JEE Main 2019 — Physics solution

JEE Main2019Physics

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 $η$ . 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 $(\frac{v _{1}}{v _{2}})$ equals :

A
$\frac{1}{27}$
B
9
C
$\frac{1}{9}$
D
27

Solution

$R^{3} = 27 r^{3}$

$\Rightarrow R = 3r$

$\frac{v _{1}}{v _{2}} = (\frac{R}{r})^{2} = 9$

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