Q73 / 75 · JEE Main 2025 (24 Jan Shift 2) · Physics
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 $\ldots\ldots$
Correct answer: 43
Solution
Since bulk modulus is given as
$\begin{aligned}
& \mathrm{B}=\frac{-\Delta \mathrm{P}}{\left(\frac{\Delta \mathrm{~V}}{\mathrm{~V}}\right)} \\
& 2.15 \times 10^9=\frac{-\Delta \mathrm{P}}{-\left(\frac{0.2}{100}\right)} \\
& \Delta \mathrm{P}=2.15 \times 10^9 \times 2 \times 10^{-3} \\
& =4.3 \times 10^6=43 \times 10^5 \mathrm{~N} / \mathrm{m}^2
\end{aligned}$
Hence, the correct answer is 43.