Question

A cylindrical vessel of height $500\, mm$ has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it upto height $H$. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being $200\, mm$. The fall in height (in mm) of water level due to opening of the orifice is [Take atmospheric pressure $= 1.0 \times10^{5} N/m^{2}$, density of water $= 1000\, kg/m^{3}$ and $g = 10 m/s^{2}$. Neglect any effect of surface tension.]

• A. $6$
• B. $4$
• C. $2$
• D. $8$

Answer 1

Answer: (A)

The water is filled up to height $H$ and then the top is closed. The pressure of air column now is atmospheric pressure, $P_{0}$.
When the tap is opened and no more water is flowing, the pressure inside $P +$ pressure due to water column is atmospheric pressure
$P+hg\rho=P_{0} $
$P+ \left(200\right)\times10^{-3}\times10\times1000=P_{0} \ldots\left(i\right)$
$\therefore P=P_{0}-hg\rho$
$\Rightarrow P=1.0 \times10^{5}-200\times10^{-3}\times10\times1000$
$P=0.98\times10^{5} N m^{2} \ldots\left(ii\right)$

$PV =$ constant
Let area of cross section of the vessel be $A$
$P_{0}\left(500-H\right)\times10^{-3}\times A=P\left(300\times10^{-3}\right)\times A$
$\Rightarrow P_{0}\left(500-H\right)=300\,P$
From equation $\left(i\right), P_{0} \frac{\left(500-H\right)}{300}+2000=P_{0}$
$\therefore H=206\, mm$
The final height of air column $= 200 \,mm $
Fall in the height $= 6 \,mm$.