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TS EAMCET · Physics · Mechanical Properties of Fluids

A copper ball of radius \(3.0 \mathrm{~mm}\) falls in an oil tank of viscosity \(1 \mathrm{~kg} / \mathrm{ms}\). Then, the terminal velocity of the copper ball will be (Density of oil \(=1.5 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\), Density of copper \(=9 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\) and \(g=10 \mathrm{~m} / \mathrm{s}^2\).)

  1. A \(18 \times 10^2 \mathrm{~m} / \mathrm{s}\)
  2. B \(25 \times 10^2 \mathrm{~m} / \mathrm{s}\)
  3. C \(15 \times 10^2 \mathrm{~m} / \mathrm{s}\)
  4. D \(20 \times 10^2 \mathrm{~m} / \mathrm{s}\)
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Answer & Solution

Correct Answer

(C) \(15 \times 10^2 \mathrm{~m} / \mathrm{s}\)

Step-by-step Solution

Detailed explanation

Given, radius of copper ball, \(r=3 \times 10^{-3} \mathrm{~m}\), viscosity of oil, \(\eta=1 \mathrm{~kg} / \mathrm{ms}\), density of oil, \(\rho=1.5 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\) and density of copper, \(\sigma=9 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\) Terminal…
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