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Physics · Newton's Laws of Motion

JEE Main 2024 — 9 April, Shift 2 — Question 44

A 1 kg mass is suspended from the ceiling by a rope of length 4 m . A horizontal force ' FF ' is applied at the mid point of the rope so that the rope makes an angle of 45∘45^{\circ} with respect to the vertical axis as shown in figure. The magnitude of F is :

Question figure
  1. Option A:

    102 N\frac{10}{\sqrt{2}} \mathrm{~N}

  2. Option B:

    1 N

  3. Option C:

    110×2 N\frac{1}{10 \times \sqrt{2}} \mathrm{~N}

  4. Option D:

    10 N

    Correct

Answer: D

Step-by-step solution

T1sin⁡45∘=F\mathrm{T}_{1} \sin 45^{\circ}=\mathrm{F}

T1cos⁡45∘=T2=1×g\mathrm{T}_{1} \cos 45^{\circ}=\mathrm{T}_{2}=1 \times \mathrm{g}

∴tan⁡45∘=Fg\therefore \tan 45^{\circ}=\frac{F}{g}

∴F=10 N\therefore \mathrm{F}=10 \mathrm{~N}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 2
Subject
Physics
Chapter
Newton's Laws of Motion
Topic
Application of NLM and Impulse
A 1 kg mass is suspended from the ceiling by a rope of length 4 m . A… | JEE Main 2024 PYQ with Solution · DhiX AI