Let \\\\(x\\\\) be the distance of centre \\\\(C\\\\) of rod from point \\\\(D.\\\\) Then\\\\(F_{2} -F_{1} =ma\\\\) or \\\\(F_{1} =3N\\\\)Further, \\\\( T_{c}=0 \\\\)\\\\(\ herefore F_{2}x=F_{1} (0.2+x) \\\\)\\\\( 5x=F_{1} (0.2 +x) \\\\) \\\\( \ herefore 5x=3 (0.2 +x) \\\\)or \\\\( x=0.3 m\\\\) \\\\(\\displaystyle \ herefore \\\\) Length of rod \\\\(=2(x+0.2) =1.0 m\\\\)