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Fig shows a block of mass 0.1 kg placed ...

Fig shows a block of mass `0.1 kg` placed on a smooth wedge of mass `1/(5sqrt(3))` kg if the block of mass m will move vertically downward with acceleration `10m//s^(2)` Then the value of tension (in newton) in the string is `(theta=30^(@))` :

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The correct Answer is:
2
Accelerating to the given problem, m is a freely falling body, i.e., contact force between M and m is zero. Under this condition, acceleration of M leftwards will be
`a=g cot alpha = g xx sqrt(3)`
Tension `T=Ma=1/5sqrt(3)xxsqrt(3)g=2N`
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