A $\sqrt{34}m$ long ladder weighing $10\mathrm{kg}$ leans on a frictionless wall. Its feet rest on the floor $3m$ 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 $\frac{{F}_{w}}{{F}_{f}}$ will be : (Use $g=10m{s}^{-2}$.) ![JEE Main 2022 Physics, Rotational Motion — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Rotational_Motion/648b5a6e417cc3fb48d67d30/question_1__q_648b5a6e417cc3fb48d67d30__cdn-question-pool.getmarks.app__6a305d3f-238a-4aa4-bece-92bd34152a39-image__c4b9097322_final_ppt_sync.png)

![JEE Main 2022 Physics, Rotational Motion — solution figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Rotational_Motion/648b5a6e417cc3fb48d67d30/solution_1__q_648b5a6e417cc3fb48d67d30__cdn-question-pool.getmarks.app__c7c1faaf-982e-474f-9f19-30ca0e6bb662-image__6b86547113_final_ppt_sync.png) Applying force balance in vertical direction ${N}_{1}=mg=100N$ Applying torque balance at the bottom contact point, $mg\times \frac{3}{2}={N}_{2}\times 5\Rightarrow {N}_{2}={F}_{w}=30N$ Applying force balance in horizontal direction ${f}_{s}={N}_{2}=30N$ Net reaction at the floor will be, ${F}_{f}=\sqrt{{{N}_{1}}^{2}+{{f}_{s}}^{2}}=\sqrt{{100}^{2}+{30}^{2}}=10\sqrt{109}N$ Therefore, $\frac{{F}_{w}}{{F}_{f}}=\frac{30}{10\sqrt{109}}=\frac{3}{\sqrt{109}}$

A √34m long ladder weighing 10… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Mechanics

2022mcqmedium

A $\sqrt{34}m$ long ladder weighing $10\mathrm{kg}$ leans on a frictionless wall. Its feet rest on the floor $3m$ 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 $\frac{{F}_{w}}{{F}_{f}}$ will be :

(Use $g=10m{s}^{-2}$ .)

JEE Main 2022 Physics, Rotational Motion — question figure

Official previous-year question

Held on 28 Jun 2022 · Verified 6 Jul 2026.

Options

A
$\frac{6}{\sqrt{110}}$
B
$\frac{3}{\sqrt{113}}$
C
$\frac{3}{\sqrt{109}}$
D
$\frac{2}{\sqrt{109}}$

Did you get this right?

Solution

JEE Main 2022 Physics, Rotational Motion — solution figure

Applying force balance in vertical direction

${N}_{1}=mg=100N$

Applying torque balance at the bottom contact point,

$mg\times \frac{3}{2}={N}_{2}\times 5\Rightarrow {N}_{2}={F}_{w}=30N$

Applying force balance in horizontal direction

${f}_{s}={N}_{2}=30N$

Net reaction at the floor will be,

${F}_{f}=\sqrt{{{N}_{1}}^{2}+{{f}_{s}}^{2}}=\sqrt{{100}^{2}+{30}^{2}}=10\sqrt{109}N$

Therefore, $\frac{{F}_{w}}{{F}_{f}}=\frac{30}{10\sqrt{109}}=\frac{3}{\sqrt{109}}$

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Solution

JEE Main 2022 Physics, Rotational Motion — solution figure

Applying force balance in vertical direction

${N}_{1}=mg=100N$

Applying torque balance at the bottom contact point,

$mg\times \frac{3}{2}={N}_{2}\times 5\Rightarrow {N}_{2}={F}_{w}=30N$

Applying force balance in horizontal direction

${f}_{s}={N}_{2}=30N$

Net reaction at the floor will be,

${F}_{f}=\sqrt{{{N}_{1}}^{2}+{{f}_{s}}^{2}}=\sqrt{{100}^{2}+{30}^{2}}=10\sqrt{109}N$

Therefore, $\frac{{F}_{w}}{{F}_{f}}=\frac{30}{10\sqrt{109}}=\frac{3}{\sqrt{109}}$

Options

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics