JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 \(F _{ a } / F _{f}\) will be: (Use \(\left.g=10\,m / s ^{2}\right)\)
- A \(\frac{6}{\sqrt{110}}\)
- B \(\frac{3}{\sqrt{113}}\)
- C \(\frac{3}{\sqrt{109}}\)
- D \(\frac{2}{\sqrt{109}}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{\sqrt{109}}\)
Step-by-step Solution
Detailed explanation
\(f = N _{2}\) \(N_{1}=m g\) \(N _{2} \times \ell \sin \theta= mg \frac{\ell}{2} \cos \theta\) \(N _{2}=\frac{ mg }{2} \cot \theta\) \(\frac{ F _{w}}{ F _{i}}=\frac{\frac{ mg }{2} \cot \theta}{\sqrt{( mg )^{2}+\left(\frac{ mg }{2} \cot \theta\right)^{2}}}\)…
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