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Question: Answered & Verified by Expert
A liquid drop placed on a horizontal plane has a near spherical shape (slightly flattened due to gravity). Let $\mathrm{R}$ be the radius of its largest horizontal section. A small disturbance causes the drop to vibrate with frequency $\mathrm{v}$ about its equilibrium shape. By dimensional analysis the ratio $\frac{\mathrm{v}}{\sqrt{\sigma / \rho \mathrm{R}^{3}}}$ can be (Here $\sigma$ is surface tension, $\rho$ is density, $\mathrm{g}$ is acceleration due to gravity, and $\mathrm{k}$ is arbitrary dimensionless constant)-
PhysicsUnits and DimensionsJEE Main
Options:
  • A $\mathrm{k} \rho \mathrm{gR}^{2} / \sigma$
  • B $\mathrm{k} \rho \mathrm{R}^{2} / \mathrm{g} \sigma$
  • C $\mathrm{k} \rho \mathrm{R}^{3} / \mathrm{g} \sigma$
  • D $\mathrm{k} \rho / \mathrm{g} \sigma$
Solution:
1388 Upvotes Verified Answer
The correct answer is: $\mathrm{k} \rho \mathrm{gR}^{2} / \sigma$
$\frac{\mathrm{v}}{\sqrt{\sigma / \rho \mathrm{R}^{3}}}$ is dimensionless $\mathrm{k} \rho \mathrm{gR}^{2} / \sigma$ is also dimensionless

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