Q107 / 150 · MHT CET 2020 (19 Oct Shift 2) · Physics

The compressibility of water is $5 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{N}$. Pressure of $15 \times 10^{6} \mathrm{~Pa}$ is applied on $100 \mathrm{ml}$ volume of water. The change in the volume of water is

  • A
    $0.75 \mathrm{ml}$ increase.
  • B
    $1.50 \mathrm{ml}$ increase.
  • Correct answer
    C
    $0.75 \mathrm{ml}$ decrease.
  • D
    zero.

Solution

(A) Given, $\begin{array}{l} \beta=\frac{1}{\mathrm{~K}}=5 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{N} \\ \Delta \mathrm{P}=15 \mathrm{MPa}=15 \times 10^{6} \mathrm{~Pa} \end{array}$ Bulk modulus, $\mathrm{K}=-\frac{\Delta \mathrm{P}}{\varepsilon_{\mathrm{V}}}$ where, $\varepsilon_{\mathrm{V}}=$ volumetric strain $\begin{array}{l} \beta=-\frac{\varepsilon_{\mathrm{V}}}{\Delta \mathrm{P}} \\ \varepsilon_{\mathrm{V}}=-\beta \Delta \mathrm{P}=-5 \times 10^{-10} \times 15 \times 10^{6} \\ \varepsilon_{\mathrm{V}}=7.5 \times 10^{-3} \end{array}$ The correct option is A.

Hence, the correct answer is option: 'C'.

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