A uniform rod of length 6 m has specify 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. If length of the part of rod which is out of water is $6/ x$ meter, find x .

Question

TestHub · Accepted Answer

Let the part of the rod out of water be 6/x m.

Balancing the weight of the rod with the buoyancy on the immersed part, together with the geometric constraint of the 5 m rope in the 10 m deep pool:

$\frac{ρ _{rod}}{ρ _{L}} = (\frac{6 - 6/ x}{6})^{2} = \frac{25}{36}$

$1 - \frac{1}{x} = \pm \frac{5}{6} \Rightarrow \frac{6}{x} = 1 or 11$

Only x = 6 is physically acceptable, so x = 6.00.

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