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

detailed solution

Correct option is C

From figure, Moment of force F about A,τ1 = F ×a, anticlockwise.Moment of weight Mg of cube about A,τ2 = Mg ×a2 , clockwiseThe cube will not exhibit any motion, if τ1 = τ2or F x a = Mg×a2 or F > Mg2 The cube will rotate only, when τ1 > τ2F ×a > Mga2 or F > Mg2If we assume that normal reaction is effective at a3 from A, then block would turn ifMg × a3 = F ×a or F = Mg3When F = Mg4 < Mg3, there will be no motion.Hence, we conclude (i) - q: (ii) - r; (iii) -p; (iv) - s.