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A string of cross-sectional area 4 mm² and length 0.5 m is attached to a rigid body of mass 2 kg. The body is whirled in a vertical circle of radius 0.5 m and reaches a speed of 5 m/s at the lowest point of the path. Find the strain produced in the string at the lowest point, expressed as a multiple of 10⁻⁵. (Take Young's modulus = 10¹¹ N/m² and g = 10 m/s².)
- 30
- 10
- 20
- 15
Correct answer: 30
Solution
At the bottom of a vertical circle the net upward force is the centripetal force, so T - mg = m*v²/r. The stress is T/A and the strain follows from Hooke's law strain = stress/Y.
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