Class 11 - Physics - Mechanical Properties of Solids

Question1.

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?

Answer:

Area of cross-section of the steel wire, A1 = (3.0 × 10–5) m2

Length of the copper wire, L2 = 3.5 m

Area of cross-section of the copper wire, A2 = (4.0 × 10–5) m2

Change in length = ΔL1 = ΔL2 = ΔL

Force applied in both the cases = F

Young’s modulus of the steel wire:

Y1 = (F1/A1) (L1/ ΔL)

    = (F x 4.7)/ (3.0 × 10–5x ΔL)    … (i)

Young’s modulus of the copper wire:

 Y2 = (F2/A2) (L2/ ΔL)

     = (F x3.5)/ (4.0 × 10–5x ΔL)    … (ii)

Dividing (i) by (ii), we get:

(Y1/Y2) = (4.7x4.0 × 10–5)/ (3.0 × 10–5x3.5)

            =1.79:1

The ratio of Young’s modulus of steel to that of copper is 1.79: 1.

Question2.

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:

It is clear from the given graph that for stress 150 × 106 N/m2, strain is 0.002.

  1. a) Young’s modulus, Y = (Stress/Strain)

= (150 x 106)/ (0.002)

= (7.5 x 1010) N/m2

Hence, Young’s modulus for the given material is 7.5 ×1010 N/m2

  1. 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

Physics Class 11 Mechanical Properties of Solids Stress & Strain Curve

Question 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?

Physics Class 11 Mechanical Properties of Solids Stress & Strain graph

Answer:

 (a) For a given strain, the stress for material A is more than it is for material B, as shown in the two graphs.

Young’s modulus = (Stress/Strain)

For a given strain, if the stress for a material is more, then Young’s modulus is also greater for that material.

Therefore, Young’s modulus for material A is greater than it is for material B.

(b) Strength of the material is determined by the amount of stress required to cause fracture.

Fracture point is the extreme point in a stress-strain curve.

It can be observed that material A can withstand more strain than material B.

Hence, material A is stronger than material B.

Question 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:

  • False. The Young’s modulus is defined as the ratio of stress to the strain within elastic limit.
  • I.e. Young’s modulus, Y = (𝑆𝑡𝑟𝑒𝑠𝑠/𝑆𝑡𝑟𝑎𝑖𝑛).For a given stretching force elongation is more in rubber as compared to steel.
  • Hence, Young’s modulus for rubber is less than it is for steel.

(b)     True. Shear modulus is the ratio of the applied stress to the change in the shape of a body.

The stretching of a coil changes its shape. Hence, shear modulus of elasticity is involved in this process.

Question 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.

Physics Class 11 Mechanical Properties of Solids Springs

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, L1 = 1.5 m

Length of the brass wire, L2 = 1.0 m

Total force exerted on the steel wire:

F1 = (4 + 6) g = (10 × 9.8) = 98 N

Young’s modulus for steel:

Y1 = (F1/A1)/ (ΔL1/L1)

Where,

ΔL1 = Change in length of the steel wire.

A1 =Area of cross section of the steel wire = π r12

Young’s modulus of steel Y1 = (2.0 x 1011) Pa

Therefore, (ΔL1) = (F1 x L1)/ (A1 x Y1)

= (F1 x L1)/ (π r12 x Y1)

= (98 x 1.5)/ (π (0.125 x 10-2)2 x 2 x 1011)

= 1.49 x 10-4 m

Total force on brass wire = F2 = (6 x 9.8)

(ΔL1) =58.8 N

 Young’s modulus for brass:

Y2 = (F2/A2)/ (ΔL2/L2)

Where,

ΔL2 = Change in length of the brass wire.

A2 =Area of cross section of the brass wire = π r22

Therefore, (ΔL2) = (F2 x L2)/ (A2 x Y2)

= (F2 x L2)/ (π r22 x Y2)

= (58.8 x 1.0)/ (π x (0.125 x 10-2)2 x (0.91 x 1011)

(ΔL2) = 1.3 x 10-4

Elongation of the steel wire = 1.49 × 10–4 m

Elongation of the brass wire = 1.3 × 10–4 m

Question 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?

Answer:

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)/ (A η) = (980x0.1)/ (10-2x (25x109))

= 3.92 × 10–7 m

 The vertical deflection of this face of the cube is 3.92 ×10–7 m.

Question7.

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.

Answer:

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 = 122500 N

Young’s modulus, Y = (𝑆𝑡𝑟𝑒𝑠𝑠/𝑆𝑡𝑟𝑎𝑖𝑛)

Where,

Area, A = π (R2 – r2)

= π ((0.6)2 – (0.3)2)

Strain = (122500/ (π [0.6) 2 – (0.3)2] x 2x1011))

= 7.22 × 10–7

Hence, the compressional strain of each column is 7.22 × 10–7

                                                                                                                                                                            

Question 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?

Answer:

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)/ (A η)

= ((44500)/ (2.9x10–4)) x (42x109)

= 3.65 × 10–3

Question 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?

Answer:

Given:

Radius of the steel cable r = 1.5cm = 0.015m

Maximum allowable stress = 108 Nm-2

Maximum Stress = (Maximum force)/ (Area of cross-section)

Therefore, (Maximum force) = (Maximum Stress) x (Area of cross-section)

= 108 x π x (0.015)2 = 7.065 x 104 N

Hence, the cable can support the maximum load of (7.065 x 104) N.

Question 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.

Answer:

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/ πd2)/Strain                Equation (i)

Where,

F = Tension force

A = Area of cross-section

D=Diameter of the wire

From equation (i) Y ∝ (1/d2)

Young’s Modulus for iron, Y1 =190x109 Pa

Diameter of the iron wire =d1

Young’s Modulus for copper, Y2 = (110x109) Pa

Diameter of the copper wire =d2

Therefore the ratios of their diameters are given as:

(d1/ d2) = √ Y1/ Y2

=√ (190x109)/ (110x109)

= √ (19/11)

=1.31:1

Question 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.

Answer:

Mass, m = 14.5 kg

Length of the steel wire, l = 1.0 m

Angular velocity, ω = 2 rev/s

Cross-sectional area of the wire, a = 0.065 cm2

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 + mlω2)

= (14.5 × 9.8) + (14.5 × 1 × (2)2) = 200.1 N

Young’s modulus = (Stress/Strain)

Y = (F/A)/ (Δl/l)

= (F l)/A Δl

Therefore Δl = (F l)/ (A Y)

Young’s modulus for steel = 2 × 1011 Pa

Therefore Δl = (200.1)/ (0.065x10-4x2x1011) = 1539.23x107

=1.539x10-4m

Hence, the elongation of the wire is 1.539 × 10–4 m.

Question 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 litres. Compare the bulk modulus of water with that of air (at constant temperature).

Explain in simple terms why the ratio is so large.

Answer:

Initial volume, V1 = 100.0l = 100.0 × 10–3 m3

Final volume, V2 = 100.5 l = 100.5 ×10–3 m3

Increase in volume, ΔV = (V2 – V1) = 0.5 × 10–3 m3

Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 105 Pa

Bulk Modulus = (Δp) / (ΔV/ V1) = (Δp) x (V1/ ΔV)

= (100x1.013 × 105x100x10-3)/ (0.5x10-3)

=2.026x106 Pa

Bulk modulus of air= 1.0x105Pa

Therefore,

Bulk modulus of water/ Bulk modulus of air

= (2.026x106)/ (1.0x105) =2.026x104

This ratio is very high because air is more compressible than water.

Question 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?

Answer:

Let the given depth be h.

Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 105 Pa

Density of water at the surface, ρ1 = 1.03 × 103 kg m–3

Let ρ2 be the density of water at the depth h.

Let V1 be the volume of water of mass m at the surface.

Let V2 be the volume of water of mass m at the depth h.

Let ΔV be the change in volume.

ΔV = (V1 - V2)

=m ((1/ ρ1) – (1/ ρ2))

Therefore, Volumetric strain = (ΔV/ V1)

=m ((1/ ρ1) – (1/ ρ2)) x (ρ1/m)

Therefore, ΔV/ V1 = (1- ρ1/ ρ2)     … (i)

Bulk modulus, B = (p V1/ ΔV)

(ΔV/ V1) = (p/B)

Compressibility of water = (1/B) =45.8x10-11 Pa-1

Therefore, (ΔV/ V1) = (80x1.013x105x45.8x10-11) = 3.71x10-3    … (ii)

For equations (i) and (ii), we get:

(1- ρ1 / ρ2) = 3.71 x10-3

ρ2 = 1.03x103/(1-(3.71x10-3)

=1.034x103kgm-3

Therefore, the density of water at the given depth (h) is 1.034 × 103 kg m–3.

Question 14.

Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Answer:

Given:

Hydraulic pressure exerted on the glass slab, p = 10 atm = (10 × 1.013 × 105) Pa

 Bulk modulus of glass, B = 37 × 109 Nm–2

Bulk modulus, B= (p/ (ΔV/V))

Where,

(ΔV/V) = Fractional change in volume

Therefore,

(ΔV/ V) = (p/B)

=10x1.013x105

=2.73x10-5

Hence, the fractional change in the volume of the glass slab is 2.73 × 10–5.

Question 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.

Answer:

Given:

Length of an edge of the solid copper cube l = 10cm =0.1m

Hydraulic pressure, p = 7.0 × 106 Pa.

Bulk modulus of copper, B = 140 x 109 Pa

Bulk modulus, B= (ρ / (ΔV/V))

Where (ΔV/V) = Fractional change in volume; ΔV = change in volume,

V = Original volume

ΔV = (pV)/ (B)

Original volume of the cube = l3

Therefore, ΔV = (p l3)/ (B)

= (7 x106 x (0.1)3)/ (140 x 109)

= 5 x 10-8 m3

=5 x 10-2 cm-2

Therefore, the volume contraction of the solid copper cube is 5 × 10–2 cm–3

 Question 16.

How much should the pressure on a litre of water be changed to compress it by 0.10%?

Answer:

Volume of water, V = 1 L

It is given that water is to be compressed by 0.10%.

Fractional change = (ΔV/V) = (0.1/100x1) = 10-3

Bulk modulus, B= (ρ/ (ΔV/V))

p= (B x ΔV/V)

Bulk Modulus of water, B = 2.2x109 Nm-2

p=2.2x109x10-3

= (2.2x106) Nm-2

Therefore, the pressure on water should be 2.2 ×106 Nm–2.

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