The increase in pressure required to decrease the volume of a water sample by 0.2 \% is P × 10 5 Nm -2 . Bulk modulus of water is 2.15 × 10 9 Nm -2 . The value of P is â€”.

Physics · Fluid Mechanics

JEE Main 2025 — 24 January, Evening Shift — Question 67

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

Answer: 43

Numerical answer — enter this value.

Step-by-step solution

Since bulk modulus is given as

$B=\frac{-\Delta P}{\left(\frac{\Delta V}{V}\right)}$

$2.15 \times 10^{9}=\frac{-\Delta \mathrm{P}}{-\left(\frac{0.2}{100}\right)}$

$\Delta \mathrm{P}=2.15 \times 10^{9} \times 2 \times 10^{-3}$

$=4.3 \times 10^{6}=43 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Fluid Mechanics
Topic: Properties of Fluids & Hydrostatic Pressure