A $\sqrt{34} m$ long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If F_f and F_w are the reaction forces of the floor and the wall, then ratio of F_w/F_f will be : (Use g = 10 m/s².)

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$\frac{2}{\sqrt{109}}$

$\frac{6}{\sqrt{110}}$

$\frac{3}{\sqrt{113}}$

$\frac{3}{\sqrt{109}}$

answer is C.

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Detailed Solution

$\begin{array}{l} f = N_{2} \\ N_{1} = mg \\ N_{2} \times ℓsin θ = mg \frac{ℓ}{2} \cos θ \\ N_{2} = \frac{mg}{2} \cot θ \\ \frac{F_{w}}{F_{f}} = \frac{\frac{mg}{2} \cot θ}{\sqrt{(mg)^{2} + {(\frac{mg}{2} \cot θ)}^{2}}} \\ = \frac{1}{\sqrt{1 + \frac{4}{\cot^{2} θ}}} \\ = \frac{3}{\sqrt{109}} \end{array}$