Q58
jee_main_2024_31_jan_morning
Bulk Modulus
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)
Numerical Answer. Answer: 18 to 18
Solution
Related Formula
β = (-Δ P)/((Δ V)/(V)) Δ P = ρ g hCore Logic
The change in pressure Δ P is the hydrostatic pressure at depth h.
Δ P = -β (Δ V)/(V) ρ g h = -β (Δ V)/(V)Step 2: Calculation
Given values: ρ = 10³ kg/m³ g = 10 m/s² β = 9 × 10⁸ N/m² (Δ V)/(V) = -0.02% = -(0.02)/(100)
Substitute into the equation:
10³ × 10 × h = - (9 × 10⁸) × (-(0.02)/(100)) 10⁴ × h = 9 × 10⁸ × 2 × 10⁻⁴ 10⁴ h = 18 × 10⁴ h = 18 mChapter Mix
Class 11 Physics: Mechanical Properties Of Solids Class 11 Physics: Mechanical Properties Of Fluids
