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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2024 · 9 April · Evening Shift · Physics · Laws of Motion

A 1 ~kg mass is suspended from the ceiling by a rope of length 4 ~m...

A 1 kg1 \mathrm{~kg} mass is suspended from the ceiling by a rope of length 4 m4 \mathrm{~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 FF is : (Assume that the system is in equilibrium and g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^2 ) A 1 ~kg mass is suspended from the ceiling by a rope of length 4 ~m . A horizontal force ' F ' is applied at the mid

Options

  1. A10 N10 \mathrm{~N}
  2. B102 N\frac{10}{\sqrt{2}} \mathrm{~N}
  3. C1 N1 \mathrm{~N}
  4. D110×2N\frac{1}{10 \times \sqrt{2}} N

Correct answer

A. 10 N10 \mathrm{~N}

Step-by-step solution

T1sin⁡45∘=FT1cos⁡45∘=T2=1×g∴tan⁡45∘=Fg∴F=10 N\begin{aligned} & \mathrm{T}_1 \sin 45^{\circ}=\mathrm{F} \\ & \mathrm{T}_1 \cos 45^{\circ}=\mathrm{T}_2=1 \times \mathrm{g} \\ & \therefore \tan 45^{\circ}=\frac{\mathrm{F}}{\mathrm{g}} \\ & \therefore \mathrm{F}=10 \mathrm{~N}\end{aligned}
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