Verified Solution • 2025–26 Session moderate

QJEEM24Ph3dad: JEE Main 2024 Physics - Class 11 JEEM24Ph3dad Solution: The depth below the surface of sea to which a rubber ball be…

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

(Take density of sea water =103kgm−3=10^3 \mathrm{kgm}^{-3} , Bulk modulus of rubber =9×108 Nm−2=9 \times 10^8 \mathrm{~Nm}^{-2} , and g=10 ms−2g=10 \mathrm{~ms}^{-2} )

Step-by-Step Verified Solution

amp;β=−ΔPΔVVamp;ΔP=−βΔVVamp;ρgh=−βΔVVamp;103×10×h=−9×108×(−0.02100)amp;⇒h=18 m\begin{aligned}& \beta=\frac{-\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}} \\& \Delta \mathrm{P}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}} \\& \rho \mathrm{gh}=-\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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