A drop of liquid of density is floating half immersed in a liquid of density and surface tension 7.5 × 10-4,N,cm -1. The radius of drop in cm will be: (Take: g =10,m / s 2)

(B) \frac{15}{\sqrt{\rho-\sigma}}

(C) \frac{3}{2 \sqrt{\rho-\sigma}}

(D) \frac{3}{20 \sqrt{2 \rho-\sigma}}

JEE Mains · Physics · STD 11 - 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 : \(g =10\,m / s ^{2}\) )

Pick an option to check your answer.

Verified Solution

Correct Answer

(A) \(\frac{15}{\sqrt{2 \rho-\sigma}}\)

Detailed explanation

Boyant force \(+\) surace tension \(= mg\)

\(\sigma \frac{ V }{2} g+2 \pi R T=\rho Vg\)

\(2 \pi RT =\frac{(2 \rho-\sigma)}{2} \frac{4}{3} \pi R ^{3} g ; \quad\left[ V =\frac{4}{3} \pi R ^{3}\right]\)

\(R ^{3}=\frac{3 T }{(2 \rho-\sigma) g} \Rightarrow R =\sqrt{\frac{3 \times 7.5 \times 10^{-2} N - m ^{-1}}{(2 \rho-\sigma) \times 10}}\)

\(R=\frac{3}{20 \sqrt{(2 \rho-\sigma)}} m =\frac{15}{\sqrt{2 \rho-\sigma}}\,\,cm\)

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LIST \(I\)	LIST \(II\)
\(A\) Intrinsic Semiconductor	\(I\) Fermi-level near conduction band
\(B\) \(n-\)type semiconductor	\(II\) Fermi-level at middle
\(C\) \(p-\) type semiconductor	\(III\) Fermi-level near valence band
\(D\) Metals	\(IV\) Fermi-level inside conduction band

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