The increase in pressure required to decrease the volume of a water sample by 0.2% is P × 10⁵ Nm⁻². Bulk modulus of water is 2.15 × 10⁹ Nm⁻². The value of P is _________. Numerical answer type. Work it out, then open the answer.

Answer: 43. Δ P = B (Δ V)/(V) Δ P = 2.15×10⁹×0.002 = 4.3×10⁶ N m⁻² = 43×10⁵ N m⁻², so P = 43.

The increase in pressure required to decrease the volume of a water… | JEE Main 2025 (24 Jan, Shift 2)

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Bulk Modulus & Compressibility

The increase in pressure required to decrease the volume of a water sample by $0.2\%$ is $\text{P} \times 10^{5}\text{ Nm}^{-2}$ . Bulk modulus of water is $2.15 \times 10^{9}\text{ Nm}^{-2}$ . The value of P is _________.

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

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Answer: 43

$\Delta P = B\,\frac{\Delta V}{V}$
$\Delta P = 2.15\times10^{9}\times0.002 = 4.3\times10^{6}\ \text{N m}^{-2}$
$= 43\times10^{5}\ \text{N m}^{-2}$ , so $P = 43$ .

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