To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%. (The bulk modulus of rubber is $9.8 \times 10^{8}$ N m$^{-2}$; and the density of sea water is $10^{3}$ kg m$^{-3}$.)

To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%. (The bulk modul

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9.8 \times 10^{8}

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Concept understanding — Bulk Modulus Application

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:

K = -\frac{\Delta P}{\Delta V / V_0}

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.

Note

A large $K$ means the material is hard to compress (like diamond or steel). A small $K$ means it's easy to compress (like air or a sponge).

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.

K = -\frac{\Delta P}{\Delta V / V_0}

Where It Shows Up in Exams

You'll typically see three types of problems:

Direct calculation: Given $\Delta P$ and $\Delta V/V_0$ , find $K$ (or vice versa).
Comparing materials: Which has higher bulk modulus? (Steel > water > air)
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.

Tip

For solids, the volume change is tiny — often given in scientific notation. For gases, the volume change can be large, so always check units carefully.

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 …

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The bulk modulus relates volumetric strain to applied pressure. At depth $h$ , hydrostatic pressure compresses the ball; equating this pressure to the product of bulk modulus and fractional volume change gives $h = 100$ m.

When a rubber ball descends into the ocean, the increasing water pressure squeezes it from all sides. The bulk modulus $B$ quantifies a material's resistance to uniform compression: it tells us how much pressure is needed to produce a given fractional change in volume. Specifically,

B = -\frac{\Delta P}{\Delta V / V}

where $\Delta P$ is the pressure increase, $\Delta V$ is the volume change (negative for compression), and $V$ is the original volume. The negative sign ensures $B$ is positive, since pressure increases cause volume to decrease.

At a depth $h$ below the sea surface, the additional pressure due to the water column is simply $\Delta P = \rho g h$ , where $\rho$ is the density of seawater and $g$ is gravitational acceleration. This hydrostatic pressure is what compresses the ball.

B = -\frac{\Delta P}{\Delta V / V} \quad \Rightarrow \quad \Delta P = -B \frac{\Delta V}{V}

Now let's find the required depth step by step.

Identify the fractional volume change. The volume decreases by 0.1%, so

\frac{\Delta V}{V} = -0.001

(negative because the volume shrinks).

Calculate the required pressure increase. Using the bulk modulus relation:

\Delta P = -B \frac{\Delta V}{V} = -(9.8 \times 10^8) \times (-0.001)

\Delta P = 9.8 \times 10^5 \text{ N m}^{-2}

Relate pressure to depth. The hydrostatic pressure at depth $h$ is …

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Q.To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%. (The bulk modulus of rubber is 9.8×1089.8 \times 10^{8}9.8×108 N m−2^{-2}−2; and the density of sea water is 10310^{3}103 kg m−3^{-3}−3.)

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Q.To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%. (The bulk modulus of rubber is $9.8 \times 10^{8}$ N m $^{-2}$ ; and the density of sea water is $10^{3}$ kg m $^{-3}$ .)

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