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Saturday, 31 October

JEE Main 2022 · PhysicsNumerical answerNumerical valueMediumMulti-step

JEE Main 27 June 2022, Shift 2, Physics Q51

Question 51 of 90 in this shift, Physics question 21 of 30, Section B.

A mass of 10 kg is suspended vertically by a rope of length 5 m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is θ=tan⁡−1(x×10−1)\theta = \tan^{-1}(x \times 10^{-1}) . The value of xx is __________. (Given, g = 10 m/s2^2)

Official answer

3

NTA final key.

Chapter
Laws of Motion
Topic
Newton's Laws of Motion
Idea tested
Equilibrium of a String Junction (Knot)

Same idea in other shifts

Asked 2× in all
  1. 2 Apr 2025, Shift 2 · Q29A body of mass 1 kg is suspended with the help of two strings making angles as shown in figure. Magnitudes of tensions T1T_1 and T2T_2,…EasySingle correctHas 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. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.