A solid sphere, of radius $R$ acquires a terminal velocity $\nu_1 $ when falling (due to gravity) through a viscous fluid having a coefficient of viscosity $\eta $. The sphere is broken into $27$ identical solid spheres. If each of these spheres acquires a terminal velocity, $\nu_2$, when falling through the same fluid, the ratio $(\nu_1/\nu_2)$ equals
JEE MAIN 2019, Diffcult
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We have
${V_T} = \frac{2}{g}\frac{{{r^2}}}{\eta }\left( {{\rho _0} - {\rho _\ell }} \right)g$
$ \Rightarrow {V_T} \propto {r^2}$
Since mass of the sphere wll be same
$\therefore \rho \frac{4}{3}\pi {R^3} = 27 \cdot \frac{4}{3}\pi {r^3}\rho $
$ \Rightarrow r = \frac{R}{3}$
$\therefore \,\frac{{{V_1}}}{{{v_2}}} = \frac{{{R_2}}}{{{r^2}}} = 9$
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