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The compressibility of water is $ 5\times {{10}^{-10}}{{m}^{2}}{{N}^{-1}} $ $ $ . The change in volume of $ 100ml $ water subjected to $ 15\times {{10}^{6}}Pa $ Pressure will
(A) No change
(B) Increase by $ 0.75\text{ }ml $
(C) Decrease by $ 1.5\text{ }ml $
(D) Decrease by $ 0.75\text{ }ml $

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Last updated date: 16th Jul 2024
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Answer
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Hint: Use the concept of Bulk Modulus of elasticity. Bulk Modulus of elasticity is defined as the ratio of normal stress to volumetric strain within the elastic limit.
Volumetric strain is defined as change of change in volume to the original volume. Since the force applied normally on the entire surface of the body and its volume decreases compressibility is reciprocal of bulk modulus of elasticity. Use the direct formula, we will get the answer.

Formula used: $ \text{ }\!\!~\!\!\text{ Bulk modulus of elasticity =}\dfrac{\text{Normal stress}}{\text{Volumetric strain}} $
 $ \text{B = }-\dfrac{\text{PV}}{\Delta \text{V}} $
Compressibility, $ \text{k = }\dfrac{1}{\text{B}} $
 $ -\Delta \text{V = pVk} $
 $ \text{P} $ is pressure , $ \text{V} $ is volume, $ \text{k} $ is compressibility.

Complete step by step solution
 $ \text{ }\!\!~\!\!\text{ Bulk modulus of elasticity =}\dfrac{\text{Normal stress}}{\text{Volumetric strain}} $
Volumetric strain $ \text{=}-\dfrac{\Delta \text{V}}{\text{V}} $ , Normal stress $ =\dfrac{\text{F}}{a} $
Also we know, $ \text{Pressure = }\dfrac{\text{Force}}{\text{Area}} $
 $ \text{B =}\dfrac{\dfrac{\text{F}}{a}}{\dfrac{-\Delta \text{V}}{\text{V}}} $ $ $
 $ \text{B = }\dfrac{\text{p}}{\dfrac{-\Delta \text{V}}{\text{V}}} $
 $ \text{B = }\dfrac{\text{pV}}{-\Delta \text{V}} $
Compressibility, $ \text{B = }\dfrac{1}{\text{k}} $
 $ \text{k =}\dfrac{1}{\text{B}}=-\dfrac{\Delta \text{V}}{\text{pV}} $
 $ -\Delta \text{V = pVk} $
Here from question $ \text{p = 15}\times \text{1}{{\text{0}}^{6}}\text{ Pa} $ [pressure is given]
Compressibility, $ \text{k = 5}\times \text{1}{{\text{0}}^{-10}}\text{ }{{\text{m}}^{2}}{{\text{N}}^{-1}} $
Volume of water $ =10\text{ ml} $
 $ \Delta \text{V} $ change in volume [negative sign shows volume is decrease]
 $ \begin{align}
  & -\Delta \text{V = 5}\times \text{1}{{\text{0}}^{-10}}\times 100\times 15\times {{10}^{6}} \\
 & -\Delta \text{V}=75\times {{10}^{-10+8}} \\
 & -\Delta \text{V}=75\times {{10}^{-2}} \\
 & -\Delta \text{V = 0}\text{.75 ml} \\
\end{align} $
Hence change in volume is $ 0.75\text{ ml} $ (Negative sign shows volume decrease.)

Additional Information
Young’s Modulus of Elasticity(y). It is defined as the ratio of normal stress to the longitudinal strain within the elastic unit.
Thus,
 $ \begin{align}
  & \text{Y = }\dfrac{\text{Normal stress}}{\text{Longitudinal strain}} \\
 & \text{Y = }\dfrac{\dfrac{\text{F}}{\text{a}}}{\dfrac{-\Delta \text{l}}{e}} \\
\end{align} $
Young’s Modulus is involved in solid only. Greater the value of Young’s Modulus, larger is elasticity.
Shearer Modulus of Elasticity: It is defined as the ratio of tangential stress to shearing strain, within the elastic limit.
 $ \begin{align}
  & \text{G = }\dfrac{\text{Tangential stress}}{\text{Shearing strain}} \\
 & \text{G = }\dfrac{\dfrac{\text{F}}{a}}{\theta } \\
 & \text{G = }\dfrac{\dfrac{\text{F}}{a}}{\dfrac{\Delta \text{L}}{\text{L}}} \\
\end{align} $
Shearer Modulus is involved with solids only.
Shear modulus is less than young’s modulus (for solids).

Note
Bulk modulus for a perfectly rigid body is infinite. The conceptual knowledge about various concepts like elasticity, Bulk’s modulus, Young’s modulus, strain and stress is required by the student.