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A man stands on a weighing machine kept ...

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 decends with the same acceleration and reading `10%` of initial. Find the acceleration of lift ?

A

`(g)/(19)m//sec^(2)`

B

`(9g)/(11)m//sec^(2)`

C

`0m//sec^(2)`

D

`gm//sec^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`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)`
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Knowledge Check

  • A body stands on a weighing machine inside a lift. When the lift is going down wit acceleration g//4 , the machine shows a reading 30 kg. When the lift goes upwards with acceleration g//4 , the reading would be

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