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A $\mathrm{\sqrt{34}}$ 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 $\mathrm{\mathrm{F}_{f}}$ and $\mathrm{\mathrm{F}_{\mathrm{w}}}$ are the reaction forces of the floor and the wall, then ratio of $\mathrm{\mathrm{F}_{\mathrm{w}} / \mathrm{F}_{f}}$ will be:

$\mathrm{\left(\text { Use } g=10 \mathrm{~m} / \mathrm{s}^{2} .\right)}$

Option: 1
$\mathrm{\frac{6}{\sqrt{110}}}$

Option: 2
$\mathrm{\frac{3}{\sqrt{113}}}$

Option: 3
$\mathrm{\frac{3}{\sqrt{109}}}$

Option: 4
$\mathrm{\frac{2}{\sqrt{109}}}$

Answers (1)

best_answer

Torque about B
$\mathrm{N_{2}}\times 5=\operatorname{mg}\times \frac{3}{2}$

$\mathrm{N_{2}=\frac{10 \times 10 \times 3}{2 \times 5}=30 \mathrm{~N}}$

$\mathrm{N_{1}=\operatorname{mg}=100 \mathrm{~N}}$

$\mathrm{N_{2}=f=30 \mathrm{~N}}$

$\mathrm{E_{f}=\sqrt{100^{2} +30^{2}}=10 \sqrt{109}}$

$\mathrm{F_{w}=30 N=N_{2} .}$

$\mathrm{\frac{F_{w}}{F_{f}}=\frac{30}{10 \sqrt{109}}=\frac{3}{\sqrt{109}}}$

Hence option 3 is the correct answer.

Posted by

HARSH KANKARIA

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