Question

Liquid is filled upto H in a container having uniform area of cross section A. A small hole of
area 'a' is made at the bottom of the container (a << A). After what time, will the container
become empty?​

Answer 1

time after which container becomes empty is t = (A/a) × √(2H/g)

Liquid is filled in container upto height H.

cross sectional area of container is A.

now a hole of area a is made at bottom of the container (a <<A).

let at certain moment height of water is h and velocity of water emerging through the hole is v.

from Bernoulli's theorem, v = √(2gh)

let dh represents decrease in water level during time interval dt when height of water is h. so, the rate of decrease in volume of water will be , -dV/dt = -A dh/dt

again the rate of flow through the hole is given by, v × a = a × √(2gh)

so, a × √(2gh) = -A dh/dt

⇒∫dt = -(A/a√2g) ∫1/√h dh

⇒[t]^t_0 = -(A/a√2g) [2√h]^0_H

⇒t = (A/a√2g) × 2√H

⇒t = (A/a) × √(2H/g)

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Answer 2

$\huge\bold\purple{Answer:-}$

Liquid is filled in container upto height H.

cross sectional area of container is A.

now a hole of area a is made at bottom of the container (a <<A).

let at certain moment height of water is h and velocity of water emerging through the hole is v.

from Bernoulli's theorem, v = √(2gh)

let dh represents decrease in water level during time interval dt when height of water is h. so, the rate of decrease in volume of water will be , -dV/dt = -A dh/dt

again the rate of flow through the hole is given by, v × a = a × √(2gh)

so, a × √(2gh) = -A dh/dt

⇒∫dt = -(A/a√2g) ∫1/√h dh

⇒[t]^t_0 = -(A/a√2g) [2√h]^0_H

⇒t = (A/a√2g) × 2√H

⇒t = (A/a) × √(2H/g)