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A rod of length 6 m has specific gravity...

A rod of length `6 m` has specific gravity `rho (= 25//36)`. One end of the rod is tied to a `5 m` long rope, which in turn is tied to the floor of a pool `10 m` deep, as shown. Find the length (in `m`) of the part of rod which is out of water.

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The correct Answer is:
1
`(6-x)^(2)Arho_(w)g.2=18Ap_(r)g`
Solving we get `x=1m`
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