The uniform lamina $$\(L\)$$, shown shaded in Figure 2, is formed by removing the square $$\(P Q R V\)$$, of side $$\(2 a\)$$, and the square $$\(R S T U\)$$, of side $$\(4 a\)$$, from a uniform square lamina $$\(A B C D\)$$, of side $$\(8 a\)$$. The lines $$\(Q R U\)$$ and $$\(V R S\)$$ are straight. The side $$\(A D\)$$ is parallel to $$\(P V\)$$ and the side $$\(A B\)$$ is parallel to $$\(P Q\)$$. The distance between $$\(A D\)$$ and $$\(P V\)$$ is $$\(a\)$$ and the distance between $$\(A B\)$$ and $$\(P Q\)$$ is $$\(a\)$$. The centre of mass of $$\(L\)$$ is at the point $$\(G\)$$. (a) Show that the distance of $$\(G\)$$ from the side $$\(A D\)$$ is $$\(\frac{42}{11} a\)$$ (5) The mass of $$\(L\)$$ is $$\(M\)$$. A particle of mass $$\(k M\)$$ is attached to $$\(L\)$$ at $$\(C\)$$. The lamina, with the attached particle, is freely suspended from $$\(B\)$$ and hangs in equilibrium with $$\(B C\)$$ making an angle of $$\(45^{\circ}\)$$ with the horizontal. (b) Find the value of $$\(k\)$$. (4)

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