Fractional Decrease in Radius Under Uniform Pressure

For a sphere, volume V = (4/3)πr³, so V depends on r³. When r changes by a small amount, take the differential: ΔV/V = 3 × (Δr/r). This means the volume changes 3 times as fast as the radius. Rearranging gives Δr/r = (1/3)(ΔV/V). The factor 3 comes only from the cube power, so it is true for any object whose volume scales as (length)³.

Bulk modulus is defined as K = P / (ΔV/V), so the fractional decrease in volume under pressure P is ΔV/V = P/K. Since Δr/r = (1/3)(ΔV/V), substitute to get Δr/r = P/(3K). The negative sign is dropped because we speak of a 'decrease'; both radius and volume get smaller under pressure.

The relation ΔV/V = 3 (ΔL/L) works for any shape where one linear dimension controls the volume as (length)³ — a sphere (radius) or a cube (side). So for a cube of side a under uniform pressure, Δa/a = P/(3K) too. The '3' is from the three dimensions, not from the sphere shape.

Yes. Uniform (hydrostatic) pressure acts equally on all sides of the body, so it is a bulk stress (volume stress). It changes volume, not shape, so bulk modulus K is the correct constant. This is different from a wire being pulled, where you use Young's modulus.

Volume and radius are linked by V ∝ r³, so you cannot shrink the volume without shrinking the radius. Uniform pressure squeezes the sphere evenly from every direction, so it stays a sphere but a smaller one — the radius drops by Δr/r = P/(3K).

Δr/r = P/(3K), where P is the uniform pressure applied and K is the bulk modulus of the material.

Because volume scales as radius cubed (V ∝ r³), so ΔV/V = 3(Δr/r). The volume changes three times faster than the radius, giving the 1/3 factor for radius.

Bulk modulus K is used, because uniform pressure is a volume (bulk) stress that changes size, not shape.

Directly Δr/r = P/(3K). For example, if P = 3 × 10⁷ Pa and K = 1 × 10¹¹ Pa, then Δr/r = 3×10⁷ / (3×10¹¹) = 10⁻⁴.