JEE Mains · Physics · STD 11 - 9.2 surface tension
A drop of liquid of density \(\rho\) is floating half immersed in a liquid of density \(\sigma\) and surface tension \(7.5 \times 10^{-4}\,N\,cm ^{-1}\). The radius of drop in \(cm\) will be : (Take : \(g =10\,m / s ^{2}\) )
- A \(\frac{15}{\sqrt{2 \rho-\sigma}}\)
- B \(\frac{15}{\sqrt{\rho-\sigma}}\)
- C \(\frac{3}{2 \sqrt{\rho-\sigma}}\)
- D \(\frac{3}{20 \sqrt{2 \rho-\sigma}}\)
Answer & Solution
Correct Answer
(A) \(\frac{15}{\sqrt{2 \rho-\sigma}}\)
Step-by-step Solution
Detailed explanation
Boyant force \(+\) surace tension \(= mg\) \(\sigma \frac{ V }{2} g+2 \pi R T=\rho Vg\) \(2 \pi RT =\frac{(2 \rho-\sigma)}{2} \frac{4}{3} \pi R ^{3} g ; \quad\left[ V =\frac{4}{3} \pi R ^{3}\right]\)…
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