NTA Abhyas JEE Main2020PhysicsMechanical Properties of FluidsPractice
U-tube moves with a constant speed v parallel to the surface of a stationary liquid. The cross-section area of the lower part of the tube lowered into the liquid, is equal to S 1 and that of the top part located over the liquid is S 2 . Friction and formation of waves should be neglected. The density of the fluid is ρ . Neglect difference in heights at both the openings of the tube. The velocity of the liquid coming
Options
- Av S 1 S 2
- Bv 1 + S 1 S 2
- Czero
- Dnone of these
Correct answer
B. v 1 + S 1 S 2
Step-by-step solution
Using the equation of continuity, we get S 1 v 1 = S 2 v 2 v 1 = v (liquid is entering in pipe with a speed v with respect to the pipe) ⇒ v 2 = S 1 S 2 v v 2 : velocity of liquid at S 2 with respect to pipe ∴ With respect to ground, the speed of liquid will be, v + S 1 S 2 v = v 1 + S 1 S 2