An open tank filled with water (density) ρ has a narrow hole at a depth of h below the water surface. The velocity of water flowing out is

Question

An open tank filled with water (density) ρ has a narrow hole at a depth of h below the water surface. The velocity of water flowing out is (A) h ρ g (B) 2 g h (C) √2 g h (D) g h

Tardigrade Admin · Accepted Answer

From Bernoulli's theorem <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/1f68a2732ad057b9b29ef2d65221f030-.png" /> p+0+ρ g H=p+(1/2) ρ v2+ρ g(H-h) (1/2) ρ v2=ρ g h ⇒ v=√2 g h

Q. An open tank filled with water (density) $ρ$ has a narrow hole at a depth of $h$ below the water surface. The velocity of water flowing out is

$h ρ g$

$2 g h$

$2 g h$

$g h$

From Bernoulli's theorem

$p + 0 + ρ g H = p + \frac{1}{2} ρ v^{2} + ρ g (H - h)$
$\frac{1}{2} ρ v^{2} = ρ g h$
$\Rightarrow v = 2 g h$

Q. An open tank filled with water (density) ρ has a narrow hole at a depth of h below the water surface. The velocity of water flowing out is

hρg

2gh

2gh​

gh

From Bernoulli's theorem p+0+ρgH=p+21​ρv2+ρg(H−h) 21​ρv2=ρgh ⇒v=2gh​

Q. An open tank filled with water (density) $ρ$ has a narrow hole at a depth of $h$ below the water surface. The velocity of water flowing out is

$h ρ g$

$2 g h$

$2 g h$

$g h$

From Bernoulli's theorem

$p + 0 + ρ g H = p + \frac{1}{2} ρ v^{2} + ρ g (H - h)$
$\frac{1}{2} ρ v^{2} = ρ g h$
$\Rightarrow v = 2 g h$