Question

The increase in pressure required to decrease the 200 L volume of a liquid by 0.004% (in kPa) is (Bulk modulus of the liquid = 2100 MPa)

• A. 8.4
• B. 84
• C. 92.4
• D. 168

Answer 1

Answer: (B)

84

Answer 2

Bulk modulus $\, \, B =\frac{normal \, stress}{volumetric \, strain }$
\hspace40mm B = \frac{? p}{- ? V/V}
Here, negative sign shows that volume is decreased,
pressure is increased.
Here \hspace30mm B = 2100 \times 10^6 \, Pa
\hspace40mm V = 200 \, L
\hspace40mm ? V = 200 \times \frac{0.004}{100} = 0.008 \, L
\therefore \hspace20mm 2100 \times 10^6 = \frac{? p}{\big(\frac{0.008}{200}\big)}
or \hspace30mm ? p = 84 \, kPa