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.
Match Column $I$ with Column $II$.
Updated On: Jul 6, 2022
$A - p, B - q, C - s, D - r$
$A - r, B - s, C - q, D - p$
$A - q , B - r , C - p , D - s$
$A - s , B - p , C - r , D - q$
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The Correct Option isC
Solution and Explanation
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 $A - q; B - r; C - p; D - s$.
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