
A cube of side \(a\) and mass \(m\) is to be tilted at point \(A\) by applying a force \(F\) as shown in…

- \(\mathrm{mg}\)
- \(\frac{2}{3} m g\)
- \(\frac{3}{2} m g\)
- \(\frac{3}{4} m g\)
Solution
Thus we get \(\operatorname{mg}\left(\frac{\mathrm{a}}{2}\right)=\mathrm{F}\left(\frac{3}{4} \mathrm{a}\right)\)
Thus we get
\(\mathrm{F}=\frac{2}{3} \mathrm{mg}\)
.Asked in: JEE Mains - Rotational Motion - Chapter Test