A solid sphere, of radius R acquires a terminal velocity nu₁ when falling (due to gravity) t

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9-1.Fluid Mechanics

hard

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

A

$27$

B

$1/27$

C

$9$

D

$1/9$

(JEE MAIN-2019)

Solution

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}$

$\therefore \,\frac{{{V_1}}}{{{v_2}}} = \frac{{{R_2}}}{{{r^2}}} = 9$

Standard 11

Physics

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