Question

A cylindrical vessel contains a liquid of density rho up to a height H. The liquid is closed by a piston of mass m and area of cross-section A. There is a small hole at the bottom of the vessel. The speed v with which the liquid comes out of the hole is

A. √3(gH+mg/rhoA)
B. √2(gH+mg/rhoA)
C. √5(gH+mg/rhoA)
D. √6(gH+mg/rhoA)

Answer 1

The speed with which the liquid comes out of the hole can be determined using Bernoulli's equation. The correct option is B) sqrt(2(gH + mg/(rhoA))).

Step-by-step explanation:

The speed with which the liquid comes out of the hole can be determined using Bernoulli's equation, which states that the sum of pressure energy, kinetic energy, and potential energy is constant along a streamline. In this case, the potential energy can be ignored as the hole is at the same level as the liquid. The pressure difference is caused by the weight of the liquid column above the hole and the weight of the piston.

The speed v can be determined using the equation:

v = sqrt(2(gH + mg/(rhoA)))

where g is the acceleration due to gravity, H is the height of the liquid column above the hole, m is the mass of the piston, rho is the density of the liquid, and A is the area of the hole.

Therefore, the correct option is B) sqrt(2(gH + mg/(rhoA))).