wide tank of cross section area a containing a liquid to height H has an orifice at its base of area a battery the initial acceleration of top surface of liquid is
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P {margin – bottom : 0.25cm; line-height 120%}
Let v be the velocity at the surface of liquid. Then from continuity principle, velocity at the orifice will be 3v
From Bernouli's principle, we have,
P+1/2pv square = pgH = p+½ p(3v) square + 0
where P = atmospheric pressure, is the density of the liquid.
Solving above equation for v, we have
V square = gh/4
differentiating both sides with respect to time, we get
2v dv/dt = g/4(dh/dt)
==> 2v dv/dt = - gv/4 as dh/dt = -v
===> a = -g/8
Acceleration is negative as velocity is decreasing, as H is decreasing.
acceleration is constant, it does not depend on the value of H
Let v be the velocity at the surface of liquid. Then from continuity principle, velocity at the orifice will be 3v
From Bernouli's principle, we have,
P+1/2pv square = pgH = p+½ p(3v) square + 0
where P = atmospheric pressure, is the density of the liquid.
Solving above equation for v, we have
V square = gh/4
differentiating both sides with respect to time, we get
2v dv/dt = g/4(dh/dt)
==> 2v dv/dt = - gv/4 as dh/dt = -v
===> a = -g/8
Acceleration is negative as velocity is decreasing, as H is decreasing.
acceleration is constant, it does not depend on the value of H
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