Question

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

Answer 1

The length of  the part of rod is 1 m.

Explanation:

Net moment about 0 = 0

FB x ⊥r x distance  = W x ⊥r x distance

FB = Vi ρ g =  A x (6 - x ) x ρw x g

τFB = A ρwg (6 - x ) x 6-x /2

      = A ρwg (6 - x )^2 /2

τw = 6A ρvg  x l/2 = 6A ρvg x 3

                              = 18A ρvg

τnet = 0

18 A ρvg = A ρwg (6 - x )^2 /2

(6 - x )^2 = 36 x ρr x g / ρw x g

             = 36 x ρr / ρw

             = 36 x 25 / 26

6 - x = 6 x 5 /6  = 5

x = 6 - 5 = 1 m

Thus the length of  the part of rod is 1 m.

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