In this illustration, we'll discuss about
floating of hollow sphere. and here we are
given that a hollow sphere of inside radius
9 centimeter and outer radius 10 centimeter
floats.
half submerged in a liquid of specific gravity
zero point 8. here we are required to calculate
the material density of sphere, and we are
required to find the density of a liquid in
which it just float in completely submerged
state.
so first here we can calculate the volume.
of sphere. which is given as 4 by 3 pie, and
outer radius is 10 centimeter this is zero
point 1 cube. so it is 4 by 3 pie multiplied
by 10 to power minus 3 meter cube. and this
is the total volume.
of sphere. and if we calculate.
the material volume of sphere. then material
volume of sphere. and we calculated by subtracting
the volume of inside cavity, so material volume
here is 4 by 3 pie. this will be zero point
1 cube minus zero point, zero 9 cube. and,
here this will be equal to 4 by, 3 pie, multiplied
by 10 to power minus 3. and inside it is.
1 minus, point 9 cube will be zero point,
7 2, 9. and that is equal to.
zero point 2 7 1 multiplied by 4 pie by 3
into 10 to power minus 3, meter cube. so this
is the material volume of the sphere.
now if we consider the first condition where,
we are given that, the sphere is half submerged
in liquid then we can write.
if sphere. is half submerged. in liquid, then,
for floatation.
we use, in this situation we can write, the
weight of, this sphere will be balancing,
the buoyant force, and buoyant force in this
situation will be v by 2 ro of liquid times
g. so this buoyant force is balancing the
weight of liquid and weight can be given by,
volume of material multiplied by density of
sphere multiplied by g, is equal to v by 2
times, ro of liquid g here g gets cancelled
out. and we can substitute.
the values for, material volume.
and the volume, as here this will give us.
zero point.
2 7 1 multiplied by 4 pie by 3 into 10 to
power minus 3.
multiplied by density of solid, is equal to,
here it is v by 2, so we can write, this is
the volume so it is half of.
4 pie by 3.
into 10 to power minus 3 multiplied by, density
of liquid here is written as zero point 8
times ro of water because the specific gravity
of liquid is point 8. here 4 pie by 3 10 to
power minus 3 gets cancelled out. and the
value of density of solid we are getting.
is equal to zero point 8 divided by point
2 7 1, into 2, times ro of water.
so this will give us, 2 point 9 5 2 divided
by 2 into 10 to power, 3. which numerically
gives us.
if we calculated it this is 1, point.
4 7, approximately we can write it 1 point
4 7, into 10 to power, 4 7, 5.
4 7 6 into 10 to power 3.
kilogram per meter cube this is 1 result of
the problem that is the density of, the material
of sphere. and further we can also write.
if, in.
another liquid. of density.
ro dash.
sphere floats, in submerged state.
in, completely submerged state.
this implies here we can write, the weight
of sphere is balanced by v. ro dash g. which
we already calculated as v by 2 times ro of,
liquid g which we have calculated from this
relation.
so this gives us the value of, ro dash, is
equal to ro of liquid by 2 which is zero point,
8 by 2 times density of water which is thousand
so this gives us, 400 kilogram, per meter
cube which is the result of this problem.
