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JEE Mains · Physics · STD 11 - 9.2 surface tension
A small spherical droplet of density \(d\) is floating exactly half immersed in a liquid of density \(\rho\) and surface tension \(\mathrm{T}\). The radius of the droplet is (take note that the surface tension applies an upward force on the droplet)
- A \(r=\sqrt{\frac{2 T}{3(d+\rho) g}}\)
- B \(r=\sqrt{\frac{3 T}{(2 d-\rho) g}}\)
- C \(r=\sqrt{\frac{T}{(d-\rho) g}}\)
- D \(r=\sqrt{\frac{\mathrm{T}}{(\mathrm{d}+\rho) \mathrm{g}}}\)
Answer & Solution
Correct Answer
(B) \(r=\sqrt{\frac{3 T}{(2 d-\rho) g}}\)
Step-by-step Solution
Detailed explanation
\(B + F = mg\) \(\mathrm{B}=\left(\frac{2}{3} \pi \mathrm{R}^{3}\right) \rho \mathrm{g}\) \(\mathrm{F}=\mathrm{T}(2 \pi \mathrm{R})\) \(\mathrm{m}=\mathrm{d}\left(\frac{4}{3} \pi \mathrm{R}^{3}\right)\)…
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