JEE Main 2022 · 28 Jun · Shift 2 · Physics Q34

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PhysicsSingle correct+4 / -1Question 34 of 90

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 $F_w/F_f$ will be :

(Use $g=10$ m/s $^2$ .)

A ladder of length √34 m leaning against a vertical wall, its foot 3 m from the wall, forming a right triangle with the floor.

\dfrac6{\sqrt{110}}

\dfrac3{\sqrt{113}}

\dfrac3{\sqrt{109}}

\dfrac2{\sqrt{109}}

Answer and worked solution

Official answer: C

Height on the wall $\sqrt{34-9}=5$ m.

Torques about the foot. $F_w\cdot5=100\cdot1.5$ , so $F_w=30$ N. The floor gives 100 N up and 30 N horizontally: $F_f=\sqrt{10900}=10\sqrt{109}$ N.

Ratio $\dfrac3{\sqrt{109}}$ .

(C)

Topics:Equilibrium of rigid bodies

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JEE Main 2022 · 28 Jun · Shift 2, Physics question 34