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Answer:
Let the Young's Modulus of steel and copper be YS and YC, respectively.
Length of the steel wire ls= 4.7 m
Length of the copper wire lc = 4.7 m
The cross-sectional area of the steel wire AS = 3.0×10−5m2
The cross-sectional area of the Copper wire AC = 4.0×10−5m2
Let the load and the change in the length be F and Δl respectively
Y=FlAΔl
FΔl=AYl
Since F and Δl is the same for both wires we have
ASYSlS=ACYClC
YSYC=AClSASlC
YSYC=4.0×10−5×4.73.0×10−5×3.5
YSYC=1.79
The ratio of Young’s modulus of steel to that of copper is 1.79.
8.2 (a) Figure 8.9 shows the strain-stress curve for a given material. What isYoung’s modulus?
Fig. 8.9
Answer:
Young’s modulus is given as the ratio of stress to strain when the body is behaving elastically.
For the given material
Y=StressStrain
Y=150×1060.002
Y=7.5×102Nm−2
8.2(b)What is the approximate yield strength for this material?
Answer:
The Yield Strength is approximately 3×108Nm−2 for the given material. We can see above this value of strain, the body stops behaving elastically.
Fig 8.10
Answer:
As we can see in the given Stress-Strain graphs that the slope is greater in the graph corresponding to material A. We conclude that A has a greater Young's Modulus.
8.3(b) Which of the two is the stronger material?
Answer: The material which fractures at higher stress is said to be stronger. As we can see in the given Stress-Strain graphs, the stress at which the material fractures is higher in A than that in B we conclude that A is the stronger material.
8.4(a) Read the following statement below carefully and state, with reasons, if it is true or false.
The Young’s modulus of rubber is greater than that of steel;
Answer:
False: Young's Modulus is defined as the ratio of the stress applied to a material and the corresponding strain that occurs. As for the same amount of pressure applied on a piece of rubber and steel, the elongation will be much less in the case of steel than that in the case of rubber and therefore the Young's Modulus of rubber is less than that of steel.
Answer:
True: As the force acts Normal to the parallel planes in which helical parts of the wire lie, the actual length of the wire would not change, but its shape would. Therefore the amount of elongation of the coil taking place for corresponding stress depends upon the Shear Modulus of elasticity.
Fig 8.11
Answer:
Tension in the steel wire is F1
F1=(4+6)×9.8F1=98N
Length of steel wire ls = 1.5 m
The diameter of the steel wire, d = 0.25 cm
A=π(d2)2
A=π×(0.25×10−22)2
A=4.9×10−6m2
Area od the steel wire, A=4.9×10−6 m2
Let the elongation in the steel wire be Δl1
Young's Modulus of steel, Y1 = 2×1011Nm−2
Y1=F1l1Δl1A
Δl1=F1l1Y1A
Δl1=98×1.52×1011×4.9×10−6
Δl1=1.5×10−4 m
Tension in the Brass wire is F2
F2=(6)×9.8F2=58.8N
Length of Brass wire l2 = 1.5 m
Area of the brass wire, A=4.9×10−6 m2
Let the elongation in the steel wire be Δl2
Young's Modulus of steel, Y2 = 0.91×1011Nm−2
Δl2=F2l2Y2A
Δl2=58.8×10.91×1011×4.9×10−6
Δl2=1.32×10−4 m
Answer:
Edge of the aluminium cube, l = 10 cm = 0.1 m
Area of the face of the Aluminium cube, A = l2 = 0.01 m2
Tangential Force is F
F=100×9.8F=980 N
Tangential Stress is F/A
FA=9800.01
FA=98000N
Shear modulus of aluminium η=2.5×1010Nm−2
Tangential Strain=FAη=980002.5×1010=3.92×10−6
Let the Vertical deflection be Δl
Δl=Tangential Stress×Side
Δl=3.92×10−6×0.1
Δl=3.92×10−7m
Answer:
Inner radii of each column, r1 = 30 cm = 0.3 m
Outer radii of each column, r2 = 60 cm = 0.6 m
Mass of the structure, m = 50000 kg
Stress on each column is P
P=50000×9.84×π×(0.62−0.32)
P=1.444×105Nm−2
Youngs Modulus of steel is Y=2×1011Nm−2
Compressional strain=PY=1.444×1052×1011=7.22×10−7
Answer:
Length of the copper piece, l = 19.1 mm
The breadth of the copper piece, b = 15.2 mm
Force acting, F = 44500 N
Modulus of Elasticity of copper, η=42×109Nm−2
Strain=FAη=Flbη=4450015.2×10−3×19.1×10−3×42×109=3.65×10−3
Answer:
Let the maximum Load the Cable Can support be T
Maximum Stress Allowed, P = 10 8 N m -2
Radius of Cable, r = 1.5 cm
T=Pπr2
T=108×π×(1.5×10−2)2
T=7.068×104N
Answer:
Each wire must support the same load and are of the same length and therefore should undergo the same extension. This, in turn, means they should undergo the same strain.
Y=FlΔlA
Y=4FlΔlπd2
Yd2=k As F, l and Δl are equal for all wires
Yiron=1.9×1011Nm−2
Ycopper=1.1×1011Nm−2
dirondcopper=YcopperYiron
dirondcopper=1.1×10111.9×1011
dirondcopper=0.761
Answer:
Mass of the body = 14.5 kg
Angular velocity, ω = 2 rev/s
ω=4π rad/s
The radius of the circle, r = 1.0 m
Tension in the wire when the body is at the lowest point is T
T=mg+mω2r
T=14.5×9.8+14.5×(4π)2
T=2431.84N
Cross-Sectional Area of wire, A = 0.065 cm2
Young's Modulus of steel, Y=2×1011Nm−2
Δl=FlAY
Δl=2431.84×10.065×10−4×2×1011
Δl=1.87mm
Answer:
Pressure Increase, P = 100.0 atm
Initial Volume = 100.0 l
Final volume = 100.5 l
Change in Volume = 0.5 l
Let the Bulk Modulus of water be B
B=StressVolumetric Strain
B=PΔVV
B=100×1.013×105×100×10−30.5×10−3
B=2.026×109Nm−2
The bulk modulus of air is Ba=1.0×105Nm−2
The Ratio of the Bulk Modulus of water to that of air is
BBa=2.026×1091.0×105
BBa=20260
This ratio is large as for the same pressure difference the strain will be much larger in the air than that in water.
Answer:
Water at the surface is under 1 atm pressure.
At the depth, the pressure is 80 atm.
Change in pressure is ΔP=79 atm
Bulk Modulus of water is B=2.2×109Nm−2
B=−PΔVVΔVV=−PBΔVV=−79×1.013×1052.2×109ΔVV=−3.638×10−3
The negative sign signifies that for the same given mass the Volume has decreased The density of water at the surface ρ=1.03×103 kg m−3
Let the density at the given depth be ρ′
Let a certain mass occupy V volume at the surface
ρ=mVρ′=mV+ΔV
Dividing the numerator and denominator of RHS by V we get
ρ′=mV1+ΔVVρ′=ρ1+ΔVVρ′=1.03×1031+(−3.638×10−3)ρ′=1.034×103 kg m−3
The density of water at a depth where pressure is 80.0 atm is 1.034×103 kg m−3 .
Answer:
Bulk's Modulus of Glass is BG=3.7×1010Nm−2
Pressure is P = 10 atm.
The fractional change in Volume would be given as
ΔVV=PB
ΔVV=10×1.013×1053.7×1010
ΔVV=2.737×10−5
The fractional change in Volume is 2.737×10−5
Answer:
The bulk modulus of copper is BC=140GPa=1.4×1011Nm−2
Edge of copper cube is s = 10 cm = 0.1 m
Volume Of copper cube is V = s3
V = (0.1)3
V = 0.001 m3
Hydraulic Pressure applies is P=7.0×106Pa
From the definition of bulk modulus
BC=PΔVVΔVV=PBCΔVV=7×1061.4×1011ΔVV=5×10−5
The volumetric strain is 5×10−5
Volume contraction will be
ΔV= Volumetric Strain × Initial Volume ΔV=5×10−5×10−3ΔV=5×10−8 m3ΔV=5×10−2 cm3
The volume contraction has such a small value even under high pressure because of the extremely large value of the bulk modulus of copper.
8.16 How much should the pressure on a litre of water be changed to compress it by 0.10%?
Answer:
Change in volume is ΔV=0.10%
Volumetric Strain=0.1100
ΔVV=0.001
Bulk modulus of water is Bw=2.2×109 Nm−2
Bw=PΔVVP=ΔVV×BwP=0.001×2.2×109P=2.2×106 Pa
A pressure of 2.2×106Pa is to be applied so that a litre of water compresses by 0.1%.
Note: The answer is independent of the volume of water taken into consideration. It only depends on the percentage change.
