid of density $\sigma$ and viscosity $\eta$ is given by the formula: $v_t = \frac{2}{9} \frac{r^2 g (ho - \sigma)}{\eta}$.Since both spheres are made of the same metal and fall through the same liquid, $ho, \sigma, g,$ and $\eta$ are constant. Thus, $v_t \propto r^2$.Given the mass $m = \frac{4}{3} \pi r^3 ho$, we have $r^3 \propto m$, which implies $r \propto m^{1/3}$.Substituting this into the proportionality for $v_t$, we get $v_t \propto (m^{1/3})^2 = m^{2/3}$.Let $v_1 = 3 \ cm s^{-1}$ for $m_1 = 8 \ g$ and $v_2$ be the terminal velocity for $m_2 = 64 \ g$.Then, $\frac{v_2}{v_1} = \left( \frac{m_2}{m_1} ight)^{2/3} = \left( \frac{64}{8} ight)^{2/3} = (8)^{2/3} = (2^3)^{2/3} = 2^2 = 4$.Therefore, $v_2 = 4 	imes v_1 = 4 	imes 3 \ cm s^{-1} = 12 \ cm s^{-1}$.

If the terminal velocity of a metal sphere of mass $8 \ g$ falling through a liquid is $3 \ cm s^{-1}$, then the terminal velocity of another sphere of mass $64 \ g$ made of the same metal falling through the same liquid is (in $cm s^{-1}$)

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