Question

A cylindrical vessel contains a liquid of density rho up to a height h a frictionless piston of mass m and area of cross section is in contact with the upper surface of the liquid there is a small hole at the bottom of the base of the speed with which liquid comes out of the holy

Answer 1

the speed of liquid is rootunder2gh

Answer 2

A cylindrical vessel contains a liquid of density rho up to a height h a frictionless piston of mass m and area of cross section is in contact with the upper surface of the liquid there is a small hole at the bottom of the base of the speed with which liquid comes out of the hole is

$\boxed{\sqrt{2(gh+\frac{mg}{A\rho})}}$

Explanation:

The diagram is attached

Applying Bernoulli's theorem on point 1 and point 2

$P_1+\rho gh+\frac{1}{2}\rho v_1^2=P_2+\rho gh+\frac{1}{2}\rho v_2^2$

At point 1 the pressure will be the pressure due to atmosphere and the pressure due to the piston and the liquid pressure

Thus,

$P_1=P_{atm}+mg/A+\rho gh$

At point 2 the pressure will be only due to atmosphere

Thus,

$P_2=P_{atm}$

Also, velocity at point 1 will be zero

Thus, the equation becomes

$P_{atm}+\frac{mg}{A}+\rho gh+\rho gh+\frac{1}{2}\rho \times 0^2=P_{atm}+\rho gh+\frac{1}{2}\rho v_2^2$

$\implies \frac{mg}{A}+\rho gh=\frac{1}{2}\rho v_2^2$

$\implies v_2^2=2(gh+\frac{mg}{A\rho})$

$\implies v_2=\sqrt{2(gh+\frac{mg}{A\rho})}$

Hope this answer is helpful.

Know More:

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