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 kg m⁻³, Bulk modulus of rubber = 9 × 10⁸ N m⁻², and g = 10 m s⁻²) Numerical answer type. Work it out, then open the answer.

Answer: 18. (Δ V)/(V) = (ρ g h)/(B) ⇒ h = (B (Δ V/V))/(ρ g) h = 9 × 10⁸ × 2 × 10⁻⁴10³ × 10 = 1.8 × 10⁵10⁴ h = 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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Density, Compressibility & Pascal's Law

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 $= 10^{3}\text{ kg m}^{-3}$ , Bulk modulus of rubber $= 9 \times 10^{8}\text{ N m}^{-2}$ , and $g = 10\text{ m s}^{-2}$ )

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

Show answer and solution

Answer: 18

$\dfrac{\Delta V}{V} = \dfrac{\rho g h}{B} \Rightarrow h = \dfrac{B\,(\Delta V/V)}{\rho g}$
$h = \dfrac{9 \times 10^{8} \times 2 \times 10^{-4}}{10^{3} \times 10} = \dfrac{1.8 \times 10^{5}}{10^{4}}$
$h = 18$ m