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 $MPa$ is)

Question

Dr. Abubacker · Accepted Answer

(b) Bulk modulus $K = \frac{\mathrm{\Delta}p}{\mathrm{\Delta}V}V$
$\mathrm{\Delta}p = \frac{K\mathrm{\Delta}V}{V}$
$\mathrm{\Delta}p = \frac{2100 	imes 10^{6} 	imes 0.008}{200} = 84\ kPa$

Q. 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 $MPa$ is)

Explanation

Reviewed by Dr. Abubacker

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