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The elastic limit of an elevator cable is 2×109N/m2. The maximum upward acceleration that an elevator of mass 2×103kg can have when supported by a cable whose cross-sectional area is 10−4m2provided the stress in cable would not exceed half of the elastic limit would be

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a

10ms−2

b

50ms−2

c

40ms−2

d

Not possible to move up

answer is C.

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Detailed Solution

From the free-body diagram of the elevator,T−mg=maT=m(g+a)Stress in cable is σ=TA=m(g+a)AFrom the given condition, σ≤σmax2m(g+a)A≤σmax2g+a≤2×1092×10−42×103=5010+a=50a=40ms-2

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