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

Correct answer: 0 Since bulk modulus is given as aligned & B= -Δ P ( Δ ~V ~V ) & 2.15 × 10^9= -Δ P -((0.2)/(100)) & Δ P=2.15 × 10^9 × 2 × 10^ -3 & =4.3 × 10^6=43 × 10^5 ~N / m^2 aligned

JEE Main 202524 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Evening Shift
Subject: Physics
Chapter: Mechanical Properties of Solids

JEE Main 2025 · 24 January · Evening Shift · Physics · Mechanical Properties of Solids

The increase in pressure required to decrease the volume of a water...

The increase in pressure required to decrease the volume of a water sample by

0.2 \%

\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

\ldots\ldots

Correct answer

Step-by-step solution

Since bulk modulus is given as

\begin{aligned} & \mathrm{B}=\frac{-\Delta \mathrm{P}}{\left(\frac{\Delta \mathrm{~V}}{\mathrm{~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 \end{aligned}

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