AP EAMCET · PHYSICS · Mechanical Properties of Fluids
A steel ball of radius \(0.05 \mathrm{~cm}\) and density \(7.8 \mathrm{~g} \mathrm{~cm}^{-3}\) is dropped into a tank of water. The terminal velocity of the steel ball is
(Density of water \(=1 \mathrm{~g} \mathrm{~cm}^{-3}\) and viscosity of water \(=\) \(0.001 \mathrm{~Pa} \mathrm{~s}\) )
- A \(3.42 \mathrm{~ms}^{-1}\)
- B \(1.81 \mathrm{~ms}^{-1}\)
- C \(5.11 \mathrm{~ms}^{-1}\)
- D \(3.77 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(A) \(3.42 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Radius of steel ball, \(\mathrm{r}=0.05 \times 10^{-2} \mathrm{~m}\) Density of steel, \(\rho=7.8 \mathrm{~g} \mathrm{~cm}^{-3}=\frac{7.8 \times 10^{-3}}{10^{-6}}\) \(=7800 \mathrm{~kg} / \mathrm{m}^3\) Density of water, \(\rho_\omega=1 \mathrm{~g} \mathrm{~cm}^{-3}\)…
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