JEE Advanced2015PhysicsMechanical Properties of FluidsActual
Two spheres P and Q of equal radii have densities ρ 1 and ρ 2 , respectively. The spheres are connected by a massless string and placed in liquids L 1 and L 2 of densities σ 1 and σ 2 and viscosities η 1 and η 2 , respectively. They float in equilibrium with the sphere P in L 1 and sphere Q in L 2 and the string being taut (see figure). If sphere P alone in L 2 has terminal velocity V → P and Q alone in L 1 has termi
Options
- AV → P V → Q = η 1 η 2
- BV → P V → Q = η 2 η 1
- CV → P . V → Q > 0
- DV → P . V → Q < 0
Correct answer
A. V → P V → Q = η 1 η 2
Step-by-step solution
Consider a body of density ρ b kept in density ρ l whose viscosity is η and terminal velocity V. then F → v i s c o u s + F → m g + F → B u o y a n c y = 0 F → v i s c o u s + ρ b 4 3 π R 3 - j ^ + ρ l 4 3 π R 3 j ^ = 0 ∴ F → v i s c o u s = ρ b - ρ l 4 3 π R 3 j ^ ⇒ 6 π η R V = ρ b - ρ l 4 3 π R 3 ∴ if ρ b > ρ 1 then F → v i s c o u s ↑ V ∝ 1 η & if ρ b ρ l F → v i s c o u s ↓ As per given diagram we can say σ 2 > σ 1 ; ρ 1 σ 1 & σ 2 > σ 2 ⇒ ρ 2 > σ 2 > σ 1 > ρ 1 ∴ if we put P in L 2 where V → P ∝ 1 η 2 when ρ 1 σ