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Physics · 2024

JEE Main · 31 January 2024, Shift 1 · Q53

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…

The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by $\displaystyle 0.02 \%$ is $\displaystyle \_\_\_\_$ $\displaystyle m$. (Take density of sea water $\displaystyle =10^3 \mathrm{kgm}^{-3}$, Bulk modulus of rubber $\displaystyle =9 \times 10^8 \mathrm{Nm}^{-2}$, and $\displaystyle g=10 \mathrm{~ms}^{-2}$ )
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JEE Main 2024 Physics question, with the answer from NTA’s final answer key. Where our answers come from.