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. `0 m//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+1))/(10)`
`10g - 10 a = g +a`
`9g = 11a rArr a = (9g)/(11)`.