A rod of length 6 m has specific gravity ρ(=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.

Assume x is the portion of rod outside waterθ is the angle made by rod with horizontalTorque acting on the rod about the point connecting rope and rod is zeroFB.l⊥=mg r⊥ (6−x)Aρωg⋅6-x2cosθ=6Aρrg.3cosθ 6-x22=18ρrρ=18×2536 ∴6-x=5 x=1 m