The bulk modulus measures a substance's resistance to uniform compression and is defined as the pressure increase needed to cause a given relative decrease in volume. It has a base unit of Pascal. For example, reducing an iron cannon ball's volume by 0.5% requires increasing the pressure by 0.8 GPa if the bulk modulus is 160 GPa. The bulk modulus is larger for solids than liquids and largest for gases, making solids the least compressible and gases the most compressible.

2. The bulk modulus (K) of a substance measures
the substance's resistance to uniform
compression. It is defined as the pressure
increase needed to cause a given relative
decrease in volume. Its base unit is Pascal.
As an example, suppose an iron cannon ball with
Bulk Modulus 160 GPa is to be reduced in
volume by 0.5%. This requires a pressure increase
of 0.005×160 GPA = 0.8 GPA.

4. THE RATIO OF HYDRAULIC STRESS TO
THE CORRESPONDING HYDRAULIC
STRAIN IS CALLED AS BULK MODULUS.
IT DONATED BY SYMBOL ‘B’.
B=-p/(∆V/V)

5. THE RECIPROCAL OF THE BULK
MODULUS IS CALLED
COMPRESSIBILITY. IT IS
DENOTED BY ‘k’. IT IS DEFINED
AS THE FRACTIONAL CHANGE IN
VOLUME PER UNIT INCREASE IN
PRESSURE.
k=(1/B)=-(1/∆p) * (∆V/V)

6. STATE MATERIAL B(10^9 N/m^2 OR GPa)
SOLIDS
ALUMINIUM 72
BRASS 61
COPPER 140
GLASS 37
IRON 100
NICKEL 260
STEEL 160
LIQUIDS
WATER 2.2
ETHANOL 0.9
CARBON DISULPHIDE 1.56
GLYCERIN 4.76
MERCURY 25
GASES AIR (at STP) 1.0 x 10^-4

7. IT CAN BE SEEN FROM THE DATA
THAT BULK MODULI FOR SOLIDS
ARE MUCH LARGER THAN FOR
LIQUIDS, WHICH ARE AGAIN
LARGER THAT BULK MODULUS OF
GASES.THUS SOLIDS ARE LEAST
COMPRESSIBLE AND GASES ARE THE
MOST COMPRESSIBLE.

8. Q: THE AVERAGE DEPTH OF INDIAN
OCEAN IS ABOUT 3000M. CALCULATE
THE FRACTIONAL COMPRESSION. ∆V/V,
OF WATER AT THE BOTTOM OF THE
OCEAN .
GIVEN: BULK MODULUS OF THE OCEAN
IS 2.2 x 10^9 N/m^2, g= 10 m/S^2

9. ANS : THE PRESSURE EXERTED BY THE
COLUMN OF WATER ON THE BOTTOM
LAYER IS
P=ρgh
= 3000m x 1000kg/m^3 x 10m/s^2
= 3 x 10^7 kg/ms^2
= 3 x 10^7 N/m^2
FRACTIONAL COMPRESSION ∆V/V, IS
∆V/V =STRESS/B
=(3 x10^7 N/m^2)/(2.2 x 10^9 N/m^2)
= 1.36 x 10^-2 or 1.36%

10. Q1.WHAT IS THE DENSITY OF OCEAN
WATER AT A DEPTH,WHERE THE
PRESSURE IS 80.0 ATM,GIVEN THAT
IT’S DENSITY AT THE SURFACE IS 1.03
x 10^3 Kg/m^3 ? COMPRESSIBILITY OF
WATER =45.8 x 10^-11 /Pa. GIVEN 1 ATM
=1.013 x 10^5.

11. ANS:
COMPRESSIBILITY =1/K=45.8 x 10^11/PA
CHANGE IN PRESSURE,P=80-1=79 ATM
=79x1.013 x 10^5 PA
DENSITY AT SURFACE,ρ= 1.03 x 10^3
Kg/m^3
K=P/ (∆v/v)
∆v/v= P/K
= 3.665 x 10^-3
∆v/v=(V-V’)/V=(M/ ρ -M/ ρ’)/(M/ ρ)
= 1-(ρ /ρ’)
ρ /ρ’=1-(∆v/v)
ρ’= 1.034 x 10^3

12. Q2. CALCULATE THE PRESSURE
REQUIRED TO STOP THE INCREASE IN
VOLUME OF A COPPER BLOCK WHEN IT IS
HEATED FROM 50˚ TO 70˚C.
COEFFICIENT OF LINEAR EXPANSION
OF COPPER = 8.0 x 10^-6/˚C AND BULK
MODULUS OF ELASTICITY = 3.6 x 10^11
N/m^2.

13. ANS: WHEN A BLOCK OF VOLUME V IS
HEATED THROUGH A TEMPERATURE OF
∆T,THE CHANGE IN VOLUME IS
∆v/v =γ ∆T
WHERE γ(=3α=24 x 10^-6) IS THE
COEFFICIENT OF CUBICAL EXPANSION.
:. VOLUME STRAIN = ∆v/v=γ∆T
BULK MODULUS = K=P/(∆v/v)=P/(γ ∆T)
PRESSURE,P = K γ ∆T
P= 1.728 x 10^8 N/m^2

14. Type Of
STRESS
TENSILE OR
COMPRESSIVE
SHEARING HYDRAULIC
STRESS Two equal & opp. forces _|
_ to opp. faces (σ=F/A)
Two equal &
opp. Forces ||
Forces _|_
everywhere to the
surface, per unit
area same
everywhere
STRAIN Elongation or compression
|| to force direction(∆v/v)
(longitudinal strain)
Pure Shear, θ Volume Change
(compression or
elongation) ∆v/v
Change
In
Shape YES YES NO
Volume NO NO YES
Elastic
modulus
Y=(FxL)/(Ax∆L) G=(Fxθ)/A B=-P/(∆V/V)
Name Of
Modulus
Young’s Modulus Shear Modulus Bulk Modulus
State Of
Matter
Solid Solid Solid, Liquid &
Gas