\\\\(\ extbf{Hint}\\\\): Draw the FBD \\\\(\ extbf{Step 1}\\\\):Given length of ladder is 5m. It is placed against the wall. Coefficient of friction is \\\\(\\mu\\\\). Let mass of the ladder is \\\\(m\\\\).From the figure: for the vertical equilibrium \\\\(N_1+\\mu N_2=mg\\\\) \\\\((1)\\\\)For horizontal equilibrium \\\\(N_2=\\mu N_1\\\\) \\\\((2)\\\\)The torque about point A is given by: \\\\(mg(2.5)cos \ heta=N_2(4)+\\mu N_2(3)\\\\)\\\\(mg(2.5 \ imes \\dfrac{3}{5})=(4+3 \\mu)N_2\\\\), since \\\\(cos \ heta=\\dfrac{3}{5}\\\\)\\\\(mg(\\dfrac{3}{2})=(4+3 \\mu)N_2\\\\) \\\\((3)\\\\)\ extbf{Step 2}\\\\(:From equation \\\\)(1)\\\\( and \\\\)(2)\\\\( \\\\)N_2(\\dfrac{1+\\mu^2}{\\mu})=mg \\implies N_2= \\dfrac{\\mu mg}{1+\\mu^2}\\\\(So, \\\\)mg \\dfrac{3}{2}=(4+3 \\mu) \\dfrac{\\mu mg}{1+\\mu^2}\\\\(\\\\)(3+3 \\mu^2)=8 \\mu+6 \\mu^2\\\\(\\\\)3 \\mu^2+8 \\mu-3=0\\\\(\\\\)\\mu=\\dfrac{-8 \\pm sqrt{64+36}}{6}=\\dfrac{-8 \\pm 10}{6}\\\\(\\\\)\\mu=\\dfrac{1}{3} Thus option C is correct.