JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2022
- Hard
- MCQ
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 \(\left.g=10\,m / s ^{2}\right)\)

Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\frac{3}{\sqrt{109}}\)
Step-by-step Solution
Detailed explanation
\(f = N _{2}\)
\(N_{1}=m g\)
\(N _{2} \times \ell \sin \theta= mg \frac{\ell}{2} \cos \theta\)
\(N _{2}=\frac{ mg }{2} \cot \theta\)
\(\frac{ F _{w}}{ F _{i}}=\frac{\frac{ mg }{2} \cot \theta}{\sqrt{( mg )^{2}+\left(\frac{ mg }{2} \cot \theta\right)^{2}}}\)
\(=\frac{1}{\sqrt{1+\frac{4}{\cot ^{2} \theta}}}\)
\(=\frac{3}{\sqrt{109}}\)
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