### Subject: What is the height of bottom of ball submerged below water surface

Date: **Sat Nov 12 20:26:26 2005**

Posted by **Azhan**

Grade level: **undergrad**
School: **University Malaya**

City: **Kuala Lumpur** State/Province: **No state entered.**
Country: **Malaysia**

Area of science: **Physics**

ID: **1131852386.Ph**

**Message:**

I have a ping pong ball of size 38 mm (r, radius = 19mm) and 2.5g weight (m,
weight = 0.0025 kg). I put the ball on water (rho=10^-6 kg/mm3) to let it float
and I want to know “h” which is the height of bottom of ball submerged below
water surface. I know that “V” is the partial volume of ball for part which
submerged below water that is V= (pi)h^2(r-0.3333h).
Could I assume that, (m divide V) = rho ?
If yes, means that r.h^2 - 0.3333.h^3 = m / (pi . rho) which solve h equals 6.9
mm.
Or, Could I assume that, (m divide A) = (h times rho) ?
If yes, means that h^2 = m / (2 (pi) r (rho)) which solve h equals 4.6 mm.
Where “A” is the surface area of ball submerged below water by taking A=2(pi)rh.

Re: What is the height of bottom of ball submerged below water surface

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