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QJEEM25Phb051: JEE Main 2025 Physics - Class 11 JEEM25Phb051 Solution: The increase in pressure required to decrease the volume of a…

The increase in pressure required to decrease the volume of a water sample by 0.2%0.2 \% is P×105Nm−2\mathrm{P} \times 10^5 \mathrm{Nm}^{-2} . Bulk modulus of water is 2.15×109Nm−22.15 \times 10^9 \mathrm{Nm}^{-2} . The value of P is _________ .

Step-by-Step Verified Solution

amp; Since bulk modulus is given as amp;amp;B=−ΔP(Δ V V)amp;2.15×109=−ΔP−(0.2100)amp;ΔP=2.15×109×2×10−3amp;=4.3×106=43×105 N/m2\begin{aligned}&\text { 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}\end{aligned}

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