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The increase in pressure required to decrease the 200 L volume of a liquid by 0.008 % in kPa is (Bulk modulus of the liquid = 2100 M Pa is )
A. 8.4
B. 84
C. 92.4
D. 168

1 Answer

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Best answer
Correct Answer - B
Bulk modulus `K = (Deltap)/(DeltaV) * V`
`Deltap = (K Delta V)/(V)`
`Deltap = (2100 xx 10^(6) xx 0.008)/(200)`
= 84kpa

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