Question

A drop of liquid of density ρ is floating half immersed in a liquid of density σ and surface tension 7.5×10−4 Ncm −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 1

Answer: (A)

$\frac{15}{\sqrt{2 \rho-\sigma}}$

Answer 2

Buoyant force + surface tension =mg
σ2V​g+2πRT=ρVg
2πRT=2(2ρ−σ)​⋅34​πR3g;[V=34​πR3]
R3=(2ρ−σ)g3T​⇒R=(2ρ−σ)×103×7.5×10−2N−m−1​​
R=20(2ρ−σ)​3​m=2ρ−σ​15​cm