4.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{1dea68fe-7916-41ed-894e-6b48f8d989bb-12_662_716_255_614}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{figure}
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 \(G\).
- Show that the distance of \(G\) from the side \(A D\) is \(\frac { 42 } { 11 } a\)
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.
- Find the value of \(k\).
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |