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201. Which property of a material is most directly measured using Hooke’s law in a tensile test?
ⓐ. Density
ⓑ. Young’s modulus
ⓒ. Volume
ⓓ. Heat capacity
Correct Answer: Young’s modulus
Explanation: Tensile testing applies Hooke’s law to measure the ratio of stress to strain in the elastic region. This gives Young’s modulus, which determines the stiffness of the material.
202. In a tensile test, why is stress–strain curve plotted?
ⓐ. To find material density
ⓑ. To determine elasticity, yield strength, and ultimate strength
ⓒ. To calculate heat expansion coefficient
ⓓ. To check conductivity of materials
Correct Answer: To determine elasticity, yield strength, and ultimate strength
Explanation: The stress–strain curve obtained in materials testing shows proportionality limit, elastic limit, yield point, and breaking point. These mechanical properties are crucial in engineering design.
203. Which of the following laboratory devices is commonly used to test Hooke’s law in wires?
ⓐ. Vernier caliper
ⓑ. Young’s modulus apparatus (Searle’s apparatus)
ⓒ. Galvanometer
ⓓ. Hydrometer
Correct Answer: Young’s modulus apparatus (Searle’s apparatus)
Explanation: Searle’s apparatus is specifically designed to measure the elongation of wires under tensile load and verify Hooke’s law.
204. In compression testing of concrete, Hooke’s law is applied up to:
ⓐ. Proportionality limit
ⓑ. Breaking point
ⓒ. Yield point
ⓓ. Ultimate strength
Correct Answer: Proportionality limit
Explanation: Concrete follows Hooke’s law up to the proportionality limit where stress is proportional to strain. Beyond this, non-linear deformation begins.
205. Which industrial test uses Hooke’s law principle to check stiffness of materials?
ⓐ. Tensile test
ⓑ. Bending test
ⓒ. Torsion test
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Hooke’s law applies in tensile, bending, and torsion tests, since all involve measurement of stress and strain in elastic region to determine elastic constants.
206. Which of the following engineering structures must be tested for Hooke’s law compliance to avoid failure?
ⓐ. Bridges
ⓑ. Aircraft wings
ⓒ. Railway tracks
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Structural components like bridges, wings, and tracks undergo elastic deformation under loads. Hooke’s law ensures that stresses remain within safe elastic limits.
207. A wire of original length $2 \, m$ elongates by $1 \, mm$ under load. If stress is $2 \times 10^{8} \, Pa$, what is Young’s modulus from testing?
ⓐ. $2 \times 10^{10} \, Pa$
ⓑ. $4 \times 10^{10} \, Pa$
ⓒ. $2 \times 10^{11} \, Pa$
ⓓ. $4 \times 10^{11} \, Pa$
Correct Answer: $4 \times 10^{11} \, Pa$
Explanation: Strain $\epsilon = \frac{\Delta L}{L} = \frac{1 \times 10^{-3}}{2} = 5 \times 10^{-4}$.
$E = \frac{\sigma}{\epsilon} = \frac{2 \times 10^{8}}{5 \times 10^{-4}} = 4 \times 10^{11} \, Pa$.
208. In torsional testing, the analog of Hooke’s law relates:
ⓐ. Torque and shear strain
ⓑ. Stress and volume
ⓒ. Force and pressure
ⓓ. Density and strain
Correct Answer: Torque and shear strain
Explanation: In torsional testing, the relation is $\tau = G \cdot \phi$, where $\tau$ is shear stress, $G$ is shear modulus, and $\phi$ is shear strain, analogous to Hooke’s law.
209. Which of the following quantities can be calculated in materials testing using Hooke’s law?
ⓐ. Stress
ⓑ. Strain
ⓒ. Modulus of elasticity
ⓓ. All of the above
Correct Answer: All of the above
Explanation: By measuring load and deformation in elastic region, one can compute stress, strain, and elastic constants such as Young’s modulus.
210. Why is it important to test materials under Hooke’s law conditions before use in construction?
ⓐ. To make sure they undergo large plastic deformation
ⓑ. To confirm they obey elastic behaviour within safe load range
ⓒ. To determine their melting point
ⓓ. To ensure they have high thermal conductivity
Correct Answer: To confirm they obey elastic behaviour within safe load range
Explanation: Materials must be tested to ensure they return to original form under expected loads (elastic region). This guarantees structural safety. Melting and conductivity are not relevant in elasticity testing.
211. Which region of the stress–strain curve obeys Hooke’s law?
ⓐ. Elastic region up to proportional limit
ⓑ. Entire plastic region
ⓒ. Beyond breaking point
ⓓ. At ultimate tensile strength
Correct Answer: Elastic region up to proportional limit
Explanation: Hooke’s law is valid only up to the proportional limit where stress and strain are directly proportional. Beyond this region, the stress–strain curve deviates from linearity.
212. What does the yield point on a stress–strain curve represent?
ⓐ. Beginning of linear proportionality
ⓑ. Beginning of plastic deformation
ⓒ. Maximum stress before fracture
ⓓ. Point of no deformation
Correct Answer: Beginning of plastic deformation
Explanation: Yield point is where a material begins to deform plastically. Beyond this, permanent deformation occurs even after stress is removed.
213. What is ultimate tensile strength (UTS) on a stress–strain curve?
ⓐ. The stress at which the material starts to yield
ⓑ. The maximum stress a material can withstand
ⓒ. The stress at which fracture begins
ⓓ. The stress at the proportionality limit
Correct Answer: The maximum stress a material can withstand
Explanation: UTS is the highest point on the stress–strain curve. Beyond this, necking begins until fracture occurs.
214. Which part of the stress–strain curve corresponds to permanent deformation?
ⓐ. Elastic region
ⓑ. Plastic region
ⓒ. Proportional region
ⓓ. Initial straight-line portion
Correct Answer: Plastic region
Explanation: The plastic region lies beyond the elastic limit. In this region, the deformation is permanent and irreversible.
215. Which characteristic point on a stress–strain curve indicates the maximum load-carrying capacity of a material?
ⓐ. Yield point
ⓑ. Ultimate tensile strength (UTS)
ⓒ. Breaking point
ⓓ. Proportional limit
Correct Answer: Ultimate tensile strength (UTS)
Explanation: UTS is the highest point on the curve and shows the maximum load a material can handle before necking and failure.
216. In the elastic region, the slope of the stress–strain curve is equal to:
ⓐ. Stress
ⓑ. Strain
ⓒ. Young’s modulus
ⓓ. Shear modulus
Correct Answer: Young’s modulus
Explanation: The linear portion of the stress–strain curve has slope = $\frac{\sigma}{\epsilon} = E$, which is Young’s modulus of the material.
217. What does the breaking point on the stress–strain curve indicate?
ⓐ. Stress up to which Hooke’s law is valid
ⓑ. Stress at which the material fails completely
ⓒ. Stress at which permanent deformation starts
ⓓ. Stress at which maximum load is borne
Correct Answer: Stress at which the material fails completely
Explanation: Breaking point is the final point on the stress–strain curve, where the material fractures and can no longer sustain any load.
218. Which of the following correctly describes the plastic region of a stress–strain curve?
ⓐ. Deformation is fully reversible
ⓑ. Deformation is permanent and irreversible
ⓒ. Stress and strain are proportional
ⓓ. Material cannot deform at all
Correct Answer: Deformation is permanent and irreversible
Explanation: In the plastic region, the material retains deformation even after the stress is removed.
219. What does the area under the stress–strain curve represent?
ⓐ. Elastic modulus of the material
ⓑ. Energy stored per unit volume (strain energy)
ⓒ. Density of the material
ⓓ. Strength of atomic bonds only
Correct Answer: Energy stored per unit volume (strain energy)
Explanation: The area under the stress–strain curve equals strain energy per unit volume absorbed by the material until that strain.
220. Which statement correctly compares ductile and brittle materials on a stress–strain curve?
ⓐ. Ductile materials have large plastic region, brittle materials have small or none
ⓑ. Brittle materials have larger plastic region than ductile
ⓒ. Both ductile and brittle show identical curves
ⓓ. Ductile materials always break at proportional limit
Correct Answer: Ductile materials have large plastic region, brittle materials have small or none
Explanation: Ductile materials (like steel, copper) show significant elongation before breaking, while brittle materials (like glass, cast iron) fracture with little or no plastic deformation.
221. Which region of the stress–strain curve corresponds to reversible deformation?
ⓐ. Elastic region
ⓑ. Plastic region
ⓒ. Yield region
ⓓ. Breaking region
Correct Answer: Elastic region
Explanation: In the elastic region, deformation is reversible. The material returns to its original dimensions once the load is removed. Beyond this, permanent changes occur.
222. What happens to a material when it is stressed just beyond the yield point?
ⓐ. It obeys Hooke’s law strictly
ⓑ. It undergoes permanent deformation
ⓒ. It returns to its exact original shape
ⓓ. It fractures immediately
Correct Answer: It undergoes permanent deformation
Explanation: The yield point marks the start of plastic deformation. Beyond this point, the material does not fully return to its original shape after unloading.
223. What type of deformation occurs in the yield region?
ⓐ. Fully reversible deformation
ⓑ. Irreversible (plastic) deformation begins
ⓒ. No deformation at all
ⓓ. Instant fracture
Correct Answer: Irreversible (plastic) deformation begins
Explanation: In the yield region, the material transitions from elastic to plastic behaviour. Deformation in this region is partially permanent.
224. Which point on the stress–strain curve marks the end of elastic behaviour?
ⓐ. Proportional limit
ⓑ. Elastic limit
ⓒ. Yield point
ⓓ. Breaking point
Correct Answer: Elastic limit
Explanation: The elastic limit is the maximum stress up to which a material can regain its original shape. Beyond this, plastic deformation begins.
225. In the plastic deformation region, what is true about stress and strain?
ⓐ. Stress is proportional to strain
ⓑ. Stress decreases with increasing strain
ⓒ. Deformation is permanent and irreversible
ⓓ. No deformation occurs
Correct Answer: Deformation is permanent and irreversible
Explanation: In plastic region, materials exhibit irreversible deformation. Even after removal of stress, the material does not return to its original form.
226. What does the yield point indicate about a material?
ⓐ. Material is perfectly rigid
ⓑ. Material begins to deform plastically
ⓒ. Material resists all stress indefinitely
ⓓ. Material instantly fractures
Correct Answer: Material begins to deform plastically
Explanation: Yield point is the stress value where plastic deformation starts. It marks the transition from elastic behaviour to plastic behaviour.
227. Which of the following materials shows a very well-defined yield point?
ⓐ. Mild steel
ⓑ. Glass
ⓒ. Rubber
ⓓ. Cast iron
Correct Answer: Mild steel
Explanation: Mild steel has a distinct yield point, easily seen in its stress–strain curve. Brittle materials like glass and cast iron fracture suddenly, while rubber shows non-linear elasticity.
228. Why is yield strength an important property in engineering design?
ⓐ. It shows how much energy a material can store
ⓑ. It indicates the maximum stress before plastic deformation starts
ⓒ. It measures the density of a material
ⓓ. It determines electrical conductivity
Correct Answer: It indicates the maximum stress before plastic deformation starts
Explanation: Yield strength ensures that structures are designed so that stresses remain below this point to avoid permanent deformation.
229. Which of the following correctly describes the elastic region?
ⓐ. Material regains shape fully after load removal
ⓑ. Material regains only part of its shape
ⓒ. Material never regains its shape
ⓓ. Material breaks instantly
Correct Answer: Material regains shape fully after load removal
Explanation: In the elastic region, deformation is reversible. When the stress is removed, the body restores its original shape and size.
230. At the breaking point, how does the material behaviour differ from plastic deformation?
ⓐ. It still regains shape
ⓑ. It fractures completely and cannot hold load
ⓒ. It continues elongating without breaking
ⓓ. It stores maximum elastic energy
Correct Answer: It fractures completely and cannot hold load
Explanation: Breaking point marks complete failure of the material, unlike plastic deformation where it still holds load but with permanent changes in shape.
231. In the elastic region, the relation between stress ($\sigma$) and strain ($\epsilon$) is given by:
ⓐ. $\sigma = k \epsilon^2$
ⓑ. $\sigma = E \cdot \epsilon$
ⓒ. $\sigma = \sqrt{\epsilon}$
ⓓ. $\sigma = \frac{1}{\epsilon}$
Correct Answer: $\sigma = E \cdot \epsilon$
Explanation: In the elastic region, Hooke’s law is valid: stress is directly proportional to strain, with $E$ (Young’s modulus) as the constant of proportionality.
232. A steel wire of length $2 \, m$ and cross-sectional area $1 \times 10^{-6} \, m^2$ is subjected to a force of $200 \, N$. If $Y = 2 \times 10^{11} \, Pa$, what is the elongation?
ⓐ. $0.1 \, mm$
ⓑ. $0.2 \, mm$
ⓒ. $0.4 \, mm$
ⓓ. $0.8 \, mm$
Correct Answer: $0.2 \, mm$
Explanation: Stress $\sigma = \frac{F}{A} = \frac{200}{1 \times 10^{-6}} = 2 \times 10^{8} \, Pa$.
Strain $\epsilon = \frac{\sigma}{Y} = \frac{2 \times 10^{8}}{2 \times 10^{11}} = 1 \times 10^{-3}$.
Elongation $ \Delta L = \epsilon L = (1 \times 10^{-3})(2) = 2 \times 10^{-3} \, m = 0.2 \, mm$.
233. Which statement is true about yield point?
ⓐ. It marks the end of the linear portion of the stress–strain curve
ⓑ. It marks the beginning of plastic deformation
ⓒ. It occurs after proportionality limit and elastic limit
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Yield point is reached after proportionality and elastic limits. Beyond it, plastic deformation starts and the material no longer obeys Hooke’s law.
234. A mild steel sample has yield stress of $ 2.5 \times 10^{8} \, Pa$. If its cross-sectional area is $2 \times 10^{-4} \, m^2$, what is the load required to reach yield point?
ⓐ. $25 \, kN$
ⓑ. $50 \, kN$
ⓒ. $75 \, kN$
ⓓ. $100 \, kN$
Correct Answer: $50 \, kN$
Explanation: Load $F = \sigma \cdot A = (2.5 \times 10^{8})(2 \times 10^{-4}) = 5 \times 10^{4} \, N = 50 \, kN$.
235. In the plastic deformation region, what happens to elongation when load is increased?
ⓐ. Elongation stops increasing
ⓑ. Elongation increases rapidly with little increase in load
ⓒ. Elongation decreases
ⓓ. Elongation is fully reversible
Correct Answer: Elongation increases rapidly with little increase in load
Explanation: In the plastic region, a small increase in stress causes large strain. This is why metals can be drawn into wires or sheets once plasticity begins.
236. Which formula is used to calculate strain energy per unit volume at yield point?
ⓐ. $U = \frac{1}{2} \sigma \epsilon$
ⓑ. $U = \sigma \epsilon$
ⓒ. $U = \frac{\sigma}{\epsilon}$
ⓓ. $U = \frac{1}{2} \sigma^2$
Correct Answer: $U = \frac{1}{2} \sigma \epsilon$
Explanation: Up to the elastic limit, the strain energy density (area under the stress–strain curve) is given by $\frac{1}{2} \sigma \epsilon$.
237. A wire of length $1.5 \, m$ and area $1 \times 10^{-6} \, m^2$ is stretched with stress of $1 \times 10^{8} \, Pa$. If $Y = 2 \times 10^{11} \, Pa$, calculate elongation.
ⓐ. $0.25 \, mm$
ⓑ. $0.50 \, mm$
ⓒ. $0.75 \, mm$
ⓓ. $1.0 \, mm$
Correct Answer: $0.75 \, mm$
Explanation: Strain = $\frac{\sigma}{Y} = \frac{1 \times 10^{8}}{2 \times 10^{11}} = 5 \times 10^{-4}$.
Elongation = $ \epsilon L = (5 \times 10^{-4})(1.5) = 7.5 \times 10^{-4} \, m = 0.75 \, mm$.
238. Which of the following is correct about elastic region and yield point?
ⓐ. Elastic region ends before yield point
ⓑ. Elastic region starts after yield point
ⓒ. Both are the same
ⓓ. Yield point lies inside proportional limit
Correct Answer: Elastic region ends before yield point
Explanation: Elastic region includes proportionality and elastic limits. Yield point comes after elastic limit, marking the start of plastic deformation.
239. What happens to Young’s modulus in the plastic region?
ⓐ. It remains constant
ⓑ. It decreases
ⓒ. It increases
ⓓ. It becomes infinite
Correct Answer: It decreases
Explanation: In the plastic region, stress and strain are no longer proportional, so the slope of stress–strain curve decreases. Thus, effective Young’s modulus reduces.
240. Which everyday example demonstrates plastic deformation after yield point?
ⓐ. Stretching a rubber band
ⓑ. Bending a steel paperclip permanently
ⓒ. Compressing a spring within limits
ⓓ. A glass rod breaking suddenly
Correct Answer: Bending a steel paperclip permanently
Explanation: When a paperclip is bent beyond its yield point, it undergoes permanent plastic deformation and does not return to its original shape.
241. What is ultimate tensile strength (UTS) of a material?
ⓐ. The stress at which permanent deformation starts
ⓑ. The maximum stress a material can withstand before necking
ⓒ. The stress at which stress and strain are proportional
ⓓ. The stress at which fracture occurs
Correct Answer: The maximum stress a material can withstand before necking
Explanation: UTS is the highest stress value on the stress–strain curve. After this point, necking occurs and the material begins to weaken until fracture.
242. What does the breaking point represent in a stress–strain curve?
ⓐ. The maximum load the material can withstand
ⓑ. The point where plastic deformation begins
ⓒ. The point where the material fractures completely
ⓓ. The end of proportional limit
Correct Answer: The point where the material fractures completely
Explanation: The breaking point is the final stage on the stress–strain curve. Here the material cannot sustain any load and breaks apart.
243. Which one of the following is true about UTS and breaking stress?
ⓐ. Breaking stress is always greater than UTS
ⓑ. UTS is always greater than breaking stress
ⓒ. Both are equal for all materials
ⓓ. They are unrelated
Correct Answer: UTS is always greater than breaking stress
Explanation: After UTS, necking reduces the effective area. The actual breaking stress is lower than UTS. Hence, UTS > breaking stress.
244. A steel wire has cross-sectional area $2 \times 10^{-6} \, m^2$. If its UTS is $5 \times 10^{8} \, Pa$, what is the maximum load it can withstand?
ⓐ. 500 N
ⓑ. 1000 N
ⓒ. 1500 N
ⓓ. 2000 N
Correct Answer: 1000 N
Explanation: Maximum load = $ \sigma \cdot A = (5 \times 10^{8})(2 \times 10^{-6}) = 1000 \, N$.
245. Why does a material fail at the breaking point?
ⓐ. Because stress is still proportional to strain
ⓑ. Because necking reduces cross-sectional area drastically
ⓒ. Because it obeys Hooke’s law perfectly
ⓓ. Because elastic limit is reached
Correct Answer: Because necking reduces cross-sectional area drastically
Explanation: At breaking point, localized thinning (necking) reduces load-bearing area, leading to fracture under reduced stress.
246. Which of the following materials has a well-defined UTS and breaking point?
ⓐ. Glass
ⓑ. Rubber
ⓒ. Mild steel
ⓓ. Wood
Correct Answer: Mild steel
Explanation: Mild steel exhibits a clear stress–strain curve with defined UTS and breaking point. Brittle materials like glass break suddenly without much plastic deformation.
247. If a material has UTS of $3 \times 10^{8} \, Pa$ and breaking stress of $2.5 \times 10^{8} \, Pa$, which statement is correct?
ⓐ. The material fractures before reaching maximum load
ⓑ. The material bears maximum load before fracture
ⓒ. UTS is irrelevant in this case
ⓓ. Breaking stress must be higher than UTS
Correct Answer: The material bears maximum load before fracture
Explanation: The material first reaches maximum load at UTS, then weakens due to necking and finally breaks at a lower breaking stress.
248. Which property does UTS measure in a material?
ⓐ. Rigidity
ⓑ. Maximum load-bearing capacity
ⓒ. Thermal stability
ⓓ. Density
Correct Answer: Maximum load-bearing capacity
Explanation: UTS indicates the maximum stress a material can sustain, directly measuring its strength under tensile loading.
249. What happens to the stress–strain curve after UTS is reached?
ⓐ. Stress continues to increase linearly
ⓑ. Stress decreases due to necking until breaking point
ⓒ. Stress remains constant until fracture
ⓓ. Material behaves elastically again
Correct Answer: Stress decreases due to necking until breaking point
Explanation: After UTS, the specimen starts thinning (necking), reducing effective area and hence stress falls until fracture.
250. A steel wire of area $1 \times 10^{-6} \, m^2$ fractures under a maximum load of $600 \, N$. What is the breaking stress?
ⓐ. $3 \times 10^{8} \, Pa$
ⓑ. $4 \times 10^{8} \, Pa$
ⓒ. $5 \times 10^{8} \, Pa$
ⓓ. $6 \times 10^{8} \, Pa$
Correct Answer: $6 \times 10^{8} \, Pa$
Explanation: Breaking stress $ \sigma = \frac{F}{A} = \frac{600}{1 \times 10^{-6}} = 6 \times 10^{8} \, Pa$.
251. Which of the following materials shows a large plastic deformation before fracture in its stress–strain curve?
ⓐ. Mild steel
ⓑ. Glass
ⓒ. Cast iron
ⓓ. Concrete
Correct Answer: Mild steel
Explanation: Mild steel is ductile and exhibits a long plastic region after yielding, allowing significant deformation before fracture. Brittle materials like glass and cast iron fracture suddenly.
252. Which of the following materials shows almost no plastic region in its stress–strain curve?
ⓐ. Steel
ⓑ. Copper
ⓒ. Glass
ⓓ. Aluminium
Correct Answer: Glass
Explanation: Glass is a brittle material. Its stress–strain curve ends suddenly at the breaking point, showing little or no plastic deformation.
253. Ductile materials can be identified from their stress–strain curve because:
ⓐ. They have a long elastic region
ⓑ. They have a wide plastic region before fracture
ⓒ. They break suddenly without warning
ⓓ. They show no elongation before breaking
Correct Answer: They have a wide plastic region before fracture
Explanation: Ductile materials like steel and copper exhibit large plastic deformation before fracture, which makes them safer for structural use.
254. Which of the following pairs correctly matches ductile and brittle material behaviour?
ⓐ. Ductile – Small plastic region; Brittle – Large plastic region
ⓑ. Ductile – Large plastic region; Brittle – Small or no plastic region
ⓒ. Ductile – No yield point; Brittle – Clear yield point
ⓓ. Ductile – Sudden fracture; Brittle – Slow fracture
Correct Answer: Ductile – Large plastic region; Brittle – Small or no plastic region
Explanation: Ductile materials like mild steel have significant plastic deformation, while brittle materials like glass fracture almost immediately after elastic region.
255. Which metal typically shows a distinct yield point on its stress–strain curve?
ⓐ. Copper
ⓑ. Aluminium
ⓒ. Mild steel
ⓓ. Glass
Correct Answer: Mild steel
Explanation: Mild steel has a very clear yield point. Copper and aluminium are ductile but don’t show a sharp yield point. Glass is brittle and has no yield point.
256. Which characteristic feature is observed in brittle material’s stress–strain curve?
ⓐ. High yield point
ⓑ. Long plastic region
ⓒ. Sudden fracture after elastic region
ⓓ. Necking before fracture
Correct Answer: Sudden fracture after elastic region
Explanation: Brittle materials like glass, ceramics, and cast iron exhibit almost no plastic deformation. They fail suddenly after elastic limit is crossed.
257. Which of the following has the highest ultimate tensile strength among the given materials?
ⓐ. Rubber
ⓑ. Glass
ⓒ. Mild steel
ⓓ. Cast iron
Correct Answer: Mild steel
Explanation: Mild steel has high tensile strength and ductility. Rubber is elastic but weak, glass and cast iron are brittle with low tensile strength.
258. A stress–strain curve with a nearly straight line up to breaking point is typical of:
ⓐ. Ductile materials
ⓑ. Brittle materials
ⓒ. Elastomers
ⓓ. Polymers
Correct Answer: Brittle materials
Explanation: Brittle materials show a linear elastic region followed by sudden fracture. They do not exhibit a large plastic deformation region.
259. Elastomers like rubber have which type of stress–strain curve?
ⓐ. Linear
ⓑ. Curved and highly non-linear
ⓒ. Sudden fracture without deformation
ⓓ. No deformation at all
Correct Answer: Curved and highly non-linear
Explanation: Rubber exhibits very large strains for small stresses. Its stress–strain curve is non-linear but reversible, showing elasticity beyond Hooke’s law.
260. Why are ductile materials preferred in construction compared to brittle ones?
ⓐ. They are always cheaper
ⓑ. They undergo plastic deformation before fracture, giving warning
ⓒ. They are lighter in density
ⓓ. They are poor conductors of heat
Correct Answer: They undergo plastic deformation before fracture, giving warning
Explanation: Ductile materials like steel deform plastically before breaking, providing visible warning signs. Brittle materials fail suddenly, making them unsafe for construction.
261. What is Young’s modulus defined as?
ⓐ. Ratio of lateral strain to longitudinal strain
ⓑ. Ratio of shear stress to shear strain
ⓒ. Ratio of tensile stress to longitudinal strain
ⓓ. Ratio of bulk stress to volumetric strain
Correct Answer: Ratio of tensile stress to longitudinal strain
Explanation: Young’s modulus is defined as
$$ Y = \frac{\sigma}{\epsilon} = \frac{\text{Tensile stress}}{\text{Longitudinal strain}} $$
It measures stiffness of a material in tension or compression.
262. What is shear modulus (also called modulus of rigidity) defined as?
ⓐ. Ratio of shear stress to shear strain
ⓑ. Ratio of tensile stress to strain
ⓒ. Ratio of lateral strain to longitudinal strain
ⓓ. Ratio of volume stress to strain
Correct Answer: Ratio of shear stress to shear strain
Explanation: Shear modulus is given by
$$ G = \frac{\tau}{\phi} $$
where $\tau$ is shear stress and $\phi$ is shear strain (angular deformation).
263. What is bulk modulus of a material?
ⓐ. Ratio of tensile stress to longitudinal strain
ⓑ. Ratio of shear stress to shear strain
ⓒ. Ratio of volumetric stress to volumetric strain
ⓓ. Ratio of lateral strain to longitudinal strain
Correct Answer: Ratio of volumetric stress to volumetric strain
Explanation: Bulk modulus is defined as
$$ K = \frac{\Delta P}{\Delta V / V} $$
It measures resistance to uniform compression.
264. Which modulus is related to compressibility of a material?
ⓐ. Young’s modulus
ⓑ. Bulk modulus
ⓒ. Shear modulus
ⓓ. Poisson’s ratio
Correct Answer: Bulk modulus
Explanation: Compressibility is the reciprocal of bulk modulus:
$$ \beta = \frac{1}{K} $$
Hence, bulk modulus determines how easily a material compresses under uniform pressure.
265. Which of the following relations is correct for Young’s modulus in terms of stress and strain?
ⓐ. $Y = \frac{\Delta L / L}{F / A}$
ⓑ. $Y = \frac{F L}{A \Delta L}$
ⓒ. $Y = \frac{A \Delta L}{F L}$
ⓓ. $Y = \frac{F}{A \Delta L}$
Correct Answer: $Y = \frac{F L}{A \Delta L}$
Explanation: From $Y = \frac{\sigma}{\epsilon}$,
$\sigma = \frac{F}{A}, \; \epsilon = \frac{\Delta L}{L} $.
So, $Y = \frac{F/A}{\Delta L/L} = \frac{F L}{A \Delta L}$.
266. Which modulus is most relevant for describing elasticity in solids under torsion?
ⓐ. Young’s modulus
ⓑ. Shear modulus
ⓒ. Bulk modulus
ⓓ. Poisson’s ratio
Correct Answer: Shear modulus
Explanation: Torsion involves twisting due to shear forces. The deformation is measured by shear modulus, $G = \frac{\tau}{\phi}$.
267. A cube of volume $0.01 \, m^3$ undergoes a volume decrease of $1 \times 10^{-5} \, m^3$ under pressure of $1 \times 10^{6} \, Pa$. What is the bulk modulus?
ⓐ. $1 \times 10^{9} \, Pa$
ⓑ. $1 \times 10^{10} \, Pa$
ⓒ. $1 \times 10^{11} \, Pa$
ⓓ. $1 \times 10^{12} \, Pa$
Correct Answer: $1 \times 10^{9} \, Pa$
Explanation: Volumetric strain $= \frac{\Delta V}{V} = \frac{1 \times 10^{-5}}{0.01} = 1 \times 10^{-3}$.
Bulk modulus $K = \frac{\Delta P}{\Delta V/V} = \frac{1 \times 10^{6}}{1 \times 10^{-3}} = 1 \times 10^{9} \, Pa$.
268. The SI unit of all elastic moduli (Y, G, K) is:
ⓐ. $N$
ⓑ. $N/m$
ⓒ. $N/m^2$ (Pascal)
ⓓ. Dimensionless
Correct Answer: $N/m^2$ (Pascal)
Explanation: All moduli are ratios of stress to strain. Since stress has unit $N/m^2$ and strain is dimensionless, all moduli share the unit Pascal.
269. If a wire of length $2 \, m$ elongates by $0.5 \, mm$ under a load of $100 \, N$ and cross-sectional area $1 \times 10^{-6} \, m^2$, 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: $2 \times 10^{11} \, Pa$
Explanation: Stress = $\frac{100}{1 \times 10^{-6}} = 1 \times 10^{8} \, Pa$.
Strain = $\frac{\Delta L}{L} = \frac{0.5 \times 10^{-3}}{2} = 2.5 \times 10^{-4}$.
$Y = \frac{1 \times 10^{8}}{2.5 \times 10^{-4}} = 4 \times 10^{11} \, Pa$.
Closer to option D.
270. Which modulus determines how a liquid resists compression?
ⓐ. Young’s modulus
ⓑ. Shear modulus
ⓒ. Bulk modulus
ⓓ. Poisson’s ratio
Correct Answer: Bulk modulus
Explanation: Liquids cannot resist shear forces but resist compression. Their compressibility is inversely related to bulk modulus.
271. Which formula correctly expresses shear modulus $G$?
ⓐ. $G = \frac{\text{Tensile stress}}{\text{Longitudinal strain}}$
ⓑ. $G = \frac{\tau}{\phi}$
ⓒ. $G = \frac{\Delta P}{\Delta V / V}$
ⓓ. $G = \frac{\epsilon}{\sigma}$
Correct Answer: $G = \frac{\tau}{\phi}$
Explanation: Shear modulus is the ratio of shear stress ($\tau$) to shear strain ($\phi$, angular deformation). It measures the rigidity of a material against shearing forces.
272. Which elastic modulus relates pressure and change in volume?
ⓐ. Young’s modulus
ⓑ. Shear modulus
ⓒ. Bulk modulus
ⓓ. Poisson’s ratio
Correct Answer: Bulk modulus
Explanation: Bulk modulus $K = \frac{\Delta P}{\Delta V/V}$ relates volume stress (pressure) to volumetric strain. It defines compressibility of solids and fluids.
273. A wire of length $2 \, m$, area $2 \times 10^{-6} \, m^2$, elongates by $0.25 \, mm$ under a load of $200 \, N$. Find Young’s modulus.
ⓐ. $7 \times 10^{10} \, Pa$
ⓑ. $3 \times 10^{11} \, Pa$
ⓒ. $5 \times 10^{11} \, Pa$
ⓓ. $8 \times 10^{11} \, Pa$
Correct Answer: $8 \times 10^{11} \, Pa$
Explanation: Stress = $200 / 2 \times 10^{-6} = 1 \times 10^{8} \, Pa$.
Strain = $0.25 \times 10^{-3} / 2 = 1.25 \times 10^{-4}$.
$Y = 1 \times 10^{8} / 1.25 \times 10^{-4} = 8 \times 10^{11} \, Pa$.
274. Which of the following materials has very high shear modulus?
ⓐ. Rubber
ⓑ. Steel
ⓒ. Glass
ⓓ. Copper
Correct Answer: Steel
Explanation: Steel resists shearing strongly, hence high shear modulus. Rubber has very low shear modulus because it deforms easily under shear.
275. A cube of side $10 \, cm$ is subjected to a tangential force of $200 \, N$ on one face. If the shear modulus $G = 8 \times 10^{10} \, Pa$, find the shear strain.
ⓐ. $2.5 \times 10^{-6}$
ⓑ. $5.0 \times 10^{-6}$
ⓒ. $1.0 \times 10^{-5}$
ⓓ. $2.5 \times 10^{-7}$
Correct Answer: $2.5 \times 10^{-7}$
Explanation: Stress = $F/A = 200 / 0.01 = 2 \times 10^{4} \, Pa$.
Shear strain = $\tau / G = 2 \times 10^{4} / 8 \times 10^{10} = 2.5 \times 10^{-7}$.
276. Which modulus is relevant in calculating the speed of longitudinal sound waves in a solid?
ⓐ. Bulk modulus
ⓑ. Young’s modulus
ⓒ. Shear modulus
ⓓ. Modulus of resilience
Correct Answer: Young’s modulus
Explanation: The velocity of longitudinal waves in a rod is $v = \sqrt{Y / \rho}$. Hence, Young’s modulus determines sound speed in solids.
277. Which modulus is relevant in calculating the speed of sound in a liquid?
ⓐ. Young’s modulus
ⓑ. Shear modulus
ⓒ. Bulk modulus
ⓓ. Elastic limit
Correct Answer: Bulk modulus
Explanation: The velocity of sound in a liquid is given by $v = \sqrt{K / \rho}$. Thus, bulk modulus determines compressibility and sound propagation.
278. If a material has bulk modulus $K = 1.6 \times 10^{11} \, Pa$ and volumetric strain = $1 \times 10^{-4}$, what pressure is applied?
ⓐ. $1.6 \times 10^{6} \, Pa$
ⓑ. $1.6 \times 10^{7} \, Pa$
ⓒ. $1.6 \times 10^{8} \, Pa$
ⓓ. $1.6 \times 10^{10} \, Pa$
Correct Answer: $1.6 \times 10^{7} \, Pa$
Explanation: $\Delta P = K \times \frac{\Delta V}{V} = 1.6 \times 10^{11} \times 1 \times 10^{-4} = 1.6 \times 10^{7} \, Pa$.
279. The resistance of a body to a change in shape without a change in volume is measured by:
ⓐ. Young’s modulus
ⓑ. Shear modulus
ⓒ. Bulk modulus
ⓓ. Poisson’s ratio
Correct Answer: Shear modulus
Explanation: Shear modulus $G$ determines how resistant a material is to shape change (shear strain) without volume change.
280. Which of the following statements is correct?
ⓐ. All three elastic moduli (Y, G, K) have the same dimensions and units
ⓑ. Only Young’s modulus has unit Pascal
ⓒ. Shear modulus is dimensionless
ⓓ. Bulk modulus has unit Joule
Correct Answer: All three elastic moduli (Y, G, K) have the same dimensions and units
Explanation: Young’s, shear, and bulk moduli are all ratios of stress to strain. Since strain is dimensionless, all three moduli have unit Pascal (Pa).
281. Which relation connects Young’s modulus $Y$, bulk modulus $K$, and Poisson’s ratio $\nu$?
ⓐ. $Y = 3K(1 – 2\nu)$
ⓑ. $Y = 2K(1 + \nu)$
ⓒ. $Y = \frac{K}{1 – \nu}$
ⓓ. $Y = K(1 + 2\nu)$
Correct Answer: $Y = 3K(1 – 2\nu)$
Explanation: The relation between $Y$, $K$, and $\nu$ is $Y = 3K(1 – 2\nu)$. It comes from combining longitudinal and volumetric strain formulas.
282. Which relation connects Young’s modulus $Y$, shear modulus $G$, and Poisson’s ratio $\nu$?
ⓐ. $Y = 3G(1 – 2\nu)$
ⓑ. $Y = 2G(1 + \nu)$
ⓒ. $Y = G(1 – \nu)$
ⓓ. $Y = \frac{G}{1 – \nu}$
Correct Answer: $Y = 2G(1 + \nu)$
Explanation: From the general elasticity relations, $Y = 2G(1 + \nu)$. This is widely used to connect shear modulus with Poisson’s ratio and Young’s modulus.
283. Which relation connects bulk modulus $K$, shear modulus $G$, and Young’s modulus $Y$?
ⓐ. $3K(1 – 2\nu) = Y$
ⓑ. $\frac{9KG}{3K + G} = Y$
ⓒ. $\frac{3K}{2G} = Y$
ⓓ. $Y = K + G$
Correct Answer: $\frac{9KG}{3K + G} = Y$
Explanation: The general relation among $Y$, $K$, and $G$ is $Y = \frac{9KG}{3K + G}$. This allows computation of one modulus if the other two are known.
284. If a material has Young’s modulus $Y = 2 \times 10^{11} \, Pa$ and Poisson’s ratio $\nu = 0.25$, what is its shear modulus $G$?
ⓐ. $7.5 \times 10^{10} \, Pa$
ⓑ. $8.0 \times 10^{10} \, Pa$
ⓒ. $9.0 \times 10^{10} \, Pa$
ⓓ. $1.0 \times 10^{11} \, Pa$
Correct Answer: $8.0 \times 10^{10} \, Pa$
Explanation: $Y = 2G(1 + \nu)$.
So, $G = \frac{Y}{2(1 + \nu)} = \frac{2 \times 10^{11}}{2(1.25)} = \frac{2 \times 10^{11}}{2.5} = 8 \times 10^{10} \, Pa$.
285. A material has bulk modulus $K = 1.6 \times 10^{11} \, Pa$ and shear modulus $G = 8 \times 10^{10} \, Pa$. Find its Young’s modulus $Y$.
ⓐ. $2.0 \times 10^{11} \, Pa$
ⓑ. $2.6 \times 10^{11} \, Pa$
ⓒ. $3.0 \times 10^{11} \, Pa$
ⓓ. $3.2 \times 10^{11} \, Pa$
Correct Answer: $2.6 \times 10^{11} \, Pa$
Explanation: $Y = \frac{9KG}{3K + G} = \frac{9 (1.6 \times 10^{11})(8 \times 10^{10})}{3(1.6 \times 10^{11}) + 8 \times 10^{10}}$.
\= $\frac{1.152 \times 10^{22}}{5.6 \times 10^{11}} \approx 2.06 \times 10^{11} \, Pa$.
286. For an incompressible material, Poisson’s ratio $\nu$ is:
ⓐ. 0
ⓑ. 0.25
ⓒ. 0.5
ⓓ. 1
Correct Answer: 0.5
Explanation: For incompressible solids (constant volume under stress), lateral strain = half of longitudinal strain, giving Poisson’s ratio $\nu = 0.5$.
287. If Young’s modulus is $2.1 \times 10^{11} \, Pa$ and shear modulus is $8.4 \times 10^{10} \, Pa$, find Poisson’s ratio.
ⓐ. 0.20
ⓑ. 0.25
ⓒ. 0.30
ⓓ. 0.35
Correct Answer: 0.25
Explanation: Relation: $Y = 2G(1 + \nu)$.
So, $\nu = \frac{Y}{2G} – 1 = \frac{2.1 \times 10^{11}}{2(8.4 \times 10^{10})} – 1 = \frac{2.1}{1.68} – 1 = 1.25 – 1 = 0.25$.
288. Which relation connects bulk modulus $K$, Young’s modulus $Y$, and Poisson’s ratio $\nu$?
ⓐ. $K = \frac{Y}{2(1 + \nu)}$
ⓑ. $K = \frac{Y}{3(1 – 2\nu)}$
ⓒ. $K = \frac{Y(1 – 2\nu)}{3(1 + \nu)}$
ⓓ. $K = \frac{3Y}{2(1 + \nu)}$
Correct Answer: $K = \frac{Y}{3(1 – 2\nu)}$
Explanation: Bulk modulus is related to Young’s modulus and Poisson’s ratio by $K = \frac{Y}{3(1 – 2\nu)}$.
289. A material has Young’s modulus $Y = 2 \times 10^{11} \, Pa$ and bulk modulus $K = 1.6 \times 10^{11} \, Pa$. Find Poisson’s ratio $\nu$.
ⓐ. 0.20
ⓑ. 0.25
ⓒ. 0.30
ⓓ. 0.35
Correct Answer: 0.20
Explanation: Relation: $Y = 3K(1 – 2\nu)$.
So, $\nu = \frac{1}{2} \left(1 – \frac{Y}{3K}\right) = \frac{1}{2}\left(1 – \frac{2 \times 10^{11}}{4.8 \times 10^{11}}\right)$.
\= $\frac{1}{2}(1 – 0.416) = 0.292$. Closest option: C. 0.30.
290. Which of the following statements is true regarding the three elastic moduli?
ⓐ. All three are independent constants
ⓑ. Any two moduli and Poisson’s ratio can determine the third
ⓒ. Only Young’s modulus is fundamental
ⓓ. Shear modulus and bulk modulus are unrelated
Correct Answer: Any two moduli and Poisson’s ratio can determine the third
Explanation: The three elastic constants $Y, K, G$ are not independent. Given any two and Poisson’s ratio, the third can be derived using the standard relations.
291. Which apparatus is commonly used to measure Young’s modulus of a wire in laboratories?
ⓐ. Searle’s apparatus
ⓑ. Barometer
ⓒ. Hydrometer
ⓓ. Vernier caliper
Correct Answer: Searle’s apparatus
Explanation: Searle’s apparatus measures elongation of a wire under load, allowing Young’s modulus to be calculated using $Y = \frac{F L}{A \Delta L}$.
292. In Searle’s experiment, why are two wires used (test wire and reference wire)?
ⓐ. To double the elongation
ⓑ. To eliminate error due to thermal expansion
ⓒ. To make the setup rigid
ⓓ. To increase the breaking load
Correct Answer: To eliminate error due to thermal expansion
Explanation: Both wires expand equally with temperature. The micrometer only records differential elongation due to the load on the test wire, not thermal effects.
293. A steel wire of length $2 \, m$, radius $0.5 \, mm$, elongates by $0.5 \, mm$ under a load of $100 \, N$. Find Young’s modulus.
ⓐ. $1.2 \times 10^{11} \, Pa$
ⓑ. $2.5 \times 10^{11} \, Pa$
ⓒ. $3.2 \times 10^{11} \, Pa$
ⓓ. $4.0 \times 10^{11} \, Pa$
Correct Answer: $1.2 \times 10^{11} \, Pa$
Explanation: Area $A = \pi r^2 = 3.14 \times (0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7} \, m^2$.
Stress = $F/A = 100 / 7.85 \times 10^{-7} = 1.27 \times 10^{8} \, Pa$.
Strain = $\Delta L / L = 0.5 \times 10^{-3} / 2 = 2.5 \times 10^{-4}$.
$Y = \sigma / \epsilon = 1.27 \times 10^{8} / 2.5 \times 10^{-4} \approx 1.2 \times 10^{11} \, Pa$.
294. Which method is generally used to measure shear modulus of a wire?
ⓐ. Young’s modulus test
ⓑ. Torsion pendulum method
ⓒ. Compression test
ⓓ. Barometer method
Correct Answer: Torsion pendulum method
Explanation: Shear modulus $G$ is measured using torsional oscillations. By suspending a disc with a wire and measuring its oscillation period, $G$ is calculated.
295. In the torsion pendulum method, the time period is given by:
ⓐ. $T = 2\pi \sqrt{\frac{I}{C}}$
ⓑ. $T = 2\pi \sqrt{\frac{C}{I}}$
ⓒ. $T = 2\pi \sqrt{\frac{Y}{\rho}}$
ⓓ. $T = 2\pi \sqrt{\frac{\rho}{Y}}$
Correct Answer: $T = 2\pi \sqrt{\frac{I}{C}}$
Explanation: $T$ is the time period of torsional oscillations, where $I$ is the moment of inertia of the disc and $C$ is the torsional constant of the wire.
296. A steel rod of length $2 \, m$, diameter $2 \, mm$, is subjected to a tensile force of $400 \, N$. It elongates by $0.4 \, mm$. Calculate Young’s modulus.
ⓐ. $1.0 \times 10^{11} \, Pa$
ⓑ. $1.5 \times 10^{11} \, Pa$
ⓒ. $2.0 \times 10^{11} \, Pa$
ⓓ. $2.5 \times 10^{11} \, Pa$
Correct Answer: $2.0 \times 10^{11} \, Pa$
Explanation: Radius = $1 \, mm = 1 \times 10^{-3} \, m$.
Area = $\pi r^2 = 3.14 \times 10^{-6} \, m^2$.
Stress = $400 / 3.14 \times 10^{-6} = 1.27 \times 10^{8} \, Pa$.
Strain = $0.4 \times 10^{-3} / 2 = 2 \times 10^{-4}$.
$Y = 1.27 \times 10^{8} / 2 \times 10^{-4} \approx 2.0 \times 10^{11} \, Pa$.
297. Bulk modulus can be measured experimentally by studying:
ⓐ. Elongation of a wire under load
ⓑ. Change in volume under hydrostatic pressure
ⓒ. Bending of a beam
ⓓ. Torsional oscillations
Correct Answer: Change in volume under hydrostatic pressure
Explanation: Bulk modulus is defined as $K = \frac{\Delta P}{\Delta V/V}$. It is measured by applying uniform fluid pressure and observing volume change.
298. A liquid has bulk modulus $K = 2 \times 10^{9} \, Pa$. If pressure of $1 \times 10^{7} \, Pa$ is applied, what is the fractional volume change?
ⓐ. $2.0 \times 10^{-2}$
ⓑ. $5.0 \times 10^{-3}$
ⓒ. $1.0 \times 10^{-2}$
ⓓ. $2.5 \times 10^{-3}$
Correct Answer: $5.0 \times 10^{-3}$
Explanation: Volumetric strain $\Delta V / V = \Delta P / K = (1 \times 10^{7}) / (2 \times 10^{9}) = 5 \times 10^{-3}$.
299. Which apparatus is commonly used to measure bulk modulus of a liquid?
ⓐ. Searle’s apparatus
ⓑ. Quincke’s apparatus
ⓒ. Hyder’s apparatus
ⓓ. Barometer
Correct Answer: Hyder’s apparatus
Explanation: Hyder’s apparatus is used to measure compressibility (inverse of bulk modulus) of liquids under high pressures.
300. Why is measurement of elastic moduli important in engineering?
ⓐ. To calculate density of materials
ⓑ. To predict elastic and plastic properties of materials in structures
ⓒ. To measure thermal conductivity
ⓓ. To calculate refractive index of materials
Correct Answer: To predict elastic and plastic properties of materials in structures
Explanation: Knowledge of elastic constants is essential in designing safe structures, selecting materials for load-bearing, and predicting deformation under stress.
The chapter Mechanical Properties of Solids plays a major role in strengthening concepts of elasticity, tensile strength, stress-strain curves, and applications of elasticity in real life. Being part of the NCERT/CBSE Class 11 syllabus, it is highly relevant for both board exams and competitive exams like JEE, NEET, and other state entrance tests. Across 6 parts, this chapter provides 560 MCQs with answers. Here in Part 3, you will practice the third set of 100 questions with explanations to test and refine your knowledge.
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