Mechanical Properties of Solids

JEE Main level, single correct, medium

95 of 118

Question 95 of 118

The bulk modulus of a liquid is 3×1010Nm−23\times {10}^{10}{Nm}^{-2}. The pressure required to reduce the volume of liquid by 22% is :

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Answer: (B) 6×108Nm−26\times {10}^{8}N{m}^{-2}

Bulk modulus B=VolumetricstressVolumetricstrain=−∆P∆VVB=\dfrac{Volumetricstress}{Volumetricstrain}=\dfrac{-∆P}{\dfrac{∆V}{V}}
Then, ∆P=B×(−∆VV)=3×1010×2100=6×108Nm−2∆P=B\times \left(-\dfrac{∆V}{V}\right)=3\times {10}^{10}\times \dfrac{2}{100}=6\times {10}^{8}N{m}^{-2}

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