A drop of mercury is present between two glass plates each of mass 50 gm. Separation between the plates is negligible as compared to spread of the mercury between the plates. What should be the mass o

Question

A drop of mercury is present between two glass plates each of mass 50 gm. Separation between the plates is negligible as compared to spread of the mercury between the plates. What should be the mass of a block (in kg) which should be placed over the plate A so that the separation between the plates becomes half of the initial separation. Assume contact angle between mercury and glass is 180°.

TestHub · Accepted Answer

Initially: m g = π R 2 Δ P , where Δ P = S ( r 1 ​ + R 1 ​ ) ≈ r S ​ (since R ≫ r ) With the block, separation halved: r ′ = 2 r ​ , so Δ P ′ = r 2 S ​ Volume conserved: π R 2 ⋅ 2 r = π R ′2 ⋅ 2 r ′ ⇒ R ′ = R 2 ​ ( M + m ) g = π R ′2 Δ P ′ = 2 π R 2 ⋅ r 2 S ​ = 4 π R 2 r S ​ = 4 m g M = 3 m = 3 × 0.050 = 0.15 kg

Physics - Fluid Question with Solution | TestHub

Doubts & Discussion