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JEE Main 2024 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 31 January 2024, Shift 1, Physics Q53

Question 53 of 90 in this shift, Physics question 23 of 30, Section B.

The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by 0.02% is ________

m

. (Take density of sea water =

10^3

kgm

^{-3}

, Bulk modulus of rubber =

9\times10^8

Nm

^{-2}

, and

g=10

ms

^{-2}

)

Official answer

18

NTA final key.

Chapter: Mechanical Properties of Solids
Topic: Elasticity
Idea tested: Compression at Liquid Depth

Same idea in other shifts

Asked 2× in all

27 Jan 2024, Shift 1 · Q53If average depth of an ocean is 4000 m and the bulk modulus of water is $2\times10^{9}$ Nm $^{-2}$ , then fractional compression…EasyNumerical 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.