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 figure. Match the Column I with Column II and mark the correct choice from the given codes.

	Column-I		Column-II
(i)	$\frac{\mathrm{Mg}}{4}<\mathrm{F}<\frac{\mathrm{Mg}}{2}$	(p)	Cube will move up
(ii)	$\mathrm{F} >\hspace{0.17em}\frac{\mathrm{Mg}}{2}$	(q)	Cube will not exhibit motion
(iii)	F > Mg	r)	Cube will begin to rotate and slip at,4.
(iv)	$\mathrm{F} = \frac{\mathrm{Mg}}{4}$	(s)	Normal reaction effectively at $\frac{\mathrm{a}}{3}$ from A, no motion

Codes

From figure, Moment of force F about A,
${\mathrm{\tau }}_1 = \mathrm{F} \times \mathrm{a}$, anticlockwise.
Moment of weight Mg of cube about A,
${\mathrm{\tau }}_2 = \mathrm{Mg} \times \frac{\mathrm{a}}{2}$ , clockwise
The cube will not exhibit any motion, if ${\mathrm{\tau }}_1 = {\mathrm{\tau }}_2$
or F x a = $\mathrm{Mg}\times \frac{\mathrm{a}}{2} \mathrm{or} \mathrm{F} > \frac{\mathrm{Mg}}{2}$ 
The cube will rotate only, when ${\mathrm{\tau }}_1 > {\mathrm{\tau }}_2$
$\mathrm{F} \times \mathrm{a} > \mathrm{Mg}\frac{\mathrm{a}}{2} \mathrm{or} \mathrm{F} > \frac{\mathrm{Mg}}{2}$
If we assume that normal reaction is effective at $\frac{\mathrm{a}}{3}$ from A, then block would turn if
$\mathrm{Mg} \times \frac{\mathrm{a}}{3} = \mathrm{F} \times \mathrm{a} \mathrm{or} \mathrm{F} = \frac{\mathrm{Mg}}{3}$

When F = $\frac{\mathrm{Mg}}{4} < \frac{\mathrm{Mg}}{3},$ there will be no motion.
Hence, we conclude (i) - q: (ii) - r; (iii) -p; (iv) - s.