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The water level in a 24 cm diameter tank is initially at a height of 10 cm. As a first grade science experiment, the students are asked to float a CD on the water. The CD has a diameter of 12 cm, a mass of 16 grams, and a density 1.2x that of water. CDs are highly hydrophobic and they float through surface tension even though they are denser than water After they float the CD, one of the students experiments by adding 4 pennies totaling 8 grams to the CD. What will be the new surface level of the water be, at points relatively far away from the CD? (3 cm is a reasonable distance)

The water level in a 24 cm diameter tank is initially at a height of 10 cm. As a first grade science experiment, the students are asked to float a CD on the water. The CD has a diameter of 12 cm, a mass of 16 grams, and a density 1.2x that of water. CDs are highly hydrophobic and they float through surface tension even though they are denser than water

After they float the CD, one of the students experiments by adding 4 pennies totaling 8 grams to the CD.

What will be the new surface level of the water be, at points relatively far away from the CD? (3 cm is a reasonable distance)

Grade:12

2 Answers

Chetan Mandayam Nayakar
312 Points
10 years ago

Let h be the height of water below and close to the CD, and H the height at far away points.

π(92)h + πH(122-92) =π(122)10

πρg(H-h)*62 = 24g

Solving the above two simultaneous equations, we get H= 10+(3/8π) = 10.12cm

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Vijay Luxmi Askiitiansexpert
357 Points
10 years ago

Dear Sohan,

Let h be the height of water below and close to the CD, and H the height at far away points.

π(92)h + πH(122-92) =π(122)10

πρg(H-h)*62 = 24g

Solving the above two simultaneous equations, we get H= 10+(3/8π) = 10.12cm

 

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