Q. 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 up 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 $200\ mm$. Find the fall in height (in $mm$) of water level due to opening of the orifice (Take atmospheric pressure $= 1.0 \times 10^{5}N/m^{2}$, density of water $= 1000\ kg/m^{3}$ and $g = 10\ m/s^{2}$. Neglect any effect of surface tension)
A) $5\ mm$
B) $6\ mm$
C) $2\ mm$
D) $1\ mm$
Explanation
(b) $P + 200 \times 10^{- 3} \times 1000 \times 10 = P_{0}$ …(i)
$P_{0}(500 - H) = P.(300mm)$
$\Rightarrow P = \frac{P_{0}(500 - H)mm}{300\ mm}$ …(ii)
From (i) and (ii)
$\frac{P_{0}(500 - H)}{300\ } + 2000 = P_{0} \Rightarrow \frac{10^{5}(500 - H)}{300} + 2000 = 10^{5}$
$\Rightarrow 5 \times 10^{7} - H \times 10^{5} + 6 \times 10^{5} = {300 \times 10}^{5}$
$\Rightarrow H = 206\ mm,$ fall in height $= 6\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.