Q.
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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a
2mgπσ
b
3mg2πσ
c
5mg2πσ
d
6mgπσ
answer is D.
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Detailed Solution
σ=FA=mg+g2πD24=3mg×42πD2=6mgπD2D=6mgπσ
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