In the shown `P_(1)` and `P_(2)` are massless pulleys `P_(1)` is fixed and `P_(2)` can move Masses of `A,B` and `C` are `(9m)/(64) 2m` amd m respectively All contacts are smooth and the string is massless `theta = tan^(-1)((3)/(4))` (Take `g = 10m//s^(2)`)
The tension in string connecting pulley `P_(2)` and block `C` is `(13)/(x)` Calculate x (Take `m =1 kg`)
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Answer

2

Explanation

`N_(1)((3)/(5))=m_(A).a_(A)rarr(1)m_(B)g-T-(4)/(5)N_(1)=m_(B)alpha_(B)rarr(2)`
`2T -m_(C) g = m_(C) a_(C)rarr(3) a_(B) = (3)/(4) a_(A)rarr(4)`
`a_(B) =2a_(C) rarr(5)` after solving above equations we get
`a_(A) =8//sec^(2), a_(B) =6m//sec^(2),a_(C) =3m//sec^(2),T=(13)/(2)N`
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