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The increase in pressure required to dec...

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

Text Solution

Verified by Experts

The correct Answer is:
b

As, `B = ("normal stress")/("volumetric strain") = (Delta P)/(-Delta V//V)`
Here, negative sign shows that volume is decreased when pressure is increased.
Here. `B = 2100 xx 10^(6) Pa, V = 200 L`
`Delta V = 200 xx (0.004)/(100) = 0.008 L`
`:. 2100 xx 10^(6) = (Delta P)/(((0.008)/(200)))`
`:. Delta P = 84 kPa`
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