JEE Advance - Physics (2009 - Paper 2 Offline - No. 15)
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 to 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.)
답변
6
설명
We have
$${P_0}{V_0} = {P_1}{V_1}$$ ..... (1)
and the pressure at equilibrium is
$$P = {P_0} - \rho gh$$
$$ = (1.0 \times {10^5}) - [({10^3})(10)(200 \times {10^{ - 3}})]$$
$$ = (98 \times {10^3})$$ N/m$$^2$$
Substituting in Eq. (1), we get
$${10^5}[A(500 - H)] = 98 \times {10^3}[A(500 - 200)]$$
where A is the cross-section of water column.
Now, if H = 206 mm, we conclude that the level of water falls down by 6 mm. Hence, the correct answer is 6.