Question

A container has two immiscible liquids of densities a capillary tube of radius r is inserted in the liquid so that its bottom reaches up to the denser liquid. The denser liquid rises in the capillary and attains a height h from the interface of the liquids which is just equal to length of the lighter liquid. Assuming angle of contact to be zero the surface tension of heavier liquid is

Answer 1

the surface tension is 6N

Total buoyant force on the block is

• B = 2/5 ρ₂ V(g+a) + 3/5 ρ₁ V(g+a)

•    = (g+a) (2/5 ρ₂ V + 3/5 ρ₁ V)

From the condition of equilibrium,

•     B = T + Vda ( g +a )

• ⇒  T = B − Vd ( g + a )
•         = ( g + a ) [ 2/5ρ₂ + 3/5ρ₁ − Vd]
• ⇒  T = B - Vd ( g + a ) = ( g + a ) V [ 2/5 ρ₂ + 3/5ρ₁ - d]
•         = 15 × 10 − 3 [ 2/5 × 1500 + 3/5 × 1000 − 800 ] = 6N