Answer:
Let the Young's Modulus of steel and copper be YS and YC, respectively.
Length of the steel wire
Length of the copper wire
The cross-sectional area of the steel wire AS =
The cross-sectional area of the Copper wire AC =
Let the load and the change in the length be F and
Since F and
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
8.2(b)What is the approximate yield strength for this material?
Answer:
The Yield Strength is approximately
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
Length of steel wire
The diameter of the steel wire, d = 0.25 cm
Area od the steel wire,
Let the elongation in the steel wire be
Young's Modulus of steel, Y1 =
Tension in the Brass wire is F2
Length of Brass wire
Area of the brass wire,
Let the elongation in the steel wire be
Young's Modulus of steel, Y2 =
Answer:
Edge of the aluminium cube,
Area of the face of the Aluminium cube, A =
Tangential Force is F
Tangential Stress is F/A
Shear modulus of aluminium
Let the Vertical deflection be
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
Youngs Modulus of steel is
Answer:
Length of the copper piece,
The breadth of the copper piece, b = 15.2 mm
Force acting, F = 44500 N
Modulus of Elasticity of copper,
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
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.
Answer:
Mass of the body = 14.5 kg
Angular velocity,
The radius of the circle, r = 1.0 m
Tension in the wire when the body is at the lowest point is T
Cross-Sectional Area of wire, A = 0.065 cm2
Young's Modulus of steel,
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
The bulk modulus of air is
The Ratio of the Bulk Modulus of water to that of air is
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
Bulk Modulus of water is
The negative sign signifies that for the same given mass the Volume has decreased The density of water at the surface
Let the density at the given depth be
Let a certain mass occupy V volume at the surface
Dividing the numerator and denominator of RHS by V we get
The density of water at a depth where pressure is 80.0 atm is
Answer:
Bulk's Modulus of Glass is
Pressure is P = 10 atm.
The fractional change in Volume would be given as
The fractional change in Volume is
Answer:
The bulk modulus of copper is
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
From the definition of bulk modulus
The volumetric strain is
Volume contraction will be
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
Bulk modulus of water is
A pressure of
Note: The answer is independent of the volume of water taken into consideration. It only depends on the percentage change.