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A rubber hose of length 50 cm and internal diameter 1 cm is stretched until its length increases by 10 cm. Taking Poisson's ratio for rubber as 0.5, find the internal diameter of the stretched hose.

1. 0.9 cm
2. 0.8 cm
3. 0.95 cm
4. 1.1 cm

Correct answer: 0.9 cm

Solution

Longitudinal strain = dL/L0 = 10/50 = 0.2. Lateral strain = -mu * longitudinal strain = -0.5 * 0.2 = -0.1. New diameter = d0*(1 + lateral strain) = 1*(1 - 0.1) = 0.9 cm.

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