The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by 0.02 % is _______ m.
(Take density of sea water =10³ kgm⁻³, Bulk modulus of rubber =9 × 10⁸ Nm⁻², and g=10 ms⁻²)

Solution

Physics sir · 4 steps

Find the extra pressure that squeezes the ball by 0.02 %, then ask how deep the water gives that pressure.

1. Volume strain. A decrease of 0.02 % means (Δ V)/V = 0.02/100 = 2 × 10⁻⁴.
2. Pressure needed. Bulk modulus is extra pressure divided by volume strain, so Δ P = B × (Δ V)/V = 9 × 10⁸ × 2 × 10⁻⁴ = 1.8 × 10⁵ N m⁻².

   hρ g = B (Δ V)/V ⇒ h = (1.8 × 10⁵)/(10³ × 10)

3. Depth. Atmospheric pressure already acts on the ball at the surface, so only the extra pressure hρ g from the water column changes its volume. This gives h = (1.8 × 10⁵)/10⁴ = 18 m.
4. So. The ball must be taken to a depth of 18 m.

The common slip: 1800. Using 0.02 as the strain without dividing by 100 gives a pressure 100 times too big and a depth of 1800 m.

The idea to keep. Bulk modulus links extra pressure to fractional volume change, Δ P = B Δ V/V. Under water that extra pressure is hρ g, so set them equal and solve for h.

Answer from the NTA final answer key. Solution by PrepWiser.