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A uniform glass tube of length 2 m is closed at the lower end, completely filled with water, and rigidly fixed at the upper end. The tube is then stretched downward. The tube length increases by 0.12 cm but the water column length increases by only 0.08 cm (treat water as incompressible). What is the value of 1/sigma, where sigma is Poisson's ratio of glass?
- 3
- 4
- 5
- 6
Correct answer: 3
Solution
Since water is incompressible, its volume A*L_water is constant; as the glass tube stretches and its radius contracts by Poisson's effect, the water column must elongate to compensate — giving sigma = 0.08 / (2 * 0.12) = 1/3, so 1/sigma = 3.
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