JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Laws of Motion

At the point of slipping the horizontal pull balances the limiting friction while the vertical forces balance. The normal reaction is reduced by the upward component of $F$: $N=mg-F\sin 30^{\circ }=100-\frac{F}{2}$. The threshold condition is $F\cos 30^{\circ }={\mu }_{s}N$, so $\frac{\sqrt{3}}{2}F=0.25\left(100-\frac{F}{2}\right)$. This gives $4\sqrt{3}F=200-F$, hence $F=\frac{200}{4\sqrt{3}+1}\approx 25.2$ N. The block just begins to move when the force is about $25.2$ N.