## Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-9 Mechanical properties of Solids

### Q.12:- Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Answer:-

Initial volume, V1 = 100.0l = 100.0 × 10 –3 m3
Final volume, V2 = 100.5 l = 100.5 ×10 –3 m3
Increase in volume, ΔV = V2V1 = 0.5 × 10–3 m3
Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 105 Pa
Bulk modulus = Δp / (ΔV/V1)  =  Δp × V1 / ΔV
= 100 × 1.013 × 105 × 100 × 10-3 / (0.5 × 10-3)
= 2.026 × 109 Pa
Bulk modulus of air = 1 × 105 Pa
∴ Bulk modulus of water / Bulk modulus of air  =  2.026 × 109 / (1 × 105)  =  2.026 × 104
This ratio is very high because air is more compressible than water.