A man stands on a weighing machine kept inside a lift . Initially , the lift is ascending with the acceleration a due to which the reading is W . Now the lift descends with the same acceleration and reading is 10 % of initial. Find the acceleration of lift.

Options

A. `(g)/(19)m//sec^(2)`
B. `(9g)/(11)m//sec^(2)`
C. `0m//sec^(2)`
D. `gm//sec^(2)`

Answer

B

Explanation

`N=m(g+a), N'=m(g-a)`
`N'=(10)/(100)xxN`
`m(g-a)=(m(g+a))/(10)`
`10g-10a=g+a` ,brgt `9g=11a`
`rArr a=(9g)/(11)`