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NCERT Class 11 Physics Chapter 8 Mechanical Properties of Solids
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Mechanical Properties of Solids
Chapter: 8
| Part – II |
EXERCISE
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?
Ans: For steel, length of wire,
Ls = 4.7 m.
Area of cross – section,
αs = 3 × 10-5 m2
For copper, length of wire
Area of cross- Section
αc = 4.0 × 10-5 m2
If F be the stretching force, I be the increase in length in each case and Ys and Yc be the young’s moduli of steel and copper respectively, then:
For steel:
𝛥Lsteel = F × Lsteel / Asteel × Ysteel
For copper:
𝛥Lcopper = F × Lcopper / Acopper × Ycopper
= 4.7 × 4 × 10-5 / 3.5 × 3 × 10-5
= 1.8.
2. Figure 8.9 shows the strain-stress curve for a given material. What are:
Ans: (a) Young’s modulus
From the Graph:
Point 1: (0.001, 100 × 106 Nm−2)
Point 2: (0.002, 200 × 106 Nm−2)
E = Stress/Strain
So,
= 200 × 106 Nm-2 – 100 × 106 Nm-2 / 0.002 – 0.001
= 100 × 106 Nm-2 / 0.001
= 100 × 109
(b) approximate yield strength for this material?
Ans: From the Graph:
The deviation from linearity starts around 250 × 106 Nm-2
Therefore the approximate yield strength is:
= 250 × 106 Nm-2
= 250 MPa.
3. The stress-strain graphs for materials A and B are shown in Fig. 8.10.
The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young’s modulus?
Ans: Young’s modulus is given by the equation, Y= stress strain At a particular strain, Young’s modulus is directly proportional to the stress. It is because the slope of the graph of material A is larger than that of B, and thus for producing the same strain, more stress is required for material B.
(b) Which of the two is the stronger material?
Ans: Material A is the stronger of the two materials. It is because it can bear greater stress before the wire of the material breaks.
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.
Ans: False. The modulus of elasticity of steel is greater than that of rubber.
(b) The stretching of a coil is determined by its shear modulus.
Ans: True. When the coil is stretched or compressed the shear stress is developed in the coil.The deformation of a coil, especially a helical spring, under stretching or compression is primarily governed by the shear modulus of the material.
5. Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. 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.
Ans: Given that:
𝜟L = FL / AY.
Diameter of both wires, d = 0.25 cm = 0.0025cm.
Radius, r = d/2 = 0.00125m
Cross – section area, A = 𝜋r2 = 𝜋 (0. 00125)2
For steel:
Unloaded length, Lsteel = 1.5m.
Mass hanging below steel wire, Mtotal,steel = 4 kg + 6 kg (mass of the brass wire and its load) = 10 kg.
Force due to load, Fsteel = mtotal,steel × g = 10 × 9.8 N.
For:
Unloaded length, Lbrass = 1.0m.
Mass hanging below brass wire, mtotal,brass = 6kg
Force due to load, Fbrass = mtotal,brass × g = 6 × 9.8 N.
Young’s modulus values:
For steel, Ysteel = 2.0 × 1011 pa.
For brass, Ybrass = 1.0 × 1011 pa.
Cross – section Area
A = 𝜋 (0.00125)2 = 𝜋 × 1.5625 × 10-6
= 4.9087 × 10-6 m2
Elongation of steel wire:
𝜟Lsteel = Fsteel × Lsteel / A × Ysteel
𝜟Lsteel = (10 × 9.8 ) × 1.5 / 4.9087 × 10-6 × 2.0 1011
𝜟Lsteel = 147 / 9.8174 × 105
= 0.00015 m
0.15m.
Elongation of brass wire:
𝜟Lbrass = Fbrass × Lbrass / A × Ybrass
𝜟Lbrass = (6 × 9.8) × 1.0 / 4.9087 × 10-6 × 1.0 × 1011
𝜟Lbrass = 58.8 / 4.9087 × 105
= 0.00012 m.
= 0.12 m.
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?
Ans: Given that:
Edge length of the cube, L = 10 cm = 0.1 m
Mass attached, m = 100 kg.
Force applied due to gravity, F= m × g = 100×9.8 N = 980 N.
Shear modulus of aluminium, G = 25 GPa = 25×109 Pa.
Area of the face, A = L2 =(0.1)2 = 0.01 m2.
𝛥L = FL / An
= 980 × 0.1 / 10-2 × (25 × 109)
= 3.92 × 10-7 m.
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 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Ans: Given that:
Total mass of structure = 50,000 kg
Gravitational force F = mass × gravity
Gravitational acceleration g = 9.8 m/s2
Number of columns = 4
Inner radius rinner = 0.3m
Outer radius router = 0.6m
Young’s modulus E = 2 × 1011 pa
Calculational Force:
F = 50, 000 kg × 9.8 m/s2 = 490,000 N
Force per column
Fcolumn = 490,000 N/4 = 122,500 N
Cross-sectional area of each column
Aouter = 𝜋 × (0.6)2 = 𝜋 × 0.36 m2
Ainner = 𝜋 × (0.3)2 = 𝜋 × 0.09 m2
A = Aouter – Ainner = 𝜋 × ( 0.36 – 0.09) = 𝜋 × 0.27 m2
A = 0.848 m2
Compressional Strain calculation
Stress = Fcolume/A = 122,500 N / 0.848m2 = 144, 433.96 N/m2
Strain = Stress/E = 144,433.96 N/m2 / 2 × 1011 pa
Strain = 7.22 × 10-7.
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?
Ans: Given that:
A = 15.2 × 19.1 × 10-6 m2
F = 44,500 N
G = 42 × 109 Nm2
Y = Stress / strain
Strain = stress / Y = F/AG
= 44500 / (15.2 × 19 .2 × 10-6) × 42 × 106
= 3.65 × 10-4.
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?
Ans: Here,
Maximum Stress = 108 N m-2
Radius of the cable, r = 1.5 cm = 0.015m
Therefore area of cross-section of the cable:
A = 𝜋r2
A = 𝜋 × (0.015)2
A = 𝜋 × 0.000225
A = 0.00070686 m2
The maximum load the cable can support:
F = maximum stress × area of cross-section
= 108 N / m2 × 0. 00070686 m2
= 76.32 N.
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 ratios of their diameters if each is to have the same tension.
Ans: Y =
= Dcopper / Diron
=
= √19/11
= 1.31.
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.
Ans: Given here:
m = 14.5 kg
I = r = 1m
ω = 2rev/ s = 4𝜋 rad./s
A = 0.065 × 10-4 m2
F = mg + mrω2
= mg + mr(4𝜋)2
= 14.5 × 9.8 + 14.5 × 1 × 4 × 4 × (3.14)2
= 142.1 + 2291.6
= 2433.9 N
Y =
𝜟I = FI / AY
=
= 1.87 × 10-3m
= 1.87 mm.
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.
Ans: Where:
K = – (ΔP × V) / ΔV
K is the bulk modulus
ΔP is the change in pressure
V is the original volume
ΔV is the change in volume
Calculating Bulk Modulus of Water
Given:
Initial volume (V) = 100 L = 0.1 m³
Pressure increase (ΔP) = 100 atm = 100 × 1.013 × 105 Pa =
1.013 × 107 Pa
Final volume = 100.5 L
Change in volume (ΔV) = 100.5 L – 100 L = 0.5 L = 0.0005 m³
K = – (1.013 × 107 Pa × 0.1 m³) / (-0.0005 m³)
K = 2.026 × 109 Pa.
This ratio is very large because water is essentially incompressible compared to air. Water molecules are closely packed and have strong intermolecular forces, making it difficult to compress. In contrast, air molecules are widely spaced and have weak intermolecular forces, making it easy to compress. It is so because gases are much more compressible than those of liquids. The molecules in gases are very poorly coupled to their neighbours as compared to those of gases.
13. What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3?
Ans: The increased in density is given by:
d p = B p q
Where “B’ is the compressibility, p is pressure and p is the density at the surface:
Here,
B = 45.8 × 10-11 pa-1
P = 80 × 1 × 105 Nm-2
P = 1.03 × 103 kg m-2
Therefore, dp = 45.8 × 10-11 × 80 × 105 × 1.03 × 103
= 3.774 kgm-3
Total density at the given depth
= P + dp
= 1.03 × 103 + 3. 744
= 1.03 × 1000 + 3.774
= 1033.774 kg m-3
= 1.034 × 103 kg m-3.
14. Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Ans: The bulk modulus of glass is approximately 5.5 × 1010 Pa.
Given:
Hydraulic pressure (ΔP) = 10 atm
= 10 × 1.013 × 105 Pa
= 1.013 × 106 Pa
Formula to solve for ΔV / V:
ΔV / V = (1.013 × 106 Pa) / (5.5 × 1010 Pa)
= 1.84 × 10-5.
15. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 106 Pa.
Ans: Bulk modulus of copper = 140 Gpa.
Given:
L = 10cm = 10m
P = 7 × 106 pa
B = 140 × 109 pa
B = pV/𝛥V
= PL3 / 𝛥V
= 𝛥V = PL3 / B
= (7 × 106) × (0.10)3 / 140 × 109
= 5 × 10-8 m3
= 5 × 10-2 cm3
16. How much should the pressure on a litre of water be changed to compress it by 0.10%? carry one quarter of the load.
Ans: Here:
ΔP = (ΔV / V) × B
ΔP = 0.001 × 2.15 × 109 Pa
ΔP = 2.15 × 106 Pa
ΔP / 4 = 2.15 × 106 Pa/4
ΔP / 4 = 5.375 × 105 Pa
This is equivalent to:
5.375 × 105 Pa / (1.013 × 105 Pa/atm)
= 5.3 atm.
