## Chapter 3 Topic D Elasticity Synopsis

The property by virtue of which the object resists the deformation in shape, size, and volume when a deforming force is applied and regains its shape, size, and volume after the withdrawal of that deforming force is called elasticity.

Cl If a body regains its original shape, size, and volume completely after the withdrawal of the deforming force of any magnitude, then that body is called a perfectly elastic body.

Elastic limit of a body is the upper limit of a deforming force acting on the body, upto which the body regains its original form if deforming force is removed, and beyond which if the force is increased, the body loses its property of elasticity and gets permanently deformed.

Due to application of an external balanced force, there is relative displacement of different parts of an elastic body. As a result, there is a change of the shape, size, and volume of the body. The ratio of change in configuration to the original configuration of the body is called strain.

**Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment**

Strain is a dimensionless physical quantity and does not have any unit.

If an elastic body is strained due to the application of an external balanced force, then there is an internal reaction force inside the body due to elasticity which resists the external force. When this external force is withdrawn, then the internal reaction force helps the body to regain its original state.

This reaction force produced per unit area of the surface of the body is called stress.

**Revised Form Of Hooke’s Law:**

Within elastic limit, stress is directly proportional to strain.

**Young’s Modulus (Y)** is defined as the ratio of longitudinal stress to longitudinal strain within elastic limit.

Where L is the length of the wire, l is the increased length of the wire, F is the ductile force and A is the cross-sectional area of the wire.

**Units of Young’s Modulus** in CGS system and SI are dyn/cm^{2} and N/m^{2}, respectively.

Force constant is defined as the required tensile force to increase the length of a spring by unity.

Units of force constant for a spring in CGS system and SI are dyn/cm and N/m, respectively.

Dimensional formula of the force constant for a spring is MT^{-2}.

## Chapter 3 Topic D Elasticity Short And Long Answer Type Questions

**Question 1. What is elasticity?**

**Answer:**

Elasticity is the property of an object or material by which it resists the deformation in its shape or volume or both due to external balanced forces acting on it. The object regains its original shape or volume when these external forces are withdrawn.

**Question 2. What is a perfectly elastic body? Also, define elastic limit.**

**Answer:**

A perfectly elastic body is that body which can regain its original shape, size and volume after the withdrawal of the external forces.

Elastic limit of a body is the upper limit of a deforming force acting on the body, upto which the body behaves as a perfectly elastic body. If the deforming force is increased beyond the elastic limit, the body loses its property of elasticity and gets permanently deformed.

**Question 3. Define a perfectly rigid body and an inelastic body.**

**Answer:**

**Perfectly Rigid Body:**

A body that has no strain in spite of the application of external force of any magnitude is called a perfectly rigid body.

**Inelastic Body:**

If a body, deformed by external forces remains in the deformed state even after the withdrawal of these deformed forces, it is called an inelastic body.

**Question 4. What is stress? How can it be measured?**

**Answer:**

If an elastic body is strained due to the application of external balanced forces, there is an internal reaction force inside the body due to elasticity which resists the external force.

When the external force are withdrawn, this internal reaction force helps the body to return to its original state. This reaction force produced per unit area of the surface of the body is called stress.

Stress is the reaction of an external applied force on the body. According to Newton’s third law of motion, for every action, there is an equal and opposite reaction.

So, stress is measured by the applied force per unit area on the surface.

∴ \(\text { stress }=\frac{\text { applied force }}{\text { area of cross section of the body }}\)

**Question 5. What is the unit of stress in CGS system and SI? Establish a relationship between these two units. Calculate the dimensional formula of stress and write down its dimension.**

**Answer:**

Units of stress in CGS system and SI are dyn/cm^{2} and N/m^{2}, respectively.

The relationship between them is given by

\(1 \mathrm{~N} / \mathrm{m}^2=\frac{10^5 \mathrm{dyn}}{10^4 \mathrm{~cm}^2}=10 \mathrm{dyn} / \mathrm{cm}^2\)Dimensional formula of stress

= \(\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)

Dimension of stress is 1 in mass, -1 in length, and -2 in time.

**Question 6. What do you mean by strain? How can it be measured? Why does it have no unit or dimension?**

**Answer:**

Due to the application of an external balanced force, there may be a change of form or shape, or both of an elastic body. This is called strain.

The measure of strain is the comparative proportional change of the initial value of length, shape, or volume of an object, i.e.,

strain = \(\frac{\text { change of form or shape }}{\text { initial form or shape }}\)

As strain is, the ratio of two quantities of the same type, it has no unit or dimension.

**Question 7. Between rubber and steel, which has greater elasticity? Why?**

**Answer:**

Between rubber and steel, steel has comparatively greater elasticity.

This is obvious from the fact that to elongate both of them to the same extent, much more force is to be applied to the steel than rubber. In scientific language, a body is said to be more elastic compared to another body if greater force has to be applied on it to bring about the same amount of strain in it.

According to this, elasticity of steel is much more than that of rubber.

**Question 8. Units of modulus of elasticity and stress are the same. Explain the authenticity of this statement.**

**Answer:**

The modulus of elasticity is the ratio of stress and strain, within the elastic limit. Strain has no unit, so the unit of modulus of elasticity should have the same unit as that of stress.

Units of modulus of elasticity in CGS system and SI are dyn/cm^{2} and N/m^{2} respectively, which are also the units of stress.

**Question 9. Write down Hooke’s law. What do you mean by modulus of elasticity?**

**Answer:**

**Revised Form Of Hooke’s Law:**

Within elastic limit, ratio of stress and strain is a constant.

So, \(\frac{stress}{strain}\) = constant

This constant is called modulus of elasticity of the material of the body. This quantity determines the extent to which a material is elastic as compared to another.

**Question 10. What do you mean by longitudinal strain arid longitudinal stress?**

**Answer:**

Suppose, a particular body is considered whose breadth and height are insignificant compared to its length. One side of this body is fixed firmly while a tensile force is applied along its length so that there is change of length only.

This type of strain is called longitudinal strain. In this case, the stress produced is called longitudinal stress. Suppose, initial length = L and increase in length = l, then longitudinal strain = \(\frac{l}{L}\).

Now if applied tensile force = F and area of cross section = A, then longitudinal stress = \(\frac{F}{A}\)

**Question 11. What is Young’s modulus? What are the units of Young’s modulus in CGS system and SI? Young’s modulus is applicable for what type of material?**

**Answer:**

- Within elastic limit, ratio between longitudinal stress and longitudinal strain is called Young’s modulus.
- Units of Young’s modulus in CGS system and SI are dyn/cm
^{2}and N/m^{2}, respectively. - Young’s modulus is applicable for solid materials only.

**Question 12. Establish a mathematical expression for Young’s modulus.**

**Answer:**

Suppose, a thin and long wire of length L and cross-sectional area A is fixed firmly at one end and the other end is loaded with a mass m and hung. Weight of the hanging body,

F = mg is the tensile force here acting along the length of the wire. As a result, l is the increase in length of the wire.

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

And longitudinal stress = \(\frac{F}{A}\)

∴ Young’s modulus, Y = \(\frac{F / A}{l / L}=\frac{F L}{A l}\)

**Question 13. What do you mean by the statement ‘Young’s modulus 1.26 x 10 ^{12} dyn/cm^{2}?**

**Answer:**

Young’s modulus 1.26 x 10^{12} dyn/cm^{2} means that a force of 1.26 x 10^{12} dyn is to be applied to every cm^{2} area of its cross section to bring about a longitudinal strain of unity in a wire made up of copper.

**Question 14. What do you understand by the force constant of a spring? Which property of a spring is measured by this constant?**

**Answer:**

If one side of a spring is firmly fixed to a support and tensile force is applied to the other end, there is an increase in length of the spring. If application of tensile force within elastic limit results in an increase of length x, then F ∝ x or, F = kx.

Here, k is a constant which is called as the force constant of a spring. If x = 1, then F = k.

So, the necessary tensile force required to increase unit length of a spring is called the force constant of the spring.

One can measure the property of stiffness of a spring by the force constant of the spring.

**Question 15. Calculate the dimensional formula of the force constant of a spring and write down the dimension of force constant.**

**Answer:**

Dimensional formula of the force constant of a spring

= \(\frac{\text { dimensional formula of force }}{\text { dimensional formula of length }}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}}=\mathrm{MT}^{-2}\)

So, dimension of the force constant is 1 in mass and -2 in time.

**Question 16. what are the units of force constant of a spring in CGS system and SI? Establish a relationship between them.**

**Answer:**

Units of force constant in CGS system and SI are dyn/cm and N/m, respectively.

\(1 \mathrm{~N} / \mathrm{m}=\frac{10^5 \mathrm{dyn}}{100 \mathrm{~cm}}=1000 \mathrm{dyn} / \mathrm{cm}\)**Question 17. With the help of a simple experiment, discuss how would you measure the force constant of a spring.**

**Answer:**

A spring is hung from a firm support. To the other end of the spring, a hook is attached and a mass hanger is hung from there. A parallel indicator is fixed with the hook. A scale is also fixed vertically by the side of the spring so that the end of the indicator touches the scale.

Now, without any weight in the hanger, reading of the scale is taken. Suppose the reading is 2.5 cm. Now a 10 g weight is placed in the mass hanger. Due to this weight, the spring is elongated along its length.

Suppose, the present reading is 2.7 cm. Applied tensile force on the spring is F= 10 gf = 10 x 980 dyn and due to this, increase of length is x = (2.7 – 2.5)cm = 0.2 cm

∴ force constant of the spring,

\(k=\frac{F}{x}=\frac{10 \times 980}{0.2}=49000 \mathrm{dyn} / \mathrm{cm}\)**Question 18. A spring with force constant k is cut into three equal portions. What is the force constant of each portion?**

**Answer:**

If x is the increase in length of a spring of force constant k when a tensile force F is applied on it, then

F = kx

or, k = \(\frac{F}{x}\)

Now, the spring is cut into three equal pieces. If a tensile force F is applied on any of these three portions, then x_{1 }= \(\frac{x}{3}\), where x_{1 }is the increase in length.

∴ force constant of each portion,

\(k_1=\frac{F}{x_1}=\frac{F}{x / 3}=\frac{3 F}{x}=3 k\)**Question 19. A spring with force constant k is cut in the ratio 1:2. What is the force constant of each portion?**

**Answer:**

If a tensile force F applied on a spring of force constant k produces an increase in length x, then

F = kx

or, k = \(\frac{F}{x}\)

Now, if the spring is cut in the ratio 1: 2 and a tensile force F is applied on each portion, increase in length of the first portion is \(x_1=\frac{x}{1+2}=\frac{x}{3}\) and increase in length of the second portion is \(x_2=\frac{2 x}{1+2}=\frac{2 x}{3}\).

∴ force constant of the first portion

\(k_1=\frac{F}{x_1}=\frac{F}{x / 3}=3 \cdot \frac{F}{x}=3 k\)and force constant of the second portion,

\(k_2=\frac{F}{x_2}=\frac{F}{2 x / 3}=\frac{3}{2} \cdot \frac{F}{x}=1.5 k\)**Question 20. It is possible to manufacture thin and fine ornaments with small amount of gold. or silver whereas it is not possible to manufacture such thin and fine ornaments with equal amount of iron or copper. Why?**

**Answer:**

If the malleability of a material is high, it can be converted into a thin sheet by simple hammering. Gold and silver have greater malleability and ductility as compared to iron or copper.

So, though it is possible to manufacture ornaments by iron or copper, it is not possible to manufacture fine and thin ornaments like those of gold and silver.

**Question 21. If a vessel of glass or ceramic drops accidentally from hand, it breaks into pieces but if a steel glass falls from the same height, it does not break. Which property of material is expressed through the above incidents?**

**Answer:**

If a vessel of glass or ceramic drops from hand accidentally, it breaks into pieces but if a steel glass falls from the same height, it does not break. It is inferred from these two incidents that glass or ceramic is more brittle than steel.

It may be mentioned that though a vessel of glass or ceramic breaks down easily, there is not any significant strain of individual parts. That is why glass or ceramic behaves like a perfectly rigid body to a great extent.

On the other hand, if a steel vessel falls from a height, it may not break but gets deformed, i.e., permanent strain may appear. That is why steel may be considered as a partially elastic body.

**Question 22. To manufacture thin chains, materials like iron, copper, gold, silver, etc. are used but lead is never used. Why?**

**Answer:**

Ductility of materials like iron, copper, gold, silver are much higher than that of lead. So these can be drawn to form sufficiently thin wires. As the ductility of lead is comparatively low, it is not possible to draw it into thin wire.

## Chapter 3 Topic D Elasticity Very Short Answer Type Questions Choose The Correct Answers

Question 1. Which of the following properties is applicable for any type of material?

- Buoyancy
- Surface tension
- Elasticity
- Viscosity

Answer: 3. Elasticity

Question 2. Two wires A and B are made up of the same material. Length of A is greater than that of B. If Young’s modulus of A and B are YA and YB respectively, then

- Y
_{A }= Y_{B} - Y
_{A >}Y_{B} - Y
_{A }< Y_{B} - Cannot be determined

Answer: 1. Y_{A}= Y_{B}

Question 3. As the value of the force constant increases, spring becomes more

- Brittle
- Ductile
- Malleable
- Stiff

Answer: 4. Stiff

Question 4. Dimensional formula of Young’s modulus is equal to which of the following quantities?

- Force
- Momentum
- Stress
- Acceleration

Answer: 3. Stress

Question 5. If temperature increases, value of Young’s modulus also

- Increases
- Decreases
- Increases at first, then decreases
- Decreases at first, then increases

Answer: 2. Decreases

Question 6. Which of the following statements is incorrect?

- Young’s modulus is a measure of the property of elasticity for any material
- Elastic property of a metal changes if some impurity is mixed with the pure metal
- Elastic limit of steel is much higher than that of rubber
- Every material is perfectly elastic upto the elastic limit

Answer: 1. Young’s modulus is a measure of the property of elasticity for any material

Question 7. If% If due to the application of a force F to a spring, there is an increase in x unit length of the spring, then

- \(\frac{F}{x}\) = constant
- Fx = constant
- Fx
^{2}= constant - \(\frac{F}{x^2}\) constant

Answer: 1. \(\frac{F}{x}\) = constant

Question 8. Dimensional formula of stress is

- ML
^{-2}T^{-2} - ML
^{-1}T^{-2} - ML
^{-1}T^{-3} - MT
^{-2}

Answer: 2. ML^{-1}T^{-2}

Question 9. Young’s modulus is the

- Characteristic of solid only
- Characteristic of liquid only
- Characteristic of gas only
- Characteristic of solid, liquid and gas

Answer: 1. Characteristic of solid only

Question 10. The force constant of a spring is 200 N/m. If the spring is divided into two parts, each part has a force constant of

- 200 N/m
- 300 N/m
- 400 N/m
- 100 N/m

Answer: 3. 400 N/m

Question 11. The force constant of a spring is 300 N/m. If the spring is divided into three parts, each part has a force constant of

- 100 N/m
- 300 N/m
- 600 N/m
- 900 N/m

Answer: 4. 900 N/m

Question 12. Dimensional formula of force constant for a spring is

- MT
^{-2} - MLT
^{-2} - ML
^{-1}T^{-2} - MT
^{-3}

Answer: 1. MT^{-2}

Question 13. The Young’s modulus of a wire is Y. If its area of cross section is unity, then the force required to double the length of the wire is

- Y
- Y2
- 2Y
- Y/2

Answer: 1. Y

Question 14. If elastic limit of a material is 10 N, then highest limit of the applied force upto which this material behaves like a perfectly elastic material is

- 5N
- 10N
- 15N
- 20N

Answer: 2. 10N

Question 15. The Young’s modulus of a wire is Y. If its area of cross section is unity, then the fore require to double the length of the wire is

- Y
- Y
^{2} - 2Y
- Y/2

Answer: 1. Y

Question 16. Young’s modulus of a perfectly rigid body is

- 0
- 1
- Infinite
- Depends on stress

Answer: 3. Infinite

Question 17. SI unit of Young’s modulus is

- N • m
^{-2} - J
- dyn
- W

Answer: 1. N • m^{-2}

Question 18. The most elastic among the following substance is

- Rubber
- Steel
- Glass
- Copper

Answer: 2. Steel

## Chapter 3 Topic D Elasticity Answer in Brief

**Question 1. How does a body behave up to its elastic limit?**

**Answer:** A body behaves like a perfectly elastic body up to its elastic limit.

**Question 2. If there is no strain of a body due to an external balanced force of any magnitude, what do we call the body?**

**Answer:** The body is called a perfectly solid body.

**Question 3. In between stress and strain, which one is fundamental?**

**Answer:** In between stress and strain, strain is fundamental.

**Question 4. Stress is not fundamental but strain is fundamental. Why?**

**Answer:** If strain is generated, then only stress is evolved. So, in between stress and strain, strain is fundamental.

**Question 5. Does liquid and gaseous materials have Young’s modulus?**

**Answer:** Only solid materials have length and as Young’s modulus is a measure of the elastic property of solid materials, so there is no existence of Young’s modulus for liquid and gaseous materials.

**Question 6. Is there any change in Young’s modulus with increasing temperature?**

**Answer:** The value of Young’s modulus is reduced if temperature is increased.

**Question 7. What is the value of Young’s modulus for a perfectly solid body?**

**Answer:** Value of Young’s modulus for a perfectly solid body is infinite.

**Question 8. It is possible to draw a thin wire of iron but it is not possible to do the same for lead due to which property of a material?**

**Answer:** This is due to the property of ductility of a material.

**Question 9. Which property of a material is more prevalent in gold than in iron so that it is possible to manufacture fine gold ornaments but not iron ornaments?**

**Answer:** Malleability of gold is more than that of iron.

**Question 10. Are the values of Young’s modulus for a thin and a thick iron wire of the same length different?**

**Answer:** No, the values of Young’s modulus of both the values are the same because both are manufactured from the same material.

**Question 11. State whether the values of Young’s moduli for thin and thick iron wires of different length will be different.**

**Answer:** Young’s moduli of two iron wires of different length and of different thickness cannot be different because Young’s modulus only depends on the nature of material of the wire.

## Chapter 3 Topic D Elasticity Fill In The Blanks

Question 1. Within the _______ limit, stress is directly proportional to strain.

Answer: Elastic

Question 2. Elasticity of steel is _______ than that of rubber.

Answer: More

Question 3. Coefficient of elasticity is defined as the ratio of the applied ________ to the change in shape of an elastic body.

Answer: Stress

Question 4. The minimum value of stress required to break a wire is called the __________ of that wire.

Answer: Breaking stress

Question 5. _________ is the reaction force that is produced per unit area of cross section of any material due to the application of an external force.

Answer: Stress

Question 6. Brittleness of glass is _______ than that of iron.

Answer: More

Question 7. More the _______ of a material, greater is the possibility of converting it to a thinner sheet by hammering.

Answer: Malleability

Question 8. k is the force constant of a spring. If the spring is cut into three equal parts, force constant of each part becomes __________

Answer: 3k

## Chapter 3 Topic D Elasticity State Whether True Or False

Question 1. Required tensile force to increase the length of spring by unity is called force constant and its unit in SI is N/m^{2}.

Answer: False

Question 2. Within elastic limit, stress is directly proportional to strain.

Answer: True

Question 3. Young’s modulus of a perfectly rigid body is finite.

Answer: False

Question 4. Rubber is more elastic than steel.

Answer: False

Question 5. In the case of an elastic body strain is more fundamental than stress.

Answer: False

Question 6. Required external force to increase the length of a spring by unity is called force constant.

Answer: True

Question 7. Glass is a brittle substance.

Answer: True

Question 8. The elasticity of a material is decreased on hammering it.

Answer: True

Question 9. Strain has no unit.

Answer: True

## Chapter 3 Topic D Elasticity Numerical Examples

**Useful information**

Stress(S) = \(\frac{\text { applied force }(F)}{\text { area of cross section }(A)}\)

Strain = \(\frac{\text { change in length }(I)}{\text { initial length }(L)}\)

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

Suppose a wire of length L and cross sectional area A suspended from a rigid support. A mass m is hung from its lower end. The increase in length of the wire is l.

∴ Young’s modulus, \(Y=\frac{F / A}{I / L}=\frac{m g L}{A l}\)

One end of a spring is fixed to a rigid support and a force F is applied at its other end.

If the spring elongates by a length x then, force constant of the spring, k = \(\frac{F}{x}\)

- If the spring is cut into two equal parts then force constant of each part is k’ = 2k.
- If the spring is cut into three equal parts then force constant of each part is k” = 3k.

If two massless springs of force constants k_{1} and k_{2} respectively are joined in series combination then the equivalent spring constant of the combination is k = \(\frac{k_1 k_2}{k_1+k_2}\)

If these two are joined in parallel combination then the equivalent spring constant of the combination is k = k_{1} + k_{2}

**Question 1. If a mass of 4 kg is put on a wire of length 1 m and cross-section 1 mm ^{2}, it extends by 0.2 mm. What is the value of Young’s modulus of the material of the wire?**

**Answer:**

Length of wire, L = 1 m = 100 cm,

area of cross-section, A = 1 mm^{2} = 1 x 10^{-2} cm^{2}

applied tensile force, F = 4 x 9.8 x 10^{5} dyn

extension of wire, l = 0.2 mm = 0.02 cm

∴ Young’s modulus of the material of wire,

Y = \(\frac{F L}{A l}=\frac{4 \times 9.8 \times 10^5 \times 100}{1 \times 10^{-2} \times 0.02}\)

= 1.96 x 10^{12} dyn/cm^{2}

**Question 2. A mass of 8 kg is hung on a metallic wire of length 2 m and cross-section 1 mm ^{2}. If Young’s modulus of the material of the metal is Y = 2 x 10^{12} dyn/cm^{2}, what is the increase of length of this wire? [ g = 10 m/s^{2}]**

**Answer:**

Initial length of the wire, L = 2 m = 200 cm.

Area of cross-section, A = 1 mm^{2} = 1 x 10^{-2} cm^{2}

Tensile force, F = 8 x 10 N = 8 x 10 x 10^{5} dyn

Suppose, increase of length of the wire =l

So, Young’s modulus of the material of the wire,

Y = \(\frac{F L}{A l}\)

∴ l = \(\frac{F L}{A Y}=\frac{8 \times 10 \times 10^5 \times 200}{1 \times 10^{-2} \times 2 \times 10^{12}}\)

= 0.08 cm = 0.8 mm

**Question 3. If a tensile force of 3 N is applied on a spring, there is an increase in length of the spring by 1 cm. Calculate the force constant of the spring.**

**Answer:**

Tensile force applied on the spring, F = 3N;

Increase of length of the spring, x = 1 cm = 0.01 m

So the force constant of the spring,

k = \(\frac{F}{x}\) = \(\frac{3}{0.01}\) = 300 N/m

**Question 4. A metallic wire of length 2 m and Young’s modulus Y = 2 x 10 ^{11} N • m^{-2} is extended by 5 mm in length by applying a longitudinal force. Find the stress produced in the wire.**

**Answer:**

Young modulus of the wire, Y = \(\frac{\text { stress }}{1 / L}\)

Here, l = 5 mm = 0.005 m and (L) = 2m

∴ The amount of stress produced in the wire

= \(Y \times\left(\frac{I}{L}\right)=2 \times 10^{11} \times \frac{0.005}{2}\)

= 5 x 10^{8} N/m^{2}

**Question 5. When a mass of 8 kg is hung from the low end of a spring. It elongate by 2 cm. Find the force constant of the spring. [ g = 10 m/s ^{2}]**

**Answer:**

Here, force applied on the spring (F) = 8 x 10 = 80 N

and elongation of the spring (x) = 2 cm = 0.02 m

∴ Force constant of the spring

(k) = \(\frac{F}{x}=\frac{80}{0.02}\) = 4000 N • m^{-1}

**Question 6. Young’s moduli of two rods of equal length and equal area of cross section are Y _{1} and Y_{2} respectively. If the rods are joined end to end, prove that the equivalent Young’s modulus of the combined rod is \(\frac{2 Y_1 Y_2}{Y_1+Y_2}\)**

**Answer:**

Suppose, length and area of cross section of each rod are L and 4 respectively. One end of the combined rod is attached to a rigid support and F force is applied to the other end. Let l_{1} and l_{2} be the increase in length of the two rods respectively.

Now, \(Y_1=\frac{F / A}{I_1 L} \text { and } Y_2=\frac{F / A}{I_2 / L}\)

or, \(l_1=\frac{F L}{A Y_1} \text { and } l_2=\frac{F L}{A Y_2}\)

∴ Total change in length of the combined rod \(l =l_1+l_2\)

= \(\frac{F L}{A Y_1}+\frac{F L}{A Y_2}=\frac{F L}{A}\left(\frac{1}{Y_1}+\frac{1}{Y_2}\right)\)

= \(\frac{F L}{A}\left(\frac{Y_1+Y_2}{Y_1 Y_2}\right)\)

For the combined rod, Young s modulus

Y = \(\frac{\frac{F}{A}}{\frac{1}{2 L}}=\frac{2 F L}{A} \times\left(\frac{Y_1 Y_2}{Y_1+Y_2}\right) \times \frac{A}{F L}=\frac{2 Y_1 Y_2}{Y_1+Y_2}\)

**Question 7. The cross-sectional area of a steel wire is 1 cm ^{2}. How much force is required to increase its length to twice its Initial length? Young’s modulus for steel is 2×10^{12} dyn • cm^{-2}.**

**Answer:**

Let us consider that the initial length of the rod is L and the applied force is F.

Here, A = 1 cm^{2} and l = L

Young’s modulus of the wire

Y = \(\frac{F / A}{I / L}\)

∴ \(2 \times 10^{12}=\frac{F \times 1}{L \times L}\)

or, F = 2 x 10^{12} dyn

∴ The required force is 2 x 10^{12} dyn.

**Match The Columns**

**1.**

**Answer:** 1. B, 2. C, 3. D, 4. A

**2.**

**Answer:** 1.D, 2. C, 3. A, 4. B