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 = 10³ kgm⁻³, Bulk modulus of rubber = 9 × 10⁸ Nm⁻², and g = 10 ms⁻²) Numerical answer type. Work it out, then open the answer.

Answer: 18. ρ g h = B (Δ V)/(V) B (Δ V)/(V) = 9×10⁸×2×10⁻⁴ = 1.8×10⁵ Pa h = 1.8×10⁵10³×10 = 18 m

The depth below the surface of sea to which a rubber ball be taken so… | JEE Main 2024 (31 Jan, Shift 1)

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Bulk Modulus & Compressibility

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 $= 10^{3}\text{ kgm}^{-3}$ , Bulk modulus of rubber $= 9 \times 10^{8}\text{ Nm}^{-2}$ , and $g = 10\text{ ms}^{-2}$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 18

$\rho g h = B\,\frac{\Delta V}{V}$
$B\,\frac{\Delta V}{V} = 9\times10^{8}\times2\times10^{-4} = 1.8\times10^{5}\ \text{Pa}$
$h = \frac{1.8\times10^{5}}{10^{3}\times10} = 18\ \text{m}$