question_answer

The apparent weight of a person inside a lift is \[{{w}_{1}}\]when lift moves up with a certain acceleration and is \[{{w}_{2}}\]when lift moves down with same acceleration. The weight of the person when lift moves up with constant speed is

A)  \[\frac{{{w}_{1}}+{{w}_{2}}}{2}\]                            

B)  \[\frac{{{w}_{1}}-{{w}_{2}}}{2}\]

C)  \[2\,{{w}_{1}}\]                                              

D)  \[2\,{{w}_{2}}\]

Correct Answer: A

Solution :

                When lift moves down with constant acceleration a, then \[mg-{{w}_{2}}=2\,mg\]                               ?(ii) From Eq. (i) and (ii), we get \[{{w}_{1}}+{{w}_{2}}=2\,mg\]                  ?(iii) When lift moves up with constant speed, its acceleration is zero. So,          \[w-mg=0\] or            \[w=mg\]                            ?(iv) From Eq. (iii) and (iv) \[{{w}_{1}}+{{w}_{2}}=2\,w\] or            \[w=\frac{{{w}_{1}}+{{w}_{2}}}{2}\]