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