A cylindrical vessel of height H and base area S is filled with water. An orifice of surface area $s \ll S$ is opened in the bottom of the vessel. Neglecting the viscosity of water, determine how soon all the water will pour out of the vessel.

**Attempt :**

Let at any instant height of water be h.

Let the speed from the bottom be $v$.

$v^2 = 2gh$

Differentating,

$2v \dfrac{dv}{dt}= 2g\dfrac{dh}{dt}$

$\implies v\dfrac{dv}{dt}= g\dfrac{dh}{dt}$

Now, $\dfrac{dh}{dt}= -v$

Substituting then integrating we get,

$v_i = gt$

$v_i = \sqrt{2gH}$

$\implies t = \sqrt{\dfrac{2H}{g}} $

**But** answer is: $t = \dfrac{S}{s}\sqrt{\dfrac{2H}{g}}$

Why am I missing that extra $S/s$ factor?

Please let me know my error. If my error is $v_2 \ne \dfrac{-dh}{dt} $, then please explain why.

Reason: -v represents the rate at which the height is decreasing i.e. dh/dt