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Friday, 9 October

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 ________ mm. (Take density of sea water = 10310^3 kgm−3^{-3}, Bulk modulus of rubber = 9×1089\times10^8 Nm−2^{-2}, and g=10g=10 ms−2^{-2})

Official answer

18

NTA final key.