Elasticity Applications of Elasticity in Daily Life

Elasticity Applications of Elasticity in Daily Life

Most materials used in our daily life undergo some kind of stress. It is therefore important to design things in a way that they continue working under the stress suffered by them. The following examples will illustrate this point.

1. In cranes: The thickness of metallic ropes used in cranes to lift and move heavy weights is decided on the basis of the elastic limit of the material of the rope and the factor of safety.

2. Indesigningabeam: When a transverse load is put across ahorizontal beam, it causes ade pression leading to bending in the beam. In various applications like bridges, buildings, etc., it is important that the beam can with stand the load or the weight and it should not bend too much or break. When the bending is not accompanied by any torsion or shear, it is said to be simple bending.

3. A bridge is declared unsafe after long use: During its long use, a bridge undergoes quick alternating strains repeatedly. It results in the loss of elastic strength of the bridge. After a long period, such a bridge starts developing large strains corresponding to the same usual value of stress and ultimately it may lead to the collapse of the bridge. To avoid such a situation a bridge is declared unsafe after its long use.

Read and Learn More: Class 11 Physics Notes

Elasticity Applications of Elasticity in Daily Life Numerical Examples

Example 1. To increase the length of an elastic string of radius 3.5 mm by 1/20 th of its initial length, within its elastic limit, a 10 N force is required. Calculate the Young’s modulus for the material of the string.
Solution:

To increase the length of an elastic string of radius 3.5 mm by 1/20 th of its initial length, within its elastic limit, a 10 N force is required.

We know that, Y = \(\frac{F L}{\pi r^2 l} .\)

Here, F = 10 N , l = 1/20 L and r = 3.5 mm = 0.0035 m

∴ Y = \(\frac{10 \times L}{3.14 \times(0.0035)^2 \times \frac{L}{20}}=5.2 \times 10^6 \mathrm{~N} \cdot \mathrm{m}^{-2}\)

Example 2. Two wires of the same length but of different materials have diameters of 1 mm and 3 mm respectively. If both of them are stretched by the same force, then the elongation of the first wire becomes thrice that of the second. Compare their Young’s moduli.
Solution:

Two wires of the same length but of different materials have diameters of 1 mm and 3 mm respectively. If both of them are stretched by the same force

Let the initial length of each wire be L and the force applied on each be F.

∴ Y = \(\frac{F L}{\pi r^2 l}\)

So, \(\frac{Y_1}{Y_2}=\left(\frac{r_2}{r_1}\right)^2 \cdot \frac{l_2}{l_1}=\left(\frac{3}{1}\right)^2 \times \frac{1}{3}=\frac{3}{1}\)

∴ \(Y_1: Y_2=3: 1\)

Example 3. If the elastic limit of a typical rock is 3 x 10⁸ N• m^-2 and its mean density is 3 x 10³ kg • m^-3, estimate the maximum height of a mountain on the earth. (g =  10 m•s^-2)
Solution:

If the elastic limit of a typical rock is 3 x 10⁸ N• m^-2 and its mean density is 3 x 10³ kg • m^-3

Let us assume that the maximum height of the mountain is h and g is nearly uniform along this height.

Then the maximum pressure at its bottom = ρg.

According to the problem, hρg = breaking stress = elastic limit.

∴ h = \(\frac{\text { elastic limit }}{\rho g}\)

Here, ρ = 3 x 10³ kg • m^-3 and the elastic limit = 3 x 10⁸ N • m^-2

∴ h = \(\frac{3 \times 10^8}{3 \times 10^3 \times 10}=10^4 \mathrm{~m}\)

Example 4. Two equal and opposite forces are applied tangentially to two mutually opposite faces of an aluminum cube of side 3 cm to produce a shear of 0.01°. If the modulus of rigidity for aluminum is 7 x 10¹⁰ N • m^-2, then calculate the force applied.
Solution:

Two equal and opposite forces are applied tangentially to two mutually opposite faces of an aluminum cube of side 3 cm to produce a shear of 0.01°. If the modulus of rigidity for aluminum is 7 x 10¹⁰ N • m^-2,

Let the magnitude of applied force be F.

We know that, n = \(\frac{F / A}{\theta} \text { or, } F=n \theta A\)

⇒ \(\theta=0.01^{\circ}=\frac{0.01 \times \pi}{180} \mathrm{rad}\)

A = \(9 \mathrm{~cm}^2=9 \times 10^{-4} \mathrm{~m}^2\)

∴ F = \(7 \times 10^{10} \times \frac{0.01 \times 3.14}{180} \times 9 \times 10^{-4}=1.1 \times 10^4 \mathrm{~N}\)

Example 5. A rubber cord of length 20 m is suspended from—a rigid support by one of Its ends and it hangs vertically. What will be the elongation of the cord due to its own weight? The density of rubber = 1.5 g • cm^-3 and Young’s modulus = 49 x 10⁷ N • m^-2.
Solution:

A rubber cord of length 20 m is suspended from—a rigid support by one of Its ends and it hangs vertically.

Here the downward force,

F = weight of the cord

= volume of the cord x density x acceleration due to gravity

= 20 x Ax 1.5 x 1000 x 9.8 N

[A = area of cross-section of the cord]

The centre of gravity of the rubber cord is at a vertical distance of 10 m from the fixed end. The weight of the cord acts through its centre of gravity.

Hence the length of the upper half, above the centre of gravity, is taken as the initial length to estimate the elongation of the cord. So, L = 10 m.

∴Y = \(\frac{F L}{A l}\)

or, \( l=\frac{F L}{A Y}=\frac{20 \times A \times 1.5 \times 1000 \times 9.8 \times 10}{A \times 49 \times 10^7}=0.006 \mathrm{~m}\)

Example 6. A steel wire of  diameter 0.8 mm and length 1 m is clamped firmly at two points A and B which are 1 m apart in the same horizontal line. A body is hung from the middle point ofthe wire such that the | middle point sags 1 cm from the original position. Calculate the mass of the body. [Y = 2 x 10¹² dyn •cm^-2]
Solution:

A steel wire of  diameter 0.8 mm and length 1 m is clamped firmly at two points A and B which are 1 m apart in the same horizontal line. A body is hung from the middle point ofthe wire such that the | middle point sags 1 cm from the original position

A steel wire of  diameter 0.8 mm and length 1 m is clamped firmly at two points A and B which are 1 m apart in the same horizontal line. A body is hung from the middle point ofthe wire such that the | middle point sags 1 cm from the original position.

C is the mid-point of the wire AB.

When a mass m is hung from C, it sags 1 cm and comes down to the point O.

AB = 1 m = 100 cm , AC = CB = 50 cm,

OC = 1 cm , OA = OB = \(\sqrt{50^2+1^2}\) = 50.01 cm,

cosθ = \(\frac{1}{50.01},\), r = 0.04 cm

Here, tension in the part CM or OB is T. The two horizontal
components of T balance each other and the vertical components balance the weight mg.

∴ 2T cosθ = mg or, T = \(\frac{m g}{2 \cos \theta}\)

The length AC of the wire changes into AO.

∴ Elongation = l = AO – Ac = 50.01-50 = 0.01

Young’s modulus, Y = \(\frac{T L}{A l}=\frac{m g L}{2 \cos \theta \cdot A l}\)

or, \(m=\frac{2 Y A l \cos \theta}{g L}\)

= \(\frac{2 \times\left(2 \times 10^{12}\right) \times 3.14 \times(0.04)^2 \times 0.01}{980 \times 50} \times \frac{1}{50.01}\) = 82.01 g

Example 7. If the work done in stretching a uniform wire, of cross section 1 mm² and length 2 m, by 1 mm is 0.05 joule, find the Young’s modulus for the material of the wire.
Solution:

If the work done in stretching a uniform wire, of cross section 1 mm² and length 2 m, by 1 mm is 0.05 joule

If the work done in stretching a uniform wire, of cross section 1 mm² and length 2 m, by 1 mm is 0.05 joule

Work done, \(W=\frac{1}{2} \frac{Y \alpha l^2}{L}\).

So, \(Y=\frac{2 W L}{\alpha l^2}\)

Here, \(W=0.05 \mathrm{~J} ; \alpha=1 \mathrm{~mm}^2=10^{-6} \mathrm{~m}^2 ; L=2 \mathrm{~m}\)

⇒ \(l=1 \mathrm{~mm}=10^{-3} \mathrm{~m}\)

∴ \(Y=\frac{2 \times 0.05 \times 2}{10^{-6} \times\left(10^{-3}\right)^2}=2 \times 10^{11} \mathrm{~N} \cdot \mathrm{m}^{-2}\)

Example 8. A cylindrical pipe of uniform cross-section and of length 120 cm is closed at one end and is completely filled with water. Kept upright, the cylinder is stretched to increase its length. It elongates by 1 cm, but the length of the water column increases by 0.7 cm. Calculate the Poisson’s ratio of the material of the pipe.
Solution:

A cylindrical pipe of uniform cross-section and of length 120 cm is closed at one end and is completely filled with water. Kept upright, the cylinder is stretched to increase its length. It elongates by 1 cm, but the length of the water column increases by 0.7 cm.

A cylindrical pipe of uniform cross-section and of length 120 cm is closed at one end and is completely filled with water. Kept upright, the cylinder is stretched to increase its length. It elongates by 1 cm, but the length of the water column increases by 0.7 cm.

Poisson’s ratio, \(\sigma=\frac{\text { lateral strain }}{\text { longitudinal strain }}\)

Here, longitudinal strain = \(\frac{l}{L}=\frac{1}{120}\)

Let the initial diameter of the pipe be D, the decrease in its diameter due to elongation be d.

Initial volume of water = \(\frac{\pi}{4} D^2 \times 120\)

and final volume of water = \(\frac{\pi}{4}(D-d)^2 \times(120+0.7)\)

= \(\frac{\pi}{4}(D-d)^2 \times 120.7\)

Since the volume of water inside the cylinder remains
unchanged,

⇒ \(\frac{\pi}{4} D^2 \times 120=\frac{\pi}{4} \times(D-d)^2 \times 120.7\)

or, \(\frac{D-d}{D}=\sqrt{\frac{120}{120.7}} \quad \text { or, } 1-\frac{d}{D}=\sqrt{0.9942}=0.997\)

or, \(\frac{d}{D}=1-0.997=0.003\)

∴ \(\sigma=\frac{d / D}{l /L}=\frac{0.003}{\frac{1}{120}}=0.003 \times 120=0.36\)

Example 9. A light bar of length 2 m is suspended horizontally by means of two wires of equal lengths connected to its two ends. One wire is of steel having a cross-sectional area of 0.1 cm², the other is of brass with a cross-sectional area of 0.2 cm². Find the point the bar from where a weight must be suspended so that both the wires experience

1. the same stress,
2. the same strain. Given, the Young’s modulus for steel =2x 10¹¹ N • m^-2 and that for brass =10 x 10¹⁰ Nm^-2.

Solution:

A light bar of length 2 m is suspended horizontally by means of two wires of equal lengths connected to its two ends. One wire is of steel having a cross-sectional area of 0.1 cm², the other is of brass with a cross-sectional area of 0.2 cm².

Let the horizontal bar be AB. A load W is hung from the point C. Let AC be x. In this condition, let the tension in the steel wire be T₁ and that in the brass wire be T₂.

Stress on the steel wire = \(\frac{T_1}{\alpha_1}\)

stress on the brass wire = \(\frac{T_2}{a_2}\)

1. If the stresses in both the wires are the same, then

⇒ \(\frac{T_1}{0.1 \times 10^{-4}}=\frac{T_2}{0.2 \times 10^{-4}} \text { or, } \frac{T_1}{T_2}=\frac{1}{2}\)…..(1)

Since the system is in equilibrium, taking moments about the point C, we get,

⇒ \(T_1 x=T_2(2-x) \quad \text { or, } \quad \frac{T_1}{T_2}=\frac{2-x}{x}\)

∴ \(\frac{1}{2}=\frac{2}{x}-1 \quad \text { or, } \frac{2}{x}=\frac{3}{2} \text { or, } x=\frac{4}{3}=1.33 \mathrm{~m}\)

So the weight should be suspended at a distance of 1.33 m from the steel wire.

2. Longitudinal strain = \(\frac{\text { longitudinal stress }}{Y}\)

= \(\frac{\frac{T}{\alpha}}{Y}=\frac{T}{\alpha Y}\)

For the same strain in the two wires, \(\frac{T_1}{0.1 \times 10^{-4} Y_1}=\frac{T_2}{0.2 \times 10^{-4} Y_2}\)

or, \(\frac{T_1}{T_2}=\frac{0.1 \times 10^{-4} \times 2 \times 10^{11}}{0.2 \times 10^{-4} \times 10 \times 10^{10}}\)=1

∴ \(T_1=T_2\)

Since, the system is in equilibrium, taking moments about the point C, we get,

T₁x = T₂(2-x) or, x = 2-x or, x= 1 m

So, the weight should be suspended from the midpoint of the horizontal bar.

Example 10. A sphere of mass 25 kg and radius 0.1 m is hung from the ceiling of a room with the help of a steel wire. The height of the ceiling from the floor is 5.21 m. When the sphere is hung just like a pendulum, its lower surface touches the floor of the room. What will be the velocity of the sphere at the lowest point of its oscillation? The Young’s modulus for steel =2x 10¹¹ N • m²; the initial length of the wire = 5 m and the radius of the wire = 5 x 10⁴ m. 
Solution:

A sphere of mass 25 kg and radius 0.1 m is hung from the ceiling of a room with the help of a steel wire. The height of the ceiling from the floor is 5.21 m. When the sphere is hung just like a pendulum, its lower surface touches the floor of the room.

According to the figure, the elongation of the wire at the lowest position of the sphere (diameter = 0.2m),

l = 5.21-(5+ 0.2) = 0.01 m

If the velocity of the sphere at the lowest point of its oscillation is v, then the tension in the wire,

⇒ \(T-m g=\frac{m v^2}{r}\)

T = \(m g+\frac{m v^2}{r}\) ….(1)

Here, m = mass of the sphere; r = distance of the center of gravity of the sphere from the point of suspension = 5.21 – 0.1 = 5.11m.

Suppose x = radius of the wire.

Then, \(Y=\frac{T}{\pi x^2} \frac{L}{l} \quad \text { or, } \quad T=\frac{Y \pi x^2 l}{L}\)

From equation (1), we get

⇒ \(mg +\frac{m v^2}{r}=\frac{Y \pi x^2 l}{L}\)

or, \(\frac{v^2}{r}=\frac{Y \pi x^2 l}{m L}-g\)

or, \(v^2=\frac{Y \pi x^2 l r}{m L}-r g\)

= \(\frac{\left(2 \times 10^{11}\right) \times 3.14 \times\left(5 \times 10^{-4}\right)^2 \times 0.01 \times 5.11}{25 \times 5}\)

= 14.1036

∴ ν = 3.76 m .s^-1

Example 11. Two blocks A and B are connected to each other by a string and a spring; the string passes over a frictionless pulley as shown. Block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C, both with the same uniform speed. The coefficient of friction between the surfaces of the blocks is 0. 2. The force constant of the spring Is 1960 N• m^-1. If the mass of the block A is 2 kg, calculate the mass of the block B and the energy stored In the spring.
Solution:

Two blocks A and B are connected to each other by a string and a spring; the string passes over a frictionless pulley as shown. Block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C, both with the same uniform speed. The coefficient of friction between the surfaces of the blocks is 0. 2. The force constant of the spring Is 1960 N• m^-1. If the mass of the block A is 2 kg

Since the block A is descending with uniform velocity, no normal reaction acts on the block A due to the block C.

Considering the motion of the blocks A and B,

T = \(m_A g=\mu m_B g\)

∴ \(m_B=\frac{m_A}{\mu}=\frac{2}{0.2}=10 \mathrm{~kg}\)

If the extension of the spring is xm, then,

T = \(k x \text { or, } x=\frac{T}{k}=\frac{m_A g}{k}=\frac{2 \times 9.8}{1960}=\frac{1}{100} \mathrm{~m}\)

Energy stored in the spring

= \(\frac{1}{2} k x^2=\frac{1}{2} \times 1960 \times\left(\frac{1}{100}\right)^2=0.098 \mathrm{~J}\)

Example 12. On application of a pressure of 21 kg • cm^-2, the volume of 1 litre of an oil decreases by 840 mm³. Calculate the bulk modulus and compressibility of the oil.
Answer:

On application of a pressure of 21 kg • cm^-2, the volume of 1 litre of an oil decreases by 840 mm³.

p = \(21 \mathrm{~kg} \cdot \mathrm{cm}^{-2}=\frac{21 \times 9.8}{(0.01)^2} \mathrm{~N} \cdot \mathrm{m}^{-2}\)

= \(21 \times 98 \times 10^3 \mathrm{~N} \cdot \mathrm{m}^{-2}\)

V = 1 litre 10^-3 m³; ν = 840 mm³ = 840×10^-9 m³

Bulk modulus of the oil,

K = \(\frac{p V}{\nu}=\frac{\left(21 \times 98 \times 10^3\right) \times 10^{-3}}{840 \times 10^{-9}}\)

= \(2.45 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^{-2}\)

= 2.45 x 109 N • m^-2

Compressibility of the oil, \(\frac{1}{K}=\frac{1}{2.45 \times 10^9}=4.1 \times 10^{-10} \mathrm{~m}^2 \cdot \mathrm{N}^{-1}\)

Example 13. A 0.5 kg block Alides from a point A on a horizontal track with an initial speed of 3 m • s^-1 towards a weightless horizontal spring of length 1 in and of force constant 2 N • m^-1. The part AB of the track is frictionless and the part BC has coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distances AB and BD are 2 m and 2.14 m respectively, find the total distance covered by the block before it comes to rest (g = 10 m • s^-2).
Solution:

A 0.5 kg block Alides from a point A on a horizontal track with an initial speed of 3 m • s^-1 towards a weightless horizontal spring of length 1 in and of force constant 2 N • m^-1. The part AB of the track is frictionless and the part BC has coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distances AB and BD are 2 m and 2.14 m respectively

Initial kinetic energy of the block

= \(\frac{1}{2} m v^2=\frac{1}{2} \times 0.5 \times(3)^2=2.25 \mathrm{~J}\)

The part AB of the horizontal track is frictionless and hence during the passage ofthe block over this path no energy is lost.

Loss in energy in the part BD = work done against friction in that path

= μmgxBD = 0.2 x 0.5 x 10×2.14 = 2.14 J

So, the kinetic energy of the block when it reaches the point

D = 2.25-2.14 = 0.11 J

Let us assume that the block compresses the spring through an amount x

According to the principle of conservation of energy,

work done against friction + work done in compressing the spring =0.11

or, \(\mu m g x+\frac{1}{2} k x^2=0.11\)

or, \(0.2 \times 0.5 \times 10 \times x+\frac{1}{2} \times 2 x^2=0.11\)

or, \(x^2+x-0.11=0\)

or, (x + 1.1)(x- 0.1) = 0

x = 0.1 m [as x cannot be -1.1 m]

∴ The distance covered by the block before it comes to rest = 2 + 2.14 + 0.1 = 4.24 m.

Example 14. Two bodies A and B of masses m and 2m respectively are put on a smooth floor. They are connected by a spring. A third body C of mass m moves with a velocity v₀ along the line joining A and B and collides elastically with A as shown in Fig. At a certain instant of time t₀ after the collision, it is found that the instantaneous velocities of A and B are the same. Further, at this instant the compression of the spring is found to be x₀. Find out

1. the common velocity of A and B at time t₀ and
2. the force constant of the spring.

Solution:

1. After elastic collision, C will come to rest and A will gain a velocity v₀ (as their masses are the same).

Let us assume that the common velocity of A and B is v, at time t₀ after the collision.

According to the principle of conservation of momentum,

⇒ \(m v_0=m \nu+2 m \nu \quad \text { or, } \quad v=\frac{v_0}{3}\)

2. According to the principle of conservation of energy,

⇒ \(\frac{1}{2} m v_0^2=\frac{1}{2} m v^2+\frac{1}{2} \times 2 m v^2+\frac{1}{2} k x_0^2\)

or, \(m v_0^2=3 m v^2+k x_0^2\)

[k = force constant of the spriong]

or, \(m v_0^2=3 m\left(\frac{v_0} {3}\right)^2+k x_0{ }^2 \quad \text { or, } k=\frac{2}{3} \frac{m v_0^2}{x_0^2}\)

Example 15. If the tension in a wire increases gradually to 6 kg, the elongation of the wire becomes 1.13 mm. Calculate the work done.
Solution:

If the tension in a wire increases gradually to 6 kg, the elongation of the wire becomes 1.13 mm.

Work done, W = \(\frac{1}{2} F l=\frac{1}{2} \times 6 \times 9.8 \times \frac{1.13}{1000}=0.033 \mathrm{~J}\)

Example 16. A body of mass 4 kg and density 2.5 g cm^-3, suspended by a metallic wire of length 1 m and diameter 2 mm, is kept completely immersed in water. What will be the increase in the length of the wire? Young’s modulus of the metal = 2x 10¹¹ N • m^-2 and g = 9.8 m • s^-2.
Solution:

A body of mass 4 kg and density 2.5 g cm^-3, suspended by a metallic wire of length 1 m and diameter 2 mm, is kept completely immersed in water.

Volume of the body, V = \(\frac{4}{2.5 \times 1000}=16 \times 10^{-4} \mathrm{~m}^3\)

Apparent weight of the body when immersed in water,

W’ = mg- V x 1000 x g

= 4 x 9.8 – 16 x 10^-4 x 1000 x 9.8

= 39.2-15.68 = 23.52 N

If the elongation of the wire is l’, then,

Y = \(\frac{F}{A} \cdot \frac{L}{l^{\prime}} \quad\left[\text { Here, } F=W^{\prime}\right]\)

or, \(l^{\prime}=\frac{F}{A} \cdot \frac{L}{Y}\)

= \(\frac{23.52 \times 1}{3.14 \times\left(10^{-3}\right)^2 \times 2 \times 10^{11}}\)

= \(3.75 \times 10^{-5} \mathrm{~m}\)

Example 17. A 0.1kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its cross sectional area is 4.9 x 10^-7 m². If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad • s^-1. If the Young’s modulus of the material of the wire is n x 10⁹N • m^-2. Calculate the value of n. 
Solution:

A 0.1kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its cross sectional area is 4.9 x 10^-7 m². If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad • s^-1. If the Young’s modulus of the material of the wire is n x 10⁹N • m^-2

Young’s modules of the material of the wire,

Y = \(\frac{\frac{F}{A}}{\frac{\Delta L}{L}}=\frac{F L}{A \Delta L} \quad \text { or, } F=\left(\frac{Y A}{L}\right) \Delta L\)….(1)

If mass m is pulled by a length ΔL then restoring force developed in the wire,

F = kΔL ……(2)

Comparing equations (1) and (2) we get,

k = \(\frac{Y A}{L}\)

Angular frequency, \(\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{Y A}{m L}}\)

or, \(140=\sqrt{\frac{n \times 10^9 \times 4.9 \times 10^{-7}}{0.1 \times 1}}=70 \sqrt{n}\)

∴ \(\sqrt{n}=2\) or, n=4

Example 18. Stress-strain graph of an elastic material is shown. Using the graph And Young’s modulus of the material.

Solution:

Young’s modulus, \(Y=\frac{\text { stress }}{\text { strain }}\)

The strain corresponding to a point P on the graph is

OA = 0.003 and stress – OB = 200 x 10⁶ N • m^-2 .

∴ Young’s modulus of the material,

Y = \(\frac{200 \times 10^6}{0.003}=6.6 \times 10^{10} \mathrm{~N} \cdot \mathrm{m}^{-2}\)

Example 19. A 3 kg mass is hanging from one end of a vertical copper wire of length 2 m and diameter 0.5 mm. Due to this the elongation produced in the wire is 2.38 mm. Find Young’s modulus of copper.
Solution:

A 3 kg mass is hanging from one end of a vertical copper wire of length 2 m and diameter 0.5 mm. Due to this the elongation produced in the wire is 2.38 mm

Young’s modulus, Y = \(Y=\frac{m g L}{\pi r^2 l} .\)

Here, m = 3 kg , g = 9.8 m • s^-2 , L = 2 m,

r = \(\frac{0.5}{2} \mathrm{~mm}=\frac{5 \times 10^{-4}}{2} \mathrm{~m}=2.5 \times 10^{-4} \mathrm{~m}\),

l = \(2.38 \mathrm{~mm}=238 \times 10^{-5} \mathrm{~m} \)

∴ Y = \(\frac{3 \times 9.8 \times 2}{3.14 \times\left(2.5 \times 10^{-4}\right)^2 \times 238 \times 10^{-5}}\)

= \(1.26 \times 10^{11} \mathrm{~N} \cdot \mathrm{m}^{-2}\)

Example 20. Six external forces, each of magnitude F, are applied on all the faces of a unit cube. Considering its elastic modulus, calculate the longitudinal strain and the volume strain on the unit cube.
Solution:

Six external forces, each of magnitude F, are applied on all the faces of a unit cube. Considering its elastic modulus

Each side of the unit cube = 1.

∴ Volume of the cube = 1, surface area of each of the six
faces = 1

Let x, y, and z axes be chosen along three mutually perpendicular sides of the cube.

Now, at first, we choose the two opposite faces perpendicular to the x-axis. Due to the forces (F, F) acting on them, the side along the x -direction will have an elongation = l (suppose)

∴ Longitudinal strain along the x -direction

= \(\frac{\text { elongation }}{\text { initial length }}=\frac{l}{1}=l\)

Also, longitudinal stress = \(\frac{\text { applied force }}{\text { surface area }}=\frac{F}{1}=F\)

∴ Young’s modulus, Y = \(\frac{\text { stress }}{\text { strain }}=\frac{F}{l} ; \text { so, } l=\frac{F}{Y}\)

Simultaneously the side along the x-direction will also suffer lateral strains due to the forces acting along y and z directions. We know,

Poisson’s ratio, \(\sigma=\frac{\text { lateral strain }}{\text { longitudinal strain }}\)

Lateral strain = σ x longitudinal strain = σl = \(\frac{\sigma F}{Y}\)

Now, an elongation is always associated with a lateral contraction. Therefore, considering the strains along all the three axes, the effective longitudinal strain along the x -axis is

⇒ \(\frac{F}{Y}-\frac{\sigma F}{Y}-\frac{\sigma F}{Y}=\frac{F}{Y}(1-2 \sigma)\)

From symmetry, the effective longitudinal strain along each
of y- and z-axes = \(\frac{F}{Y}(1-2 \sigma)\)

As a result, the final length of each of the three sides of the cube = \(1+\frac{F}{Y}(1-2 \sigma)\)

∴ Final volume of the unit cube

= \(\left[1+\frac{F}{Y}(1-2 \sigma)\right]^3=1+\frac{3 F}{Y}(1-2 \sigma)\) [neglecting higher order terms]

∴ Volume expansion = \(1+\frac{3 F}{Y}(1-2 \sigma)-1\)

= \(\frac{3 F}{Y}(1-2 \sigma)\)

∴ Volume strain = \(\frac{\frac{3 F}{Y}(1-2 \sigma)}{1}=\frac{3 F}{Y}(1-2 \sigma)\)

Example 21. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, find the ratio of the elongation in the thin wire to that in the thick wire.
Solution:

One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends,

Let the elongation produced in wire MN be l₁ and that in PQ be l₂

Since both the wires are made of copper, Young’s moduli will be same for both.

∴ From \(Y=\frac{F L}{A l}\) we have,

⇒ \(\frac{F}{\frac{\pi(2 R)^2}{\frac{l_2}{2 L}}}=\frac{\frac{F}{\pi R^2}}{\frac{l_1}{L}}\)

or, \(\frac{2 F L}{4 \pi R^2 l_2}=\frac{F L}{\pi R^2 l_1}\)

or, \(\frac{l_1}{l_2}=2\)

∴ \(l_1: l_2=2: 1\)