Ratio of heights of a sphere, over and under water When a sphere is floating in the water, only 10 % of the volume is above the surface, while the rest is below. I need to calculate the relationship between height for the part above, and below the water.
The solution in the textbook is 0.244


Thank you for all the feedback, however, I am still getting wrong answers as for H and h. Is there something else that I am doing wrong here
 A: From the spherical cap of unit radius the volume formula is
$$ \pi h^2 (R-h/3)= ( 0.1 ) \frac43. \pi.1^3 \to \frac{0.4}{3}=h^2(1-h/3) $$
The cubic equation solves to ( two other real solutions are unacceptable) $$h\approx 0.3916,\; H= 2-h\approx 1.6084, \; \frac{h}{H}\approx 0.243472 ;$$
A: The Part above the water is known as Spherical Cap with known formulae for the volume.
The total Diameter is $D=h+H$ & $r=(h+H)/2$
Plug this into the Spherical Cap Volume formula and equate it to 10% of total volume. Solve this Equation in terms of $h/H$.
If the solution in the textbook is correct, you should get $h/H=0.244$ , else Post a comment here.
Alternatively, you can verify your answer yourself.
UPDATE:
Working out the Solution:
Volume of Spherical Cap : $(1/3) \Pi h^2 (3r-h)$ = $(1/3) \Pi h^2 (3(h+H)/2-h)$ = $(1/3) \Pi h^2 (3h/2+3H/2-2h/2)$ = $(1/3) \Pi h^2 (h/2+3H/2)$
Volume of Sphere is: $(4/3) \Pi r^3$ = $(4/3) \Pi (h+H)^3/8$ = $(1/3) \Pi (h+H)^3/2$
Hence:
$(1/3)\Pi h^2 (h/2+3H/2) = (0.1) (1/3) \Pi (h+H)^3/2$
$(1/3) h^2 (h/2+3H/2) = (0.1) (1/3) (h+H)^3/2$
$h^2 (h/2+3H/2) = (0.1) (h+H)^3/2$
$h^2 (h+3H) = (0.1) (h+H)^3$
Divide though-out by $H^3$
$h^2 (h+3H)/H^3 = (0.1) (h+H)^3/H^3$
$(h/H)^2 (h/H+3H/H) = (0.1) (h/H+H/H)^3$
$(h/H)^2 (h/H+3) = (0.1) (h/H+1)^3$
Let $h/H=X$
$X^2 (X+3) = (0.1) (X+1)^3$
Solve this.
Check whether $h/H = X = 0.244$ is valid.
$(0.244)^2 (0.244+3) = (0.1) (0.244+1)^3$
We get $ (0.193134784)$ & $(0.1) (1.925134784) $
