Question:
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 :
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 :
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Terminal velocity of a sphere in a viscous fluid is proportional to \(r^2\); find the new radius from volume conservation.
Updated On: Nov 11, 2026
- \(\frac{1}{9}\)
- \(\frac{1}{27}\)
- 27
- 9
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The Correct Option is D
Solution and Explanation
Volume conservation: \(27\cdot\frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3\), so \(r = R/3\).
Terminal velocity: \(v_t = \frac{2r^2(\rho-\sigma)g}{9\eta}\), so \(v_t \propto r^2\).
\(\frac{v_1}{v_2} = \frac{R^2}{r^2} = 3^2 = 9\).
Answer: (4) 9
Terminal velocity: \(v_t = \frac{2r^2(\rho-\sigma)g}{9\eta}\), so \(v_t \propto r^2\).
\(\frac{v_1}{v_2} = \frac{R^2}{r^2} = 3^2 = 9\).
Answer: (4) 9
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