Maximum stress that can be applied to a wire which supports an elevator is σ . Mass of elevator is m and it is moved upwards with an acceleration of g2. Minimum diameter of wire (Neglecting weight of wire) must be

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2mgπσ

3mg2πσ

5mg2πσ

6mgπσ

answer is D.

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Detailed Solution

σ=FA=mg+g2πD24=3mg×42πD2=6mgπD2D=6mgπσ

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