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)

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) (15/√2 ρ-σ) (B) (15/√ρ-σ) (C) (3/2 √ρ-σ) (D) (3/20 √2 ρ-σ)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/82efac8007830de95f9a47e28eec0360-.png" /> Buoyant force + surface tension = mg σ ( V /2) g +2 π R T=ρ V g 2 π R T=((2 ρ-σ)/2) ⋅ (4/3) π R 3 g ; [ V =(4/3) π R 3] R3=(3 T/(2 ρ-σ) g) ⇒ R=√(3 × 7.5 × 10-2 N - m -1/(2 ρ-σ) × 10) R =(3/20 √(2 ρ-σ)) m =(15/√2 ρ-σ) cm

Q. A drop of liquid of density $ρ$ is floating half immersed in a liquid of density $σ$ and surface tension $7.5 \times 1 0^{- 4} N c m^{- 1}$ . The radius of drop in $c m$ will be : $($ Take : $g = 10 m / s^{2})$

$\frac{15}{2 ρ - σ}$

$\frac{15}{ρ - σ}$

$\frac{3}{2 ρ - σ}$

$\frac{3}{20 2 ρ - σ}$

Q. A drop of liquid of density ρ is floating half immersed in a liquid of density σ and surface tension 7.5×10−4Ncm−1. The radius of drop in cm will be : ( Take : g=10m/s2)

2ρ−σ​15​

ρ−σ​15​

2ρ−σ​3​

202ρ−σ​3​

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

Q. A drop of liquid of density $ρ$ is floating half immersed in a liquid of density $σ$ and surface tension $7.5 \times 1 0^{- 4} N c m^{- 1}$ . The radius of drop in $c m$ will be : $($ Take : $g = 10 m / s^{2})$

$\frac{15}{2 ρ - σ}$

$\frac{15}{ρ - σ}$

$\frac{3}{2 ρ - σ}$

$\frac{3}{20 2 ρ - σ}$