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Saturday, 31 October

JEE Main 2023 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 11 April 2023, Shift 2, Physics Q52

Question 52 of 90 in this shift, Physics question 22 of 30, Section B.

The surface tension of soap solution is 3.5×10−2 Nm−13.5\times10^{-2}\ \mathrm{Nm^{-1}}. The amount of work done required to increase the radius of soap bubble from 10 cm10\ \mathrm{cm} to 20 cm20\ \mathrm{cm} is ____________ ×10−4 J\times10^{-4}\ \mathrm{J}. (take π=22/7\pi = 22/7)

Official answer

264

NTA final key.

Chapter
Mechanical Properties of Fluids
Topic
Surface Tension
Idea tested
Surface Energy and Work

Same idea in other shifts

Asked 2× in all
  1. 4 Apr 2024, Shift 1 · Q53A soap bubble is blown to a diameter of 7 cm. 36960 erg of work is done in blowing it further. If surface tension of soap solution is 40…MediumNumerical value

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.