Ncert Solutions for Class 11 Physics Mechanical Properties of Solids
NCERT Solutions for Class 11th: Ch 9 Mechanical properties of Solids Physics Science
Page No: 242
Excercises
9.1. A steel wire of length 4.7 m and cross-sectional area 3.0 × 10–5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2 under a given load. What is the ratio of the Young's modulus of steel to that of copper?
Length of the steel wire, L 1 = 4.7 m
Area of cross-section of the steel wire, A 1 = 3.0 × 10–5 m2
Length of the copper wire, L 2 = 3.5 m
Area of cross-section of the copper wire, A 2 = 4.0 × 10–5 m2
Change in length = ΔL 1 = ΔL 2 = ΔL
Force applied in both the cases = F
Young's modulus of the steel wire:
Y 1 = (F 1 / A 1) (L 1 / ΔL1)
= (F / 3 X 10-5) (4.7 / ΔL) ....(i)
Young's modulus of the copper wire:
Y 2 = (F 2 / A 2) (L 2 / ΔL 2)
= (F / 4 × 10-5) (3.5 / ΔL) ....(ii)
Dividing (i) by (ii), we get:
Y 1 / Y 2 = (4.7 × 4 × 10-5) / (3 × 10-5 × 3.5)
= 1.79 : 1
The ratio of Young's modulus of steel to that of copper is 1.79 : 1.
9.2. Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for this material?
Answer
(a) It is clear from the given graph that for stress 150 × 106 N/m2, strain is 0.002.
∴Young's modulus, Y = Stress / Strain
= 150 × 106 / 0.002 = 7.5 × 1010 Nm-2
Hence, Young's modulus for the given material is 7.5 ×1010 N/m2.
(b) The yield strength of a material is the maximum stress that the material can sustain without crossing the elastic limit.
It is clear from the given graph that the approximate yield strength of this material is 300 × 106 Nm/2 or 3 × 108 N/m2.
Page No: 243
9.3. The stress-strain graphs for materials A and B are shown in Fig. 9.12.
The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young's modulus?
(b) Which of the two is the stronger material?
(a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence, Young's modulus (=stress/strain) is greater for A than that of B.
(b) A is stronger than B. Strength of a material is measured by the amount of stress required to cause fracture, corresponding to the point of fracture.
9.4. Read the following two statements below carefully and state, with reasons, if it is true or false.(a) The Young's modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.
Answer
(a) False, because for given stress there is more strain in rubber than steel and modulus of elasticity is inversely proportional to strain.
(b) True, because the stretching of coil simply changes its shape without any change in the length of the wire used in the coil due to which shear modulus of elsticity is involved.
9.5. Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.
Answer
Elongation of the steel wire = 1.49 × 10–4 m
Elongation of the brass wire = 1.3 × 10–4 m
Diameter of the wires, d = 0.25 m
Hence, the radius of the wires, r = d/2 = 0.125 cm
Length of the steel wire, L 1 = 1.5 m
Length of the brass wire, L 2 = 1.0 m
Total force exerted on the steel wire:
F 1 = (4 + 6) g = 10 × 9.8 = 98 N
Young's modulus for steel:
Y1 = (F1/A1) / (ΔL 1 / L 1)
Where,
ΔL 1 = Change in the length of the steel wire
A 1 = Area of cross-section of the steel wire = πr 1 2
Young's modulus of steel, Y 1 = 2.0 × 1011 Pa
∴ ΔL 1 = F 1 ×L 1 / (A 1 ×Y 1)
= (98 × 1.5)/ [ π(0.125 × 10-2)2 × 2 × 1011 ] = 1.49 × 10-4 m
Total force on the brass wire:
F 2 = 6 × 9.8 = 58.8 N
Young's modulus for brass:
Y 2 = (F 2/A 2) / (ΔL 2 / L 2)
Where,
ΔL 2 = Change in the length of the brass wire
A 1 = Area of cross-section of the brass wire = πr1 2
∴ ΔL 2 = F 2 ×L 2 / (A 2 ×Y 2)
= (58.8 X 1)/ [ (π × (0.125 × 10-2)2 × (0.91 × 1011) ] = 1.3 × 10-4 m
Elongation of the steel wire = 1.49 × 10–4 m
Elongation of the brass wire = 1.3 × 10–4 m.
Page No: 244
9.6. The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Edge of the aluminium cube, L = 10 cm = 0.1 m
The mass attached to the cube, m = 100 kg
Shear modulus (η) of aluminium = 25 GPa = 25 × 109 Pa
Shear modulus, η = Shear stress / Shear strain = (F/A) / (L/ΔL)
Where,
F = Applied force = mg = 100 × 9.8 = 980 N
A = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01 m2
ΔL = Vertical deflection of the cube
∴ ΔL = FL / Aη
= 980 × 0.1 / [ 10-2 × (25 × 109) ]
= 3.92 × 10–7 m
The vertical deflection of this face of the cube is 3.92 ×10–7 m.
9.7. Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Mass of the big structure, M = 50,000 kg
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young's modulus of steel, Y = 2 × 1011 Pa
Total force exerted, F = Mg = 50000 × 9.8 N
Stress = Force exerted on a single column = 50000 × 9.8 / 4 = 122500 N
Young's modulus, Y = Stress / Strain
Strain = (F/A) / Y
Where,
Area, A = π (R 2 – r 2) = π ((0.6)2 – (0.3)2)
Strain = 122500 / [ π ((0.6)2 – (0.3)2) × 2 × 1011 ] = 7.22 × 10-7
Hence, the compressional strain of each column is 7.22 × 10–7.
9.8. A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Length of the piece of copper, l = 19.1 mm = 19.1 × 10–3 m
Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10–3 m
Area of the copper piece:
A = l × b
= 19.1 × 10–3 × 15.2 × 10–3
= 2.9 × 10–4 m2
Tension force applied on the piece of copper, F = 44500 N
Modulus of elasticity of copper, η = 42 × 109 N/m2
Modulus of elasticity, η = Stress / Strain
= (F/A) / Strain
∴ Strain = F / Aη
= 44500 / (2.9 × 10-4 × 42 × 109)
= 3.65 × 10–3.
9.9. A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 N m–2, what is the maximum load the cable can support?
Radius of the steel cable, r = 1.5 cm = 0.015 m
Maximum allowable stress = 108 N m–2
Maximum stress = Maximum force / Area of cross-section
∴ Maximum force = Maximum stress × Area of cross-section
= 108 × π (0.015)2
= 7.065 × 104 N
Hence, the cable can support the maximum load of 7.065 × 104 N.
9.10. A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.
The tension force acting on each wire is the same. Thus, the extension in each case is the same. Since the wires are of the same length, the strain will also be the same.
The relation for Young's modulus is given as:
Y = Stress /Strain
= (F/A) / Strain = (4F/πd 2) / Strain ....(i)
Where,
F = Tension force
A = Area of cross-section
d = Diameter of the wire
It can be inferred from equation (i) that Y ∝ (1/d2)
Young's modulus for iron, Y 1 = 190 × 109 Pa
Diameter of the iron wire = d 1
Young's modulus for copper, Y 2 = 120 × 109 Pa
Diameter of the copper wire = d 2
Therefore, the ratio of their diameters is given as:
9.11. A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.
Mass, m = 14.5 kg
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s
Cross-sectional area of the wire, a = 0.065 cm2 = 0.065 × 10-4 m2
Let Δl be the elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass is:
F = mg + mlω 2
= 14.5 × 9.8 + 14.5 × 1 × (12.56)2
= 2429.53 N
Young's modulus = Strss / Strain
Y = (F/A) / (∆ l/l)
∴ ∆ l = Fl / AY
Young's modulus for steel = 2 × 1011 Pa
∆ l = 2429.53× 1 / (0.065× 10-4× 2× 10 11 ) = 1.87× 10 -3 m
Hence, the elongation of the wire is 1.87 × 10–3 m.
9.12. 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.
Ncert Solutions for Class 11 Physics Mechanical Properties of Solids
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