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.

Answer

1

Explanation

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