JEE Mains · Physics · STD 11 - Fluid mechanics
- JEE Mains 2019
- Hard
- Numerical
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
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Answer & Solution
Correct Answer
(C) \(9\)
Step-by-step Solution
Detailed explanation
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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