A) \[1\times {{10}^{-6}}\,N/{{m}^{2}}\]
B) \[1\times {{10}^{8}}\,N/{{m}^{2}}\]
C) \[1\times {{10}^{7}}\,N/{{m}^{2}}\]
D) \[1\times {{10}^{9}}\,N/{{m}^{2}}\]
Correct Answer: D
Solution :
\[\beta =\frac{-\Delta P}{\Delta V/V},\,\,here\,\,\Delta P=hdg\]
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