Q93 / 125 · BITSAT 2009 · Physics
A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force $\mathrm{F}$ is applied normal to one of the faces at a point that is directly above the centre of the face, at a height $3 \mathrm{a} / 4$ above the base. The minimum value of $\mathrm{F}$ for which the cube begins to topple an edge is (assume that cube does not slide)
- A$\frac{\mathrm{mg}}{3}$
- B$\frac{\mathrm{mg}}{2}$
- Correct answerC$\frac{2 \mathrm{mg}}{3}$
- D$\frac{3 \mathrm{mg}}{4}$
Solution
For toppling about edge $x$ x At the moment of toppling the normal force pass through axis $\mathrm{xx}$ '.
$$
\mathrm{F}_{\min } \frac{3 \mathrm{a}}{4}=\mathrm{mg} \frac{\mathrm{a}}{2} \text { or } \mathrm{F}_{\min }=\frac{2 \mathrm{mg}}{3}
$$
Hence, the correct answer is option: 'C'.