In the situation given, all surfaces are frictionless. Pulley is ideal and string is light if `F=mg//2` the acceleration of the big block is `g//x` then x is:

Answer

4

Explanation

First consider both the blocks as system force that we apply one end of string is tension in the string
For system block `(A+B) (Mg)/(2) =2Ma a =g//4`
Thus `a = g//hati`
For system block `A: Mg - (Mg)/(2) =Ma_(r)`
`vecar =g//2hatj`
Thus, `vecaA =g//4hati + g// 2hatj`
`veca3 = g//4hatj`
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