JEE Main 2024 Physics question
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)
Answer: 18
Solution
B = ΔP/(ΔV/V), so ρgh = B(ΔV/V) and h = (B/ρg)(ΔV/V) = (9 × 10⁸/(10³ × 10))(0.02 × 10⁻²) = 18 m.