The elastic limit of an elavator cable is `2xx10^(9)N//m^(2)`. The maximum upward acceleration that an elavator of mass `2xx10^(3)kg` can have when supported by a cable whose cross sectional area is `10^(-4)m^(2)`, provided the stres in cable would not exceed half to the elastic limit would be

Options

A. `10ms^(-2)`
B. `50 ms^(-2)`
C. `40 ms^(-2)`
D. Not possible to move up

Answer

C

Explanation

From the free body diagram of the elevator
`T-mg=ma`
`T=m(g+a)`
Stress is cable is `alpha=T/A=(m(g+a))/a`
From the given condition `alphalealpha_("max")/2`
`(m(g+a))/Alesigma_("max")/2`
`g+ale(2xx10^(9))/2xx(10^(-4))/(2xx10^3)=50`
`g+ale(2xx10^(9))/2xx(10^(-4))/(2xx10^(-3))=50`