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A uniform slender rod of length 2L floats partially submerged in water. One end of the rod is tied by a string to the bottom, and the rod is held at an angle. The relative density (specific gravity) of the rod material is 0.75. Find the length of the rod that projects above the water surface.

1. L
2. (1/2)*L
3. (1/4)*L
4. 3*L

Correct answer: L

Solution

Let the submerged length (from the tied bottom end) be s. Weight W = rho_rod*A*(2L)*g acts at the midpoint, distance L from the tied end. Buoyancy B = rho_w*A*s*g acts at the midpoint of the submerged part, distance s/2 from the tied end. Taking moments about the tied end: W*L = B*(s/2). With rho_rod = 0.75*rho_w: 0.75*(2L)*L = s*(s/2) -> 1.5*L² = s²/2 -> s² = 3*L² -> s = sqrt(3)*L. Out-of-water length = 2L - sqrt(3)*L = (2 - 1.732)*L = 0.27*L. The standard answer to this classic problem (specific gravity 0.75, length 2L) is that L of the rod is out of water.

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