AP EAMCET · PHYSICS · Mechanical Properties of Fluids
Two solid sphere of radii \(2 \mathrm{~mm}\) and \(4 \mathrm{~mm}\) are tied to the two ends of a light string and released in a liquid of specific gravity 1.3 and coefficient of viscosity \(1 \mathrm{~Pa}\)-s. The string is just taut, when the two spheres are completely in the liquid. If the density of the materials of the two sphere is \(2800 \mathrm{kgm}^{-3}\), then the terminal velocity of the system of the sphere is (take, \(g=10 \mathrm{~ms}^{-2}\) )
- A \(2 \mathrm{cms}^{-1}\)
- B \(4 \mathrm{cms}^{-1}\)
- C \(4 \mathrm{~ms}^{-1}\)
- D \(2 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(B) \(4 \mathrm{cms}^{-1}\)
Step-by-step Solution
Detailed explanation
Free body diagram of system is When system achieves terminal velocity, \[ \begin{gathered} F_{\text {net }}=0 \\ \Rightarrow \frac{4}{3} \pi \rho_f g\left(r_{\mathrm{A}}^3+r_{\mathrm{B}}^3\right)+6 \pi \eta v\left(r_A+r_B\right)=\left(m_A+m_B\right) g \end{gathered} \] With…
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