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(C) The terminal velocity $v_t$ of a sphere of radius $r$ falling in a viscous liquid is given by $v_t = \frac{2r^2g(\rho - \sigma)}{9\eta}$, where $\

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$4$

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(C) The terminal velocity $v_t$ of a sphere of radius $r$ falling in a viscous liquid is given by $v_t = \frac{2r^2g(\rho - \sigma)}{9\eta}$, where $\rho$ is the density of the sphere and $\sigma$ is the density of the liquid.Since the spheres are of the same metal, $\rho$ is constant, so $v_t \propto r^2$.The mass $M$ of a sphere is given by $M = \frac{4}{3}\pi r^3 \rho$, which implies $M \propto r^3$, or $r \propto M^{1/3}$.Substituting this into the proportionality for terminal velocity, we get $v_t \propto (M^{1/3})^2 = M^{2/3}$.Given the masses $M_1 = M$ and $M_2 = 8M$, the ratio of their terminal velocities is $\frac{v_1}{v_2} = \left(\frac{M_1}{M_2}\right)^{2/3}$.Substituting the values, $\frac{v}{nv} = \left(\frac{M}{8M}\right)^{2/3} = \left(\frac{1}{8}\right)^{2/3} = \left(\left(\frac{1}{2}\right)^3\right)^{2/3} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.Thus, $\frac{1}{n} = \frac{1}{4}$, which gives $n = 4$.

Two solid spheres of the same metal but of mass $M$ and $8M$ fall simultaneously in a viscous liquid. If their terminal velocities are $v$ and $nv$, then the value of $n$ is:

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