Question:

The terminal velocity of a small steel ball of radius \(r\) falling in a fluid is proportional to

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By Stokes’ law, terminal velocity of a small sphere in a viscous fluid varies as: \[ v_t\propto r^2 \]
Updated On: Apr 27, 2026
  • \(r^8\)

  • \(r^7\)

  • \(r^2\)

  • \(r^1\)

  • \(r^5\)

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The Correct Option is C

Solution and Explanation

For a small sphere falling through a viscous fluid, terminal velocity is given by Stokes’ law: \[ v_t=\frac{2r^2(\rho-\sigma)g}{9\eta} \] where:
  • \(r\) is the radius of the sphere,
  • \(\rho\) is density of the sphere,
  • \(\sigma\) is density of the fluid,
  • \(\eta\) is coefficient of viscosity.
From this formula: \[ v_t \propto r^2 \]
Hence, the correct answer is the option corresponding to: \[ \boxed{r^2} \]
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