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JEE MAINS 2024
31.01.24 S1
Question
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.
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Solution
Sol. $$ \begin{aligned} & \beta=\frac{-\Delta \mathrm{P}}{\frac{\Delta \mathrm{~V}}{\mathrm{~V}}} \\ & \Delta \mathrm{P}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{~V}} \\ & \rho g h=-\beta \frac{\Delta \mathrm{V}}{\mathrm{~V}} \\ & 10^3 \times 10 \times \mathrm{h}=-9 \times 10^8 \times\left(-\frac{0.02}{100}\right) \\ & \Rightarrow \mathrm{h}=18 \mathrm{~m} \end{aligned} $$
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