The apparent weight of a man in a lift is `W_(1)` when lift moves upwards with some acceleration and is `W_(2)` when it is accerating down with same acceleration. Find the true weight of the man and acceleration of lift .

Explanation

(a) `W_(1) =m(g +a),W_(2) =m(g -a)`
`W_(1) +W_(2) =2mg rArr W_(1) +W_(2) =2W(":' W =mg)`
`rArr (W_(1)+W_(2))/(2) =W`
(b) `(W_(1))/(W_(2)) =(m(g+a))/(m(g-a)) =(g+a)/(g-a)`
`(g)/(a) =(W_(1)+W_(2))/(W_(1)+W_(2)) rArra =g ((W_(1)-W_(2))/(W_(1)+W_(2)))` .