A uniform rod of density ρ and length ℓ is having square cross-section of side a. It is slowly lowered in a liquid of same density ρ vertically along its length in a tank having sufficient height of liquid. The surface tension of liquid is T and angle of contact is 120°. Then :

Engineering

Physics

Surface Tension

Question

Rod will sink to bottom of tank.

Rod will float completely immersed inside the liquid.

Rod will float partially submerged with height $\frac{4 T}{aρg}$ above liquid.

Rod will float partially submerged with height $\frac{2 T}{a ρg}$ above liquid.

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$f_{B} = Vρg + 4 Tcos θa$

$ℓ^{1} a^{2} ρg = {ℓa}^{2} ρg + 4 T (- 1 / 2) a$

$ℓ^{1} = \frac{{ℓa}^{2} ρg - \frac{4 T}{2} a}{a^{2} eg}$

$= ℓ - \frac{4 T}{2 aρg}$

$ℓ^{1} = ℓ - \frac{2 T}{aρg}$ ] submerged length

outside length $= ℓ - (ℓ - \frac{2 T}{aρg}) = \frac{2 T}{aρg}$

Please subscribe our Youtube channel to unlock this solution.