A rod P of length 1 m is hinged at one end A | vedclass

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(B) Let the distance between the hinges $A$ and $B$ be $l$. The ring moves in a circular path of radius $R = 1 \ m$ centered at $A$.When the angle $\p

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$2$

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(B) Let the distance between the hinges $A$ and $B$ be $l$. The ring moves in a circular path of radius $R = 1 \ m$ centered at $A$.When the angle $\phi$ is at its maximum value,the rod $Q$ is tangent to the circular path of the ring.In this configuration,the triangle formed by the hinges $A, B$ and the position of the ring is a right-angled triangle with the right angle at the ring's position.Given that the angle at $A$ is $	heta = 60^{\circ}$ at the maximum value of $\phi$,we have:$\cos 	heta = \frac{	ext{Adjacent side}}{	ext{Hypotenuse}} = \frac{R}{l}$$\cos 60^{\circ} = \frac{1}{l}$$\frac{1}{2} = \frac{1}{l}$$l = 2 \ m$Thus,the distance between the hinges $A$ and $B$ is $2 \ m$.

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