JEE Main · 2006 · 30 Apr 2006 · Offline · Laws of Motion

The mass hangs on a string and a slowly increasing horizontal force draws it aside until the string is at $45^{\circ }$ to the vertical. Writing the string length as $L$, the bob rises through $L(1-\cos 45^{\circ })$ and moves sideways through $L\sin 45^{\circ }$. Because the force is built up quasi-statically, the work it does equals the gain in gravitational potential energy: $F(L\sin 45^{\circ })=MgL(1-\cos 45^{\circ })$. Cancelling $L$ and using $\sin 45^{\circ }=\cos 45^{\circ }=\frac{1}{\sqrt{2}}$ gives $F=Mg\frac{1-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=Mg(\sqrt{2}-1)$. So the horizontal force needed is $Mg(\sqrt{2}-1)$.