A container with square base of side a is filled up to a height H with a liquid. A hole is made at a depth h from the free surface of water. With what acceleration the container must be accelerated, so that the water does not come out?

Question

A container with square base of side a is filled up to a height H with a liquid. A hole is made at a depth h from the free surface of water. With what acceleration the container must be accelerated, so that the water does not come out? (A)  g  (B)  (g/2)  (C)  (2gH/2)  (D)  (2gh/a)

Tardigrade Admin · Accepted Answer

Bernoulli's theorem is a form of conservation of energy, hence we have

H_{0} + h ρ g = H_{0} + \frac{1}{2} ρ V^{2}

When vessel is accelerated down with an acceleration

g

(free fall), then pseudo acceleration

g

will act vertically upwards and effective value of

g

is zero.
Hence, water will not flow.

Q. A container with square base of side a is filled up to a height $H$ with a liquid. A hole is made at a depth $h$ from the free surface of water. With what acceleration the container must be accelerated, so that the water does not come out?

$g$

$\frac{g}{2}$

$\frac{2 g H}{2}$

$\frac{2 g h}{a}$

Q. A container with square base of side a is filled up to a height H with a liquid. A hole is made at a depth h from the free surface of water. With what acceleration the container must be accelerated, so that the water does not come out?

g

2g​

22gH​

a2gh​

Bernoulli's theorem is a form of conservation of energy, hence we have H0​+hρg=H0​+21​ρV2 When vessel is accelerated down with an acceleration g (free fall), then pseudo acceleration g will act vertically upwards and effective value of g is zero. Hence, water will not flow.

Q. A container with square base of side a is filled up to a height $H$ with a liquid. A hole is made at a depth $h$ from the free surface of water. With what acceleration the container must be accelerated, so that the water does not come out?

$g$

$\frac{g}{2}$

$\frac{2 g H}{2}$

$\frac{2 g h}{a}$