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Exercises · 8.16

Q.How much should the pressure on a litre of water be changed to compress it by 0.10%?

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To compress water by 0.10%, the pressure must be increased by an amount determined by water's bulk modulus, resulting in a pressure change of 2.2×106 Pa\boxed{2.2 \times 10^6 \text{ Pa}}.

The problem asks us to determine the change in pressure required to achieve a specific percentage compression of water. This directly involves the concept of Bulk Modulus, which is a measure of a substance's resistance to compression under pressure.

Concept and Intuition: Bulk Modulus

Imagine trying to squeeze a balloon filled with air versus a balloon filled with water. The air-filled balloon is much easier to compress. This difference in "squeezability" is quantified by the bulk modulus.

  • Bulk Modulus (BB) is defined as the ratio of the change in pressure (ΔP\Delta P) to the resulting fractional change in volume (ΔV/V\Delta V / V). It tells us how much pressure is needed to cause a certain relative change in volume.
  • A high bulk modulus means the material is difficult to compress (like water or steel). A low bulk modulus means it's easy to compress (like air).
  • When you increase pressure on a substance, its volume decreases. This means if ΔP\Delta P is positive (pressure increases), ΔV\Delta V will be negative (volume decreases). To ensure the bulk modulus BB is always a positive value, a negative sign is included in its definition.

The formula for Bulk Modulus is:

B=−ΔPΔV/VB = - \frac{\Delta P}{\Delta V / V}

where:

  • BB is the Bulk Modulus (in Pascals, Pa)
  • ΔP\Delta P is the change in pressure (in Pascals, Pa)
  • ΔV\Delta V is the change in volume (in m3^3 or litres)
  • VV is the original volume (in m3^3 or litres) The ratio ΔV/V\Delta V / V is the fractional change in volume.

Now, let's apply this concept to solve the problem.

Step-by-Step Solution

  1. Identify the given information and the unknown.
    • The initial volume of water is V=1 litreV = 1 \text{ litre}. While the absolute volume is given, notice that the formula uses the fractional change in volume (ΔV/V\Delta V / V), so the specific value of VV will cancel out.
    • The water is to be compressed by 0.10%0.10\%. This means the fractional change in volume is ΔVV=−0.10%\frac{\Delta V}{V} = -0.10\%. The negative sign indicates compression (decrease in volume).

ΔVV=−0.10100=−0.0010\frac{\Delta V}{V} = - \frac{0.10}{100} = -0.0010

*   We need to find the change in pressure, $\Delta P$. …

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