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# Physics/What would happen in this scenario?

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Question
We have a circle shaped hole 150 meters deep, 4 meters in diameter. Half of it is filled with water. We drop a a rod 75 meters long and 4 meters in diameter into the hole. The rod is exactly as wide as the hole so there is no space between the edges of the rod and the surrounding hole. What happens to the water?

Answer
Nothing happens to the water. The rod sits on top of the water with 25 meters of the rod sticking out of the hole. Water is for all intensive purposes incompressible and so it's volume essentially remains unchanged. Since the diameter of he hole is identical to the diameter of the rod no water can slip past the rod and escape leaving all of the water below the rod.
From Wikipedia!
"The compressibility of water is a function of pressure and temperature. At 0 °C, at the limit of zero pressure, the compressibility is 5.1×10−10 Pa−1.[30] At the zero-pressure limit, the compressibility reaches a minimum of 4.4×10−10 Pa−1 around 45 °C before increasing again with increasing temperature. As the pressure is increased, the compressibility decreases, being 3.9×10−10 Pa−1 at 0 °C and 100 MPa.
The bulk modulus of water is 2.2 GPa.[31] The low compressibility of non-gases, and of water in particular, leads to their often being assumed as incompressible. The low compressibility of water means that even in the deep oceans at 4 km depth, where pressures are 40 MPa, there is only a 1.8% decrease in volume.[31]"

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#### James J. Kovalcin

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

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