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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. - Physics

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.

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Solution

Initial volume, V= 100.0l = 100.0 × 10 –3 m3

Final volume, V= 100.5 l = 100.5 ×10 –3 m3

Increase in volume, ΔV = V2 – V= 0.5 × 10–3 m3

Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 105 Pa

Bulk modulus = `((trianglep)/(triangleV))/V_1 = (trianglep xx V_1)/(triangleV)`

`= (100 xx 1.013 xx 10^5 xx 100 xx 10^(-3))/(0.5xx 10^(-3))`

`= 2.026 xx 10^9 Pa`

Bulk modulus of air =`1.0 xx 10^5 Pa`

`:. "Bulk modulus of water"/"Bulk modulus of air" = (2.026 xx 10^9)/(1.0 xx 10^5) = 2.026 xx 10^4`

This ratio is very high because air is more compressible than water.

Concept: Elastic Moduli - Bulk Modulus
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APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 9 Mechanical Properties of Solids
Q 12 | Page 244
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