A cylindrical vessel contains a liquid of density ρ 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

Question

A cylindrical vessel contains a liquid of density ρ 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 <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-5mosxxrifocb.png" /> (A) √2 g h (B) √2 (g h + (m g/ρ A)) (C) √2 (g h + (m g/A)) (D) √2 g h + (m g/A)

Tardigrade Admin · Accepted Answer

Applying Bernoulli's theorem at 1 and 2: difference in pressure energy between 1 and 2 = difference in kinetic energy between 1 and 2 <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-5mosxxrieb1f.png" /> or ρ gh + (m g/A) = (1/2)ρ v2 or v=√2 g h + (2 m g/ρ A)= √2 (g h + (m g/ρ A))

Q. A cylindrical vessel contains a liquid of density $ρ$ 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

$2 g h$

$2 (g h + \frac{m g}{ρ A})$

$2 (g h + \frac{m g}{A})$

$2 g h + \frac{m g}{A}$

Applying Bernoulli's theorem at 1 and 2 : difference in pressure energy between 1 and $2 =$ difference in kinetic energy between 1 and 2

or $ρ g h + \frac{m g}{A} = \frac{1}{2} ρ v^{2}$
or $v = 2 g h + \frac{2 m g}{ρ A} = 2 (g h + \frac{m g}{ρ A})$

Q. A cylindrical vessel contains a liquid of density ρ 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

2gh​

2(gh+ρAmg​)​

2(gh+Amg​)​

2gh+Amg​​