JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Laws of Motion

At the midpoint three pulls balance. The lower half of the rope holds up the $10$ kg load, so its tension is vertical and equal to $100$ N; the applied force $F$ is horizontal; and the upper half pulls at $45^{\circ }$ to the vertical with tension $T$. Vertical balance gives $T\cos 45^{\circ }=100$ N, so $T=100\sqrt{2}$ N, and horizontal balance gives $F=T\sin 45^{\circ }=100\sqrt{2}\times \frac{1}{\sqrt{2}}=100$ N.