When a rubber ball is taken to a depth of __________m in deep sea, its volume decreases by 0.5%.
(The bulk modulus of rubber = 9.8 × 10⁸ Nm⁻², Density of sea water = 10³ kgm⁻³, g = 9.8 m/s²)

Solution

Physics sir · 3 steps

The extra water pressure does the squeezing, so set hρ g equal to B times the volume strain.

1. Pressure from strain. Bulk modulus links the extra pressure to the fractional change in volume: B = (Δ P)/(Δ V/V). The volume strain here is 0.5% = 0.005.

   Δ P = B (Δ V)/V = 9.8 × 10⁸ × 0.005 = 4.9 × 10⁶ N m⁻²

2. Pressure from depth. At depth h the water adds pressure Δ P = hρ g. So h = (Δ P)/(ρ g) = (4.9 × 10⁶)/(10³ × 9.8).
3. So. h = (4.9 × 10⁶)/(9.8 × 10³) = 500 m, so the answer is 500.

The common slip: 50000. Putting the strain in as 0.5 instead of 0.005 gives 50 000 m, a hundred times too deep. Convert the percentage to a fraction first.

The idea to keep. Bulk modulus questions always come down to Δ P = B (Δ V)/V. Find the pressure from depth with hρ g, and turn percentages into fractions first.

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