Correct Answer - c
From the figure, Moment of force F about a,
`tau_(1)= F xx a`, anticlockwise.
Moment of weight Mg of cube about A,
`tau_(2)= Mg xx a/2`, clockwise.
The cube will not exhibit any motion, if `tau_(1)=tau_(2)`
or `F xx a = Mg xx a/2`or `F=(Mg)/(2)`
The cube will rotate only, when `tau_(1) gt tau_(2)`
`F xx a gt (Mga)/2` or `F gt (Mg)/(2)`
If we assume that normal reaction is effective at `a//3` from A, then block would turn if
`Mg xx a/3 =F xx a` or `F = (Mg)/(3)`
When `F=(Mg)/(4) lt (Mg)/3`, there will no motion.
Hence we conclude A-q, B-r, C-p, D-s.