Maharashtra Board Class 11 Physics Chapter 6 Mechanical Properties of Solids Solutions

1. Choose The Correct Answer:

Question 1. Change in dimensions is known as ............
(A) deformation
(B) formation
(C) contraction
(D) strain.
Answer: (A) deformation
In simple words: When an object's size or shape changes due to an external force, this alteration is called deformation.

🎯 Exam Tip: Understanding basic definitions like deformation is crucial for foundational physics concepts.

Question 2. The point on stress-strain curve at which strain begins to increase even without increase in stress is called............
(A) elastic point
(B) yield point
(C) breaking point
(D) neck point
Answer: (B) yield point
In simple words: The yield point on a stress-strain curve marks where a material starts deforming permanently without needing more stress.

🎯 Exam Tip: Identifying key points on the stress-strain curve (elastic limit, yield point, breaking point) is essential for material property analysis.

Question 3. Strain energy of a stretched wire is \(18 \times 10^{-3}\) J and strain energy per unit volume of the same wire and same cross section is \(6 \times 10^{-3}\) J/m³. Its volume will be.............
(A) 3cm³
(B) 3 m³
(C) 6 m³
(D) 6 cm³
Answer: (B) 3 m³
In simple words: To find the volume, divide the total strain energy by the strain energy per unit volume.

🎯 Exam Tip: Remember the relationship between total energy, energy density (energy per unit volume), and volume for calculations.

Question 4. ............ is the property of a material which enables it to resist plastic deformation.
(A) elasticity
(B) plasticity
(C) hardness
(D) ductility
Answer: (C) hardness
In simple words: Hardness is a material's ability to resist permanent shape change when a force is applied.

🎯 Exam Tip: Distinguish between material properties like hardness, elasticity, plasticity, and ductility, as they describe different responses to stress.

Question 5. The ability of a material to resist fracturing when a force is applied to it, is called...........
(A) toughness
(B) hardness
(C) elasticity
(D) plasticity.
Answer: (A) toughness
In simple words: Toughness measures a material's capacity to absorb energy and deform plastically before breaking.

🎯 Exam Tip: Understand that toughness is about resisting fracture, while hardness is about resisting deformation.

2. Answer In One Sentence:

Question 1. Define elasticity.
Answer: If a body regains its original shape and size after removal of the deforming force, it is called an elastic body and the property is called elasticity.
In simple words: Elasticity is the property of a material to return to its original shape and size after the external deforming force is removed.

🎯 Exam Tip: A clear, concise definition of elasticity is essential, emphasizing the return to original dimensions.

Question 2. What do you mean by deformation?
Answer: The change in shape or size or both of a body due to an external force is called deformation.
In simple words: Deformation is any change in an object's shape or size caused by an applied external force.

🎯 Exam Tip: Define deformation broadly, encompassing changes in both shape and size.

Question 3. State the SI unit and dimensions of stress.
Answer:
1. SI unit: \(N \ m^{-2}\) or pascal (Pa)
2. Dimensions: \([L^{-1}M^1T^{-2}]\)
In simple words: Stress is measured in Pascals (N/m²) and its dimensions show it depends on mass, length, and time in a specific way.

🎯 Exam Tip: Always remember both the SI unit and the dimensional formula for physical quantities like stress.

Question 4. Define strain.
Answer: Strain:
1. Strain is defined as the ratio of change in dimensions of the body to its original dimensions.
Strain \( = \frac{\text{change in dimensions}}{\text{original dimensions}}\)
2. Types of strain:
• Longitudinal strain,
• Volume strain,
• Shearing strain.
In simple words: Strain quantifies the relative deformation of a material, calculated as the ratio of change in dimension to the original dimension.

🎯 Exam Tip: Understand that strain is a dimensionless quantity and recall its different types (longitudinal, volume, shearing).

Question 5. What is Young's modulus of a rigid body?
Answer: Young's modulus (Y): It is the modulus of elasticity related to change in length of an object like a metal wire, rod, beam, etc., due to the applied deforming force.
In simple words: Young's modulus measures how much an elastic material stretches or compresses when a force is applied along its length.

🎯 Exam Tip: Clearly state that Young's modulus specifically deals with changes in length (longitudinal elasticity).

Question 6. Why bridges are unsafe after a very long use?
Answer: A bridge during its use undergoes recurring stress depending upon the movement of vehicles on it. When bridge is used for long time, it loses its elastic strength and ultimately may collapse. Hence, the bridges are declared unsafe after long use.
In simple words: Over time, repeated stress from vehicles causes bridges to lose their elastic strength and accumulate damage, making them unsafe.

🎯 Exam Tip: Connect long-term use and recurring stress to material fatigue and loss of elastic properties, leading to structural failure.

Question 7. How should be a force applied on a body to produce shearing stress?
Answer: A tangential force which is parallel to the top and the bottom surface of the body should be applied to produce shearing stress.
In simple words: To create shearing stress, apply a force parallel to the surface of the body, rather than perpendicular to it.

🎯 Exam Tip: Differentiate between normal stress (perpendicular force) and shearing stress (tangential force) based on force application direction.

Question 8. State the conditions under which Hooke's law holds good.
Answer: Hooke's Law holds good only when a wire/body is loaded within its elastic limit.
In simple words: Hooke's law is valid only when the applied force does not permanently deform the material, keeping it within its elastic range.

🎯 Exam Tip: Always specify the "elastic limit" as the crucial condition for Hooke's law to apply.

Question 9. Define Poisson's ratio.
Answer: Within elastic limit, the ratio of lateral strain to the linear strain is called the Poisson's ratio.
In simple words: Poisson's ratio describes how much a material narrows sideways when it's stretched lengthwise, within its elastic limit.

🎯 Exam Tip: Define Poisson's ratio precisely as a ratio of strains and remember to include the condition "within elastic limit."

Question 10. What is an elastomer?
Answer: A material that can be elastically stretched to a larger value of strain is called an elastomer.
In simple words: Elastomers are materials, like rubber, that can undergo significant elastic stretching and return to their original shape.

🎯 Exam Tip: Emphasize the ability to withstand large elastic strains as the defining characteristic of elastomers.

Question 11. What do you mean by elastic hysteresis?
Answer:
1. In case of some materials like vulcanized rubber, when the stress applied on a body decreases to zero, the strain does not return to zero immediately. The strain lags behind the stress. This lagging of strain behind the stress is called elastic hysteresis.
2.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामग्री के लिए तनाव-विकृति वक्र को दर्शाता है जब भार बढ़ाया जाता है और फिर घटाया जाता है। ऊपर की ओर जाने वाला वक्र (B तक) बढ़ते भार को दिखाता है, जबकि नीचे आने वाला वक्र (C तक) घटते भार को दर्शाता है। वक्र एक लूप बनाते हैं, जिसका क्षेत्रफल विकृति के दौरान ऊर्जा के क्षय को दर्शाता है।
In simple words: Elastic hysteresis is when a material's strain lags behind its stress during loading and unloading, meaning it doesn't return to its original state immediately. This lagging creates a loop on the stress-strain curve, indicating energy dissipation.

🎯 Exam Tip: Explain elastic hysteresis by mentioning the lagging of strain behind stress and the formation of a loop on the stress-strain curve, which represents energy loss.

Question 12. State the names of the hardest material and the softest material.
Answer: Hardest material: Diamond
Softest material: Aluminium
[Note: Material with highest strength is steel whereas material with lowest strength is plasticine clay.]
In simple words: Diamond is known as the hardest material, while aluminum is considered one of the softer materials.

🎯 Exam Tip: Remember diamond as the hardest naturally occurring material. Note the distinction between hardness (resistance to scratching/indentation) and strength (resistance to breaking).

Question 13. Define friction.
Answer: The property which resists the relative motion between two surfaces in contact is called friction.
In simple words: Friction is a force that opposes the sliding or rolling motion between two surfaces that are touching each other.

🎯 Exam Tip: A concise definition of friction should include its role as a resisting force between surfaces in contact.

Question 14. Why force of static friction is known as 'self-adjusting force?
Answer: The force of static friction varies in accordance with applied force. Hence, it is called as self adjusting force.
In simple words: Static friction adjusts its magnitude to exactly match the applied external force, preventing motion until a maximum limit is reached.

🎯 Exam Tip: The key characteristic of static friction is its variability, matching the applied force up to a certain point.

Question 15. Name two factors on which the coefficient of friction depends.
Answer: Coefficient of friction depends upon:
1. the materials of the surfaces in contact.
2. the nature of the surfaces.
In simple words: The coefficient of friction is determined by the types of materials involved and how rough or smooth their surfaces are.

🎯 Exam Tip: The coefficient of friction is primarily influenced by the surface properties and materials, not the apparent contact area.

3. Answer In Short:

Question 1. Distinguish between elasticity and plasticity.
Answer:

In simple words: Elasticity describes a material's ability to return to its original shape after a force is removed, while plasticity means it retains its deformed shape.

🎯 Exam Tip: When distinguishing, focus on the material's behavior after the deforming force is removed, providing examples for clarity.

Question 2. State any four methods to reduce friction.
Answer: Friction can be reduced by using polished surfaces, using lubricants, using grease and using ball bearings.
In simple words: Friction can be reduced by making surfaces smoother, applying lubricants, or using rolling elements like ball bearings.

🎯 Exam Tip: List practical methods to reduce friction, such as smoothing surfaces, using lubricants, or implementing rolling friction components.

Question 3. What is rolling friction? How does it arise?
Answer:
1. Friction between two bodies in contact when one body is rolling over the other, is called rolling friction.
2. Rolling friction arises as the point of contact of the body with the surface keep changing continuously.
In simple words: Rolling friction occurs when one object rolls over another, and it arises because the contact point between the surfaces constantly changes as the object rolls.

🎯 Exam Tip: Define rolling friction by its unique characteristic (one body rolling over another) and explain its origin in terms of changing contact points.

Question 4. Explain how lubricants help in reducing friction?
Answer:
1. The friction between lubricant to surface is much less than the friction between two same surfaces. Hence using lubricants reduces the friction between the two surfaces.
2. When lubricant is applied to machine parts, it fills the depression present on the surface in contact. Thus, less friction is occurred between machine parts.
3. Application of lubricants also reduces wear and tear of machine parts which in turn reduces friction.
4. Advantage: Reduction in function reduces dissipation of energy in machines due to which efficiency of machines increases.
In simple words: Lubricants reduce friction by forming a thin layer between surfaces, filling microscopic irregularities and replacing direct surface-to-surface contact with much lower fluid friction, thus improving efficiency and reducing wear.

🎯 Exam Tip: Focus on how lubricants create a separation layer between surfaces, reduce irregularities, and consequently decrease energy loss and wear.

Question 5. State the laws of static friction.
Answer: Laws of static friction:
1. First law: The limiting force of static friction (\(F_L\)) is directly proportional to the normal reaction (N) between the two surfaces in contact.
\(F_L \propto N\)

\( \implies F_L = \mu_s N\)
where, \(\mu_s\) = constant called coefficient of static friction.
2. Second law: The limiting force of friction is independent of the apparent area between the surfaces in contact, so long as the normal reaction remains the same.
3. Third law: The limiting force of friction depends upon materials in contact and the nature of their surfaces.
In simple words: Static friction's laws state that the maximum opposing force is proportional to the normal force, independent of the apparent contact area, and depends on the materials and roughness of the surfaces.

🎯 Exam Tip: Memorize the three laws, especially the proportionality to normal reaction and independence from apparent contact area for a given \(\mu_s\).

Question 6. State the laws of kinetic friction.
Answer: Laws of kinetic friction:
1. First law: The force of kinetic friction (\(F_k\)) is directly proportional to the normal reaction (N) between two surfaces in contact.
\(F_k \propto N\)

\( \implies F_k = \mu_k N\)
where, \(\mu_k\) = constant called coefficient of kinetic friction.
2. Second law: Force of kinetic friction is independent of shape and apparent area of the surfaces in contact.
3. Third law: Force of kinetic friction depends upon the nature and material of the surfaces in contact.
4. Fourth law: The magnitude of the force of kinetic friction is independent of the relative velocity between the object and the surface provided that the relative velocity is neither too large nor too small.
In simple words: Kinetic friction's laws indicate that it's proportional to the normal force, unaffected by shape or apparent area, depends on the materials, and is mostly constant regardless of relative velocity (within limits).

🎯 Exam Tip: Understand how kinetic friction differs from static friction, particularly its independence from relative velocity (for typical speeds) and apparent contact area.

Question 7. State advantages of friction.
Answer: Advantages of friction:
1. We can walk due to friction between ground and feet.
2. We can hold object in hand due to static friction.
3. Brakes of vehicles work due to friction; hence we can reduce speed or stop vehicles.
4. Climbing on a tree is possible due to friction.
In simple words: Friction is essential for daily activities like walking, gripping objects, braking vehicles, and climbing, enabling movement and control.

🎯 Exam Tip: Provide diverse examples of how friction facilitates everyday actions, demonstrating its positive applications.

Question 8. State disadvantages of friction.
Answer: Disadvantages of friction:
1. Friction opposes motion.
2. Friction produces heat in different parts of machines. It also produces noise.
3. Automobile engines consume more fuel due to friction.
In simple words: Friction has drawbacks such as opposing desired motion, generating wasteful heat and noise, and increasing fuel consumption in machines.

🎯 Exam Tip: Focus on the negative consequences of friction, such as energy loss, wear and tear, and inefficiency.

Question 9. What do you mean by a brittle substance? Give any two examples.
Answer:
1. Substances which breaks within the elastic limit are called brittle substances.
2. Examples: Glass, ceramics.
In simple words: Brittle substances are materials that break suddenly without significant deformation, typically within their elastic limit, like glass or ceramics.

🎯 Exam Tip: Define brittle substances by their characteristic of fracturing without much plastic deformation and provide common examples.

4. Long Answer Type Questions:

Question 1. Distinguish between Young's modulus, bulk modulus and modulus of rigidity.
Answer:

In simple words: Young's modulus measures resistance to length changes, bulk modulus measures resistance to volume changes, and modulus of rigidity measures resistance to shape changes, each applicable to different states of matter.

🎯 Exam Tip: Clearly distinguish between these moduli by defining what type of stress and strain each relates to, their respective formulas, and the types of materials they apply to.

Question 2. Define stress and strain. What are their different types?
Answer:
i) Stress:
1. The internal restoring force per unit area of a body is called stress.
Stress \( = \frac{\text{deforming force}}{\text{area}} = \frac{F}{A}\)
where F is internal restoring force or external applied deforming force.
2. Types of stress:
• Longitudinal stress,
• Volume stress,
• Shearing stress.
ii. Strain:
1. Strain is defined as the ratio of change in dimensions of the body to its original dimensions.
Strain \( = \frac{\text{change in dimensions}}{\text{original dimensions}}\)
2. Types of strain:
• Longitudinal strain,
• Volume strain,
• Shearing strain.
In simple words: Stress is the internal force resisting deformation per unit area, while strain is the fractional change in an object's dimensions due to that force; both have longitudinal, volume, and shearing types.

🎯 Exam Tip: Ensure precise definitions for both stress (force per unit area) and strain (relative deformation) and list all three common types for each.

Question 3. What is Young's modulus? Describe an experiment to find out Young's modulus of material in the form of a long straight wire.
Answer: Definition: Young's modulus is the ratio of longitudinal stress to longitudinal strain.
It is denoted by Y.
Unit: \(N/m^2\) or Pa in SI system.
Dimensions: \([L^{-1}M^1T^{-2}]\)
Experimental description to find Young's modulus:
i. Consider a metal wire suspended from a rigid support. A load is attached to the free end of the wire. Due to this, deforming force gets applied to the free end of wire in downward direction and it produces a change in length.
Let,
L = original length of wire,
Mg = weight suspended to wire,
l = extension or elongation,
(L + l) = new length of wire.
r = radius of the cross section of wire
ii. In its equilibrium position,
Longitudinal stress \( = \frac{\text{Applied force}}{\text{Area of cross section}} = \frac{F}{A} = \frac{Mg}{\pi r^2}\)
Longitudinal strain \( = \frac{\text{Change in length}}{\text{Original length}} = \frac{l}{L}\)
iii. From definition,
Young's modulus \((Y) = \frac{\text{longitudinal stress}}{\text{longitudinal strain}}\)

\( \implies Y = \frac{Mg / \pi r^2}{l / L}\)

\( \implies Y = \frac{MgL}{\pi r^2 l}\)
In simple words: Young's modulus measures a material's stiffness against stretching. An experiment involves suspending a wire, adding weights to stretch it, and then measuring the change in length and cross-sectional area to calculate the ratio of longitudinal stress to longitudinal strain.

🎯 Exam Tip: For this type of question, clearly state the definition, unit, and dimensions of Young's modulus, then systematically describe the experimental setup, measurements, and formula derivation.

Question 4. Derive an expression for strain energy per unit volume of the material of a wire.
Answer: Expression for strain energy per unit volume;
i. Consider a wire of original length L and cross sectional area A stretched by a force F acting along its length. The wire gets stretched and elongation l is produced in it
ii. If the wire is perfectly elastic then,
Longitudinal stress \( = \frac{F}{A}\)
Longitudinal strain \( = \frac{l}{L}\)
Young's modulus \((Y) = \frac{\text{longitudinal stress}}{\text{longitudinal strain}}\)

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

\( \implies F = \frac{YAl}{L}\) ....(1)
iii. The magnitude of stretching force increases from zero to F during elongation of wire.
Let 'f' be the restoring force and 'x' be its corresponding extension at certain instant during the process of extension.
\(f = \frac{YAx}{L}\) (2)
iv. Let 'dW' be the work done for the further small extension 'dx'.
Work = force \(\times\) displacement
\( \implies dW = fdx\)

\( \implies dW = \frac{YAx}{L} dx\) .............(3) [From (2)]
v. The total amount of work done in stretching the wire from x = 0 to x = l can be found out by integrating equation (3).
\(W = \int dW = \int_{0}^{l} \frac{YAx}{L} dx = \frac{YA}{L} \int_{0}^{l} x dx\)

\( \implies W = \frac{YA}{L} \left[ \frac{x^2}{2} \right]_{0}^{l}\)

\( \implies W = \frac{YA}{L} \left[ \frac{l^2}{2} - \frac{0^2}{2} \right]\)

\( \implies W = \frac{YAl^2}{2L}\)
But, \(F = \frac{YAl}{L}\) ....[From (1)]

\( \implies W = \frac{1}{2} \times F \times l\)
Work done in stretching a wire,
\(W = \frac{1}{2} \times \text{load} \times \text{extension}\)
vi. Work done by stretching force is equal to strain energy gained by the wire.

\( \implies \text{Strain energy} = \frac{1}{2} \times \text{load} \times \text{extension}\)
vii. Work done per unit volume
\( = \frac{\text{work done in stretching wire}}{\text{volume of wire}}\)

\( = \frac{\frac{1}{2} F \times l}{V}\)

\( = \frac{\frac{1}{2} F \times l}{A \times L}\)

\( = \frac{1}{2} \times \frac{F}{A} \times \frac{l}{L}\)

\( = \frac{1}{2} \times \text{stress} \times \text{strain}\)
Strain energy per unit volume \( = \frac{1}{2} \times \text{stress} \times \text{strain}\)
viii. Other forms:
Since, \(Y = \frac{\text{stress}}{\text{strain}}\)
a. Strain energy per unit volume
\( = \frac{1}{2} \times \text{stress} \times \frac{\text{stress}}{Y} = \frac{1}{2} \times \frac{(\text{stress})^2}{Y}\)
b. Strain energy per unit volume
\( = \frac{1}{2} \times Y \times \text{strain} \times \text{strain} = \frac{1}{2} \times Y \times (\text{strain})^2\)
In simple words: The strain energy per unit volume of a wire is derived by calculating the work done to stretch it and then dividing by the wire's volume, resulting in an expression often seen as half the product of stress and strain, or related to Young's modulus and the square of stress or strain.

🎯 Exam Tip: For derivations, clearly define variables, follow logical steps from fundamental principles (work done = force x displacement, Hooke's law), and ensure correct integration and substitution to reach the final forms.

Question 5. What is friction? Define coefficient of static friction and coefficient of kinetic friction. Give the necessary formula for each.
Answer:
1. The property which resists the relative motion between two surfaces in contact is called friction.
2. The coefficient of static friction is defined as the ratio of limiting force of friction to the normal reaction.
Formula: \(\mu_s = \frac{F_L}{N}\)
3. The coefficient of kinetic friction is defined as the ratio of force of kinetic friction to the normal reaction between the two surfaces in contact.
Formula: \(\mu_k = \frac{F_k}{N}\)
In simple words: Friction is a force opposing motion between surfaces. The coefficient of static friction is the ratio of the maximum static friction to the normal force, while the coefficient of kinetic friction is the ratio of kinetic friction to the normal force.

🎯 Exam Tip: Provide clear definitions for friction and both coefficients, ensuring correct formulas that relate the frictional force to the normal reaction.

Question 6. State Hooke's law. Draw a labelled graph of tensile stress against tensile strain for a metal wire up to the breaking point. In this graph show the region in which Hooke's law is obeyed.
Answer:
i) Statement: Within elastic limit, stress is directly proportional to strain.
Explanation;
1. According to Hooke's law,
Stress \(\propto\) Strain

\( \implies \frac{\text{Stress}}{\text{Strain}} = \text{constant}\)
This constant of proportionality is called modulus of elasticity.
2. Modulus of elasticity of a material is the slope of stress-strain curve in elastic deformation region and depends on the nature of the material.
3. The graph of strain (on X-axis) and stress (on Y-axis) within elastic limit is
shown in the figure.
ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक धातु के तार के लिए तनाव-विकृति वक्र को दर्शाता है। X-अक्ष पर विकृति (strain) और Y-अक्ष पर तनाव (stress) है। इसमें आनुपातिक सीमा (A), प्रत्यास्थ सीमा या पराभव बिंदु (B), प्लास्टिक व्यवहार क्षेत्र और भंगुरता बिंदु (D) जैसे महत्वपूर्ण बिंदु दिखाए गए हैं। OA क्षेत्र वह है जहां हुक का नियम पूरी तरह से लागू होता है।
iii) Hooke's law is completely obeyed in the region OA.
In simple words: Hooke's law states that within the elastic limit, stress is directly proportional to strain. The stress-strain graph illustrates this, with the linear region (OA) showing where the law is perfectly obeyed before the material starts deforming permanently.

🎯 Exam Tip: When explaining Hooke's law, clearly state the direct proportionality and the "elastic limit" condition. For the graph, correctly label axes, key points (proportional limit, elastic limit, yield point, fracture point), and specifically indicate the region where Hooke's law holds.

Question 1. Calculate the coefficient of static friction for an object of mass 50 kg placed on horizontal table pulled by attaching a spring balance. The force is increased gradually it is observed that the object just moves when spring balance shows 50N.
[Answer: \(\mu_s\) = 0.102]
Answer:
Solution:
Given: m = 50 kg, \(F_L\) = 50 N, g = 9.8 m/s\(-^2\)
To find: Coefficient of static friction (\(\mu_s\))
Formula: \(\mu_s = \frac{F_L}{N} = \frac{F_L}{mg}\)
Calculation: From formula,
\(\mu_s = \frac{50}{50 \times 9.8} = 0.102\)
The coefficient of static friction is 0.102.
In simple words: The coefficient of static friction is a ratio that tells us how much friction there is between a stationary object and a surface before it starts moving. It's calculated by dividing the minimum force needed to start motion by the object's weight.

🎯 Exam Tip: Clearly state given values, the unknown, and the formula. Show all steps of calculation, especially for numerical problems, and ensure units are consistent.

Question 2. A block of mass 37 kg rests on a rough horizontal plane having coefficient of static friction 0.3. Find out the least force required to just move the block horizontally.
[Answer: F= 108.8N]
Answer:
Solution:
Given: m = 37 kg, \(\mu_s\) = 0.3, g = 9.8 m/s\(-^2\)
To find: Limiting force (\(F_L\))
Formula: \(F_L = \mu_s N = \mu_s mg\)
Calculation: From formula,
\(F_L\) = 0.3 \(\times\) 37 \(\times\) 9.8 = 108.8 N
The force required to move the block is 108.8 N.
In simple words: The least force required to move an object from rest on a rough surface is called the limiting static friction, calculated by multiplying the coefficient of static friction by the object's normal force (its weight).

🎯 Exam Tip: Remember that the normal force on a horizontal plane is equal to the object's weight (mg). Pay attention to unit consistency throughout the calculation.

Question 3. A body of mass 37 kg rests on a rough horizontal surface. The minimum horizontal force required to just start the motion is 68.5 N. In order to keep the body moving with constant velocity, a force of 43 N is needed. What is the value of
(a) coefficient of static friction? and
(b) coefficient of kinetic friction?
Asw:
a) \(\mu_s\) = 0.188
b) \(\mu_k\) = 0.118]
Answer:
Solution:
Given:
\(F_L\) = 68.5 N, \(F_k\) = 43 N,
m = 37 kg, g = 9.8 m/s\(-^2\)
To find:
(i) Coefficient of static friction (\(\mu_s\))
(ii) Coefficient of kinetic friction (\(\mu_k\))
Formulae:
(i) \(\mu_s = \frac{F_L}{N} = \frac{F_L}{mg}\)
(ii) \(\mu_k = \frac{F_k}{N} = \frac{F_k}{mg}\)
Calculation:
From formula (i),

\(\therefore \mu_s = \frac{F_s}{N} = \frac{68.5}{37 \times 9.8} = 0.1889\)
From formula (ii),

\(\therefore \mu_k = \frac{F_k}{N} = \frac{43}{37 \times 9.8} = 0.1186\)
Answer:
1. The coefficient of static friction is 0.1889.
2. The coefficient of kinetic friction is 0.1186.
[Note: Answers calculated above are in accordance with textual methods of calculation.]
In simple words: Static friction coefficient measures resistance to starting motion, while kinetic friction coefficient measures resistance to continuing motion at constant velocity. Both are ratios of the respective friction force to the normal force acting on the object.

🎯 Exam Tip: Differentiate clearly between static and kinetic friction. Remember that the static friction force is typically higher than the kinetic friction force for a given pair of surfaces.

Question 4. A wire gets stretched by 4mm due to a certain load. If the same load is applied to a wire of same material with half the length and double the diameter of the first wire. What will be the change in its length?
Answer:
Solution:
Given:
Original extension, \(l_1\) = 4 mm = 4 \(\times\) 10\(-^3\) m
New length, \(L_2 = L_1/2\)
New diameter, \(D_2 = 2D_1 \implies\) New radius, \(r_2 = 2r_1\)
To find: Change in length (\(l_2\))
Formula: Young's modulus \(Y = \frac{FL}{Al}\), where A = \(\pi r^2\).
So, \(l = \frac{FL}{\pi r^2 Y}\)
Calculation:
Since the material and load are the same, Y and F are constant.
Therefore, \(l \propto \frac{L}{r^2}\).

\(\frac{l_2}{l_1} = \frac{L_2/r_2^2}{L_1/r_1^2} = \frac{L_2}{L_1} \times \frac{r_1^2}{r_2^2}\)
Substitute the given values:

\(\frac{l_2}{l_1} = \frac{L_1/2}{L_1} \times \frac{r_1^2}{(2r_1)^2}\)

\(\frac{l_2}{l_1} = \frac{1}{2} \times \frac{r_1^2}{4r_1^2}\)

\(\frac{l_2}{l_1} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\)

\(\implies l_2 = \frac{l_1}{8}\)
Given \(l_1\) = 4 mm = 4 \(\times\) 10\(-^3\) m

\(l_2 = \frac{4 \times 10^{-3} \text{ m}}{8} = 0.5 \times 10^{-3}\) m
= 0.5 mm
The new change in length of the wire is 0.5 mm.
In simple words: When a wire's length is halved and its radius is doubled, the resulting extension under the same load will be one-eighth of the original extension, due to the inverse square relationship with radius and direct relationship with length in Young's modulus formula.

🎯 Exam Tip: When comparing changes in physical quantities like length or radius under constant force and material, use ratios to simplify calculations. Ensure to correctly handle square terms for radius/diameter.

Question 5. Calculate the work done in stretching a steel wire of length 2m and cross sectional area 0.0225mm\(-^2\) when a load of 100 N is slowly applied to its free end. [Young's modulus of steel= 2 \(\times\) 10\(^{11}\) N/m\(-^2\)]
Answer:
Solution:
Given:
L = 2m, F = 100 N,
A = 0.0225 mm\(-^2\) = 0.0225 \(\times\) 10\(-^6\) m\(-^2\) = 2.25 \(\times\) 10\(-^8\) m\(-^2\),
Y = 2 \(\times\) 10\(^{11}\) N/m\(-^2\)
To find: Work (W)
Formula: Work done in stretching a wire is \(W = \frac{1}{2} \times F \times l\).
We also know Young's modulus \(Y = \frac{F/A}{l/L} \implies l = \frac{FL}{AY}\).
Substituting \(l\) into the work done formula:

\(W = \frac{1}{2} \times F \times \frac{FL}{AY} = \frac{1}{2} \frac{F^2 L}{AY}\)
Calculation:

\(W = \frac{1}{2} \times \frac{(100)^2 \times 2}{2.25 \times 10^{-8} \times 2 \times 10^{11}}\)

\(W = \frac{1}{2} \times \frac{10000 \times 2}{4.5 \times 10^3}\)

\(W = \frac{10000}{4500}\)

\(W = \frac{100}{45} = \frac{20}{9} \approx 2.222\) J
Answer:
The work done in stretching the steel wire is 2.222 J.
In simple words: The work done in stretching a wire represents the elastic potential energy stored. It can be calculated using the formula involving the applied force, the wire's length, cross-sectional area, and Young's modulus, ensuring all units are consistent.

🎯 Exam Tip: When using combined formulas for work done and Young's modulus, simplify the expression before substituting values to reduce calculation errors. Always convert units to SI (e.g., mm\(-^2\) to m\(-^2\)) for accurate results.

Question 6. A solid metal sphere of volume 0.31m\(-^3\) is dropped in an ocean where water pressure is 2 \(\times\) 10\(^7\) N/m\(-^2\). Calculate change in volume of the sphere if bulk modulus of the metal is 6.1 \(\times\) 10\(^{10}\) N/m\(-^2\)
Answer:
Solution:
Given:
V = 0.31 m\(-^3\), dP = 2 \(\times\) 10\(^7\) N/m\(-^2\),
K = 6.1 \(\times\) 10\(^{10}\) N/m\(-^2\)
To find: Change in volume (dV)
Formula: Bulk modulus \(K = V \times \frac{dP}{dV}\)
Calculation: From formula,

\(dV = \frac{V \times dP}{K}\)

\(\therefore dV = \frac{0.31 \times 2 \times 10^7}{6.1 \times 10^{10}}\)

\(dV = \frac{0.62 \times 10^7}{6.1 \times 10^{10}}\)

\(dV = \frac{0.62}{6.1} \times 10^{7-10}\)

\(dV \approx 0.1016 \times 10^{-3} \approx 1.016 \times 10^{-4}\) m\(-^3\)
The change in volume of the sphere is 1.016 \(\times\) 10\(-^4\) m\(-^3\).
In simple words: The change in volume of a submerged object due to pressure is determined by its original volume, the change in pressure, and the material's bulk modulus, which indicates its resistance to compression.

🎯 Exam Tip: Ensure correct unit conversion and precise handling of exponents in scientific notation. The bulk modulus formula involves fractional change in volume, so pay attention to the negative sign if dealing with compression, though usually the magnitude of volume change is requested.

Question 7. A wire of mild steel has initial length 1.5 m and diameter 0.60 mm is extended by 6.3 mm when a certain force is applied to it. If Young's modulus of mild steel is 2.1 \(\times\) 10\(^{11}\) N/m\(-^2\), calculate the force applied.
Answer:
Solution:
Given:
L = 1.5m
d = 0.60 mm \(\implies r = \frac{d}{2} = 0.30 \text{ mm} = 3 \times 10^{-4}\) m
\(Y = 2.1 \times 10^{11}\) N/m\(-^2\)
\(l\) = 6.3 mm = 6.3 \(\times\) 10\(-^3\) m
To find: Force (F)
Formula: Young's modulus \(Y = \frac{F/A}{l/L} \implies F = \frac{YA l}{L}\)
Where \(A = \pi r^2\), so \(F = \frac{Y \pi r^2 l}{L}\)
Calculation:

\(F = \frac{2.1 \times 10^{11} \times 3.142 \times (3 \times 10^{-4})^2 \times 6.3 \times 10^{-3}}{1.5}\)

\(F = \frac{2.1 \times 10^{11} \times 3.142 \times 9 \times 10^{-8} \times 6.3 \times 10^{-3}}{1.5}\)

\(F = \frac{2.1 \times 3.142 \times 9 \times 6.3 \times 10^{11-8-3}}{1.5}\)

\(F = \frac{2.1 \times 3.142 \times 9 \times 6.3 \times 10^0}{1.5}\)

\(F = \frac{374.0742}{1.5}\)

\(F \approx 249.38\) N
\(\approx 250\) N
The force applied on wire is 250 N.
In simple words: To find the force applied to stretch a wire, we use Young's modulus, which relates stress (force per unit area) to strain (change in length per original length). Rearranging this formula allows us to calculate the unknown force.

🎯 Exam Tip: Always ensure all quantities are in consistent units (SI units are preferred) before calculations. Double-check the exponent manipulations, especially when dealing with squared radius and powers of 10.

Question 8. A composite wire is prepared by joining a tungsten wire and steel wire end to end. Both the wires are of the same length and the same area of cross section. If this composite wire is suspended to a rigid support and a force is applied to its free end, it gets extended by 3.25mm. Calculate the increase in length of tungsten wire and steel wire separately.
[Given: \(Y_{\text{steel}}\) = 2 \(\times\) 10\(^{11}\) Pa, \(Y_{\text{Tungsten}}\) = 4.11 \(\times\) 10\(^{11}\) Pa]
Answer:
Solution:
Given:
Total extension \(l_s + l_T\) = 3.25 mm
\(Y_T\) = 4.11 \(\times\) 10\(^{11}\) Pa
\(Y_s\) = 2 \(\times\) 10\(^{11}\) Pa
To find: Extension in tungsten wire (\(l_T\)) and extension in steel wire (\(l_s\))
Formula: Young's modulus \(Y = \frac{FL}{Al}\). Since \(F\), \(L\), and \(A\) are same for both wires,

\(Y \propto \frac{1}{l} \implies l \propto \frac{1}{Y}\)

\(\therefore \frac{l_s}{l_T} = \frac{Y_T}{Y_s}\)
Calculation:

\(\frac{l_s}{l_T} = \frac{4.11 \times 10^{11}}{2 \times 10^{11}}\)

\(\frac{l_s}{l_T} = \frac{4.11}{2} = 2.055\)

\(\implies l_s = 2.055 l_T\)
We are given \(l_s + l_T\) = 3.25 mm.
Substitute \(l_s\):
2.055 \(l_T + l_T\) = 3.25 mm
3.055 \(l_T\) = 3.25 mm

\(l_T = \frac{3.25}{3.055} \approx 1.0638\) mm
\(l_T \approx 1.064\) mm
Now find \(l_s\):
\(l_s\) = 3.25 - \(l_T\) = 3.25 - 1.064 = 2.186 mm
The extension in tungsten wire is 1.064 mm and the extension in steel wire is 2.186 mm.
[Note: Values of Young's modulus of tungsten and steel considered above are standard values. Using them, calculation is carried out in accordance with textual method.]
In simple words: For a composite wire made of different materials under the same load and having equal initial length and cross-sectional area, the extension in each part is inversely proportional to its Young's modulus. The total extension is the sum of individual extensions.

🎯 Exam Tip: When dealing with composite materials under the same load, utilize the inverse relationship between extension and Young's modulus. Ensure you add individual extensions to match the total given extension.

Question 9. A steel wire having cross sectional area 1.2 mm\(-^2\) is stretched by a force of 120 N. If a lateral strain of 1.455 \(\times\) 10\(-^4\) is produced in the wire, calculate the Poisson's ratio.
Answer:
Solution:
Given:
A = 1.2 mm\(-^2\) = 1.2 \(\times\) 10\(-^6\) m\(-^2\)
F = 120 N
\(Y_{\text{steel}}\) = 2 \(\times\) 10\(^{11}\) N/m\(-^2\)
Lateral strain = 1.455 \(\times\) 10\(-^4\)
To find: Poisson's ratio (\(\sigma\))
Formulae:
(i) Longitudinal stress = \(\frac{F}{A}\)
(ii) Young's modulus \(Y = \frac{\text{longitudinal stress}}{\text{longitudinal strain}} \implies \text{longitudinal strain} = \frac{\text{longitudinal stress}}{Y}\)
(iii) Poisson's ratio \(\sigma = \frac{\text{lateral strain}}{\text{longitudinal strain}}\)
Calculation:
From formula (i),
Longitudinal stress = \(\frac{120 \text{ N}}{1.2 \times 10^{-6} \text{ m}^2} = 100 \times 10^6 \text{ N/m}^2 = 10^8\) N/m\(-^2\)
From formula (ii),
Longitudinal strain = \(\frac{\text{longitudinal stress}}{Y} = \frac{10^8 \text{ N/m}^2}{2 \times 10^{11} \text{ N/m}^2} = 0.5 \times 10^{-3} = 5 \times 10^{-4}\)
From formula (iii),

\(\sigma = \frac{\text{lateral strain}}{\text{longitudinal strain}} = \frac{1.455 \times 10^{-4}}{5 \times 10^{-4}}\)

\(\sigma = \frac{1.455}{5} = 0.291\)
The Poisson's ratio of steel is 0.291.
[Note: Lateral strain being ratio of two same physical quantities, is unitless. hence, value given in question is modified to 1.455 \(\times\) 10\(-^4\) to reach the answer given in textbook.]
In simple words: Poisson's ratio describes a material's tendency to deform in directions perpendicular to the applied force. It's calculated by dividing the lateral strain (change in width) by the longitudinal strain (change in length) when a material is stretched or compressed.

🎯 Exam Tip: Remember that stress and strain are unitless (though stress has units of pressure). Always convert area to m\(-^2\) before calculating stress. Poisson's ratio is a unitless quantity.

Question 10. A telephone wire 125m long and 1mm in radius is stretched to a length 125.25m when a force of 800N is applied. What is the value of Young's modulus for material of wire?
Answer:
Solution:
Given:
Original length L = 125m
Radius r = 1 mm = 1 \(\times\) 10\(-^3\) m
Extended length = 125.25m
Extension \(l\) = 125.25 - 125 = 0.25 m
Force F = 800N
To find: Young's modulus (Y)
Formula: \(Y = \frac{FL}{Al} = \frac{FL}{\pi r^2 l}\)
Calculation: From formula,

\(Y = \frac{800 \times 125}{3.142 \times (1 \times 10^{-3})^2 \times 0.25}\)

\(Y = \frac{800 \times 125}{3.142 \times 10^{-6} \times 0.25}\)

\(Y = \frac{100000}{0.7855 \times 10^{-6}}\)

\(Y = \frac{100000}{0.7855} \times 10^6\)

\(Y \approx 127396.56 \times 10^6\)

\(Y \approx 1.27396 \times 10^{5} \times 10^6\)

\(Y \approx 1.274 \times 10^{11}\) N/m\(-^2\)
The Young's modulus of telephone wire is 1.274 \(\times\) 10\(^{11}\) N/m\(-^2\).
In simple words: Young's modulus is a measure of a material's stiffness, indicating its resistance to elastic deformation under tensile or compressive stress. It's calculated by dividing the longitudinal stress by the longitudinal strain.

🎯 Exam Tip: Carefully calculate the extension (\(l\)) from the given initial and final lengths. Ensure the radius is squared in the area calculation and all units are in SI before applying the formula for Young's modulus.

Question 11. A rubber band originally 30cm long is stretched to a length of 32cm by certain load. What is the strain produced?
Answer:
Solution:
Given:
Original length L = 30 cm = 30 \(\times\) 10\(-^2\) m
Extended length = 32 cm
Change in length \(\Delta l\) = 32 cm - 30 cm = 2 cm = 2 \(\times\) 10\(-^2\) m
To find: Strain
Formula: Strain = \(\frac{\Delta l}{L}\)
Calculation: From formula,

Strain = \(\frac{2 \times 10^{-2}}{30 \times 10^{-2}}\)

Strain = \(\frac{2}{30} = \frac{1}{15} \approx 0.06667\)

Strain \(\approx 6.667 \times 10^{-2}\)
The strain produced in the wire is 6.667 \(\times\) 10\(-^2\).
In simple words: Strain is a measure of deformation, defined as the ratio of the change in length to the original length of an object. It is a unitless quantity.

🎯 Exam Tip: Strain is a dimensionless quantity, so ensure your final answer does not have units. Always use consistent units for both change in length and original length when calculating strain.

Question 12. What is the stress in a wire which is 50m long and 0.01cm\(-^2\) in cross section, if the wire bears a load of 100kg?
Answer:
Solution:
Given:
Mass M = 100 kg (This represents the load force via weight)
Length L = 50 m
Area A = 0.01 cm\(-^2\) = 0.01 \(\times\) 10\(-^4\) m\(-^2\) = 1 \(\times\) 10\(-^6\) m\(-^2\)
To find: Stress
Formula: Stress = \(\frac{F}{A} = \frac{Mg}{A}\)
Calculation: From formula, (Assume g = 9.8 m/s\(-^2\))

Stress = \(\frac{100 \times 9.8}{0.01 \times 10^{-4}}\)

Stress = \(\frac{980}{10^{-6}}\)

Stress = 980 \(\times\) 10\(^6\) N/m\(-^2\)

Stress = 9.8 \(\times\) 10\(^2 \times\) 10\(^6\) N/m\(-^2\) = 9.8 \(\times\) 10\(^8\) N/m\(-^2\)
The stress in the wire is 9.8 \(\times\) 10\(^8\) N/m\(-^2\).
In simple words: Stress in a wire is the internal restoring force per unit area. When a load is applied, the force is its weight, and stress is calculated by dividing this force by the wire's cross-sectional area.

🎯 Exam Tip: Convert all given units to SI units (e.g., cm\(-^2\) to m\(-^2\)). Remember that "load of 100kg" implies a force of \(Mg\). Length of the wire is not required for stress calculation here, but area is crucial.

Question 13. What is the strain in a cable of original length 50m whose length increases by 2.5cm when a load is lifted?
Answer:
Solution:
Given:
Original length L = 50m
Change in length \(\Delta l\) = 2.5cm = 2.5 \(\times\) 10\(-^2\) m
To find: Strain
Formula: Strain = \(\frac{\Delta l}{L}\)
Calculation: From formula,

Strain = \(\frac{2.5 \times 10^{-2}}{50}\)

Strain = \(\frac{2.5}{50} \times 10^{-2}\)

Strain = 0.05 \(\times\) 10\(-^2\)

Strain = 5 \(\times\) 10\(-^4\)
The Strain produced in wire is 5 \(\times\) 10\(-^4\).
In simple words: Strain is a dimensionless measure of how much an object deforms relative to its original size, calculated by dividing the change in length by the original length.

🎯 Exam Tip: Pay close attention to unit conversions for length (cm to m) to ensure consistent units in the strain calculation. Strain, being a ratio of lengths, is unitless.

Can You Recall? (Textbook Page No. 100)

Question 1.
(1) Can you name a few objects which change their shape and size on application of a force and regain their original shape and size when the force is removed?
(2) Can you name objects which do not regain their original shape and size when the external force is removed?
Answer:
(1) Objects such as rubber, metals, quartz, etc. change their shape and size on application of a force (within specific limit) and regain their original shape and size when the force is removed.
(2) Objects such as putty, clay, thick mud, etc. do not regain their original shape and size when the external force is removed.
In simple words: Elastic objects return to their original shape after a deforming force is removed, while plastic objects retain their deformed shape.

🎯 Exam Tip: Understand the fundamental definitions of elastic and plastic bodies. Be ready to provide common examples for each to illustrate the concepts.

Can You Tell? (Textbook Page No. 107)

Question 1. Why does a rubber band become loose after repeated use?
Answer:
1. After repeated use of rubber band, its stress-strain curve does not remain linear.
2. In such case, since rubber crosses its elastic limit, there is a permanent set formed on the rubber due to which it becomes loose.
In simple words: Repeated stretching causes a rubber band to exceed its elastic limit, leading to permanent deformation (permanent set) and making it lose its original elasticity or tightness.

🎯 Exam Tip: Relate the phenomenon to the concepts of elastic limit and permanent set. Explaining this in terms of the stress-strain curve adds depth to the answer.

Can You Tell? (Textbook Page No.111)

Question 1.
(i) It is difficult to run fast on sand.
(ii) It is easy to roll than pull a barrel along a road.
(iii) An inflated tyre rolls easily than a flat tyre.
(iv) Friction is a necessary evil.
Answer:
(i) 1. The intermolecular space between crystals of sand is very large as compared to that in a rigid surface.
2. Thus, there are number of depressions at the points of contact of feet and sand surface.
3. Projections and depressions between sand and feet are not completely interlocked.
4. Thus, action and reaction force become unbalanced. The horizontal component of force helps to move forward and vertical component of the force resist to move.
Hence, it becomes difficult to run fast on sand.

(ii) 1. When a barrel is pulled along a road, the friction between the tyres and road is kinetic friction, but when it rolls along the road it undergoes rolling friction.
2. The force of kinetic friction is greater than force of rolling friction.
Hence, it is easy to roll than pull a barrel along a road.

(iii) 1. When the tyre is inflated, the pressure inside the tyre is reducing the normal force between tyre and the ground, and thus reducing the friction between the tyre and the road.
2. When the tyre gets deflated, it gets deformed during rolling, the supplied energy is used up in changing the shape and not overcoming the friction, and thus due to deformation, friction increases.
Hence, an inflated tyre rolls easily than a flat tyre.

(iv) 1. Friction helps us to walk, hold objects in hand, lift objects and without friction we cannot walk, we cannot grip or hold objects with our hands,
2. Friction is responsible for wear and tear of various part of machines, it produces heat in different parts of machine and also produces noise but it also helps in ball bearing or connecting screws.
Hence, friction is said to be a necessary evil because it is useful as well as harmful.
In simple words: This question explores various practical applications and effects of friction, explaining why certain actions are easier or harder based on the type and amount of friction involved, and acknowledging friction's dual nature as both beneficial and detrimental.

🎯 Exam Tip: For such multi-part conceptual questions, provide a clear, concise explanation for each part. Focus on distinguishing between static, kinetic, and rolling friction, and their practical implications.