A uniform L-shaped tube has one vertical arm and one horizontal arm. The liquid occupies a total length 2L between the two free surfaces, and both free surfaces are exposed to the same pressure. A small displacement of the liquid along the tube is produced and the liquid is released. Neglect viscosity. What is the time period?
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Question

A uniform L-shaped tube has one vertical arm and one horizontal arm. The liquid occupies a total length 2L between the two free surfaces, and both free surfaces are exposed to the same pressure. A small displacement of the liquid along the tube is produced and the liquid is released. Neglect viscosity. What is the time period?

Options

G1 Physics · Accepted Answer

If the liquid shifts by x along the tube, only the free surface in the vertical arm changes its gravitational height by x. Thus, ΔP = ρgx. Restoring force: F = −ρgAx. The total moving liquid length is 2L, so its mass is m = 2ρAL. Hence, 2ρAL ẍ = −ρgAx. Thus, ω² = g/(2L). Therefore, T = 2π√(2L/g). Correct option: B.