In figure shown if `m_(A) = 20` kg and `m_(B) = 80` kg. The acceleration of block A if the system is set free to move is `beta m//s^(2)` . Find the value of `beta` . (Take `g = 10 ms^(-2)`)

Answer

A, B, D

Explanation

Let `x_(1), x_(2)` and `x_(3)` be the displacement of the three bodies as shown, measured downwards from the ceiling:
Also, let us assume acceleration of each body in the downward direction.
`(a_(1), a_(2), a_(3))`
From newton 's law
(1) `20 g - 4T = 20 a_(1)` (2) `80 g- 2T = 80 a_(2)`
From constraint equation
`x_(1) + x_(1) - x_(2)` = constant `implies 2a_(1) - a_(2) = 0`
`2x_(2) + 2x_(3)` = constant `implies a_(2) + a_(3) = 0`
combining the equation to get (3) `a_(3) = - 2a_(1)`
Solve to get `a_(1) = (-7 g)/(17)` and `a_(2) = (14 g)/(17)`
Acceleraion of 20 kg, `(7 g)/(17) m//s^(2)` upwards acceleration of 80 kg, `(14 g)/(17) m//s^(2)` downwards