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JEE Main 2019 (12 Apr, Shift 2) · Terminal Velocity

Terminal Velocity question from JEE Main 2019 (12 Apr, Shift 2)

Q51JEE Main 2019 · 12 APR · SHIFT 2Medium

Terminal Velocity

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:

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Answer: (d)

  1. Volume conserved: R3=27r3⇒r=R/3R^3 = 27r^3 \Rightarrow r = R/3.
  2. v∝r2⇒v1v2=(Rr)2=9v \propto r^2 \Rightarrow \dfrac{v_1}{v_2} = \left(\dfrac{R}{r}\right)^2 = 9
  3. So the answer is (d) 9.
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