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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. 8.4B. 84C. 92.4D. 168 |
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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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