In this illustration we'll analyze the water
in an accelerated container.
we are given that an open cubical tank completely
filled with water is kept on a horizontal
surface.
and its acceleration is then slowly increase
to 2 meter per second square as shown in the
figure, we are required to find the mass of
water that would spill out of the tank. here
we can consider that tank is completely filled
with water upto its length. and in solution
here we can say.
if, acceleration of container.
is ay, then, angle of, free surface.
of liquid.
in container.
is given as, this we have, already studied
in concept videos that if the container is
accelerated, with acceleration ay, then the
liquid level will be, inclined at an angle
theta where the value of tan theta is given
as ay by g, we already studied it, and the
free surfaces of the liquid, are all inclined.
that means in this situation, the volume which
will spill over is the volume in this triangular
cross section.
so here we can say if, this depth to which
on the right wall the liquid level exist is
x. and this length we are already provided
with 1 meter.
the length of this vessel.
then here we can directly write the value
of tan theta which is ay by g. this angle
is also theta so this will be x by 1. and
value of ay, we are provided with 2 meter
per second and g we can take as 10, so in
this situation the value of x we are getting,
is 2 by 10. and, that is equal to 0 point
2 meter.
so this is the value of x to which, the level
will get depressed on the right wall.
so here we can now calculate, the volume of
water, that is spilled out. is given as. this
volume of this triangular cross section we
can write.
is half multiplied by, x multiplied by 1 this
is the cross sectional area, and as the tank
is cubical its depth is also 1 so we multiplied
with 1.
so this will be x divided by 2. and the value
of x we have calculated as point 2 meter so
this will be, zero point 2 by 2 so the volume
will be zero point 1.
meter cube. so, the mass. of water, spilled
out can be given as density of water multiplied
by this volume.
which is thousand multiplied by zero point
1, that is equal to hundred kilogram which
is the result of this problem.
