JEE PYQ: Mechanical Properties Of Fluids Question 29
Question 29 - 2019 (12 Apr 2019 Shift 2)
A solid sphere, of radius $R$ acquires a terminal velocity $v_1$ 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, $v_2$, when falling through the same fluid, the ratio ($v_1/v_2$) equals:
(1) 9
(2) 1/27
(3) 1/9
(4) 27
Show Answer
Answer: (3) 1/9
Solution
$27 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$ or $r = \frac{R}{3}$ Terminal velocity, $v \propto r^2$ $\therefore \frac{v_1}{v_2} = \frac{r_1^2}{r_2^2}$ or $\frac{v_1}{v_2} = \left(\frac{r_2}{r_1}\right)^2 v_1 = \left(\frac{R/3}{R}\right)^2 v_1 = \frac{1}{9}$ or $\frac{v_1}{v_2} = \frac{1}{9}$
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