Q.What is the density of ocean water at a depth where the pressure is 100 atm. Given that its density at the surface is 1.03×10^3 kg m^-3? Compressibility 45.8×10^-11 pa^-1, 1atm =1.013×10^5 Pa. OR A 10kW drilling machine is used to drill a bore in a small aluminium block of mass 10 kg. How much is rise in the temperature of the block in 3 minutes assuming 50% of power is used up in heating the machine itself or lost to the surrounding. Specific heat of aluminium = 0.91Jg^-1 °C^-1.

Bulk Modulus: The Resistance to Squeezing

Imagine you have a sponge. When you squeeze it from all sides — say, by pushing it into a smaller space — it gets compressed. Now imagine a block of steel. If you try to squeeze it from all sides, it barely changes size. The bulk modulus is the number that tells you how hard it is to compress a material when you apply pressure evenly from every direction.

This is different from stretching or bending. Here, the force is uniform all around — like the pressure deep underwater, where water pushes on every surface of an object.

The Intuition: Pressure vs. Volume Change

Take a cube of material. If you increase the pressure on it (push harder from all sides), its volume decreases. The bulk modulus $K$ is defined as:

Bulk modulus = $\frac{\text{pressure applied}}{\text{fractional change in volume}}$

In symbols:

Where:

• $\Delta P$ = change in pressure (force per area)
• $\Delta V$ = change in volume (final minus initial)
• $V_0$ = original volume

The minus sign is there because when pressure increases ($\Delta P > 0$), volume decreases ($\Delta V < 0$), so the ratio comes out positive.

The Precise Statement

The bulk modulus is a material property. It tells you how much the volume of a substance changes when you apply a uniform pressure. The reciprocal of bulk modulus is called compressibility ($\beta = 1/K$), which is often used for gases and liquids.

For a solid, $K$ is usually very large — a few hundred gigapascals for metals. For water, $K \approx 2.2 \times 10^9$ Pa (about 2.2 GPa). For air at room temperature, $K \approx 1.4 \times 10^5$ Pa — much smaller, which is why you can easily squeeze a balloon.

Where It Shows Up in Exams

You'll typically see three types of problems:

1. Direct calculation: Given $\Delta P$ and $\Delta V/V_0$, find $K$ (or vice versa).
2. Comparing materials: Which has higher bulk modulus? (Steel > water > air)
3. Applications: Why does a submarine's hull need to be strong? Because at depth, $\Delta P$ is huge, and a small $K$ would mean dangerous compression.

A Common Mistake

Students often forget the negative sign in the formula. Remember: when pressure goes up, volume goes down. The ratio $\Delta V / V_0$ is negative, so $-\Delta P / (\Delta V/V_0)$ gives a positive $K$. If you drop the minus sign, you'll get a negative bulk modulus — which is physically meaningless.

Real-World Example …