JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2025
- Medium
- Numerical
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 ________.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 43
Step-by-step Solution
Detailed explanation
Since bulk modulus is given as
\(\begin{aligned}
& \mathrm{B}=\frac{-\Delta \mathrm{P}}{\left(\frac{\Delta \mathrm{~V}}{\mathrm{~V}}\right)} \\
& 2.15 \times 10^9=\frac{-\Delta \mathrm{P}}{-\left(\frac{0.2}{100}\right)} \\
& \Delta \mathrm{P}=2.15 \times 10^9 \times 2 \times 10^{-3} \\
& =4.3 \times 10^6=43 \times 10^5 \mathrm{~N} / \mathrm{m}^2
\end{aligned}\)
\(\begin{aligned}
& \mathrm{B}=\frac{-\Delta \mathrm{P}}{\left(\frac{\Delta \mathrm{~V}}{\mathrm{~V}}\right)} \\
& 2.15 \times 10^9=\frac{-\Delta \mathrm{P}}{-\left(\frac{0.2}{100}\right)} \\
& \Delta \mathrm{P}=2.15 \times 10^9 \times 2 \times 10^{-3} \\
& =4.3 \times 10^6=43 \times 10^5 \mathrm{~N} / \mathrm{m}^2
\end{aligned}\)
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