JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 6)
A solid sphere, of radius R acquires a terminal velocity v1 when falling (due to gravity) through a viscous
fluid having a coefficient of viscosity . The sphere is broken into 27 identical solid spheres. If each of these
spheres acquires a terminal velocity, v2, when falling through the same fluid, the ratio (v1/v2) equals :
$${1 \over 9}$$
$${1 \over {27}}$$
27
9
Explanation
$${4 \over 3}$$$$\pi $$R3 = 27 $$ \times $$ $${4 \over 3}$$$$\pi $$r3
$$ \Rightarrow $$ r = $${R \over 3}$$
Terminal velocity, $${V_T} = {2 \over 9}{{{r^2}} \over \eta }\left( {{\sigma _s} - {\rho _l}} \right)g$$
$$ \therefore $$ VT $$ \propto $$ r2
$$ \therefore $$ $${{{v_1}} \over {{v_2}}} = {{{R^2}} \over {{r^2}}}$$ = $${{{R^2}} \over {{{{R^2}} \over 9}}}$$ = 9
$$ \Rightarrow $$ r = $${R \over 3}$$
Terminal velocity, $${V_T} = {2 \over 9}{{{r^2}} \over \eta }\left( {{\sigma _s} - {\rho _l}} \right)g$$
$$ \therefore $$ VT $$ \propto $$ r2
$$ \therefore $$ $${{{v_1}} \over {{v_2}}} = {{{R^2}} \over {{r^2}}}$$ = $${{{R^2}} \over {{{{R^2}} \over 9}}}$$ = 9
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