In the arrangement shown in fig the mass of the ball 1 is `eta=1.8` times as great as that of rod 2 The length of the later is `l=100` cm The masses of the pulleys and the threads as well as the friction are negligible The ball is set on the same level as the lower end of the rod and then released How soon will the ball be opposite to the other end of the rod

Explanation

Let the acceleration of the mass `m_(1)` is a upward, then the acceleration of `m_(2)` will be 2a downward

From FBD we have
`T-m_(1)g=m_(1)a` …(i)
`m_(2)g-T' = m_(2)(2a)` …(ii)
and `T=2T'` …(iii)
Solving above equations we get
`a=[(2m_(2)-m_(1)]/(m_(1)+4m_(2))]g`
Given `m_(1)=1.8m_(2)`
`:. a=g/29m//s^(2)`
Acceleration of rod relative to ball `=2a-(-a)=3a`
Displacement `l=100 cm =1m`
By second equation of motion, we have
`l=ut+1/2(3a)t^(2)`
or `1=0+1/2 3xxg/29 xxt^(2)`
`:. t=1.40s`