A uniform cube of mass `m` and side a is placed on a frictionless horizontal surface. A vertical force `F` is applied to the edge as shown in Fig. Match the following (most appropriate choice) :
(a)`mg//4 lt F lt mg//2` (i) Cube will move up.
(b) `F gt mg//2` (ii) Cube will not exhibit motion.
(c ) `F gt mg` (iii) Cube will begin to rotate and slip at `A`.
(d) `F = mg//4` (iv) Normal reaction effectively at `a//3` from A, no motion.

Explanation

Consider the below diagram
Moment of the force F about point A `tau_(1)=Fxxa` (anti - clockwise )
moment of weight mg of the cube about point A ,
`tau_(2)=mg xx(a)/(2)` (clock wise )

Cube will not exhibit motion ,if ` tau_(1)= ta_(2)`
(`:'` in this case , both the torque will cancel the effect the effect of each other )
`thetefore "" fxxa= mg xx(a)/(2)implies F=(mg)/(2)`
Cube will roate only when ,`tau_(1)gt tau_(2)`
`Fxxa gt mgxx(a)/(2)implies Fgt(mg)/(2)`
Let normal reaction is acting at `(a)/(3)` from point A , then
`mgxx(a)/(3)=Fxxa or F=(mg)/(3)`
when `F=(mg)/(4)` which is less than `(mg)/(3)` ""`(Flt(mg)/(3))`
there will be no motion .
`therefore (a) to (ii) (b) to (iii) (c ) to (i) (d) to (iv)`