AP EAMCET · PHYSICS · Mechanical Properties of Fluids
A spherical body of density \(\rho\) is floating half immersed in a liquid of density \(d\). If \(\sigma\) is the surface tension of the liquid, then the diameter of the body is
- A \(\sqrt{\frac{3 \sigma}{g(2 \rho-d)}}\)
- B \(\sqrt{\frac{6 \sigma}{g(2 \rho-d)}}\)
- C \(\sqrt{\frac{4 \sigma}{g(2 \rho-d)}}\)
- D \(\sqrt{\frac{12 \sigma}{g(2 \rho-d)}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{\frac{3 \sigma}{g(2 \rho-d)}}\)
Step-by-step Solution
Detailed explanation
Weight of body \(=\) Buoyant force + Force of surface tension \( \begin{aligned} \frac{4}{3} \pi r^3 \rho \times g & =\frac{2}{3} \pi r^3 d g+2 \pi r \sigma \\ \frac{2}{3} \pi r^3 g(2 \rho-d) & =2 \pi r \sigma \end{aligned} \) So, \(r^2=\frac{3 \sigma}{g(2 \rho-d)}\) So,…
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