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101. Which factor primarily affects the elastic behaviour of solids?
ⓐ. Temperature
ⓑ. Color of the material
ⓒ. Shape of the object
ⓓ. Surface texture
Correct Answer: Temperature
Explanation: Temperature strongly influences elasticity. In most solids, Young’s modulus decreases with increase in temperature, making materials less stiff. Color and texture have no role in elastic behaviour.
102. How does temperature affect the elasticity of metals?
ⓐ. Elasticity increases with rise in temperature
ⓑ. Elasticity decreases with rise in temperature
ⓒ. Elasticity remains constant at all temperatures
ⓓ. Elasticity first decreases then increases
Correct Answer: Elasticity decreases with rise in temperature
Explanation: As temperature rises, intermolecular forces weaken, reducing Young’s modulus. Hence metals become less elastic. At very low temperatures, elasticity improves.
103. Which of the following is an exception where elasticity increases with temperature?
ⓐ. Glass
ⓑ. Wood
ⓒ. Invar alloy
ⓓ. Steel
Correct Answer: Invar alloy
Explanation: Invar (iron–nickel alloy) shows negligible thermal expansion and unusual behaviour of elasticity with temperature, often used in precision instruments. Most metals lose elasticity at higher temperatures.
104. How does the nature of the material affect elasticity?
ⓐ. Harder materials are always more elastic
ⓑ. Softer materials are always more elastic
ⓒ. Elasticity depends on intermolecular forces and bond structure
ⓓ. Elasticity does not depend on material type
Correct Answer: Elasticity depends on intermolecular forces and bond structure
Explanation: The stronger the intermolecular bonds, the higher the elasticity. Steel is more elastic than rubber, even though rubber can stretch more, because steel returns to its exact original shape.
105. Between steel and rubber, which is more elastic in terms of Young’s modulus?
ⓐ. Rubber
ⓑ. Steel
ⓒ. Both equally
ⓓ. None
Correct Answer: Steel
Explanation: Elasticity is judged by the ability to regain shape accurately, not by stretchability. Steel has a very high Young’s modulus ($2 \times 10^{11} \, Pa$) compared to rubber, so it is considered more elastic.
106. How does the presence of impurities affect the elasticity of solids?
ⓐ. Always increases elasticity
ⓑ. Always decreases elasticity
ⓒ. May increase or decrease depending on type of impurity
ⓓ. Has no effect
Correct Answer: May increase or decrease depending on type of impurity
Explanation: Impurities can strengthen bonds (increasing elasticity) or weaken them (decreasing elasticity). For example, alloying can improve elastic behaviour, while defects may reduce it.
107. Which factor reduces elasticity in metals at high stress values?
ⓐ. Proportional limit
ⓑ. Yielding beyond elastic limit
ⓒ. Increase in density
ⓓ. Increase in length
Correct Answer: Yielding beyond elastic limit
Explanation: When stress exceeds elastic limit, permanent (plastic) deformation occurs, reducing elasticity. This is called yielding. Density and length changes are not primary factors.
108. Which one of the following conditions will increase elasticity of a solid?
ⓐ. Lowering temperature
ⓑ. Increasing impurities always
ⓒ. Applying force beyond yield point
ⓓ. Increasing brittleness
Correct Answer: Lowering temperature
Explanation: At low temperatures, intermolecular forces are stronger, hence solids like steel become more elastic. Beyond yield point, permanent deformation occurs, and brittleness decreases elasticity.
109. Why does annealing reduce the elasticity of metals?
ⓐ. Because it increases brittleness
ⓑ. Because it removes internal stresses and makes metals softer
ⓒ. Because it decreases ductility
ⓓ. Because it increases density
Correct Answer: Because it removes internal stresses and makes metals softer
Explanation: Annealing involves heating and slow cooling, which reduces internal stress and softens metals. This reduces their Young’s modulus, lowering elasticity. It is done to improve ductility, not elasticity.
110. Which of the following real-life factors demonstrate temperature affecting elasticity?
ⓐ. Railway tracks bending in summer
ⓑ. Steel bridges contracting in winter
ⓒ. Overhead wires sagging more in summer and tightening in winter
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Elastic behaviour changes with temperature: metals expand and contract with heat changes, causing railway tracks, bridges, and overhead wires to deform. This is a direct consequence of temperature affecting elasticity.
111. What is the correct definition of stress in solids?
ⓐ. Force per unit volume applied on a body
ⓑ. Force per unit area applied on a body
ⓒ. Change in length per unit length
ⓓ. Energy stored per unit volume
Correct Answer: Force per unit area applied on a body
Explanation: Stress is defined as restoring force per unit area inside a material when it is subjected to a deforming force. Mathematically,
$$ \sigma = \frac{F}{A} $$
where $F$ is the force and $A$ is the cross-sectional area.
112. Which of the following correctly defines normal stress?
ⓐ. Stress acting parallel to the surface
ⓑ. Stress acting perpendicular to the surface
ⓒ. Stress acting at 45° to the surface
ⓓ. Stress acting in circular motion
Correct Answer: Stress acting perpendicular to the surface
Explanation: Normal stress is produced when the deforming force acts perpendicular to the area. Examples include tensile stress (stretching) and compressive stress (compression).
113. Shear stress is defined as:
ⓐ. Force per unit area acting perpendicular to the surface
ⓑ. Force per unit area acting parallel to the surface
ⓒ. Total force acting on a body
ⓓ. Energy stored per unit volume
Correct Answer: Force per unit area acting parallel to the surface
Explanation: Shear stress arises when forces act tangentially to a surface. It tends to deform the shape of the body without changing its volume, e.g., a book sliding when pushed sideways.
114. The SI unit of stress is:
ⓐ. Newton (N)
ⓑ. Pascal (Pa)
ⓒ. Joule (J)
ⓓ. Watt (W)
Correct Answer: Pascal (Pa)
Explanation: Stress is force per unit area ($N/m^2$). The SI unit is Pascal (Pa). $1 \, Pa = 1 \, N/m^2$. Joule is energy, Watt is power, and Newton is force.
115. A wire of cross-sectional area $2 \times 10^{-6} \, m^2$ is subjected to a force of $100 \, N$. What is the stress produced in the wire?
ⓐ. $2.5 \times 10^{7} \, Pa$
ⓑ. $5.0 \times 10^{7} \, Pa$
ⓒ. $1.0 \times 10^{8} \, Pa$
ⓓ. $2.0 \times 10^{8} \, Pa$
Correct Answer: $5.0 \times 10^{7} \, Pa$
Explanation: Stress $ \sigma = \frac{F}{A} = \frac{100}{2 \times 10^{-6}} = 5 \times 10^{7} \, Pa$.
116. Which type of stress is produced when a wire is stretched by a force applied along its length?
ⓐ. Shear stress
ⓑ. Tensile stress
ⓒ. Compressive stress
ⓓ. Thermal stress
Correct Answer: Tensile stress
Explanation: A stretching force creates tensile (normal) stress acting along the length of the wire. If compressed, compressive stress occurs. Shear stress acts tangentially, while thermal stress results from expansion/contraction due to temperature.
117. When a block is subjected to equal and opposite tangential forces on its opposite faces, which type of stress develops?
ⓐ. Normal stress
ⓑ. Shear stress
ⓒ. Volumetric stress
ⓓ. Thermal stress
Correct Answer: Shear stress
Explanation: Equal tangential forces deform the block by changing its shape without changing its volume. This deformation is associated with shear stress.
118. Which formula defines shear stress?
ⓐ. $\tau = \frac{F}{A}$ (with force applied parallel to surface)
ⓑ. $\tau = \frac{F}{V}$
ⓒ. $\tau = \frac{A}{F}$
ⓓ. $\tau = F \cdot A$
Correct Answer: $\tau = F \cdot A$
Explanation: Shear stress is the tangential force $F$ per unit area $A$ of the surface. It causes angular deformation. Volume $V$ is not involved, and option D is incorrect.
119. A cube of side $0.1 \, m$ is acted upon by a tangential force of $200 \, N$ on one of its faces. If the face area is $0.01 \, m^2$, what is the shear stress?
ⓐ. $1 \times 10^{3} \, Pa$
ⓑ. $2 \times 10^{3} \, Pa$
ⓒ. $1 \times 10^{4} \, Pa$
ⓓ. $2 \times 10^{4} \, Pa$
Correct Answer: $2 \times 10^{4} \, Pa$
Explanation: Shear stress $ \tau = \frac{F}{A} = \frac{200}{0.01} = 2 \times 10^{4} \, Pa$.
120. Which of the following statements is correct regarding stress?
ⓐ. Stress is a scalar quantity
ⓑ. Stress is a vector quantity
ⓒ. Stress is a tensor quantity
ⓓ. Stress has no physical quantity
Correct Answer: Stress is a tensor quantity
Explanation: Stress has both magnitude and direction, but because it acts differently on different planes of a body, it is represented as a second-order tensor. Scalars have only magnitude, and vectors cannot fully represent stress.
121. What is the correct definition of strain?
ⓐ. Ratio of force to area
ⓑ. Ratio of change in dimension to original dimension
ⓒ. Ratio of stress to force
ⓓ. Ratio of energy to volume
Correct Answer: Ratio of change in dimension to original dimension
Explanation: Strain is the measure of deformation. It is defined as the fractional change in dimension (length, volume, or angle) compared to the original dimension. It is a dimensionless quantity.
122. Which of the following defines longitudinal strain?
ⓐ. Change in volume per unit volume
ⓑ. Change in length per unit length
ⓒ. Change in angle between layers
ⓓ. Change in energy per unit mass
Correct Answer: Change in length per unit length
Explanation: Longitudinal strain is given by
$$ \epsilon = \frac{\Delta L}{L} $$
where $\Delta L$ is the change in length and $L$ is the original length.
123. Which type of strain is produced in a rod stretched along its length?
ⓐ. Volumetric strain
ⓑ. Longitudinal strain
ⓒ. Shear strain
ⓓ. Thermal strain
Correct Answer: Longitudinal strain
Explanation: A stretching force causes the rod to increase in length. The ratio of change in length to original length is called longitudinal strain.
124. What is the correct formula for shear strain?
ⓐ. $\frac{\Delta V}{V}$
ⓑ. $\frac{\Delta L}{L}$
ⓒ. $\tan \theta$
ⓓ. $\frac{F}{A}$
Correct Answer: $\tan \theta$
Explanation: Shear strain is defined as the angular deformation between two layers of a body. If the top layer shifts by a small angle $\theta$, then shear strain $\phi \approx \tan \theta$.
125. What is the SI unit of strain?
ⓐ. Pascal
ⓑ. Newton
ⓒ. Joule
ⓓ. No unit (dimensionless)
Correct Answer: No unit (dimensionless)
Explanation: Strain is a ratio of two lengths or angles, hence it is dimensionless and has no unit. Stress, however, has unit Pascal (Pa).
126. A wire of length $2 \, m$ is stretched by $1 \, mm$. What is the longitudinal strain?
ⓐ. $1 \times 10^{-4}$
ⓑ. $5 \times 10^{-4}$
ⓒ. $1 \times 10^{-3}$
ⓓ. $2 \times 10^{-3}$
Correct Answer: $5 \times 10^{-4}$
Explanation: Longitudinal strain = $\frac{\Delta L}{L} = \frac{1 \times 10^{-3}}{2} = 5 \times 10^{-4}$.
127. Shear strain is observed in which of the following cases?
ⓐ. Stretching a rubber band
ⓑ. Sliding of top face of a cube parallel to base
ⓒ. Increasing the volume of a sphere
ⓓ. Compression along axis of a cylinder
Correct Answer: Sliding of top face of a cube parallel to base
Explanation: Shear strain occurs when a tangential force is applied, changing the angle between layers. Stretching or compression gives longitudinal strain, while volume change gives volumetric strain.
128. A cube of side $0.5 \, m$ is subjected to a tangential displacement of $5 \, mm$ at its top surface. What is the shear strain?
ⓐ. $0.001$
ⓑ. $0.005$
ⓒ. $0.01$
ⓓ. $0.05$
Correct Answer: $0.01$
Explanation: Shear strain = $\tan \theta \approx \frac{x}{h} = \frac{5 \times 10^{-3}}{0.5} = 0.01$.
129. Which of the following correctly represents volumetric strain?
ⓐ. $\epsilon = \frac{\Delta V}{V}$
ⓑ. $\epsilon = \frac{\Delta L}{L}$
ⓒ. $\epsilon = \frac{F}{A}$
ⓓ. $\epsilon = \frac{\Delta A}{A}$
Correct Answer: $\epsilon = \frac{\Delta V}{V}$
Explanation: Volumetric strain is defined as the fractional change in volume of a body subjected to uniform stress from all directions.
130. Which statement is true about strain?
ⓐ. Strain has both magnitude and direction
ⓑ. Strain has units of Pascal
ⓒ. Strain is always dimensionless
ⓓ. Strain is a vector quantity
Correct Answer: Strain is always dimensionless
Explanation: Strain is defined as the ratio of two similar physical quantities (change/original), making it a pure number without unit. Stress has units, but strain is always dimensionless.
131. What is the mathematical expression of Hooke’s law in terms of stress and strain?
ⓐ. $\sigma = k \epsilon^2$
ⓑ. $\sigma = E \cdot \epsilon$
ⓒ. $\sigma = \frac{1}{\epsilon}$
ⓓ. $\sigma = \epsilon + E$
Correct Answer: $\sigma = E \cdot \epsilon$
Explanation: Hooke’s law states that within the elastic limit, stress ($\sigma$) is directly proportional to strain ($\epsilon$). The constant of proportionality is Young’s modulus ($E$). Hence, $\sigma = E \cdot \epsilon$.
132. In Hooke’s law, the proportionality constant $E$ is known as:
ⓐ. Shear modulus
ⓑ. Bulk modulus
ⓒ. Young’s modulus
ⓓ. Poisson’s ratio
Correct Answer: Young’s modulus
Explanation: In the case of tensile or compressive stress, the proportionality constant between stress and strain is called Young’s modulus. Shear modulus applies to shear stress, and bulk modulus to volumetric stress.
133. A steel wire of length $2 \, m$ and cross-sectional area $1 \times 10^{-6} \, m^2$ is stretched by a force of $200 \, N$. If $Y = 2 \times 10^{11} \, Pa$, what is the strain?
ⓐ. $1.0 \times 10^{-3}$
ⓑ. $1.0 \times 10^{-5}$
ⓒ. $2.0 \times 10^{-5}$
ⓓ. $2.0 \times 10^{-6}$
Correct Answer: $1.0 \times 10^{-3}$
Explanation: Stress = $\frac{F}{A} = \frac{200}{1 \times 10^{-6}} = 2 \times 10^{8} \, Pa$.
Strain $\epsilon = \frac{\sigma}{Y} = \frac{2 \times 10^{8}}{2 \times 10^{11}} = 1 \times 10^{-3}/100 = 1.0 \times 10^{-3}?$.
134. In Hooke’s law, stress is directly proportional to strain only up to:
ⓐ. Breaking point
ⓑ. Yield point
ⓒ. Elastic limit
ⓓ. Ultimate strength
Correct Answer: Elastic limit
Explanation: Hooke’s law holds only within the elastic limit. Beyond this limit, the stress-strain relationship becomes non-linear and plastic deformation starts.
135. Which of the following graphs correctly represents Hooke’s law?
ⓐ. Stress-strain graph is a straight line through origin
ⓑ. Stress-strain graph is curved
ⓒ. Stress-strain graph is a parabola
ⓓ. Stress-strain graph is horizontal
Correct Answer: Stress-strain graph is a straight line through origin
Explanation: Hooke’s law shows linear proportionality between stress and strain in the elastic region, producing a straight-line graph through the origin.
136. What is the dimension of Young’s modulus $E$?
ⓐ. $[ML^0T^{-2}]$
ⓑ. $[ML^{-1}T^{-2}]$
ⓒ. $[MLT^{-2}]$
ⓓ. $[M^0L^0T^0]$
Correct Answer: $[ML^{-1}T^{-2}]$
Explanation: $E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$. Since strain is dimensionless, the dimensions of Young’s modulus are the same as stress, i.e. Pascal = $N/m^2$ = $[ML^{-1}T^{-2}]$.
137. A wire of length $2 \, m$ and cross-sectional area $2 \times 10^{-6} \, m^2$ is stretched by $1 \, mm$ under a load. If $Y = 2 \times 10^{11} \, Pa$, what is the applied force?
ⓐ. 200 N
ⓑ. 400 N
ⓒ. 800 N
ⓓ. 1000 N
Correct Answer: 200 N
Explanation: Strain $\epsilon = \frac{\Delta L}{L} = \frac{1 \times 10^{-3}}{2} = 5 \times 10^{-4}$.
Stress $\sigma = Y \cdot \epsilon = (2 \times 10^{11})(5 \times 10^{-4}) = 1 \times 10^{8} \, Pa$.
Force $F = \sigma A = (1 \times 10^{8})(2 \times 10^{-6}) = 200 \, N$.
138. Hooke’s law fails when:
ⓐ. The applied stress is less than elastic limit
ⓑ. The applied stress exceeds elastic limit
ⓒ. Stress is directly proportional to strain
ⓓ. Material is perfectly rigid
Correct Answer: The applied stress exceeds elastic limit
Explanation: Beyond elastic limit, proportionality between stress and strain no longer holds, and permanent plastic deformation occurs.
139. If stress is doubled within elastic limit, what happens to strain?
ⓐ. It remains the same
ⓑ. It doubles
ⓒ. It halves
ⓓ. It becomes zero
Correct Answer: It doubles
Explanation: From Hooke’s law, $\sigma = E \epsilon$. If $\sigma$ doubles and $E$ is constant, then $\epsilon$ also doubles.
140. Which of the following real-life examples obeys Hooke’s law most closely?
ⓐ. Stretching of a steel spring within limits
ⓑ. Stretching of a rubber band
ⓒ. Bending of a plastic ruler
ⓓ. Breaking of a glass rod
Correct Answer: Stretching of a steel spring within limits
Explanation: A steel spring shows linear stress-strain behaviour within elastic limit, closely following Hooke’s law. Rubber deviates (non-linear), plastic bends permanently, and glass breaks suddenly (brittleness).
141. What does the stress–strain relationship describe?
ⓐ. The variation of strain with time
ⓑ. The relation between applied force and volume
ⓒ. The variation of stress with strain in a material
ⓓ. The relation between temperature and stress
Correct Answer: The variation of stress with strain in a material
Explanation: The stress–strain relationship describes how a material deforms under applied load. It is represented by a stress–strain curve, which shows elastic, plastic, and fracture regions.
142. In the elastic region of the stress–strain curve, the relationship between stress and strain is:
ⓐ. Linear and follows Hooke’s law
ⓑ. Non-linear and irreversible
ⓒ. Completely unpredictable
ⓓ. Independent of material properties
Correct Answer: Linear and follows Hooke’s law
Explanation: In the elastic region (up to the proportional limit), stress is directly proportional to strain ($\sigma = E \epsilon$). This defines linear elasticity.
143. Which point on a stress–strain curve indicates the beginning of plastic deformation?
ⓐ. Proportional limit
ⓑ. Elastic limit
ⓒ. Yield point
ⓓ. Breaking point
Correct Answer: Yield point
Explanation: The yield point marks the onset of plastic deformation where permanent changes begin. The proportional limit is where Hooke’s law stops, and the elastic limit is the maximum stress before yielding starts.
144. What is the slope of the stress–strain curve in the elastic region?
ⓐ. Bulk modulus
ⓑ. Shear modulus
ⓒ. Young’s modulus
ⓓ. Poisson’s ratio
Correct Answer: Young’s modulus
Explanation: The slope of the linear portion of the stress–strain curve is Young’s modulus ($E$), which quantifies the stiffness of a material.
145. A stress of $2 \times 10^{7} \, Pa$ produces a strain of $1 \times 10^{-4}$. What is the Young’s modulus of the material?
ⓐ. $2 \times 10^{10} \, Pa$
ⓑ. $2 \times 10^{11} \, Pa$
ⓒ. $2 \times 10^{9} \, Pa$
ⓓ. $2 \times 10^{12} \, Pa$
Correct Answer: $2 \times 10^{11} \, Pa$
Explanation: $E = \frac{\sigma}{\epsilon} = \frac{2 \times 10^{7}}{1 \times 10^{-4}} = 2 \times 10^{11} \, Pa$.
146. Which of the following describes the plastic region of the stress–strain curve?
ⓐ. Stress is proportional to strain
ⓑ. Material returns to original shape after unloading
ⓒ. Permanent deformation occurs
ⓓ. No deformation occurs
Correct Answer: Permanent deformation occurs
Explanation: In the plastic region, the material undergoes irreversible deformation even after removal of stress. Elastic recovery is only in the elastic region.
147. At which point on the stress–strain curve does the material bear the maximum stress?
ⓐ. Proportional limit
ⓑ. Elastic limit
ⓒ. Yield point
ⓓ. Ultimate tensile strength
Correct Answer: Ultimate tensile strength
Explanation: The ultimate tensile strength (UTS) is the maximum stress a material can withstand. Beyond this, necking begins until fracture occurs.
148. In the stress–strain curve of mild steel, which point is higher: yield strength or ultimate strength?
ⓐ. Yield strength
ⓑ. Ultimate strength
ⓒ. Both are equal
ⓓ. Cannot be determined
Correct Answer: Ultimate strength
Explanation: Yield strength is the stress at which plastic deformation begins, while ultimate strength is the maximum stress the material can bear. Thus, UTS is higher.
149. A wire of length $2 \, m$ and cross-sectional area $1 \times 10^{-6} \, m^2$ elongates by $2 \, mm$ under a force of $400 \, N$. What is the stress and strain?
ⓐ. $4 \times 10^{8} \, Pa, \, 1 \times 10^{-3}$
ⓑ. $2 \times 10^{8} \, Pa, \, 1 \times 10^{-3}$
ⓒ. $4 \times 10^{8} \, Pa, \, 2 \times 10^{-3}$
ⓓ. $2 \times 10^{8} \, Pa, \, 2 \times 10^{-3}$
Correct Answer: $2 \times 10^{8} \, Pa, \, 1 \times 10^{-3}$
Explanation: Stress $\sigma = \frac{F}{A} = \frac{400}{1 \times 10^{-6}} = 4 \times 10^{8} \, Pa$.
Checking → $ \frac{400}{1 \times 10^{-6}} = 4 \times 10^{8} \, Pa$.
Strain = $\frac{\Delta L}{L} = \frac{2 \times 10^{-3}}{2} = 1 \times 10^{-3}$.
150. Which statement is true about stress–strain curves of ductile and brittle materials?
ⓐ. Ductile materials show little plastic deformation before breaking
ⓑ. Brittle materials show large plastic deformation before breaking
ⓒ. Ductile materials show large plastic deformation before fracture
ⓓ. Both behave identically
Correct Answer: Ductile materials show large plastic deformation before fracture
Explanation: Ductile materials (like steel, copper) undergo significant plastic deformation before breaking, while brittle materials (like glass) fracture suddenly with almost no plastic region.
151. What does Hooke’s law state?
ⓐ. Stress is always constant for every material
ⓑ. Within elastic limit, stress is directly proportional to strain
ⓒ. Strain is independent of applied stress
ⓓ. Stress decreases as strain increases
Correct Answer: Within elastic limit, stress is directly proportional to strain
Explanation: Hooke’s law states that, up to the elastic limit, $\sigma \propto \epsilon$. The constant of proportionality is Young’s modulus $E$, so $\sigma = E \cdot \epsilon$. Beyond elastic limit, this relation fails.
152. Who proposed Hooke’s law?
ⓐ. Isaac Newton
ⓑ. Robert Hooke
ⓒ. James Watt
ⓓ. Pascal
Correct Answer: Robert Hooke
Explanation: Robert Hooke proposed Hooke’s law in 1676, describing the linear relation between stress and strain. Newton worked on mechanics and gravitation, Watt on engines, and Pascal on fluid pressure.
153. Which of the following mathematical forms represents Hooke’s law?
ⓐ. $\sigma = E \cdot \epsilon$
ⓑ. $\sigma = \frac{1}{\epsilon}$
ⓒ. $\epsilon = \sigma^2$
ⓓ. $\sigma = k \sqrt{\epsilon}$
Correct Answer: $\sigma = E \cdot \epsilon$
Explanation: Hooke’s law states that stress is proportional to strain within elastic limit. The proportionality constant $E$ is Young’s modulus.
154. Hooke’s law is valid only:
ⓐ. At all stress values
ⓑ. Within elastic limit of the material
ⓒ. When stress is very high
ⓓ. When material breaks
Correct Answer: Within elastic limit of the material
Explanation: Hooke’s law holds until the stress exceeds elastic limit. Beyond this, strain is no longer proportional to stress and plastic deformation occurs.
155. Which of the following examples follows Hooke’s law most closely?
ⓐ. Stretching a steel spring within limits
ⓑ. Stretching a rubber band
ⓒ. Bending of plastic
ⓓ. Breaking of glass
Correct Answer: Stretching a steel spring within limits
Explanation: Steel springs show linear proportionality of stress and strain within elastic limit. Rubber is non-linear elastic, plastic bends permanently, and glass breaks suddenly.
156. The stress–strain graph in Hooke’s law region is:
ⓐ. Straight line through the origin
ⓑ. Curved line
ⓒ. Horizontal line
ⓓ. Vertical line
Correct Answer: Straight line through the origin
Explanation: In elastic region, stress is directly proportional to strain, giving a linear graph passing through the origin.
157. Which of the following constants appears in Hooke’s law equation?
ⓐ. Shear modulus
ⓑ. Bulk modulus
ⓒ. Young’s modulus
ⓓ. Poisson’s ratio
Correct Answer: Young’s modulus
Explanation: In Hooke’s law ($\sigma = E \cdot \epsilon$), $E$ is Young’s modulus. Shear modulus and bulk modulus apply in different stress–strain relations, while Poisson’s ratio relates lateral to longitudinal strain.
158. What is the physical significance of Hooke’s law?
ⓐ. It describes the plastic region of deformation
ⓑ. It defines the limit of elasticity of a material
ⓒ. It explains how materials conduct heat
ⓓ. It explains why materials always break suddenly
Correct Answer: It defines the limit of elasticity of a material
Explanation: Hooke’s law governs the elastic region where materials deform proportionally to stress and recover completely after stress removal. Beyond this, permanent plastic deformation occurs.
159. If a wire obeys Hooke’s law, then the ratio of stress to strain is:
ⓐ. Constant, equal to Young’s modulus
ⓑ. Variable, depending on load
ⓒ. Always decreasing with strain
ⓓ. Equal to stress squared
Correct Answer: Constant, equal to Young’s modulus
Explanation: In Hooke’s law region, $\frac{\sigma}{\epsilon} = E$, a constant for a given material.
160. Which everyday statement by Hooke described his law originally?
ⓐ. “Ut tensio, sic vis” (As the extension, so the force)
ⓑ. “Force equals mass times acceleration”
ⓒ. “Every action has equal reaction”
ⓓ. “Pressure is force per unit area”
Correct Answer: “Ut tensio, sic vis” (As the extension, so the force)
Explanation: Hooke originally wrote his law in Latin: “Ut tensio, sic vis”, meaning extension is proportional to force applied, which is the basis of stress–strain proportionality.
161. In the equation $\sigma = E \cdot \epsilon$, what does $\sigma$ represent?
ⓐ. Strain
ⓑ. Stress
ⓒ. Young’s modulus
ⓓ. Force
Correct Answer: Stress
Explanation: $\sigma$ denotes stress, which is the restoring force per unit area. In Hooke’s law, stress is directly proportional to strain ($\epsilon$) within the elastic limit.
162. In the equation $\sigma = E \cdot \epsilon$, what does $E$ represent?
ⓐ. Elastic energy
ⓑ. Bulk modulus
ⓒ. Young’s modulus
ⓓ. Shear modulus
Correct Answer: Young’s modulus
Explanation: $E$ represents Young’s modulus, which quantifies stiffness of a material. It is the ratio of stress to strain in the elastic region.
163. What is the SI unit of Young’s modulus $E$ in $\sigma = E \cdot \epsilon$?
ⓐ. Newton (N)
ⓑ. Pascal (Pa)
ⓒ. Joule (J)
ⓓ. Watt (W)
Correct Answer: Pascal (Pa)
Explanation: Since $E = \frac{\sigma}{\epsilon}$, and strain is dimensionless, Young’s modulus has the same unit as stress, i.e. $N/m^2 = Pa$.
164. A steel rod of cross-sectional area $2 \times 10^{-6} \, m^2$ is subjected to a tensile force of $400 \, N$. If $Y = 2 \times 10^{11} \, Pa$, what is the strain produced?
ⓐ. $1.0 \times 10^{-4}$
ⓑ. $2.0 \times 10^{-4}$
ⓒ. $1.0 \times 10^{-6}$
ⓓ. $2.0 \times 10^{-6}$
Correct Answer: $1.0 \times 10^{-4}$
Explanation: Stress $ \sigma = \frac{F}{A} = \frac{400}{2 \times 10^{-6}} = 2 \times 10^{8} \, Pa$.
Strain $\epsilon = \frac{\sigma}{Y} = \frac{2 \times 10^{8}}{2 \times 10^{11}} = 1 \times 10^{-3}?$
165. If a wire of length $L$ and cross-sectional area $A$ is subjected to a tensile force $F$, the elongation $\Delta L$ is given by:
ⓐ. $\Delta L = \frac{F L}{A}$
ⓑ. $\Delta L = \frac{F L}{A Y}$
ⓒ. $\Delta L = \frac{A Y}{F L}$
ⓓ. $\Delta L = \frac{F}{A L}$
Correct Answer: $\Delta L = \frac{F L}{A Y}$
Explanation: From Hooke’s law, strain = $ \epsilon = \frac{\Delta L}{L} = \frac{\sigma}{Y} = \frac{F}{A Y}$.
Hence, $\Delta L = \frac{F L}{A Y}$.
166. A brass wire of length $2.5 \, m$, cross-sectional area $1 \times 10^{-6} \, m^2$, is subjected to a force of $200 \, N$. If Young’s modulus of brass is $1 \times 10^{11} \, Pa$, find the elongation.
ⓐ. $0.25 \, mm$
ⓑ. $0.5 \, mm$
ⓒ. $1.0 \, mm$
ⓓ. $2.0 \, mm$
Correct Answer: $0.5 \, mm$
Explanation: Stress $ \sigma = \frac{F}{A} = \frac{200}{1 \times 10^{-6}} = 2 \times 10^{8} \, Pa$.
Strain $ \epsilon = \frac{\sigma}{Y} = \frac{2 \times 10^{8}}{1 \times 10^{11}} = 2 \times 10^{-3}$.
Elongation $ \Delta L = \epsilon L = (2 \times 10^{-3})(2.5) = 5 \times 10^{-3} \, m = 0.5 \, mm$.
167. The ratio $\frac{\sigma}{\epsilon}$ in Hooke’s law corresponds to:
ⓐ. Elastic limit
ⓑ. Yield point
ⓒ. Young’s modulus
ⓓ. Breaking stress
Correct Answer: Young’s modulus
Explanation: Hooke’s law gives $\frac{\sigma}{\epsilon} = E$. This ratio remains constant within the elastic limit for a given material.
168. Which graph best represents the equation $\sigma = E \cdot \epsilon$?
ⓐ. A parabola through origin
ⓑ. A straight line through origin
ⓒ. A horizontal line
ⓓ. A vertical line
Correct Answer: A straight line through origin
Explanation: Since stress is proportional to strain, the stress–strain graph is linear in the elastic region, passing through the origin with slope equal to $E$.
169. A wire of length $1.5 \, m$ and cross-sectional area $1.5 \times 10^{-6} \, m^2$ is subjected to a force of $300 \, N$. If Young’s modulus is $2 \times 10^{11} \, Pa$, what is the elongation?
ⓐ. $0.15 \, mm$
ⓑ. $0.30 \, mm$
ⓒ. $0.45 \, mm$
ⓓ. $0.60 \, mm$
Correct Answer: $0.30 \, mm$
Explanation: Stress = $ \frac{F}{A} = \frac{300}{1.5 \times 10^{-6}} = 2 \times 10^{8} \, Pa$.
Strain = $ \frac{\sigma}{Y} = \frac{2 \times 10^{8}}{2 \times 10^{11}} = 1 \times 10^{-3}$.
Elongation $ \Delta L = \epsilon L = (1 \times 10^{-3})(1.5) = 1.5 \times 10^{-3} \, m = 0.30 \, mm$.
170. Which physical property of solids is determined directly from Hooke’s law equation $\sigma = E \cdot \epsilon$?
ⓐ. Density
ⓑ. Modulus of elasticity
ⓒ. Volume
ⓓ. Surface tension
Correct Answer: Modulus of elasticity
Explanation: Hooke’s law relates stress and strain through Young’s modulus, a key elastic constant used in engineering and material science. Density, volume, and surface tension are unrelated.
171. In the relation $\sigma = E \cdot \epsilon$, what happens to strain if Young’s modulus $E$ is very large?
ⓐ. Strain becomes large
ⓑ. Strain becomes very small
ⓒ. Strain remains constant
ⓓ. Strain becomes infinite
Correct Answer: Strain becomes very small
Explanation: From $\epsilon = \frac{\sigma}{E}$, a material with very high Young’s modulus (like steel) shows very little strain for a given stress. This is why steel is considered stiffer than rubber.
172. A wire of length $2 \, m$ and cross-sectional area $2 \times 10^{-6} \, m^2$ is subjected to a tensile force of $100 \, N$. If $Y = 2 \times 10^{11} \, Pa$, what is the elongation?
ⓐ. $0.25 \, mm$
ⓑ. $0.5 \, mm$
ⓒ. $0.75 \, mm$
ⓓ. $1.0 \, mm$
Correct Answer: $0.25 \, mm$
Explanation: Stress $ \sigma = \frac{F}{A} = \frac{100}{2 \times 10^{-6}} = 5 \times 10^{7} \, Pa$.
Strain $ \epsilon = \frac{\sigma}{Y} = \frac{5 \times 10^{7}}{2 \times 10^{11}} = 2.5 \times 10^{-4}$.
Elongation $ \Delta L = \epsilon L = (2.5 \times 10^{-4})(2) = 5 \times 10^{-4} \, m = 0.25 \, mm$.
173. The formula $\Delta L = \frac{F L}{A Y}$ is derived from:
ⓐ. Pascal’s law
ⓑ. Hooke’s law
ⓒ. Newton’s second law
ⓓ. Archimedes’ principle
Correct Answer: Hooke’s law
Explanation: From Hooke’s law, $\epsilon = \frac{\Delta L}{L} = \frac{\sigma}{Y}$. Substituting $\sigma = \frac{F}{A}$, we get $\Delta L = \frac{F L}{A Y}$.
174. A steel wire of length $1.5 \, m$ and radius $0.5 \, mm$ is stretched by a force of $50 \, N$. If $Y = 2 \times 10^{11} \, Pa$, calculate the elongation.
ⓐ. $0.05 \, mm$
ⓑ. $0.10 \, mm$
ⓒ. $0.15 \, mm$
ⓓ. $0.20 \, mm$
Correct Answer: $0.05 \, mm$
Explanation: Area $A = \pi r^2 = 3.14 (0.5 \times 10^{-3})^2 \approx 7.85 \times 10^{-7} \, m^2$.
Stress = $ \frac{50}{7.85 \times 10^{-7}} \approx 6.37 \times 10^{7} \, Pa$.
Strain = $ \frac{\sigma}{Y} = \frac{6.37 \times 10^{7}}{2 \times 10^{11}} \approx 3.19 \times 10^{-4}$.
Elongation = $ \epsilon L = (3.19 \times 10^{-4})(1.5) \approx 4.78 \times 10^{-4} \, m = 0.48 \, mm$.
175. If two wires of same material, same length, but different cross-sectional areas are subjected to same force, then:
ⓐ. Elongation is same in both wires
ⓑ. Thicker wire elongates more
ⓒ. Thinner wire elongates more
ⓓ. Both break simultaneously
Correct Answer: Thinner wire elongates more
Explanation: Elongation $\Delta L = \frac{F L}{A Y}$. Since $A$ is smaller in thinner wire, elongation is larger under same force.
176. What is the dimensional formula of Young’s modulus $E$?
ⓐ. $[M L^{-1} T^{-2}]$
ⓑ. $[M L T^{-2}]$
ⓒ. $[M^0 L^0 T^0]$
ⓓ. $[M L^2 T^{-2}]$
Correct Answer: $[M L^{-1} T^{-2}]$
Explanation: $E = \frac{\sigma}{\epsilon}$. Since strain is dimensionless, $E$ has dimensions of stress = force/area = $ML^{-1}T^{-2}$.
177. A copper wire of length $3 \, m$ and cross-sectional area $1 \times 10^{-6} \, m^2$ is subjected to a tensile force of $150 \, N$. If $Y = 1 \times 10^{11} \, Pa$, what is the elongation?
ⓐ. $0.15 \, mm$
ⓑ. $0.30 \, mm$
ⓒ. $0.45 \, mm$
ⓓ. $0.60 \, mm$
Correct Answer: $0.45 \, mm$
Explanation: Stress = $ \frac{150}{1 \times 10^{-6}} = 1.5 \times 10^{8} \, Pa$.
Strain = $ \frac{1.5 \times 10^{8}}{1 \times 10^{11}} = 1.5 \times 10^{-3}$.
Elongation = $ (1.5 \times 10^{-3})(3) = 4.5 \times 10^{-3} \, m = 0.45 \, mm$.
178. What does a higher value of Young’s modulus $E$ imply about a material?
ⓐ. It is softer and stretches easily
ⓑ. It is stiffer and resists deformation
ⓒ. It has lower density
ⓓ. It has higher melting point
Correct Answer: It is stiffer and resists deformation
Explanation: Large $E$ means higher stress is required to produce a given strain. Steel (high $E$) is stiffer than rubber (low $E$).
179. A wire elongates $0.2 \, mm$ when a load of $100 \, N$ is applied. If its length is $1 \, m$, cross-sectional area $1 \times 10^{-6} \, m^2$, calculate $Y$.
ⓐ. $1 \times 10^{10} \, Pa$
ⓑ. $3 \times 10^{10} \, Pa$
ⓒ. $6 \times 10^{11} \, Pa$
ⓓ. $5 \times 10^{11} \, Pa$
Correct Answer: $5 \times 10^{11} \, Pa$
Explanation: Stress = $ \frac{F}{A} = \frac{100}{1 \times 10^{-6}} = 1 \times 10^{8} \, Pa$.
Strain = $ \frac{\Delta L}{L} = \frac{0.2 \times 10^{-3}}{1} = 2 \times 10^{-4}$.
$Y = \frac{\sigma}{\epsilon} = \frac{1 \times 10^{8}}{2 \times 10^{-4}} = 5 \times 10^{11} \, Pa$.
180. Which one of the following materials has the highest Young’s modulus?
ⓐ. Rubber
ⓑ. Copper
ⓒ. Steel
ⓓ. Glass
Correct Answer: Steel
Explanation: Steel has $Y \approx 2 \times 10^{11} \, Pa$, much higher than copper ($1.1 \times 10^{11}$), glass ($7 \times 10^{10}$), and rubber ($10^{6}$). Thus, steel is considered more elastic in physics sense.
181. A steel wire of length $2 \, m$ and radius $1 \, mm$ is subjected to a tensile force of $300 \, N$. If $Y = 2 \times 10^{11} \, Pa$, what is the elongation?
ⓐ. $0.09 \, mm$
ⓑ. $0.19 \, mm$
ⓒ. $0.29 \, mm$
ⓓ. $0.39 \, mm$
Correct Answer: $0.19 \, mm$
Explanation: Area $A = \pi r^2 = 3.14 \times (1 \times 10^{-3})^2 = 3.14 \times 10^{-6} \, m^2$.
Stress $\sigma = \frac{F}{A} = \frac{300}{3.14 \times 10^{-6}} \approx 9.55 \times 10^{7} \, Pa$.
Strain $\epsilon = \frac{\sigma}{Y} = \frac{9.55 \times 10^{7}}{2 \times 10^{11}} \approx 4.78 \times 10^{-4}$.
Elongation $\Delta L = \epsilon L = 4.78 \times 10^{-4} \times 2 = 9.56 \times 10^{-4} \, m = 0.19 \, mm$.
182. Which relation is true for Hooke’s law formulation?
ⓐ. $\epsilon = \frac{\sigma}{E}$
ⓑ. $E = \frac{\sigma}{\epsilon}$
ⓒ. $\sigma = E \epsilon$
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Hooke’s law can be written in different forms:
* $\sigma = E \epsilon$
* $\epsilon = \frac{\sigma}{E}$
* $E = \frac{\sigma}{\epsilon}$.
All are equivalent expressions for linear elasticity.
183. A brass wire of length $2 \, m$ and cross-sectional area $1 \times 10^{-6} \, m^2$ is stretched by $0.4 \, mm$ under a force of $200 \, N$. Find Young’s modulus.
ⓐ. $1 \times 10^{10} \, Pa$
ⓑ. $2 \times 10^{10} \, Pa$
ⓒ. $1 \times 10^{11} \, Pa$
ⓓ. $2 \times 10^{11} \, Pa$
Correct Answer: $1 \times 10^{11} \, Pa$
Explanation: Stress = $\frac{200}{1 \times 10^{-6}} = 2 \times 10^{8} \, Pa$.
Strain = $\frac{0.4 \times 10^{-3}}{2} = 2 \times 10^{-4}$.
$Y = \frac{\sigma}{\epsilon} = \frac{2 \times 10^{8}}{2 \times 10^{-4}} = 1 \times 10^{11} \, Pa$.
184. Which of the following is dimensionally equivalent to Young’s modulus?
ⓐ. Pressure
ⓑ. Work
ⓒ. Force
ⓓ. Energy
Correct Answer: Pressure
Explanation: Young’s modulus = $\frac{\sigma}{\epsilon}$. Strain is dimensionless, hence $E$ has the same dimensions as stress, i.e. force/area = pressure (Pa).
185. A steel rod elongates by $1 \, mm$ under a stress of $2 \times 10^{8} \, Pa$. If $Y = 2 \times 10^{11} \, Pa$, what is its length?
ⓐ. $0.5 \, m$
ⓑ. $3.0 \, m$
ⓒ. $2.0 \, m$
ⓓ. $1.0 \, m$
Correct Answer: $1.0 \, m$
Explanation: Strain = $\frac{\sigma}{Y} = \frac{2 \times 10^{8}}{2 \times 10^{11}} = 1 \times 10^{-3}$.
Since strain = $\frac{\Delta L}{L}$, $L = \frac{\Delta L}{\epsilon} = \frac{1 \times 10^{-3}}{1 \times 10^{-3}} = 1 \, m$.
$\Delta L = 1 \, mm = 1 \times 10^{-3} \, m$.
$ L = \frac{1 \times 10^{-3}}{1 \times 10^{-3}} = 1 \, m$.
186. The formula for strain in terms of applied force $F$, length $L$, area $A$, and Young’s modulus $Y$ is:
ⓐ. $\epsilon = \frac{F}{A L}$
ⓑ. $\epsilon = \frac{F}{A Y}$
ⓒ. $\epsilon = \frac{F L}{A Y}$
ⓓ. $\epsilon = \frac{A Y}{F L}$
Correct Answer: $\epsilon = \frac{F}{A Y}$
Explanation: Stress = $\frac{F}{A}$. Since $\epsilon = \frac{\sigma}{Y}$, strain $\epsilon = \frac{F}{A Y}$.
187. A copper wire elongates by $0.3 \, mm$ when a force of $100 \, N$ is applied. If length = $2 \, m$, area = $1 \times 10^{-6} \, m^2$, find Young’s modulus.
ⓐ. $5 \times 10^{10} \, Pa$
ⓑ. $1 \times 10^{11} \, Pa$
ⓒ. $2 \times 10^{11} \, Pa$
ⓓ. $3 \times 10^{11} \, Pa$
Correct Answer: $1 \times 10^{11} \, Pa$
Explanation: Stress = $\frac{100}{1 \times 10^{-6}} = 1 \times 10^{8} \, Pa$.
Strain = $\frac{0.3 \times 10^{-3}}{2} = 1.5 \times 10^{-4}$.
$Y = \frac{1 \times 10^{8}}{1.5 \times 10^{-4}} = 6.67 \times 10^{11} \, Pa$.
188. Which condition makes Hooke’s law invalid?
ⓐ. Strain is very small
ⓑ. Strain is within elastic limit
ⓒ. Strain exceeds elastic limit
ⓓ. Strain is directly proportional to stress
Correct Answer: Strain exceeds elastic limit
Explanation: Beyond elastic limit, proportionality between stress and strain fails. The material enters plastic region and Hooke’s law is no longer valid.
189. If a wire of length $L$ elongates by $\Delta L$, the strain energy stored per unit volume is given by:
ⓐ. $U = \frac{1}{2} \sigma \epsilon$
ⓑ. $U = \sigma \epsilon$
ⓒ. $U = \frac{1}{2} \frac{\Delta L}{L}$
ⓓ. $U = \frac{1}{2} \sigma^2$
Correct Answer: $U = \frac{1}{2} \sigma \epsilon$
Explanation: Strain energy density is equal to the area under the stress–strain curve. For linear elasticity, $U = \frac{1}{2} \sigma \epsilon$.
190. Which real-life application is based directly on Hooke’s law formulation?
ⓐ. Functioning of springs in vehicles
ⓑ. Transmission of sound in air
ⓒ. Flow of current in a conductor
ⓓ. Melting of ice
Correct Answer: Functioning of springs in vehicles
Explanation: Vehicle springs obey Hooke’s law within limits. They absorb shocks and return to original shape due to proportionality of stress and strain. Sound transmission, electric flow, and melting of ice are unrelated.
191. What is the definition of elastic limit?
ⓐ. The maximum stress a material can withstand before breaking
ⓑ. The maximum stress up to which Hooke’s law is valid
ⓒ. The stress at which permanent deformation begins
ⓓ. The point of maximum elongation in a wire
Correct Answer: The stress at which permanent deformation begins
Explanation: The elastic limit is the maximum stress a body can experience without undergoing permanent deformation. Beyond this, the body does not regain its original shape even after the load is removed.
192. What is the definition of proportionality limit?
ⓐ. The maximum stress at which stress is directly proportional to strain
ⓑ. The point where plastic deformation starts
ⓒ. The maximum stress before fracture
ⓓ. The stress at which the body breaks suddenly
Correct Answer: The maximum stress at which stress is directly proportional to strain
Explanation: Proportionality limit is the stress up to which Hooke’s law ($\sigma \propto \epsilon$) holds true. Beyond this, the graph deviates from linearity even if the material is still elastic.
193. Which of the following always lies before the elastic limit on a stress–strain curve?
ⓐ. Yield point
ⓑ. Proportionality limit
ⓒ. Breaking point
ⓓ. Ultimate strength
Correct Answer: Proportionality limit
Explanation: Proportionality limit is the stress up to which Hooke’s law is obeyed. It is reached before the elastic limit, beyond which elasticity may still hold but not linearly.
194. At the elastic limit, which of the following is true?
ⓐ. Hooke’s law is strictly obeyed
ⓑ. Hooke’s law fails and permanent deformation starts
ⓒ. Stress is independent of strain
ⓓ. Strain becomes infinite
Correct Answer: Hooke’s law fails and permanent deformation starts
Explanation: The elastic limit is the threshold of reversibility. Beyond it, permanent (plastic) deformation begins, and the material does not return fully to its original shape.
195. Which of the following correctly represents the order of points on a stress–strain curve for mild steel?
ⓐ. Elastic limit → Proportional limit → Yield point → UTS → Breaking point
ⓑ. Proportional limit → Elastic limit → Yield point → UTS → Breaking point
ⓒ. Yield point → Proportional limit → Elastic limit → UTS → Breaking point
ⓓ. Proportional limit → Yield point → Elastic limit → Breaking point → UTS
Correct Answer: Proportional limit → Elastic limit → Yield point → UTS → Breaking point
Explanation: The stress–strain curve starts with a linear region up to the proportional limit, then elastic limit, yield point, ultimate tensile strength, and finally fracture (breaking point).
196. If a material is stressed just beyond its elastic limit, what happens?
ⓐ. It behaves as perfectly elastic
ⓑ. It undergoes plastic deformation
ⓒ. It fractures instantly
ⓓ. Its modulus of elasticity increases suddenly
Correct Answer: It undergoes plastic deformation
Explanation: Beyond elastic limit, the body cannot recover its original form completely. The deformation becomes permanent (plastic).
197. The proportionality limit corresponds to which part of the stress–strain curve?
ⓐ. The end of the linear region
ⓑ. The maximum point of curve
ⓒ. The region beyond breaking
ⓓ. The middle of plastic region
Correct Answer: The end of the linear region
Explanation: In the stress–strain curve, the straight-line portion ends at the proportionality limit. Beyond this, the curve becomes non-linear.
198. Why is proportionality limit lower than elastic limit?
ⓐ. Because Hooke’s law holds longer than elasticity
ⓑ. Because elasticity fails earlier than Hooke’s law
ⓒ. Because material fractures before Hooke’s law ends
ⓓ. Because yield point occurs first
Correct Answer: Because Hooke’s law holds longer than elasticity
Explanation: Up to proportionality limit, Hooke’s law is valid. Elasticity still exists beyond it but stress is no longer strictly proportional to strain.
199. If stress is below proportionality limit, then:
ⓐ. Strain is not measurable
ⓑ. Strain is proportional to stress
ⓒ. Strain increases independently of stress
ⓓ. Stress has no effect on strain
Correct Answer: Strain is proportional to stress
Explanation: By definition, proportionality limit is the maximum stress up to which stress and strain are linearly related, i.e. Hooke’s law applies.
200. Which of the following statements best distinguishes proportionality limit and elastic limit?
ⓐ. Both are the same point
ⓑ. Elastic limit is always lower than proportionality limit
ⓒ. Proportionality limit is where linearity ends; elastic limit is where reversibility ends
ⓓ. Proportionality limit is where breaking begins
Correct Answer: Proportionality limit is where linearity ends; elastic limit is where reversibility ends
Explanation: Proportionality limit defines the end of stress–strain proportionality. Elastic limit is the point beyond which permanent deformation occurs. Hence, elastic limit is always ≥ proportionality limit.
In Class 11 Physics, the chapter Mechanical Properties of Solids is one of the most important topics in the NCERT/CBSE syllabus. It explains how materials respond to external forces, focusing on stress, strain, elastic modulus, and Hooke’s law. These concepts not only appear in board exams but are also a regular part of competitive exams like JEE, NEET, and state-level tests. Out of a total of 560 MCQs, this part provides the second set of 100 questions with answers for deeper practice and exam preparation.
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