A solid cone is kept in three immiscible liquids of density ρ , 2 ρ and 3 ρ inside a container. The radius of cone is ' R ' and its height is 4 R . Container is accelerated with a constant a

Question

A solid cone is kept in three immiscible liquids of density $ρ, 2 ρ$ and $3 ρ$ inside a container. The radius of cone is ' $R$ ' and its height is $4 R$ . Container is accelerated with a constant acceleration $g /2$ in vertically upward direction as shown. If cone is in equilibrium relative to the frame of container then density of the cone is

TestHub · Accepted Answer

Tan θ = 4 1 ​ v 1 ​ = 3 π ​ [ x 2 2 ​ ⋅ 2 R − x 1 2 ​ ⋅ R ] = 48 7 π ​ R 3 v 2 ​ = 3 π ​ [ x 3 2 ​ ⋅ 3 R − x 2 2 ​ ⋅ 2 R ] = 48 19 π ​ R 3 v 3 ​ = 3 π ​ [ R 2 ⋅ 4 R − x 3 2 ​ ⋅ 3 R ] = 48 37 π ​ R 3 So, m = v 1 ρ ​ + v 2 ​ ( 2 ρ ) + v 3 ​ ( 3 ρ ) ρ 0 ​ 3 π R 2 ​ × 4 R ( 2 3 g ​ ) = ρ 48 π R 3 ​ ( 156 ) ( 2 3 g ​ ) Density of cone, ρ 0 ​ = 16 39 ​ ρ

Physics - Fluid Question with Solution | TestHub

Options:

Doubts & Discussion

JEE Main 2027 (Class 12)