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6.System of Particles and Rotational Motion
hard
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
$\frac{2}{\sqrt{109}}$
(JEE MAIN-2022)
Solution
$f = N _{2}$
$N_{1}=m g$
$N _{2} \times \ell \sin \theta= mg \frac{\ell}{2} \cos \theta$
$N _{2}=\frac{ mg }{2} \cot \theta$
$\frac{ F _{w}}{ F _{i}}=\frac{\frac{ mg }{2} \cot \theta}{\sqrt{( mg )^{2}+\left(\frac{ mg }{2} \cot \theta\right)^{2}}}$
$=\frac{1}{\sqrt{1+\frac{4}{\cot ^{2} \theta}}}$
$=\frac{3}{\sqrt{109}}$
Standard 11
Physics
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