Elastic moduli MCQ Quiz - Objective Question with Answer for Elastic moduli - Download Free PDF

Calculation: 

Let's consider the steel ring of radius r" id="MathJax-Element-2295-Frame" role="presentation" style="position: relative;" tabindex="0">rr $r$ and cross-sectional area A" id="MathJax-Element-2296-Frame" role="presentation" style="position: relative;" tabindex="0">AA $A$ fitted onto a wooden disc of radius R" id="MathJax-Element-2297-Frame" role="presentation" style="position: relative;" tabindex="0">RR $R$ (where R>r" id="MathJax-Element-2298-Frame" role="presentation" style="position: relative;" tabindex="0">R>rR>r $R > r$ ).

The ring needs to be expanded from its natural radius r" id="MathJax-Element-2299-Frame" role="presentation" style="position: relative;" tabindex="0">rr $r$ to the radius R" id="MathJax-Element-2300-Frame" role="presentation" style="position: relative;" tabindex="0">RR $R$ to fit onto the disc.

The strain in the ring is given by:

Strain=Δrr=R−rr" id="MathJax-Element-2301-Frame" role="presentation" style="position: relative;" tabindex="0">Strain=Δrr=R−rrStrain=Δrr=R−rr $\text{Strain} = \frac{\Delta r}{r} = \frac{R - r}{r}$

According to Hooke's Law, the stress is given by:

Stress=Y×Strain" id="MathJax-Element-2302-Frame" role="presentation" style="position: relative;" tabindex="0">Stress=Y×StrainStress=Y×Strain $\text{Stress} = Y \times \text{Strain}$

Stress=Y×R−rr" id="MathJax-Element-2303-Frame" role="presentation" style="position: relative;" tabindex="0">Stress=Y×R−rrStress=Y×R−rr $\text{Stress} = Y \times \frac{R - r}{r}$

The force with which the steel ring is expanded can be found using the relation:

Force=Stress×Area" id="MathJax-Element-2304-Frame" role="presentation" style="position: relative;" tabindex="0">Force=Stress×AreaForce=Stress×Area $\text{Force} = \text{Stress} \times \text{Area}$

Force=Y×R−rr×A" id="MathJax-Element-2305-Frame" role="presentation" style="position: relative;" tabindex="0">Force=Y×R−rr×AForce=Y×R−rr×A $\text{Force} = Y \times \frac{R - r}{r} \times A$

Simplifying the expression, we get:

Force=AY(R−rr)" id="MathJax-Element-2306-Frame" role="presentation" style="position: relative;" tabindex="0">Force=AY(R−rr)Force=AY(R−rr) $\text{Force} = A Y \left( \frac{R - r}{r} \right)$

Therefore, the correct answer is option 4: 

AY(R−rR)" id="MathJax-Element-2307-Frame" role="presentation" style="position: relative;" tabindex="0">AY(R−rR)AY(R−rR) $A Y \left( \frac{R - r}{R} \right)$

Calculation:

\(\mathrm{B}=\frac{\rho \mathrm{gh}}{\left(\frac{\Delta \mathrm{v}}{\mathrm{v}}\right)}\)

\(\frac{\Delta \mathrm{v}}{\mathrm{v}} \times 100=\frac{\rho \mathrm{gh}}{\mathrm{B}} \times 100 \)

\(\frac{1000 \times 10 \times 2.5 \times 10^{3}}{2 \times 10^{9}} \times 100 \%\)

= 1.25 %

Calculation: 

The extension versus load graph for a wire of length 'L' and radius 'r' is a straight line passing through the origin. The slope of this line is tan θ.

According to Hooke's Law, the extension (ΔL) of a wire is directly proportional to the load (F) applied:

F = kΔL

Where k is the force constant (stiffness) of the wire. From the graph, the slope (tan θ) is given by:

tan θ = ΔL / F

We also know that the Young's modulus (Y) of the material of the wire is given by:

Y = (F / A) / (ΔL / L)

Where A is the cross-sectional area of the wire. For a wire of radius 'r', the cross-sectional area A is:

A = πr²

Substituting the value of A and rearranging the equation, we get:

Y = (F × L) / (πr² × ΔL)

Since F = kΔL, we can substitute k = F / ΔL into the equation:

Y = (kΔL × L) / (πr² × ΔL)

Y = kL / πr²

From the graph, k = 1 / tan θ, thus:

Y = (L / πr²) × (1 / tan θ)

Therefore, the Young's modulus Y of the material of the wire can be calculated using the formula:

Y = L / (πr² × tan θ)

∴ The correct answer is option 1.

We have change in length due to the change in temperature (considering no obstruction)

\(\Delta L = \alpha L \Delta T\)

or

\(\Delta L = 1.1 \times 10^{-5} \times 0.10 \times 100 = 1.1 \times 10^{-4} \, \text{m}\)

Now pressure which needs to be applied would be given using

\(\frac{F}{A} = \text{Pressure} = \frac{\Delta L \times Y}{L} = \frac{1.1 \times 10^{-4} \times 2 \times 10^{11}}{0.1}\)

(Here \(\Delta L\) is the expansion prevented)

\(= 2.2 \times 10^8 \, \text{Pa}\)

Calculation:

Energy stored per unit volume can be given as:

\(E = \dfrac{1}{2} \times \text{stress} \times \text{strain}\) -----------(1)

From Hooke's law:

Young's modulus, \(Y = \dfrac{\text{Stress}}{\text{Strain}}\)

\(\implies \text{Strain} = \dfrac{\text{Stress}}{Y} = \dfrac{S}{Y}\) -----------(2)

From equation (1) and (2):

\(\therefore E = \dfrac{1}{2} \times S \times \dfrac{S}{Y}\)

\(\Rightarrow E = \dfrac{1}{2} \dfrac{S^2}{Y}\)

Hence, the correct option is (B)

CONCEPT:

• If a wire is under force, there will be an increase or decrease in length due to the developed stress and strain in it.
  • Stress is calculated by Force (F) divided by the cross-section area (A) on which it is applied.
  • Stress = F/A
  • The strain is the change in length of wire (x) divided by the original length (L) of the wire.
  • Strain = x / L
• Hook's Law of elasticity Young's modulus = stress/strain

\( \Rightarrow Y = \frac{{Fl}}{{A{\rm{Δ }}l}}\)

Where F = Force applied, A = area of cross-section, l = original length and Δl = change in length

​

Given - Area (A) = 1 cm² = 10^-4 m², L₁ = l, L₂ = 2l 

Δl = L₂ – L₁ = l 

As we know, the Young modulus is

\( \Rightarrow Y = \frac{{Fl}}{{A{\rm{Δ }}l}}\)

\( \Rightarrow F = \frac{{YA{\rm{Δ }}l}}{l}\)

\( \Rightarrow F = \frac{{2 \times {{10}^{11}} \times {{10}^{ - 4}} \times l}}{l} = 2 \times {10^7}N\)

CONCEPT:

• Elasticity: The ability of a deformed body to return to its original shape and size when the forces causing the deformation are removed is called Elasticity.
  • An elastic variable is one that response more than proportionally to changes in other variables.
• Elastic stress: When the body is deformed by the application of external forces, forces within the body are brought into play.
  • Elastic bodies regain their original shape due to internal restoring forces.
  • The internal forces and external forces are opposite in direction. 
  • If a force F is applied uniformly over a surface of area A then the stress is defined as the force per unit area.
  • S.I unit for stress is Nm-2

Stress = Force/Area

• Strain: The ratio of charge of any dimension to its original dimension is called strain. 
  • The fractional change in the dimension of a body produced by the external stress acting on is called strain.
  •  A body under stress gets deformed.
  • Since strain is the ratio of two identical physical quantities, it is just a number.
  • It has no unit and dimension.

Strain = change in Dimension/Original Dimension

• Elastic Limit: It is the greatest stress that can be applied to it without causing plastic (permanent) deformation.
  • When a material is stressed to a point below its elastic limit, it will return to its original length once the stress is removed.

• Young's Modulus: It is defined as the relationship between stress (force per unit area) and strain (proportional deformation in an object. 
  • A solid object deforms when a particular load is applied to it.
  • If the object is elastic, the body regains its original shape when the pressure is removed.
  • Many materials are not linear and elastic beyond a small amount of deformation

EXPLANATION:

• So, it can be said that for a given amount of stress, the strain produced in the steel is comparatively smaller than the strain produced in the rubber. Therefore, with the help of Young's modulus, it can be concluded that steel has greater elasticity than rubber, glass, and silver.

 The correct option is Steel.

CONCEPT:

• Hooke’s law and Modulus of Elasticity: According to this law, within the elastic limit, stress is proportional to the strain.

i.e. stress ∝ strain

or,

\(\frac{{stress}}{{strain}} = constant = E\)

Where E  = modulus of elasticity.

EXPLANATION:

• From the above relation, it is very much clear that Hook’s law gives information about the relationship between stress and strain. It does not define stress or strain. Thus, option 1 and 2 is incorrect. 
• From the above relation, it is very much clear that Hook's law defines the modulus of elasticity. Hence option 3 is correct.

Stress: Stress is the ratio of the load or force to the cross-sectional area of the material to which the load is applied.

The standard unit of stress is N/m².

Strain: Strain is a measure of the deformation of the material as a result of the force applied.

The strain is a unitless quantity.

Hooke's law states that within the elastic limit the stress applied on a body is directly proportional to strain produced.

⇒ Stress ∝ Strain

⇒  Stress = E × Strain  (Where E = modulus of elasticity).

Hooke's law is important because it helps us understand how a stretchy object will behave when it is stretched or compacted.

This law was named after Robert Hooke. 

​EXPLANATION:

From above it is clear that, In the relation, stress = k × strain, k is the modulus of elasticity.

CONCEPT:

Young’s modulus: 

• Young's modulus a modulus of elasticity, applicable to the stretching of wire, etc., equal to the ratio of the applied load per unit area of the cross-section to the increase in length per unit length.
• It is denoted as E or Y.
• The unit of Young’s modulus is N m-2.

\(\text{Y}=\frac{\text{ }\!\!\sigma\!\!\text{ }}{\epsilon }\)

Where σ = stress, ϵ = strain in wire.

• Young’s Modulus Formula by using other quantities:

\({\rm{Y}} = \frac{{{\rm{F}}{{\rm{L}}_0}}}{{{\rm{A\Delta L}}}}\)

Where F = force exerted under tension, A = actual cross-sectional area, L0 = actual length, ΔL = change in length.

EXPLANATION:

• The value of Young’s Modulus depends upon the nature of the material of the body and the manner in which the body is deformed. Therefore option  1 is correct.
• Its value depends upon the temperature of the body.
• Its value is independent of the dimensions (length, radius, volume, etc.) of the body.
• As we know young's modulus of elasticity is a material's property. Therefore, Young's modulus does not depend upon the length of the radius of the wire. It is constant for the material.

CONCEPT:

• Bulk modulus: The ratio of hydrostatic stress to the volumetric strain within the elastic range is called bulk modulus.
• It is denoted by K or B.

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

Where K = bulk modulus, ΔP = change in pressure, ΔV = change in volume

• SI unit of the bulk modulus is the same as that of pressure i.e., N m–2 or Pa.

• It tells how resistant a substance (solid or fluid) is to compression.

EXPLANATION:

• From above it is clear that bulk modulus can be mathematically written as 

\(\Rightarrow K=-\frac{\Delta P}{\frac{\Delta V}{V}}=-\frac{\Delta pV}{\Delta V}\)

• Therefore option 2 is correct.

The correct answer is Bulk modulus.

 Key Points

• Bulk modulus is the correct term for the ratio of hydraulic stress to volumetric strain.
• It quantifies the substance's resistance to uniform compression, a fundamental property in understanding how materials deform under stress.

Additional Information 

• Equivalence - This term does not directly relate to the concept of the ratio between hydraulic stress and volumetric strain. Equivalence generally refers to a state of being equal or equivalent in value, function, meaning, etc., but not in this specific physical context.
• Compression - While compression is related to the concept of applying stress and causing strain, it is a process or effect rather than a measure or ratio. Compression refers to the act of pressing together or the state of being compressed.
• Semi Modulus - This term is not standard in the context of material science or physics related to stress and strain. There isn't a defined physical quantity known as ""Semi modulus"" in the study of materials' responses to external forces. -
• Summary - Bulk modulus is the appropriate term for describing the ratio of hydraulic stress to volumetric strain, highlighting the material's resistance to compression. Other options either refer to different concepts or are not directly related to the specific ratio in question."

Concept-

The force/load per unit cross-sectional area is called stress.

\(Stress\;\left( S \right) = \frac{{Force\left( F \right)}}{{Area\;\left( A \right)}}\)

Explanation-

Given that:

Force /load (F) on both the wires is the same.

Let the cross-sectional area of wire B is equal to A.

Cross-sectional area of wire A (A₁) = 2 × Cross-sectional area of wire B (A₂) = 2 A

\(Stress\;on\;wire\;A\left( S \right) = \frac{{Force\left( F \right)}}{{Area\;of\;wires\;A\;\left( {{A_1}} \right)}} = \frac{F}{{2A}}\)   ------- (1)

\(Stress\;on\;wire\;B\left( {S'} \right) = \frac{{Force\left( F \right)}}{{Area\;of\;wires\;B\;\left( {{A_2}} \right)}} = \frac{F}{A} \)   ------- (2)

On dividing equations 1 and 2, we get

\(⇒ \frac{Stress\;on\;wire\;A\left( S \right)}{Stress\;on\;wire\;B\left( S' \right)}=\frac{1}{2}\)

⇒ Stress on wire B(S') = 2 × Stress on wire A(S)

So option 2 is correct.

CONCEPT:

• Stress: Stress is the ratio of the load or force to the cross-sectional area of the material to which the load is applied. It is denoted by σ.
  •  The standard unit of stress is N/m².
• The maximum stress that a material can bear before fracture is called Ultimate stress.
• Strain: Strain is a measure of the deformation of the material as a result of the force applied. The strain is a unit less quantity. It is denoted by ϵ.
• Hooke's Law: The force needed to extend or compress a spring by some distance is proportional to that distance.
• Stress-strain curve: Stress-strain curve is a graphical representation of the behavior of a material when it's subjected to a load or force. The two characteristics that are plotted are stress on the y-axis and strain on the x-axis.

EXPLANATION:

• The stress at which any material transform from elastic to plastic behaviour is called the yield strength. So option 2 is correct.
• The Yield strength of a material is a constant that represents the maximum limit of elastic behaviour of that material.

CONCEPT:

• Linear strain: If the deforming force produces a change in length alone, the strain produced in the body is called linear strain or tensile strain.

\(Linear\;strain = \frac{{change\;in\;length\;\left( {{\rm{\Delta }}l} \right)\;}}{{Original\;length\;\left( l \right)\;}}\)

• Stress: The internal restoring force acting per unit area of cross-section of the deformed body is called stress

\(Stress = \frac{{Force\;\left( F \right)}}{{Area\;\left( A \right)}}\)

• Young's Modulus (Y): It is defined as the ratio of normal stress to longitudinal strain within the limit of proportionality.

\(Y = \frac{{Normal\;stress}}{{longitudinal\;\;strain}} = \frac{{\frac{F}{A}}}{{\frac{{{\rm{\Delta }}l}}{l}}} = \frac{{Fl}}{{A{\rm{\Delta }}l}}\)

EXPLANATION:

• Its value depends upon the nature of the material of the body and the manner in which the body is deformed.
• Its value depends upon the temperature of the body.
• Its value is independent of the dimensions (length, radius, volume, etc.) of the body.
• As we know young's modulus of elasticity is a material's property. Therefore, Young's modulus does not depend upon the length of the radius of the wire. It is constant for the material.