RBSE Solutions Class 11 Physics Chapter 10 Properties of Material Substances

Class 11 Physics Chapter 10 Textbook Exercises With Solutions

Class 11 Physics Chapter 10 Very Short Answer Type Questions

 

Question 1. What is the force called by applying which the body changes its shape and size?
Answer: The force that causes a body to change its shape and size is known as a deforming force. This force makes an object deform from its original state.
In simple words: A force that makes something change its shape or size is called a deforming force.

🎯 Exam Tip: Remember that a deforming force is responsible for changing an object's physical dimensions or form, distinguishing it from forces that only cause motion.

 

Question 2. What is the property called by which the body regains its original shape and size after the removal of deforming force.
Answer: The property that allows a body to return to its original shape and size once the deforming force is removed is called elasticity. This is a fundamental property of many materials.
In simple words: The ability of a body to return to its first shape after a force is taken away is called elasticity.

🎯 Exam Tip: Always associate elasticity with the ability of a material to recover its original form after being stretched or compressed.

 

Question 4. Write the dimension formula of strain.
Answer: Strain is a ratio of two similar physical quantities, which means it has the same dimension in both the numerator and denominator. Because of this, strain is a dimensionless quantity. Therefore, its dimensional formula is \( [M^0L^0T^0] \). It does not depend on mass, length, or time.
In simple words: Strain has no dimensions because it is a ratio of two things that have the same unit. So, its formula is \( [M^0L^0T^0] \).

🎯 Exam Tip: When writing dimensional formulas, always ensure that dimensionless quantities are correctly represented with zero powers for mass, length, and time.

 

Question 5. Write down the modulus of elasticity of copper, steel, glass and rubber in ascending order.
Answer: To arrange the modulus of elasticity for rubber, glass, copper, and steel in ascending order (from smallest to largest), we consider their stiffness. Rubber is the least stiff, followed by glass, then copper, and finally steel, which is the stiffest among them. Therefore, the ascending order is: Rubber, Glass, Copper, Steel.
In simple words: From least elastic to most elastic, the order is rubber, glass, copper, steel.

🎯 Exam Tip: Remember that a higher modulus of elasticity means a material is stiffer and requires more force to deform, while a lower modulus means it's more easily deformable.

 

Question 6. What is the ratio of stress to strain called within the limit of elasticity?
Answer: Within the elastic limit, the ratio of stress to strain for any specific material is known as the modulus of elasticity. This value indicates how much a material resists deformation when a force is applied. It is a fundamental property used to characterize material stiffness.
In simple words: The modulus of elasticity is what we call the ratio of stress to strain when a material is still elastic.

🎯 Exam Tip: Make sure to specify "within the limit of elasticity" as Hooke's Law and the modulus of elasticity only apply up to this point.

 

Question 7. What is the unit of modulus of elasticity?
Answer: The modulus of elasticity is given by the formula \( \text{Modulus of elasticity} = \frac{\text{Stress}}{\text{Strain}} \). Since strain is a dimensionless quantity (it has no units), the unit of modulus of elasticity is the same as the unit of stress. Stress is defined as force per unit area, so its unit is Newton per square meter \( (\text{Nm}^{-2}) \).
In simple words: The unit for modulus of elasticity is the same as for stress, which is Newtons per square meter \( (\text{Nm}^{-2}) \).

🎯 Exam Tip: When determining units, always break down the formula to its base quantities and remember that dimensionless quantities do not affect the overall unit.

 

Question 8. What is the ratio of lateral strain to longitudinal strain called?
Answer: The ratio of lateral strain to longitudinal strain is known as Poisson's ratio. This ratio describes how a material expands or contracts perpendicular to the direction of an applied force. It is a very important material property that shows how materials respond to stretching or compressing forces.
In simple words: The ratio of how much a material changes width compared to how much it changes length is called Poisson's ratio.

🎯 Exam Tip: Clearly define "lateral strain" (change perpendicular to force) and "longitudinal strain" (change parallel to force) when discussing Poisson's ratio.

 

Question 11. When we stretch a wire, why do we have to do work?
Answer: When a wire is stretched, internal reactive forces come into play that oppose the stretching action. We have to do work to overcome these opposing internal forces. This work is not lost but is stored within the wire as elastic potential energy. This stored energy allows the wire to return to its original state once the stretching force is removed.
In simple words: We do work to stretch a wire because internal forces in the wire pull back. This work turns into stored elastic energy.

🎯 Exam Tip: Emphasize the concept of "internal reactive force" and "elastic energy storage" to explain why work is required for stretching.

 

Question 12. What is the dimensional formula of modulus of rigidity?
Answer: The modulus of rigidity, like other moduli of elasticity, represents the ratio of stress to strain. Specifically, it is the ratio of shearing stress to shearing strain. Since strain is dimensionless, the dimensional formula for modulus of rigidity is the same as that for stress, which is \( M^1 L^{-1} T^{-2} \).
In simple words: The dimensional formula for modulus of rigidity is \( M^1 L^{-1} T^{-2} \), just like for stress.

🎯 Exam Tip: Remember that all moduli of elasticity (Young's, Bulk, Rigidity) share the same dimensional formula as stress.

Class 11 Physics Chapter 10 Short Answer Type Questions

 

Question 1. What do you understand by elasticity?
Answer: Elasticity is the property of a body that allows it to resist changes in its shape or size caused by a deforming force, and to return to its original shape and size once that force is removed.
Perfectly elastic bodies are those that regain their original shape and size immediately and completely after the deforming force is removed. For example, quartz fibers are nearly perfectly elastic.
Perfectly plastic bodies, on the other hand, do not regain their original shape or size at all after the deforming force is removed. Putty and wax are examples of nearly perfectly plastic bodies.
In reality, most bodies fall somewhere between these two extremes, exhibiting both elastic and plastic properties to varying degrees.
In simple words: Elasticity is the ability of a material to bounce back to its original shape after being stretched or squeezed. If it bounces back completely, it's perfectly elastic; if it doesn't at all, it's perfectly plastic.

🎯 Exam Tip: Clearly distinguish between perfectly elastic and perfectly plastic bodies, giving examples for each, and state that most real-world materials show a mix of both properties.

 

Question 2. What are restitution forces?
Answer: A substance naturally stays in its original, balanced state due to internal forces between its atoms. When an outside force is applied, the substance changes shape or size, or both. This change creates internal reactive forces within the substance. When the outside forces are removed, these internal reactive forces cause the substance to return to its original shape and size. These internal reactive forces are called restoring forces or restitution forces. They always act in the opposite direction to the deforming force. When the substance is stable again, the restoring force per unit area, caused by strain, is called stress.
In simple words: Restitution forces are the internal forces inside a material that try to bring it back to its original shape and size after an outside force has changed it.

🎯 Exam Tip: Focus on explaining that restitution forces are internal, reactive forces that oppose deformation and work to restore the material's original configuration.

 

Question 4. What is strain?
Answer: Strain is the measure of deformation an object undergoes when a deforming force is applied. When an external force acts on an object, its length, volume, or shape may change. The proportional change in the object's shape and size due to this force is called strain. It is a dimensionless quantity because it is a ratio of two similar lengths or volumes.
Based on how an object is deformed, there are three main types of strain:
(a) Longitudinal strain (change in length)
(b) Volume strain (change in volume)
(c) Shearing strain (change in shape or angle)
In simple words: Strain is how much an object changes its shape or size compared to its original shape when a force pushes or pulls it. It has no units and can be longitudinal, volume, or shearing.

🎯 Exam Tip: Remember to define strain as a proportional change and mention its dimensionless nature, then list and briefly explain the three main types.

 

Question 5. What do you understand by limit of elasticity?
Answer: The limit of elasticity is the maximum deforming force a material can withstand while still behaving as a perfectly elastic body. This means that if the applied force is within this limit, the object will fully return to its original shape and size once the force is removed. However, if the deforming force goes beyond this elastic limit, the material will undergo permanent deformation, meaning it will not completely return to its original shape, even after the force is gone. This limit is crucial for engineering and material science applications.
In simple words: The limit of elasticity is the biggest force a material can handle and still go back to its original shape perfectly. If the force is stronger, the material changes forever.

🎯 Exam Tip: Clearly state that beyond the elastic limit, a material experiences permanent deformation, which is a key concept in material behavior.

 

Question 6. What is Poisson's ratio?
Answer: Poisson's ratio, represented by \( \sigma \), is defined as the ratio between the lateral strain and the longitudinal strain within a material's elasticity limit. When a material is stretched (longitudinal strain), it tends to become thinner (lateral strain). This ratio quantifies that lateral contraction relative to the longitudinal extension. It's an important property that describes the bulk behavior of a material.
\( \text{Poisson's ratio } \sigma = \frac{\text{Lateral strain}}{\text{Longitudinal strain}} = \frac{\beta}{\alpha} = \frac{d/D}{l/L} = \frac{Ld}{lD} \)
In simple words: Poisson's ratio tells us how much a material gets thinner when it's stretched longer. It's the ratio of how much it changes in width to how much it changes in length.

🎯 Exam Tip: Always remember that Poisson's ratio describes the material's response perpendicular to the applied stress, and it is a dimensionless quantity.

 

Question 8. What do you understand by the rigidity of a solid?
Answer: The rigidity of a solid refers to its resistance to changes in shape when a tangential or shearing force is applied. Materials with high rigidity are stiff and do not easily deform their shape. For solids, the modulus of elasticity (which includes Young's modulus, bulk modulus, and shear modulus) is generally very high, making them less compressible and more rigid. This means a greater force is needed to cause even a small change in their shape or length. For instance, steel is known to be more rigid (and elastic) than materials like copper, brass, or aluminum, meaning it resists shape changes more effectively. Rigidity is a measure of a solid's stiffness against shearing forces.
In simple words: Rigidity is how much a solid material resists changing its shape when pushed sideways. Stiff materials like steel have high rigidity, meaning they don't easily deform.

🎯 Exam Tip: When explaining rigidity, focus on resistance to *shape* change (shear deformation) rather than just length or volume change, and connect it to the shear modulus.

 

Question 9. What do you understand by the perfectly elastic, plastic and rigid bodies? Discuss their limits.
Answer: We can understand perfectly elastic, plastic, and rigid bodies as follows:
* **Perfectly Elastic Bodies:** These are bodies that completely and instantly return to their original shape and size once the deforming force is removed. An ideal rubber band or a quartz fiber can behave almost like a perfectly elastic body. The "limit" here refers to the elastic limit, beyond which they would deform permanently.
* **Perfectly Plastic Bodies:** These bodies do not regain any of their original shape or size after the deforming force is removed. They undergo permanent deformation. Putty, clay, or wax are examples of materials that are nearly perfectly plastic. They do not have an elastic limit in the same sense as elastic bodies because they deform permanently from the start.
* **Rigid Bodies:** These bodies are theoretical constructs that do not deform at all, or show very minimal deformation, when an external force is applied. They maintain their shape and size perfectly. In reality, no body is perfectly rigid, but very hard materials like diamond or steel can be considered nearly rigid under typical forces.
Most real-world materials exist on a spectrum between perfectly elastic and perfectly plastic, with various degrees of elasticity and plasticity. For example, some materials might be highly elastic up to a point, then become plastic.
In simple words: Perfectly elastic bodies snap back to their original shape, perfectly plastic bodies stay deformed, and rigid bodies don't deform at all. Real objects are usually a mix of elastic and plastic, and no object is truly rigid.

🎯 Exam Tip: Define each type of body clearly and provide common examples. Highlight that ideal perfectly elastic, plastic, or rigid bodies are theoretical and real materials exhibit a combination of these properties.

Class 11 Physics Chapter 10 Long Answer Type Questions

 

Question 1. Explain stress, strain, limit of elasticity and describe various strains produced in a material.
Answer:
**Stress:** When an external force is applied to a substance, it causes an internal reactive force within the material. This internal reactive force per unit cross-sectional area, which arises due to the deformation (strain), is called stress. It opposes the external deforming force and is equal in magnitude but opposite in direction to the external force when the body is in equilibrium.
The formula for stress is: \( \text{Stress} = F/A \), where F is the applied force and A is the cross-sectional area.
The unit of stress in the M.K.S. system is \( \text{N.m}^{-2} \) (Newton per square meter), and its dimensional formula is \( [M^1L^{-1}T^{-2}] \).
There are three main types of stress, corresponding to the types of deformation:
(a) **Longitudinal Stress:** This occurs when a deforming force is applied along the length of an object, like a cylinder. The reactive force acting on the longitudinal cross-sectional area is called longitudinal stress. If the length increases (due to pulling), it's called tensile stress. If the length decreases (due to pushing), it's called compressive stress.
\( \text{Longitudinal stress} = \frac{\text{Force}}{\text{Area}} = \frac{F}{A} \) \( [\text{Nm}^{-2}] \)
(b) **Normal (Volume) Stress:** This occurs when a uniform force is applied perpendicularly to all surfaces of an object. This causes a change in the object's volume but not its shape. The reactive normal force per unit area is called normal stress.
(c) **Tangential (Shearing) Stress:** This occurs when one surface of a body is held fixed, and a tangential force is applied to an opposite surface. This causes the body to deform in shape, resulting in a shearing angle. The tangential reactive force applied per unit area is called tangential stress.
\( \text{Tangential Stress} = \frac{F}{A} \) \( [\text{N.m}^{-2}] \)
**Strain:** Strain is the measure of the deformation an object undergoes relative to its original dimensions when a deforming force is applied. It is a dimensionless quantity.
There are three primary types of strain:
(a) **Longitudinal Strain:** This is the change in unit length or the proportional change in the length of an object due to an external force.
\[ \text{Longitudinal strain} = \frac{\text{Change in length}}{\text{Original length}} = \frac{l}{L} \]
If the length increases, it is called tensile strain; if it decreases, it is called compressive strain.
(b) **Volume Strain:** This is the change in unit volume or the proportional change in the volume of an object due to an external force applied uniformly over its surface.
\[ \text{Volume Strain} = \frac{\text{Change in volume}}{\text{Original volume}} = \frac{\Delta V}{V} \]
Volume strain is generally less in solids and liquids compared to gases.
(c) **Shearing Strain:** This occurs when a tangential force causes a change only in the shape (or angle) of an object, while its volume remains unchanged. The shearing angle, \( \Phi \), is the measure of this strain.
\[ \text{Shearing strain} = \text{Shearing angle} = \Phi = \frac{l}{L} \]
**Limit of Elasticity:** This is the maximum value of the deforming force a material can withstand and still fully return to its original shape and size once the force is removed. If the applied force exceeds this limit, the material will undergo permanent deformation and will not recover its original state.
In simple words: Stress is the internal force per area trying to resist change. Strain is how much a material changes its shape or size compared to its original. The limit of elasticity is the point up to which a material can return to its original shape. Strains can be about changing length (longitudinal), volume (volume), or shape (shearing).

🎯 Exam Tip: When explaining these terms, define each clearly with its formula and units (where applicable). For strains, differentiate between their types and provide the relevant formulas.

 

Question 2. Explain elasticity, limit of elasticity, yield point and breaking point.
Answer: To understand elasticity, limit of elasticity, yield point, and breaking point, we can look at a stress-strain curve for a material under tensile stress (Figure 10.6, which is a graph showing how stress changes with strain).
When stress is applied to a material, it produces strain.
* **Elasticity and Elastic Limit:** From point O to E on the stress-strain curve, the relationship between stress and strain is linear, meaning the material follows Hooke's Law. In this region, the material behaves elastically, and if the applied force is removed, it returns to its original position. The material's ability to return to its original shape is its elasticity. The elastic limit (also called the proportional limit) is the maximum stress a material can endure without permanent deformation.
* **Yield Point:** Beyond point E, the curve becomes non-linear, meaning stress is no longer directly proportional to strain. However, the material still possesses elasticity up to a point Y, known as the yield point or elastic limit. If the stress exceeds the yield point, the material begins to deform permanently, even after the load is removed.
* **Plastic Deformation:** After the yield point (Y), if the weight or force on the wire increases, its length grows rapidly. Even if the weight is removed, the wire does not return to its original length, showing permanent deformation, also called plastic deformation. This occurs in the section of the curve between Y and U.
* **Breaking Point (Fracture Point):** Point U on the graph represents the maximum tensile strength the material can withstand. After this point, any additional strain, even with reduced applied force, leads to a rapid increase in deformation, and the material eventually breaks at point B. This point B is called the breaking point or fracture point. If the maximum tensile limit (U) and breaking point (B) are close, the material is considered brittle; if they are far apart, it is ductile.
Sometimes, a material might not return to its original position immediately after the elastic limit but does so after some time. This time delay is known as elastic relaxation time. Also, sometimes the strain lags behind the applied force, a phenomenon called Elastic Hysteresis, which leads to energy loss as heat.
In simple words: Elasticity is how a material returns to its shape. The elastic limit is the maximum force it can take and still go back to normal. The yield point is when it starts to change shape permanently. The breaking point is when it finally snaps or breaks.

🎯 Exam Tip: When describing these points, refer to the stress-strain curve and explain what happens to the material at each stage, especially the difference between elastic and plastic deformation.

 

Question 3. Explain stress, strain and modulus of elasticity and define Young's modulus of elasticity.
Answer:
**Stress:** A substance normally stays in its original, balanced state because of forces between its atoms. When an external force is applied, the substance changes in shape, size, or both. This change creates an internal reactive force. When the external force is removed, these internal reactive forces help the substance regain its original shape. These internal reactive forces are equal in strength but opposite in direction to the external forces. In a balanced state, the internal reactive force (or restoring force) per unit of cross-sectional area, caused by strain, is called stress. Its unit is \( \text{Nm}^{-2} \) and dimensional formula is \( [M^1L^{-1}T^{-2}] \).
**Strain:** Strain is a measure of how much an object deforms when a force is applied, relative to its original size. It is the ratio of the change in dimension to the original dimension. Strain is a dimensionless quantity.
\[ \text{Longitudinal Strain} = \frac{\text{Change in length}}{\text{Original length}} = \frac{l}{L} \]
**Modulus of Elasticity:** The modulus of elasticity is a measure of a material's stiffness or resistance to elastic deformation. It is defined as the ratio of stress to strain within the elastic limit. All moduli of elasticity (Young's, Bulk, and Shear) share the same unit as stress, which is \( \text{Nm}^{-2} \).
**Young's Modulus of Elasticity (\(Y\)):** Young's modulus specifically measures the stiffness of a solid material when stretched or compressed. It is defined as the ratio of longitudinal stress to longitudinal strain within the elastic limit. It is represented by \( Y \).
\[ Y = \frac{\text{Longitudinal stress}}{\text{Longitudinal strain}} \]
If a tensile force \( F \) is applied to a wire with initial length \( L \) and cross-sectional area \( A \), causing a change in length \( l \), then:
\[ \text{Longitudinal Strain} = \frac{\text{Change in length}}{\text{Original length}} = \frac{l}{L} \]
\[ \text{and Longitudinal stress} = \frac{\text{Deforming force}}{\text{Area of cross section}} = \frac{F}{A} \]
Therefore, Young's modulus of elasticity is:
\[ Y = \frac{F/A}{l/L} = \frac{FL}{Al} \]
If \( r \) is the radius of the wire and the mass \( M \) is hanging on it (so force \( F=Mg \)), and the cross-sectional area \( A=\pi r^2 \), then:
\[ Y = \frac{MgL}{\pi r^2 l} \]
In the MKS system, the unit of Young's modulus is \( \text{Nm}^{-2} \). Young's modulus is a value calculated only for solids and is a key characteristic of solid materials.
In simple words: Stress is the internal force a material uses to resist outside forces. Strain is how much a material stretches or changes shape. Modulus of elasticity shows how stiff a material is by comparing stress to strain. Young's modulus specifically measures this stiffness for stretching or compressing a solid material.

🎯 Exam Tip: Provide clear definitions and formulas for stress, strain, and Young's modulus. Highlight that Young's modulus is specific to longitudinal deformation and is a property of solids.

 

Question 4. What is Poisson's ratio? Define its limits.
Answer: When a force is applied along the length of a wire, causing it to stretch (increase in length), it also causes the wire to become thinner in the perpendicular direction (decrease in width).
**Poisson's ratio**, denoted by \( \sigma \), is defined as the ratio between the lateral strain (change in width per original width) and the longitudinal strain (change in length per original length), within the elastic limit of the material.
\[ \text{Poisson's ratio } \sigma = \frac{\text{Lateral strain}}{\text{Longitudinal strain}} = \frac{\beta}{\alpha} \]
Here, \( \alpha \) represents the longitudinal strain and \( \beta \) represents the transverse or lateral strain.
\[ \alpha = \frac{\text{Change in length in the direction of force}}{\text{Original length in the direction of force}} \]
\[ \beta = \frac{\text{Change in width in the perpendicular direction of force}}{\text{Original width in the perpendicular direction of force}} \]
Since Poisson's ratio is a ratio of two strains, it is a dimensionless quantity.
**Limits of Poisson's Ratio:**
Generally, the theoretical values of Poisson's ratio range from -1 to +0.5. However, for most common materials, its practical value typically falls between 0.2 and 0.4. For example, steel has a Poisson's ratio of approximately 0.19, copper is around 0.32, and brass is about 0.26. It is a critical material property that shows how much a material will change in width when stretched or compressed.
In simple words: Poisson's ratio shows how much a material gets thinner sideways when you stretch it lengthwise. It's the ratio of side change to length change. This ratio is usually between 0.2 and 0.4 for most real materials.

🎯 Exam Tip: Clearly state the definition and formula for Poisson's ratio. Emphasize that it is dimensionless and explain the typical range of values for common materials.

 

Question 5. Define Young's modulus of elasticity, modulus of rigidity and Bulk modulus of elasticity. Calculate Young's; modulus of elasticity by Searle's experiment method.
Answer:
**Young's Modulus of Elasticity (\(Y\)):** This measures a material's resistance to change in length under tensile or compressive stress. It is defined as the ratio of longitudinal stress to longitudinal strain within the elastic limit.
\[ Y = \frac{\text{Longitudinal stress}}{\text{Longitudinal strain}} \]
If a tensile force \( F \) is applied to a wire with initial length \( L \) and cross-sectional area \( A \), causing a change in length \( l \), then:
\[ \text{Longitudinal Strain} = \frac{l}{L} \quad \text{and} \quad \text{Longitudinal stress} = \frac{F}{A} \]
Therefore, Young's modulus is \( Y = \frac{F/A}{l/L} = \frac{FL}{Al} \).
If \( r \) is the radius of the wire and a mass \( M \) is hanging on it (so \( F = Mg \)), and \( A = \pi r^2 \), then \( Y = \frac{MgL}{\pi r^2 l} \). The unit for \( Y \) is \( \text{Nm}^{-2} \).
**Bulk Modulus of Elasticity (\(K\)):** This measures a material's resistance to change in volume under uniform hydrostatic pressure. It is defined as the ratio of normal stress (pressure) to volume strain. When pressure is applied uniformly on all points of an object's surface, its volume changes, but its shape does not.
\[ K = \frac{\text{Normal stress}}{\text{Volume strain}} \]
If \( F \) is the normal deforming force applied uniformly on surface area \( A \), causing a change in volume \( \Delta V \) from an original volume \( V \), then \( K = \frac{F/A}{\Delta V/V} = \frac{PV}{\Delta V} \). The inverse of bulk modulus is called compressibility (\( \frac{1}{K} \)).
**Modulus of Rigidity (\( \eta \) or \( G \)):** This measures a material's resistance to shearing deformation (change in shape) under tangential stress. It is defined as the ratio of tangential stress to shearing strain within the elastic limit.
\[ \eta = \frac{\text{Tangential stress}}{\text{Shearing strain}} \]
If a tangential force \( F \) is applied to an area \( A \) causing a shearing angle \( \Phi \), then \( \eta = \frac{F/A}{\Phi} \). The unit for \( \eta \) is also \( \text{Nm}^{-2} \).
**Searle's Experiment to Calculate Young's Modulus:**
Searle's apparatus uses two wires, A and B, of the same material and length, hung from a rigid base. Wire A is called the reference wire, and wire B is the experimental wire. A spirit level is attached to a frame that connects both wires. A dead load is attached to wire A to keep it taut. A hanger is attached to wire B, onto which increasing weights can be added.
Initially, the spherometer reading is taken when the spirit level is horizontal. As weights are added to wire B, it stretches, and the spirit level tilts. The spherometer screw is then rotated to bring the spirit level back to a horizontal position. The difference in spherometer readings gives the increase in length (\( l \)) of wire B.
By taking different weights (\( M \)) and measuring the corresponding increase in length (\( l \)), a graph of \( M \) versus \( l \) can be plotted (Fig. 10.9). This graph is typically a straight line within the elastic limit.
If \( L \) is the initial length of wire B, \( r \) is its radius, and \( M \) is the applied mass (so \( F = Mg \)), then Young's modulus \( Y \) can be calculated as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{l/L} = \frac{MgL}{\pi r^2 l} \]
The values for \( M/l \) can be found from the slope of the graph, \( r \) can be measured using a screw gauge, and \( L \) using a meter scale. Substituting these values into the formula gives Young's modulus. This experiment is a reliable way to determine Young's modulus for a solid material.
In simple words: Young's modulus shows how much a material stretches or compresses. Bulk modulus shows how much its volume changes under pressure. Modulus of rigidity shows how much its shape changes when pushed sideways. Searle's experiment uses two wires and a scale to measure how much a wire stretches under weight to find Young's modulus.

🎯 Exam Tip: When defining each modulus, clearly state what type of stress and strain it relates to. For Searle's experiment, describe the setup, how measurements are taken, and how the Young's modulus is calculated from the collected data.

Class 11 Physics Chapter 10 Numerical Questions

 

Question 1. Tn a wire by producing \( 4 \times 10^{-4} \) linear strain a \( 4.8 \times 10^7 \text{ Nm}^{-2} \) stress is produced. Calculate the Young's modulus of elasticity of the material.
Answer:
Given:
Linear strain \( = 4 \times 10^{-4} \)
Stress \( = 4.8 \times 10^7 \text{ Nm}^{-2} \)
To calculate Young's modulus of elasticity, we use the formula:
\[ Y = \frac{\text{Longitudinal stress}}{\text{Longitudinal strain}} \]
Now, substitute the given values into the formula:
\[ Y = \frac{4.8 \times 10^7}{4 \times 10^{-4}} \]
Perform the division:
\[ Y = \frac{4.8}{4} \times \frac{10^7}{10^{-4}} \]
\[ Y = 1.2 \times 10^{7 - (-4)} \]
\[ Y = 1.2 \times 10^{11} \text{ Nm}^{-2} \]
The Young's modulus of elasticity for the material is \( 1.2 \times 10^{11} \text{ Nm}^{-2} \). This high value indicates a stiff material.
In simple words: We are given how much stress (force per area) is applied and how much the material stretches (strain). To find Young's modulus, we divide the stress by the strain. The answer is \( 1.2 \times 10^{11} \text{ Nm}^{-2} \).

🎯 Exam Tip: Always state the formula clearly before substituting values and remember to include the correct units in the final answer.

 

Question 2. The average depth of Indian Ocean is 3 km. Calculate \( \frac{\Delta V}{V} \) given that the Bulk modulus of elasticity for water is \( 2.2 \times 10^9 \text{ Nm}^{-2} \), (g = 10 m/s²).
Answer:
Given values:
Average depth of Indian Ocean, \( h = 3 \text{ km} = 3 \times 10^3 \text{ m} \)
Density of water, \( \rho = 1.0 \times 10^3 \text{ kg/m}^3 \)
Bulk modulus of elasticity for water, \( K = 2.2 \times 10^9 \text{ Nm}^{-2} \)
Acceleration due to gravity, \( g = 10 \text{ m/s}^2 \)
First, calculate the pressure at the given depth:
\( \text{Pressure } P = h \rho g \)
\( P = (3 \times 10^3 \text{ m}) \times (1.0 \times 10^3 \text{ kg/m}^3) \times (10 \text{ m/s}^2) \)
\( P = 3 \times 10^7 \text{ Nm}^{-2} \)
Next, use the formula for Bulk modulus:
\[ K = \frac{\text{Pressure}}{\text{Volume strain}} = \frac{P}{\Delta V/V} \]
We need to find the fractional change in volume, \( \frac{\Delta V}{V} \). Rearrange the formula:
\[ \frac{\Delta V}{V} = \frac{P}{K} \]
Substitute the calculated pressure and the given bulk modulus:
\[ \frac{\Delta V}{V} = \frac{3 \times 10^7 \text{ Nm}^{-2}}{2.2 \times 10^9 \text{ Nm}^{-2}} \]
\[ \frac{\Delta V}{V} = \frac{3}{2.2} \times 10^{7-9} \]
\[ \frac{\Delta V}{V} = 1.3636 \times 10^{-2} \]
Rounding to two decimal places:
\[ \frac{\Delta V}{V} \approx 1.36 \times 10^{-2} \]
This means the volume reduces by approximately 1.36%. Water is not very compressible, so the change is small.
In simple words: We first find the pressure at the ocean's depth using the depth, water density, and gravity. Then, we use the formula for Bulk modulus (pressure divided by volume strain) to find the fractional change in volume. The answer is about \( 1.36 \times 10^{-2} \).

🎯 Exam Tip: Remember to convert all units to SI (meters, kilograms, seconds) before calculation. Ensure you correctly apply the formula for hydrostatic pressure and bulk modulus.

 

Question 3. A wire of length 0.5 m and diameter 2 mm is tied at one end and is twisted at the other end by 0.8 rad. Calculate the shearing strain on the surface of the wire.
Answer:
Given values:
Length of the wire, \( L = 0.5 \text{ m} \)
Diameter of the wire \( = 2 \text{ mm} \)
Radius of the wire, \( r = \frac{\text{Diameter}}{2} = \frac{2 \text{ mm}}{2} = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \)
Twist angle at the other end, \( \theta = 0.8 \text{ rad} \)
Shearing strain \( \Phi \) at the surface of the wire is given by the formula:
\[ \Phi = \frac{r\theta}{L} \]
Substitute the given values:
\[ \Phi = \frac{(1 \times 10^{-3} \text{ m}) \times (0.8 \text{ rad})}{0.5 \text{ m}} \]
\[ \Phi = \frac{0.8 \times 10^{-3}}{0.5} \]
\[ \Phi = 1.6 \times 10^{-3} \text{ rad} \]
To convert this to degrees (though typically strain is left in radians for physics):
\( \Phi = 1.6 \times 10^{-3} \times \frac{180^\circ}{\pi} \)
\( \Phi = 1.6 \times 10^{-3} \times \frac{180}{3.14} \)
\( \Phi \approx 1.6 \times 10^{-3} \times 57.32^\circ \)
\( \Phi \approx 0.0917^\circ \)
The shearing strain on the surface of the wire is \( 1.6 \times 10^{-3} \text{ radians} \) or approximately \( 0.092^\circ \). This shows the small angular deformation.
In simple words: We use the formula for shearing strain which involves the wire's radius, the angle it was twisted, and its length. By plugging in the numbers, we find the shearing strain is \( 1.6 \times 10^{-3} \) radians.

🎯 Exam Tip: Ensure all units are consistent (meters for length and radius) and that the twist angle is in radians when using the shearing strain formula.

Factors Affecting Elasticity

Elasticity in a material is influenced by several factors:
(i) Effect of annealing: When a material is heated and then cooled slowly, it forms larger crystal grains, which makes the material less elastic.
(ii) Effect of the impurities present in the material: Adding different types of impurities can either increase or decrease the material's elasticity.
(iii) Effect of hammering and rolling: These processes break the crystal grains into smaller pieces, which increases the elasticity of the materials.
(iv) Effect of temperature: Generally, the elasticity of materials decreases as their temperature increases.

In Searle's experiment, a metal frame has two wires, A and B, of the same material hanging from a strong base. Wire A has a hook P, and wire B has a hook Q. A spirit level is connected to hook P and a measuring tool called a spherometer. When a weight is hung on wire A, it stretches and gets longer. The spherometer is used to measure how much the wire stretches.

Scientists measure how much a wire stretches by hanging different weights on it. They draw a graph with the weight (M) on the X-axis and how much the wire lengthens (l) on the Y-axis. If the wire's initial length is L, the applied mass is M, its radius is r, and it stretches by 'l', then these values are used in calculations.
Stress \( = \frac{F}{A} = \frac{Mg}{\pi r^2} \)
Strain \( = \frac{l}{L} \)
Young's modulus \( Y = \frac{\text{Stress}}{\text{Strain}} = \frac{MgL}{\pi r^2 l} \)

The value of Y (Young's modulus) is determined from this graph and by measuring the wire. This modulus helps us understand how much a material can resist being stretched or compressed.

 

Class 11 Physics Chapter 10 Numerical Questions

 

Question 1. Tn a wire by producing 4 × 10-4 linear strain a 4.8 × 107 Nm-2 stress is produced. Calculate the Young's modulus of elasticity of the material.
Answer: The problem provides the linear strain and the stress produced in a wire. We need to find the Young's modulus of elasticity. Young's modulus is the ratio of stress to strain.
Given linear strain \( = 4 \times 10^{-4} \).
Given stress \( = 4.8 \times 10^7 \, \text{Nm}^{-2} \).
We calculate Young's modulus (Y) using the formula:
\( Y = \frac{\text{Stress}}{\text{Strain}} \)
\( Y = \frac{4.8 \times 10^7 \, \text{Nm}^{-2}}{4 \times 10^{-4}} \)
\( Y = 1.2 \times 10^{11} \, \text{Nm}^{-2} \)
The Young's modulus tells us how much a material can resist being stretched or compressed.
In simple words: To find Young's modulus, we divide the stress by the strain. This shows how stiff the material is.

🎯 Exam Tip: Remember that Young's modulus is a measure of a material's stiffness, so a higher value means the material is harder to deform.

 

Question 2. The average depth of Indian Ocean is 3 km. Calculate AV/V given that the Bulk modulus of elasticity for water is 2.2 × 10⁹ Nm¯², (g = 10 m/s²).
Answer: We are given the average depth of the Indian Ocean, the bulk modulus of water, and gravity. We need to find the fractional change in volume (\( \frac{\Delta V}{V} \)). First, calculate the pressure at that depth.
Given average depth of Indian Ocean, \( h = 3 \, \text{km} = 3 \times 10^3 \, \text{m} \).
Density of water \( \rho = 1.0 \times 10^3 \, \text{kg/m}^3 \).
Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).
Bulk modulus of elasticity of water \( K = 2.2 \times 10^9 \, \text{Nm}^{-2} \).
Pressure exerted by water \( P = h \rho g \)
\( P = (3 \times 10^3 \, \text{m}) \times (1.0 \times 10^3 \, \text{kg/m}^3) \times (10 \, \text{m/s}^2) \)
\( P = 3 \times 10^7 \, \text{Nm}^{-2} \)
The bulk modulus is defined as \( K = \frac{P}{(\Delta V / V)} \). We need to find \( \frac{\Delta V}{V} \).
\( \frac{\Delta V}{V} = \frac{P}{K} \)
\( \frac{\Delta V}{V} = \frac{3 \times 10^7 \, \text{Nm}^{-2}}{2.2 \times 10^9 \, \text{Nm}^{-2}} \)
\( \frac{\Delta V}{V} \approx 1.3636 \times 10^{-2} \)
So, the fractional change in volume is about \( 1.36 \times 10^{-2} \). This change is relatively small, showing water's incompressibility.
In simple words: We first calculate the pressure at the ocean's depth. Then, we divide this pressure by the bulk modulus of water to find how much its volume changes compared to its original volume.

🎯 Exam Tip: Remember to convert all units to SI units (km to m) before calculations. The bulk modulus shows how resistant a fluid is to compression.

 

Question 3. A wire of length 0-5 m and diameter 2 mm is tied at one end and is twisted at the other end by 0.8 rad. Calculate the shearing strain on the surface of the wire.
Answer: We need to calculate the shearing strain on the surface of a twisted wire. We are given the wire's length, diameter, and the twisting angle.
Given length of the wire \( L = 0.5 \, \text{m} \).
Diameter \( D = 2 \, \text{mm} \), so radius \( r = D/2 = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \).
Angle of twist \( \theta = 0.8 \, \text{rad} \).
The shearing strain \( \phi \) at the surface of a twisted wire is given by the formula:
\( \phi = \frac{r \theta}{L} \)
\( \phi = \frac{(1 \times 10^{-3} \, \text{m}) \times (0.8 \, \text{rad})}{0.5 \, \text{m}} \)
\( \phi = 1.6 \times 10^{-3} \, \text{rad} \)
To convert this to degrees, we multiply by \( \frac{180}{\pi} \):
\( \phi = 1.6 \times 10^{-3} \times \frac{180}{\pi} \approx 1.6 \times 10^{-3} \times 57.295 \approx 0.0917^\circ \)
The shearing strain measures the deformation due to twisting forces.
In simple words: Shearing strain for a twisted wire is found by multiplying the wire's radius by the twist angle (in radians) and then dividing by the wire's length. This tells us how much the wire is distorted.

🎯 Exam Tip: Ensure that the angle of twist is in radians when using the formula \( \phi = \frac{r \theta}{L} \). Converting to degrees is often done for easier interpretation but not strictly required unless specified.

 

Question 5. A 5 m long steel rod is tied between two rigid bases. Steel's linear modulus of elasticity = 12 × 10-6 l°C and Y = 2 × 1011 Nm-2. Calculate the stress produced in the rod if the temperature is increased by 40°C.
Answer: We need to find the stress produced in a steel rod when its temperature increases. We are given its length, the coefficient of linear thermal expansion for steel, its Young's modulus, and the temperature change.
Given length of the steel rod \( L = 5 \, \text{m} \).
Coefficient of linear thermal expansion \( \alpha = 12 \times 10^{-6} \, \text{/}^\circ\text{C} \).
Young's modulus of steel \( Y = 2 \times 10^{11} \, \text{Nm}^{-2} \).
Increase in temperature \( \Delta T = 40^\circ\text{C} \).
When a rod's temperature changes, it tries to expand. If it's held rigidly, stress is created. The thermal strain produced is given by \( \text{Strain} = \alpha \Delta T \). The stress produced is given by Hooke's Law: \( \text{Stress} = Y \times \text{Strain} \). So, \( \text{Stress} = Y \alpha \Delta T \)
\( \text{Stress} = (2 \times 10^{11} \, \text{Nm}^{-2}) \times (12 \times 10^{-6} \, \text{/}^\circ\text{C}) \times (40^\circ\text{C}) \)
\( \text{Stress} = 96 \times 10^6 \, \text{Nm}^{-2} \)
\( \text{Stress} = 9.6 \times 10^7 \, \text{Nm}^{-2} \)
This stress is a compressive force that tries to prevent the rod from expanding.
In simple words: When a metal rod heats up and can't expand, it creates an internal push or pull called stress. We find this stress by multiplying the material's Young's modulus, its expansion rate, and how much the temperature changed.

🎯 Exam Tip: Be careful to identify the correct physical constant; 'linear modulus of elasticity' with units of /°C typically refers to the coefficient of linear thermal expansion (\( \alpha \)).

 

Question 6. How much will be the work done in stretching a 2 m long wire of area 1 mm² by 0.1 nun?
Answer: Given length of the wire \( L = 2 \, \text{m} \).
Cross-sectional area \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \).
Increase in length \( l = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} = 1 \times 10^{-4} \, \text{m} \).
Since the material is not specified, we will assume it is steel, for which Young's modulus \( Y = 21 \times 10^{10} \, \text{Nm}^{-2} \) (from common values for steel).
The work done in stretching a wire (elastic potential energy stored) is given by the formula:
\( \text{Work Done} = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume} \)
Which can also be written as:
\( \text{Work Done} = \frac{1}{2} \times Y \times A \times \frac{l^2}{L} \)
\( \text{Work Done} = \frac{1}{2} \times (21 \times 10^{10} \, \text{Nm}^{-2}) \times (1 \times 10^{-6} \, \text{m}^2) \times \frac{(1 \times 10^{-4} \, \text{m})^2}{2 \, \text{m}} \)
\( \text{Work Done} = \frac{21 \times 10^{10} \times 10^{-6} \times 10^{-8}}{4} \)
\( \text{Work Done} = \frac{21}{4} \times 10^{-4} \, \text{J} \)
\( \text{Work Done} = 5.25 \times 10^{-4} \, \text{J} \)
This small amount of work represents the energy stored in the wire due to its elastic deformation.
In simple words: We calculate the energy stored in the wire when it is stretched. We use a specific formula that considers how stiff the material is, its size, and how much it stretched.

🎯 Exam Tip: When the material isn't specified, but Young's modulus is needed for a numerical answer, check if typical values are given elsewhere in the chapter or consider the context for a plausible material like steel.

 

Question 7. Calculate the Poisson's ratio when a rubber string is stretched and in that situation the change in its volume is negligible in comparision to the change in shape.
Answer: We need to find Poisson's ratio for a rubber string when its volume change is negligible during stretching.
Poisson's ratio (\( \sigma \)) is related to Young's modulus (Y) and Bulk modulus (K) by the formula:
\( Y = 3K(1-2\sigma) \)
When the change in volume is negligible, it means the material is practically incompressible. This implies that the Bulk modulus \( K \) is very large, tending towards infinity (\( K \rightarrow \infty \)).
For \( Y \) to remain a finite value, the term \( (1-2\sigma) \) must approach zero.
\( 1 - 2\sigma = 0 \)
\( 2\sigma = 1 \)
\( \sigma = 0.5 \)
This value is the theoretical maximum for Poisson's ratio for an isotropic material.
In simple words: When something like rubber is stretched and its total volume stays almost the same, its Poisson's ratio becomes 0.5. This means it expands sideways as much as possible for its length change.

🎯 Exam Tip: Remember that for an incompressible material (where volume change is negligible), Poisson's ratio is always 0.5. This is the theoretical upper limit for most materials.

 

Question 8. A solid rubber ball when taken from the surface to the bottom of a 200 m deep lake the volume of the ball reduces by 0.1%. Density of water of the lake = 1.0 × 10³ kg/m³. Calculate the Bulk modulus of elasticity of rubber, (g = 10 m/s²)
Answer: We need to calculate the bulk modulus of elasticity for a rubber ball that changes volume when submerged in a deep lake.
Given depth of lake \( h = 200 \, \text{m} \).
Fractional reduction in volume \( \frac{\Delta V}{V} = 0.1\% = \frac{0.1}{100} = 0.001 \).
Density of water \( \rho = 1.0 \times 10^3 \, \text{kg/m}^3 \).
Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).
First, calculate the pressure at the bottom of the lake:
\( P = h \rho g \)
\( P = (200 \, \text{m}) \times (1.0 \times 10^3 \, \text{kg/m}^3) \times (10 \, \text{m/s}^2) \)
\( P = 2 \times 10^6 \, \text{Nm}^{-2} \)
The bulk modulus of elasticity \( K \) is defined as the ratio of pressure (stress) to the fractional change in volume (volume strain):
\( K = \frac{P}{(\Delta V / V)} \)
\( K = \frac{2 \times 10^6 \, \text{Nm}^{-2}}{0.001} \)
\( K = 2 \times 10^9 \, \text{Nm}^{-2} \)
The bulk modulus shows how much a material resists changes in its volume when pressure is applied.
In simple words: First, we find the water pressure at the bottom of the lake. Then, we divide this pressure by the percentage change in the rubber ball's volume to get its bulk modulus, which tells us how hard it is to squeeze.

🎯 Exam Tip: Ensure that the percentage volume change is converted to a fractional value (e.g., 0.1% = 0.001) before using it in the bulk modulus formula.

 

Question 9. A shearing force of 9.0 × 104 N is applied on a thin surface of a square glass slab of 50 cm side and width 10 cm. Second thin surface is stick to the floor. How much will the upper surface get displaced?
Answer: We need to calculate the displacement of the upper surface of a glass slab when a shearing force is applied.
Given shearing force \( F = 9.0 \times 10^4 \, \text{N} \).
The slab has a base area where the force is applied. Assuming the base dimensions are 50 cm by 10 cm, the area \( A = (0.5 \, \text{m}) \times (0.1 \, \text{m}) = 0.05 \, \text{m}^2 \).
The height of the slab (the distance over which the shear occurs) is assumed to be 50 cm, so \( L = 0.5 \, \text{m} \).
Modulus of rigidity for glass \( \eta = 5.6 \times 10^9 \, \text{Nm}^{-2} \) (as given in the problem statement).
Shearing stress \( = F/A \).
Shearing strain \( = x/L \), where \( x \) is the displacement of the upper surface.
The modulus of rigidity \( \eta \) is defined as:
\( \eta = \frac{\text{Shearing stress}}{\text{Shearing strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax} \)
We can rearrange this formula to solve for the displacement \( x \):
\( x = \frac{FL}{A\eta} \)
\( x = \frac{(9.0 \times 10^4 \, \text{N}) \times (0.5 \, \text{m})}{(0.05 \, \text{m}^2) \times (5.6 \times 10^9 \, \text{Nm}^{-2})} \)
\( x = \frac{4.5 \times 10^4}{0.28 \times 10^9} \)
\( x = 16.0714 \times 10^{-5} \, \text{m} \)
\( x \approx 1.61 \times 10^{-4} \, \text{m} \)
This displacement shows how much the top layer shifts compared to the fixed bottom layer due to the applied force.
In simple words: We find how much the top of the glass slab moves sideways by using the force, the slab's size, and how stiff the glass is against twisting.

🎯 Exam Tip: When dealing with geometric contradictions in a problem (like "square slab... and width"), choose the most consistent interpretation that allows for calculation, usually guided by how area and length are used in the formula. Convert all units to SI before calculation.

 

Question 10. For a steel wire of 4.7 m length and transverse cross section of 3.0 × 10-5 m² and a copper wire of length 3.5 m and 4.0 × 10-5 m², the increase in length of both is same when same value of weights are hanged from the wires. What is the ratio of steel and copper's Young's modulus of elasticity?
Answer: We need to find the ratio of Young's modulus for a steel wire and a copper wire, given their dimensions and the condition that they stretch by the same amount under the same weight.
For the steel wire (subscript 's'):
Length \( L_s = 4.7 \, \text{m} \).
Cross-sectional area \( A_s = 3.0 \times 10^{-5} \, \text{m}^2 \).
For the copper wire (subscript 'c'):
Length \( L_c = 3.5 \, \text{m} \).
Cross-sectional area \( A_c = 4.0 \times 10^{-5} \, \text{m}^2 \).
We are given that the increase in length is the same (\( l_s = l_c = l \)) and the force (weight) applied is the same (\( F_s = F_c = F \)).
Young's modulus \( Y \) is defined as:
\( Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{l/L} = \frac{FL}{Al} \)
So, for steel: \( Y_s = \frac{FL_s}{A_s l} \)
And for copper: \( Y_c = \frac{FL_c}{A_c l} \)
Now, we find the ratio of their Young's moduli:
\( \frac{Y_s}{Y_c} = \frac{FL_s / (A_s l)}{FL_c / (A_c l)} \)
Since \( F \) and \( l \) are the same for both wires, they cancel out:
\( \frac{Y_s}{Y_c} = \frac{L_s / A_s}{L_c / A_c} = \frac{L_s A_c}{L_c A_s} \)
Substitute the given values:
\( \frac{Y_s}{Y_c} = \frac{4.7 \, \text{m} \times (4.0 \times 10^{-5} \, \text{m}^2)}{3.5 \, \text{m} \times (3.0 \times 10^{-5} \, \text{m}^2)} \)
\( \frac{Y_s}{Y_c} = \frac{4.7 \times 4.0}{3.5 \times 3.0} = \frac{18.8}{10.5} \)
\( \frac{Y_s}{Y_c} \approx 1.79 \)
This ratio indicates that steel is significantly stiffer than copper, as expected.
In simple words: We compare how much steel and copper wires stretch under the same weight. Since Young's modulus tells us how stiff a material is, we use a formula to find the ratio of stiffness between steel and copper.

🎯 Exam Tip: When comparing properties of two materials under similar conditions, set up a ratio of their formulas. Common terms (like force or elongation) will often cancel out, simplifying the calculation.

 

Question 11. Two wires are made up of the same material (metal). Length of the first wire is half the length of the second wire, and its diameter is double the diameter of the second wire. If same weights are hanged from the wires then what would be the ratio of increase in the length of the wires?
Answer: We need to find the ratio of how much two wires stretch, given their different lengths and diameters, but made of the same material and with the same weight hanging from them.
Let's denote the first wire with subscript '1' and the second wire with subscript '2'.
Given:
Materials are the same, so Young's modulus \( Y_1 = Y_2 = Y \).
Length of the first wire is half the length of the second: \( L_1 = \frac{L_2}{2} \implies L_2 = 2L_1 \).
Diameter of the first wire is double the diameter of the second: \( D_1 = 2D_2 \implies r_1 = 2r_2 \).
Same weights are hanged, so force \( F_1 = F_2 = F \).
The formula for the increase in length \( l \) is derived from Young's modulus \( Y = \frac{FL}{Al} \):
\( l = \frac{FL}{AY} \)
Since the cross-sectional area \( A = \pi r^2 \), we can write:
\( l = \frac{FL}{\pi r^2 Y} \)
Now, let's find the ratio of the increase in length for the two wires:
\( \frac{l_1}{l_2} = \frac{FL_1 / (\pi r_1^2 Y)}{FL_2 / (\pi r_2^2 Y)} \)
Since \( F \), \( \pi \), and \( Y \) are the same for both wires, they cancel out:
\( \frac{l_1}{l_2} = \frac{L_1 / r_1^2}{L_2 / r_2^2} = \frac{L_1 r_2^2}{L_2 r_1^2} \)
Substitute the given relationships \( L_2 = 2L_1 \) and \( r_1 = 2r_2 \):
\( \frac{l_1}{l_2} = \frac{L_1 r_2^2}{(2L_1) (2r_2)^2} \)
\( \frac{l_1}{l_2} = \frac{L_1 r_2^2}{2L_1 \times 4r_2^2} \)
\( \frac{l_1}{l_2} = \frac{1}{8} \)
So, the ratio of the increase in length of the wires is \( 1:8 \). This means the first wire stretches much less than the second wire, mainly because it is thicker.
In simple words: We compare how much two wires stretch using a formula. Even though they are made of the same material and have the same weight, the thicker and shorter wire stretches less.

🎯 Exam Tip: Pay close attention to inverse and direct relationships in formulas. Here, length \( l \) is directly proportional to \( L \) but inversely proportional to \( r^2 \), making diameter a very strong factor in stretching.

 

Question 12. A material breaks by the stress of 109 N.m¯²; if the density of the material is 3 × 103 kg m⁻³ then calculate that length of the wire; which on hanging by its own weight breaks.
Answer: We need to find the maximum length of a wire that can hang by its own weight before it breaks, given its breaking stress and density.
Given breaking stress \( \sigma_b = 10^9 \, \text{Nm}^{-2} \).
Density of the material \( \rho = 3 \times 10^3 \, \text{kg m}^{-3} \).
Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \).
When a wire hangs by its own weight, the stress at its top end is caused by the total weight of the wire below that point. This stress is given by:
\( \text{Stress} = \frac{\text{Weight of wire}}{\text{Cross-sectional Area}} \)
The weight of the wire can be expressed as:
\( \text{Weight} = \text{Volume} \times \text{Density} \times g = (A \times L) \times \rho \times g \)
Where \( A \) is the cross-sectional area and \( L \) is the length of the wire.
So, the stress is:
\( \sigma = \frac{(A L \rho g)}{A} = L \rho g \)
For the wire to break, this stress must be equal to the breaking stress (\( \sigma_b \)):
\( \sigma_b = L \rho g \)
We can rearrange this to find the maximum length \( L \):
\( L = \frac{\sigma_b}{\rho g} \)
\( L = \frac{10^9 \, \text{Nm}^{-2}}{(3 \times 10^3 \, \text{kg m}^{-3}) \times (9.8 \, \text{m/s}^2)} \)
\( L = \frac{10^9}{29.4 \times 10^3} \)
\( L \approx 3.401 \times 10^4 \, \text{m} \)
\( L \approx 34 \, \text{km} \)
This very long length shows that materials can be quite strong even under their own massive weight.
In simple words: We find the maximum length a wire can be before it breaks from its own weight. We use the wire's breaking strength, its density, and the force of gravity to figure this out.

🎯 Exam Tip: Remember that stress due to self-weight is maximum at the support point and increases linearly with length. Use consistent units (SI) for all values.

 

Question 13. The ratio of radius of two wires of same material is 2 : 1 If they are stretched by applying similar force then what would be the ratio of the stress generated?
Answer: We need to find the ratio of stress generated in two wires that have different radii but are made of the same material and stretched by the same force.
Let's denote the first wire with subscript '1' and the second wire with subscript '2'.
Given:
Ratio of radii \( r_1 : r_2 = 2 : 1 \). This means \( \frac{r_1}{r_2} = \frac{2}{1} \), so \( \frac{r_2}{r_1} = \frac{1}{2} \).
Similar force applied \( F_1 = F_2 = F \).
Stress \( \sigma \) is defined as force per unit area:
\( \sigma = \frac{\text{Force}}{\text{Area}} = \frac{F}{A} \)
Since the cross-sectional area \( A = \pi r^2 \), we can write:
\( \sigma = \frac{F}{\pi r^2} \)
Now, we find the ratio of the stress generated in the two wires:
\( \frac{\sigma_1}{\sigma_2} = \frac{F / (\pi r_1^2)}{F / (\pi r_2^2)} \)
Since \( F \) and \( \pi \) are the same for both wires, they cancel out:
\( \frac{\sigma_1}{\sigma_2} = \frac{r_2^2}{r_1^2} = \left( \frac{r_2}{r_1} \right)^2 \)
Substitute the ratio of radii:
\( \frac{\sigma_1}{\sigma_2} = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \)
So, the ratio of the stress generated in the wires is \( 1:4 \). This means the thinner wire experiences much higher stress for the same force.
In simple words: We find the stress in two wires with different thicknesses but pulled by the same force. The thinner wire will always feel more stress because the force is spread over a smaller area.

🎯 Exam Tip: Remember that stress is inversely proportional to the square of the radius. A small change in radius can lead to a large change in stress for the same force.

 

Question 14. From any rigid base a steel wire of length 2 m and weight 3g a weight of 2.5 kg is hanged. The increase in length of the steel wire is 2.5 mm. If the density of the steel wire is 7.8 × 103 kg/m³ then calculate the Young's modulus of elasticity.
Answer: We need to calculate the Young's modulus of a steel wire, given its dimensions, the mass hanging from it, its density, and how much it stretched.
Given length of steel wire \( L = 2 \, \text{m} \).
Mass of the wire \( m_{\text{wire}} = 3 \, \text{g} = 3 \times 10^{-3} \, \text{kg} \).
Hanging load \( M = 2.5 \, \text{kg} \).
Increase in length \( l = 2.5 \, \text{mm} = 2.5 \times 10^{-3} \, \text{m} \).
Density of steel \( \rho = 7.8 \times 10^3 \, \text{kg/m}^3 \).
Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \).
First, we calculate the cross-sectional area \( A \) of the wire using its mass, length, and density:
Volume of wire \( V = A \times L \)
Mass of wire \( m_{\text{wire}} = \rho \times V = \rho \times A \times L \)
So, \( A = \frac{m_{\text{wire}}}{\rho L} \)
\( A = \frac{3 \times 10^{-3} \, \text{kg}}{(7.8 \times 10^3 \, \text{kg/m}^3) \times (2 \, \text{m})} \)
\( A = \frac{3 \times 10^{-3}}{15.6 \times 10^3} = 1.923 \times 10^{-7} \, \text{m}^2 \) (approximately)
Young's modulus \( Y \) is calculated using the formula \( Y = \frac{MgL}{Al} \), where \( M \) is the hanging load:
\( Y = \frac{(2.5 \, \text{kg}) \times (9.8 \, \text{m/s}^2) \times (2 \, \text{m})}{(1.923 \times 10^{-7} \, \text{m}^2) \times (2.5 \times 10^{-3} \, \text{m})} \)
\( Y = \frac{49}{4.8075 \times 10^{-10}} \)
\( Y \approx 10.1923 \times 10^{10} \, \text{Nm}^{-2} \)
\( Y \approx 1.02 \times 10^{11} \, \text{Nm}^{-2} \)
This value is typical for steel, confirming its high stiffness.
In simple words: We find the thickness of the wire using its small weight and density. Then, we use the amount it stretched under the hanging load, along with its length, to calculate its Young's modulus, which measures how much it resists stretching.

🎯 Exam Tip: Be careful with units (grams to kilograms, mm to meters) and correctly identify which mass (wire's own vs. hanging load) is used for calculating cross-sectional area and for Young's modulus, respectively. Assume standard 'g' if not provided.

 

Question 15. A copper wire of transverse cross section of 0.8 cm² is tied at both the ends rigidly. If the temperature of the wire is reduced by 20°C then calculate the tension in the wire. Young's modulus of elasticity of copper is 11 × 1010 Nm-2 and coefficient of expansion is 17 × 10-6/°C.
Answer: We need to calculate the tension (force) created in a copper wire when it is cooled and held firmly at both ends, preventing it from shrinking.
Given cross-sectional area \( A = 0.8 \, \text{cm}^2 = 0.8 \times 10^{-4} \, \text{m}^2 \).
Reduction in temperature \( \Delta T = 20^\circ\text{C} \).
Young's modulus of copper \( Y = 11 \times 10^{10} \, \text{Nm}^{-2} \).
Coefficient of linear thermal expansion \( \alpha = 17 \times 10^{-6} \, \text{/}^\circ\text{C} \).
When a material is cooled but not allowed to contract, it experiences a thermal stress. This stress creates tension in the wire. The thermal strain produced is given by:
\( \text{Strain} = \alpha \Delta T \)
From Hooke's Law, stress is related to strain and Young's modulus:
\( \text{Stress} = Y \times \text{Strain} = Y \alpha \Delta T \)
The tension (force \( F \)) in the wire is the stress multiplied by the cross-sectional area:
\( F = \text{Stress} \times A = Y \alpha \Delta T A \)
Substitute the given values into the formula:
\( F = (11 \times 10^{10} \, \text{Nm}^{-2}) \times (17 \times 10^{-6} \, \text{/}^\circ\text{C}) \times (20^\circ\text{C}) \times (0.8 \times 10^{-4} \, \text{m}^2) \)
\( F = (11 \times 17 \times 20 \times 0.8) \times 10^{(10 - 6 - 4)} \)
\( F = (2992) \times 10^0 \)
\( F = 2992 \, \text{N} \)
This tension is a significant force, showing how materials resist thermal changes when constrained.
In simple words: When a copper wire cools down but cannot shrink, it pulls inward, creating tension. We calculate this tension using how much the wire tries to shrink, how stiff it is, and its thickness.

🎯 Exam Tip: Remember to use consistent units (SI) and that for thermal stress, the strain is \( \alpha \Delta T \). Ensure careful calculation of powers of ten to avoid errors.

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