Mechanical Properties Of Solids

Given Data,

Length of the steel wire, L_s = 4.7 m

Area of cross-section of the steel wire, A_s = 3.0 × 10^-5 m²

Length of the copper wire, L_c = 3.5 m

Area of cross-section of the steel wire, A_c = 4.0 × 10^-5 m²

Consider the force applied in both the cases is = F_s = F_c = F

Change in Length = ΔL_s = ΔL_c = ΔL (since both the wires stretch by the same amount)

Young's modulus ( E ) describes tensile elasticity, or the tendency of an object to deform along an axis when opposing forces are applied along that axis; it is defined as the ratio of tensile stress to tensile strain. It is often referred to simply as the elastic modulus.

Formula for young’s modulus for steel:

…………… (1)

Formula for young’s modulus for steel:

Y_c =

Y_c = ………… (2)

Dividing (1) by (2)

=

Cancelling the common terms F andΔL, the equation changes to,

=

⇒

Hence, the ratio of Young’s modulus of steel to that of copper is 1.79:1.

(a) Considering the graph from the question, highlight the part which corresponds to stress @ 150 × 10⁶ N/m²

We can clearly observe that the strain at the stress point 150 × 10⁶ N/m², is 0.002.

By definition, Young’s modulus, Y =

⇒ Y =

⇒ Y = 7.5 × 10¹⁰ N/m²

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 above graph that the approximate yield strength of the material is 300 × 10⁶ Nm^-2.

(a) From the graphs given in the question, we note that for a given strain, stress for A is more than that of B.

Hence, young’s modulus, which is defined as the ratio of tensile stress to tensile strain, is greater for A than that of B.

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

On observing the graphs mentioned in the question, we can see that material A is stronger than B since the stress curve in Graph for A fails at a larger distance compared to the curve in graph for B.

(a) False.

Please be informed that the strain factor in rubber material is higher than the strain in the steel material. The young’s modulus is inversely proportional to the strain.

Hence, the Young’s modulus of steel is greater than that of rubber.

(b) True.

In material science, shear modulus or modulus of rigidity, denoted by G, or sometime S or μ, is defined as the ratio of shear stress to the shear strain.

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 elasticity is involved.

Given data,

Diameter of both the wires, d = 0.25 cm

Hence, radius of both the wires = r = d/2

⇒ r = 0.25/2 cm

⇒ r = 0.125 cm.

Length of the steel wire, L_s = 1.5 m

Length of the brass wire, L_b = 1.0 m

From the given figure, the total force exerted on the steel wire:

F_s = (4 kg + 6 kg) × 9.8 ms^-2

Here, 9.8 ms^-2 is the acceleration due to gravity which acts along with the force attached to the wires.

⇒ F_s = 10 × 9.8 = 98 N.

Young’s modulus for steel:

Y_s = ………………… (1)

Where,

= Change in the length of the steel wire.

= Area of cross-section of the steel wire = πr²

⇒ A_s = π×(0.125× 10^-2)² m²

Young’s modulus for steel from standard table is, 2.0× 10¹¹ Pa

Substituting the value is (1)

2.0× 10¹¹ Pa =

ΔL_s =

⇒ ΔL_s = 1.49 × 10^-4 m

Hence, elongation of the steel wire is 1.49 × 10^-4 m.

From the given figure, the total force exerted on the brass wire:

F_b = 6 kg × 9.8 ms^-2

Here, 9.8 ms^-2 is the acceleration due to gravity which acts along with the force attached to the wires.

⇒ F_b = 6 kg × 9.8 ms^-2 = 58.8 N.

Young’s modulus for steel:

Y_b = ………………… (1)

Where,

= Change in the length of the brass wire.

= Area of cross-section of the brass wire = πr²

⇒ A_b = π×(0.125× 10^-2)² m²

Young’s modulus for brass from standard table is, 0.91× 10¹¹ Pa

Substituting the value is (1)

0.91× 10¹¹ Pa =

=

⇒ ΔL_b = 1.3 × 10^-4 m

Hence, elongation of the steel wire is 1.3 × 10^-4 m.

Given data,

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 =

Where,

F = applied force = mg = 100kg × 9.8ms^-2 = 980 N

A = Area of one of the faces of the cube = L² = 0.1×0.1 = 0.01m²

ΔL = vertical deflection of the cube.

Modifying the formula of Shear modulus,

ΔL =

Substituting all the values mentioned above, we get,

⇒ ΔL =

⇒ ΔL = 3.92 × 10^-7m

Hence, the vertical deflection of this face of the cube is

3.92× 10^-7 m.

Given Data,

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_s, from the standard chart is 2×10¹¹ Pa

Total force exerted on all the columns, F = Mg = 5000kg × 9.8 N

Amount of force exerted on each column, F₁ = (5000Kg× 9.8N)/4

⇒ F₁ = 122500 N

Young’s modulus, Y = Stress/ Strain

⇒ Y = (/Strain

⇒ Strain = …………………………… (1)

Where, A = area of each column = π (R²-r²)

⇒ A = π ((0.6)² - (0.3)²)

Substituting value of Area in equation (1), we get,

Strain = 122500 N/ π ((0.6)²-(0.3)²) × 2×10¹¹ Pa

⇒ Strain = 7.22× 10^-7

Hence, the compressional strain on each column is 7.22 × 10^-7.

Given Data,

Length of the piece of copper = l = 19.1 mm = 19.1 × 10^-3m

Breadth of the piece of copper = b = 15.2 mm = 15.2× 10^-3m

Tension force applied on the piece of cooper, F = 44500N

Area of rectangular cross section of copper piece,

Area = l× b

⇒ Area = (19.1 × 10^-3m) × (15.2× 10^-3m)

⇒ Area = 2.9 × 10^-4 m²

Modulus of elasticity of copper from standard list, η = 42× 10⁹ N/m²

By definition, Modulus of elasticity, η = stress/strain

⇒

⇒ Strain = F/Aη

⇒

⇒ Strain = 3.65 × 10^-3

Hence, the resulting strain is 3.65 × 10^-3

Given data,

Radius of the steel cable. R = 1.5 cm = 0.015 m

Maximum allowable stress = 10⁸ Nm^-2

Maximum stress = Maximum force/Area of cross-section

Here, Area of steel cable, A = π r² = π (0.015)²

⇒ Maximum force = Maximum stress × Area of cross-section.

⇒ Maximum force = 10⁸ Nm^-2× π (0.015 m)²

⇒ Maximum force = 7.065× 10⁴ N

Hence, the cable can support maximum load of 7.065× 10⁴ N

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 formula for young’s modulus is given as:

Y = Stress/Strain

⇒ Y = (F/A)/Strain

Here, A = area of wire = π r² = π (d/2)² =

F = Tension force

d = Diameter of the wire

⇒ Y =

From the above equation, we can infer that young’s modulus,

Y α (1/d²) …………….. (1)

Young’s modulus for iron from standard table, Y₁ = 190 × 10⁹ Pa

Young’s modulus for copper from standard table, Y₂ = 120× 10⁹ Pa

Let,

Diameter of the iron wire be d₁

Diameter of the copper wire be d₂

From (1), the ratios of young’s modulus is,

⇒

⇒

⇒

⇒

Therefore, the ratio of their diameters is 1.25: 1

Given data,

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^-4m²

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

⇒ F = (14.5 × 9.8) + (14.5× 1 × (12.56)²)

⇒ F = 2429.53 N

Young’s modulus = Stress/Strain

Y = (F/A)/(Δl/l)

∴ Δl = Fl/AY

Young’s modulus for steel = 2× 10¹¹ Pa

Δl = 2429.53× 1 / (0.065× 10^-4× 2× 10¹¹)

⇒ Δl = 1.87× 10^-3 m

Hence, the elongation of the wire is 1.87× 10^-3m.

Given data,

Initial volume, V₁ = 100.0 L = 100.0 × 10^-3 m ³.

Final volume, V₂ = 100.5 L = 100.5 × 10^-3 m ³.

Increase in volume, Δv = v₂-v₁ = 0.5 × 10^-3 m ³.

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

The bulk modulus (or) of a substance is a measure of how incompressible/resistant to compressibility that substance is. It is defined as the ratio of the infinitesimal pressure increase to the resulting relative decrease of the volume.

Bulk modulus of water = Δp/(Δv/v₁) = Δp×v₁/ Δv

⇒ Bulk modulus = 100× 1.013× 10⁵× 100× 10^-3/⇒ (0.5× 10^-3)

⇒ Bulk modulus = 2.026× 10⁹ Pa

⇒ Bulk modulus of air = 1 × 10⁵ Pa.

∴ Bulk modulus of water/ Bulk modulus of air = 2.026× 10⁹/(1× 10⁵) = 2.026× 10⁴

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

Given data,

Pressure at the given depth, p = 80.0 atm

⇒ Pressure = 80× 1.01 × 10⁵ Pa.

Density of water at the surface, ρ₁ = 1.03 × 10³ kgm^-3

Let the given depth be h.

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

⇒ V = m[(1/ρ₁)-(1/ρ₂)]

∴ Volumetric strain = ΔV/V₁

⇒ Volumteric Strain = m [(1/ρ₁)-(1/ρ₂)] × (ρ₁/m)

⇒ ΔV/V₁ = 1-(ρ₁/ρ₂) ………… (1)

Bulk modulus, B = pV₁/ ΔV

ΔV/V₁ = p/B

Compressibility of water = (1/B) = 45.8× 10^-11 Pa^-1

Compressibility is the fractional change in volume per unit increase in pressure. For each atmosphere increase in pressure, the volume of water would decrease 46.4 parts per million.

∴ ΔV/V₁ = 80× 1.013× 10⁵× 45.8× 10^-11 = 3.71 × 10^-3 ……. (2)

From equations 1 & 2, we get:

1-(ρ₁/ρ₂) = 3.71 × 10^-3

⇒ ρ₂ = 1.03× 10³ / [1-(3.71 × 10^-3)]

⇒ ρ₂ = 1.034× 10³ kgm^-3

Therefore, the density of water at the given depth (h) is 1.034× 10³ kgm^-3.

Given data,

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

The bulk modulus (or) of a substance is a measure of how incompressible/resistant to compressibility that substance is. It is defined as the ratio of the infinitesimal pressure increase to the resulting relative decrease of the volume.

Bulk modulus of glass by standards, B = 37 × 10⁹ Nm^-2.

Bulk modulus, B = p/ (Δv/v)

Where,

Δv/v = Fractional change in volume.

∴ Δv/v = ρ / B

⇒ 10× 1.013 × 10⁵ / (37 × 10⁹)

= 2.73 × 10^-5

Hence, the fractional change in the volume of the glass slab is 2.73 × 10^-5

Given data,

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

Hydraulic pressure, p = 7.0 × 10⁶ Pa.

Bulk modulus of copper, B = 140 × 10⁹ Pa

Bulk modulus, B = p/(Δv/v)

Where,

Δv/v = volumetric strain

Δv = change in volume

v = original volume

Δv = pv/B

Original volume of the cube, v = l³

∴ Δv = pl³/B

⇒ 7× 10⁶ × (0.1)³ / (140× 10⁹)

⇒ 5× 10^-8m³ = 5× 10^-2 cm^-3

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

Given data,

Volume of water, V = 1 L

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

∴ Fractional change, ΔV/V = 0.1/ (100× 1) = 10^-3.

Bulk modulus, B = ρ/ (ΔV/V)

ρ = B × (ΔV/V)

Bulk modulus of water, B = 2.2 × 10⁹ Nm^-2

ρ = 2.2× 10⁹× 10^-3 = 2.2× 10⁶ Nm^-2

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

Here the Compressive force will be transmitted from the wide ends to the narrow end so there will be pressure at the Narrow end or the tip, which is also Experiencing force in opposite direction, from the lower end in order to keep it stationary, since forces must be balanced i.e. equal and opposite on each member of anvil the free body diagram disassembling each member of anvil has been shown in the figure

We know Pressure is force acting per unit area and is given by

P = F/A

Where P is the pressure, F is the force, and A is the surface area in contact

Now here the force on the Tip of anvil is

F = 50000 N = 5 × 10⁴ N

We are given diameter of the tip

d = 0.50 mm

so the radius of the tip is

r = d/2 = 0.50mm/2 = 0.25 mm

i.e. r = 2.5 × 10^-4 m

now since the tip is a circular area of a circle is given by

A = πr²

Where A is the area of circle,

r is the radius

Putting value of r we get area of the Tip as

A = π × (2.5 × 10^-4 m)²

Solving we get area

A = 1.96 × 10^-7 m²

So putting value of F and A we get pressure

So the pressure at the tip of anvil is 2.55 × 10¹¹ Pa

Now the rod is held by two strings and weight is suspended it from it.

Now the tension in the strings will be in vertically upward direction and the weight of block is acting in vertically downward direction now since rod is held stationary it must be in equilibrium i.e. all the forces on it must cancel each other and also net resultant torque on it due to all the forces must also be zero

Now tension in the strings is the force acting due to them on rod and same amount of force is also acting on them in opposite direction let the Force on Rod due to tension in string A and B be T_A and T_Brespectively now let the block be at a distance d from string end tied to String A, let the weight of block of mass m be w

Now we know weight is given by

w = m × g

Where m is mass of block and g is acceleration due to gravity

All the forces on the rod is represented by free body diagram as shown in the figure below

So Balancing all the forces for equilibrium condition i.e. net magnitude of forces in vertically upward direction must be equal to magnitude of force in downward direction so we get

T_A + T_B = w

Now rod is in rotational equilibrium also since it is not rotating so sum of torque due to all the forces acting on rod must be zero around any point

We know magnitude torque on body due to a force with respect to any point is given by

Where r_⊥ is the perpendicular distance of point from line of force

And F is the magnitude of force

Now let us consider o point where mass is attached for rotation equilibrium, here forces are perpendicular to rod so the perpendicular distance from any point to line of force is same as the distance from point o now force due to point

We can see torque due to force T_A will be in clockwise direction denoted by and due to T_Bin anticlockwise direction denoted by and there will be no torque to weight of mass m since it is acting on the same point about we are calculating torques, so r_⊥ or the perpendicular distance is zero so torque is also zero since

Now the direction of torques are shown in figure below

Now for force T_A

r_⊥ = d

(Distance from o to left most end where T_A is acting)

F = T_A

So torque is

Now for force T_B

r_⊥ = 1.05 – d

(Distance from o to Right most end where T_B is acting since total length of rod is 1.05m)

F = T_B

So torque is

Now both torque must cancel each other i.e. should be equal in magnitude

i.e.

or

so we get the relation

(a) Now we know stress in a wire is restoring force per unit area , restoring force is same as the Tension in the wire or the force applied by the wire

We know stress in in a wire is given as

𝝈 = F/A

Where 𝝈 is the stress in wire, F is the restoring force applied by wire same as applied on it, and A is the area of cross section of the wire

Now here we are given area of cross section of wire A as

A_A = 1.0 mm²

Force on wire A is T_A

So stress on wire A is given by

𝝈_A = T_A/A_A

And we are given area of cross section of wire B as

A_B = 2.0 mm²

Force on wire B is T_B

So stress on wire B is given by

𝝈_B = T_B/A_B

Now stress in both the wires is equal

i.e. 𝝈_A = 𝝈_B

or T_A/A_A = T_B/A_B

re arranging we get

T_A/T_B = A_A/A_B

i.e. Putting the value of A_A and A_B we get

or

Note :We have not converted area of cross section of wires to m² from mm² because we were taking their ratio or were dividing them so units would ultimately cancel out , even if we would have converted result would have been same .

we already know

Where d is the distance of leftmost point to point o where the mass m is attached

Equating both the equation we get

Solving we get

d = 2.1m - 2d

Or 3d = 2.1m

So we have

that means wire should be tied at a distance of 0.7m from the left end of the rod

(b) Now we are given that strain in both the wires A and B must be equal we know strain is ratio of change in length to original length of wire , S = Δl/l where l is original length of wire and Δl is change in length of wire due to application of strain , to find strain in wire we use young modulus which is the ratio of stress to strain and is different for different materials of wire and is given by

Y = Stress/Strain

Or we can write

Strain = Stress/Y

Or we can say

S = 𝝈/Y

Where S is the strain in wire, 𝝈 is the stress and Y is the young’s modulus of wire

Now we know stress in in a wire is given as

𝝈 = F/A

Where 𝝈 is the stress in wire,

F is the restoring force applied by wire same as applied on it, and

A is the area of cross section of the wire

Putting value in S = 𝝈/Y

We get stress , as S = F/AY

Now for wire A

Its made up of steel so young’s modulus of wire is

⇒ Y_A = 2 × 10¹¹ Nm^-2

And stress is ⇒ S_A = T_A/A_AY_A

Where S_A is the stress in wire A, A_A is area of cross section of wire , T_A is the force acting on it and A_A is the area of cross section of wire

Now for wire B

Its made up of Aluminium so young’s modulus of wire is

Y_B = 7 × 10¹⁰ Nm^-2

And stress is

S_B = F_B/A_BY_B

Where S_B is the stress in wire B, A_B is area of cross section of wire , T_B is the force acting on it and A_B is the area of cross section of wire

Since strain in both the wires must be equal

So we have

S_A = S_B

T_A/A_AY_A = T_B/A_BY_B

On rearranging equation we get ratio of forces in two wires as

T_A/T_B = A_AY_A/A_BY_B

Putting values of A_A, A_B, Y_A, Y_B We get

Note :We have not converted area of cross section of wires to m² from mm² because we were taking their ratio or were dividing them so units would ultimately cancel out , even if we would have converted result would have been same .

we already know

Where d is the distance of leftmost point to point o where the mass m is attached

Equating both the equation we get

Solving we get

10d = 7.35m - 7d

Or, 17d = 7.35m

So we get

so the mass m must be tied at a distance of 0.43m from the leftmost portion of rod in order to have an equal strain in both the wires

Now here when a mass is suspended at midpoint of the stretched wire, the wire will elongate i.e. its size will increase due to which a depression will be there on the midpoint, and originally horizontal wire will make some angle with the horizontal

let us say the original length of wire is

l = 1m

Mass suspended is

m = 100g = 100 × 10^-3 Kg = 0.1 Kg

area of cross section of wire is

A = 0.50 × 10^-2 cm² = 0.50 × 10^-6 m² (since 1 cm² = 10^-4 m²)

Let the depression in the middle of wire be x

After elongation let the new length of wire be l’

And angle made by wire with horizontal on both sides be 𝜽

As has been shown in the figure

now elongation in the wire changes in its length of wire let it be

Δl = l’ – l

Now the mass m has been held at rest i.e. in equilibrium position by elongated string or tension due to it ,there are two forces acting on block its weight ‘w’ acting in vertically downward direction and the tension ‘T’ of both part of string AD and BD at an angle 𝜽 with the horizontal , the vertical component of Tension of string of the two parts balances the weight of block , and horizontal component balances each other

As shown in free body diagram

The horizontal component of tension in BD is

T_x = Tcos𝜽 in right direction

The horizontal component of tension in AD is

T_x = Tcos𝜽 in left direction

So both cancel out

As shown in figure

The vertical component of tension in BD is

T_y = Tsin𝜽 in vertically upward direction

The horizontal component of tension in AD is

T_y = Tsin𝜽 in vertically upward direction

So net force in vertically upward direction is

T_y + T_y = 2 T_y = 2Tsin𝜽

The force in vertically downward direction is weight of block

w = mg

where m is mass of block and g is acceleration due to gravity

since the block is in equilibrium both must be equal and opposite

i.e. w = 2T_y

or mg = 2Tsin𝜽

as shown in figure

So rearranging mg = 2Tsin𝜽 ,we get tension in the string as

T = mg/2sin𝜽

Now the depression is x using Pythagoras theorem in the ΔBCD

We have

CD² + BC² = BD²

CD = x

BC = l/2 (half of original length of wire)

BD = l’/2 (half of elongated length of wire)

So we have

Or

So elongated length is

(since )

We know increase in length Δl = l’ – l

So we have

Or

(putting (1/4)^1/2=1/2 , value of l = 1 ,and 4x² = (2x)² )

We get

We know the Binomial expansion

Using the above expansion Here we have n = 1/2 and y = (2x)²

So we have

Since depression will be very small so x<<l or x<<1

So neglecting terms with powers more than 1 like x⁴,x⁶ etc.

We get

Now finding value of sin𝜽

we know sin𝜽 = Perpendicular/Hypotenuse

in the ΔBCD

Perpendicular = CD = x

Hypotenuse = BD = l’/2

So we get

Sin𝜽 = x/(l’/2) = 2x/l’

We already know using Pythagoras theorem elongated length is

So we get

(solving as done earlier)

(putting l =1)

Now since x² << 1 neglecting it

i.e. putting

we get

Sin𝜽 = 2x

the tension in the string as

T = mg/2sin𝜽

So putting sin𝜽 = 2x we get

Tension in the string as

T = mg/4x

Now we know stress in a wire is restoring force per unit area, restoring force is same as the Tension in the wire or the force applied by the wire

We know stress in in a wire is given as

𝝈 = F/A

Where 𝝈 is the stress in wire, F is the restoring force applied by wire same as applied on it, and A is the area of cross section of the wire

So we get stress in the wire as

𝝈 = T/A = mg/4Ax

Now we know strain is ratio of change in length to original length of wire,

S = Δl/l

where l is original length of wire and Δl is change in length of the wire due to the application of strain

here Δl = 2x²

so we get strain in the wire as

S = 2x²/l

Original length of wire is l = 1 m,

so we have

S = 2x²

we use young modulus which is the ratio of stress to strain and is different for different materials of wire and is given by

Y = Stress/Strain

Or we can write

Y = 𝝈/S

Where Y is the young’s modulus of material of wire for steel

Y= 2 × 10¹¹ Nm^-2

Putting value of 𝝈 and S we get

Y = (mg/4Ax)/2x² = mg/8Ax³

Solving we get

X³ = mg/8AY

Or we get depression in the wire as

X = (mg/8AY)^1/3

Putting m = 0.1 kg

g = 9.8 ms^-2

A = 0.50 × 10^-6 m²

Y= 2 × 10¹¹ Nm^-2

We get

Or

So the depression in wire is 0.01m

Now here the metal strips are riveted together so the stress in the rivet would be shearing stress since force would be perpendicular to surface area of rivet and stress is the restoring force per unit area, restoring force is same as applied force in equilibrium condition, the rivet joints and force on them are as shown in the figure

Now we know shearing stress is given as

𝝈_s = F/A

Where 𝝈_s is the shearing stress F is the Shearing Force and A is the surface area

So, we get maximum force that can be applied to a rivet as

F = 𝝈_s/A

There are four rivets and We are given diameter of each rivet as d = 6.0 mm

So, radius of each rivet is

r = d/2 = 6.0mm/2 = 3.0 mm

or r = 3 × 10^-3m

since rivet is circular in shape area is given as

A = πr²

So, area of each rivet is

A = π × 3 × 10^-3m×3 × 10^-3m = 28.27 × 10^-6 m²

So, area of each rivet is

A = 2.827 × 10^-5m²

Now we know shearing stress is given as

𝝈_s = F/A

Where 𝝈_s is the shearing stress F is the Shearing Force and A is the surface area

So, we get maximum force that can be applied to a rivet as

F = 𝝈_s × A

We are given maximum shearing stress that can be applied to a rivet as

𝝈_s = 6.9 × 10⁷ Pa

Putting value of 𝝈_s and A we get maximum shearing force each rivet can with stand as

F = 6.9 × 10⁷Nm^-2 × 2.827 × 10^-5m² = 19.50 × 10² N

So each rivet can withstand a force of

F = 1.95 × 10³ N = 1.95 KN

Now each rivet is to carry one-quarter of the load i.e 1/4 th of the total load so total load that can be applied or maximum tension that can be exerted is four timed of that can be applied on each rivet i.e. the maximum tension

T_max = 4 × F

So T_max = 4 × 1.95 KN = 78 KN

So, the maximum tension that can be exerted by riveted strip is 78 ×10³ N or 78 KN

Here the solid Steel Ball placed in the water under high pressure is compressed uniformly at all Points. The force applied by the water acts in a perpendicular direction at each point on the surface of sphere This leads to decrease in its volume. The body develops internal restoring forces that are equal and opposite to the forces applied to it by water. The internal restoring force per unit is known as hydraulic stress and in magnitude is equal to the hydraulic pressure, since stress is restoring force per unit area. The strain produced by the pressure is called volume strain and is defined as the ratio of change in volume to the original volume. Since the strain is a ratio of change in dimension to the original dimension

The Force on Sphere, Restoring force by it and volume before and after change due to pressure has been shown in the following figures

Here the volume of ball will change and will decrease due to external pressure of water as there is always change in dimensions of a material upon application of an external force which depends upon the elastic property of that element for volumetric change we consider bulk modulus, now ball is made up of steel we have bulk modulus for steel is

B = 1.6 × 10¹¹ N/m²

The force on the ball will be due to pressure on the surface of sphere and on every point on surface of sphere pressure will be

P = 1.1 × 10⁸ Pa

Now the original volume of sphere without shrinking due to external pressure is

V = 0.32 m³

Not let us say change in volume of the ball be ΔV

So the strain on ball or the ratio of change in volume to original volume is given by ΔV/V

Now as explained earlier the external hydraulic pressure on the Ball is same as the stress on its surface

Now bulk modulus is the ratio of stress to strain so bulk modulus is given by

B = -P/(ΔV)/V

Where B is the bulk modulus, P is the excess pressure or the pressure on the surface of body due to restoring forces which are actually the volumetric stress, ΔV is the change in volume of the body initially having original volume V

So re-arranging we get

B = -(P × V)/(ΔV)

Or we can say change in volume

ΔV = -(P × V)/B

Putting values of P, V, and B we get

1Pa = 1 N/m²

Minus sign suggest that volume has decreased as compared to original So we get change in volume of sphere is 2.2 × 10^-4 m³