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A cylindrical tin 8cm in diameter contains water to a depth of 4cm.
If a cylindrical wooden rod 4cm in diameter and 6cm long is placed in the tin,
it floats exactly half submerged. What is the new depth of water?
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The volume of the water in the tin is (= was originally)
= = = 201.0619 cm^3 (rounded).
The volume of the half of the cylindrical wooden rod is
= = 37.6991 cm^3 (rounded).
The combined volume occupied by the water in the tin and the half of the cylindrical wooden rod
submerged in the water is this sum
201.0619 + 37.6991 = 238.761 cm^3 (rounded).
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| Notice that from the numbers given in the problem, it is clear |
| that the submerged rod does not touch the bottom of the tin. |
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Let x be the level of the water in the tin after submerging half of the cylindrical
wooden rod (= the final level).
Then we have this equation to find x
= 238.761 cm^3,
or
= 238.761.
It gives
x = = 4.75 cm (rounded).
ANSWER. The new depth of the water in the tin is 4.75 cm.
Solved.