Question

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 $II$.

• A. $A - p, B - q, C - s, D - r$
• B. $A - r, B - s, C - q, D - p$
• C. $A - q , B - r , C - p , D - s$
• D. $A - s , B - p , C - r , D - q$

Answer 1

Answer: (C)

$A - q , B - r , C - p , D - s$

Answer 2

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 $A - q; B - r; C - p; D - s$.