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Q. A uniform cube of mass M and side a is placed on a frictionless horizontal surface. A vertical force F is applied to edge as shown in figure.
image
Match Column $I$ with Column $II$.
Column I Column I
A $\frac{Mg}{4} < F < \frac{Mg}{2}$ p Cube will move up
B $F > \frac{Mg}{2}$ q Cube will not exhibit motion
C $F > Mg$ r Cube will begin to rotate and slip at A
D $F = \frac{Mg}{4}$ s Normal reaction effectively at $a/3$ from $A$ no motion

System of Particles and Rotational Motion

Solution:

image
From figure, Moment of force $F$ about $A$,
$\tau_1 = F\times a$, anticlockwise.
Moment of weight $Mg$ of cube about $A$,
$\tau_2 = Mg \times \frac {a}{2}$, clockwise.
The cube will not exhibit any motion,
if $ \tau_1 = \tau_2$
or $ F \times a = Mg \times \frac{a}{2}$ or $F = \frac{Mg}{2}$
The cube will rotate only, when $ \tau_1 > \tau_2$
$F \times a > Mg \frac{a}{2}$ or $F > \frac{Mg}{2}$
If we assume that normal reaction is effective at $a/3$ from $A$, then block would turn if
$Mg \times \frac{a}{3}= F \times a$ or $F =\frac{Mg}{3}$.
When $F= \frac{Mg}{4} < \frac{Mg}{3}$, there will be no motion.
Hence, we conclude $A - q; B - r; C - p; D - s$.