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Saturday, 10 October

JEE Main 2024 · PhysicsMultiple choiceSingle correctEasyApplication

JEE Main 9 April 2024, Shift 2, Physics Q34

Question 34 of 90 in this shift, Physics question 4 of 30, Section A.

A 11 kg mass is suspended from the ceiling by a rope of length 44 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/s^2})

The figure, in words

Rope hangs from ceiling at 45 deg to the vertical (tension T1); a 1 kg mass hangs below on a vertical rope segment (tension T2); horizontal force F pulls to the left at the junction (mid point of the rope).
  1. (1)1010 NOfficial answer
  2. (2)102\frac{10}{\sqrt{2}} N
  3. (3)110×2\frac{1}{10\times\sqrt{2}} N
  4. (4)11 N

Official answer

Option 1

NTA final key.

Chapter
Laws of Motion
Topic
Newton's Laws of Motion
Idea tested
Static Equilibrium

Same idea in other shifts

Asked 2× in all
  1. 25 Jul 2022, Shift 1 · Q52Four forces are acting at a point P in equilibrium as shown in figure. The ratio of force F1F_1 to F2F_2 is 1:x1:x where x=x= ______.EasyNumerical valueHas a figure

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.