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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.
THE RATIO OF HYDRAULIC STRESS TO THE CORRESPONDING HYDRAULIC STRAIN IS CALLED AS BULK MODULUS. IT DONATED BY SYMBOL ‘B’. B=-p/(∆V/V)
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)
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
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.
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
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%
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.
ANS: COMPRESSIBILITY =1/K=45.8 x 10^11/PACHANGE IN PRESSURE,P=80-1=79 ATM
=79x1.013 x 10^5 PADENSITY 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
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.
ANS: WHEN A BLOCK OF VOLUME V IS HEATED THROUGH A TEMPERATURE OF ∆T,THE CHANGE IN VOLUME IS ∆v/v =γ ∆TWHERE 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
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