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)
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)
Updated On: Mar 28, 2026
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The Correct Option is B
Solution and Explanation
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 $\hspace30mm ? p = 84 \, kPa$
$\hspace40mm B = \frac{? p}{- ? V/V}$
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 $\hspace30mm ? p = 84 \, kPa$
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