In the above problem, given that $M_{2} = 2 M_{1} a n d M_{2}$ moves vertically downwards with acceleration a. If the position of the masses are reversed the acceleration of $M_{2}$ down the inclined plane will be

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Solution

The correct option is D

If $m_{2} = 2 m_{1}, t h e n m_{2}$ moves vertically downward with acceleration

$a = \frac{m_{2} - m_{1} s i n θ}{m_{1} + m_{2}} g = \frac{2 m_{1} - m_{1} s i n 30}{m_{1} + 2 m_{1}} g = \frac{g}{2}$

If the position of masses are reversed then $m_{2}$ moves downward with acceleration

$a^{'} = \frac{m_{2} s i n θ - m_{1}}{m_{1} + m_{2}} g = \frac{2 m_{1} s i n 30 - m_{1}}{m_{1} + 2 m_{1}} . g = 0 [A s m_{2} = 2 m_{1}]$

i.e. the $m_{2}$ will not move.

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