Equilibrium and Elasticity MCQ Quiz - Objective Question with Answer for Equilibrium and Elasticity - Download Free PDF

Concept:

The potential energy U is given as a function of the interatomic distance r, and the equation is expressed as:

U = A/r¹ - B/r², where A and B are positive constants and r is the interatomic distance. The equilibrium distance between the two atoms is the value of r where the potential energy is minimized, i.e., the force is zero.

Explanation:

To find the equilibrium distance, we differentiate the potential energy with respect to r and set the result equal to zero. The condition for equilibrium is given by:

dU/dr = 0

On differentiating the equation U = A/r¹ - B/r², we get:

dU/dr = -A/r² + 2B/r³

Setting dU/dr = 0 for equilibrium:

-A/r² + 2B/r³ = 0

Solving for r, we get:

r = (2A/B)^1/3

Correct option: Option 3: (2A/B)^1/3

Given:

We are asked to determine the dimensions of the elasticity constant.

Concept:

• The elasticity constant (or modulus of elasticity) E is related to stress and strain. The stress is force per unit area, and strain is a dimensionless quantity.
• Mathematically, elasticity is given by the relation \(\sigma = E \epsilon\) , where:
  • \(\sigma\) is stress (force per unit area),
  • \(\epsilon\) is strain (dimensionless, ratio of change in length to original length),
  • E is the modulus of elasticity.
• Stress has dimensions of force per unit area:\( [\sigma] = \frac{\text{MLT}^{-2}}{L^2} = \text{ML}^{-1}\text{T}^{-2}\) .
• Since strain is dimensionless, the dimensions of elasticity constant E are the same as that of stress, i.e., \([E] = \text{ML}^{-1}\text{T}^{-2} .\)

Calculation:

The dimensions of elasticity constant are\( \text{ML}^{-1}\text{T}^{-2}\) , which corresponds to option 2.

∴ The correct answer is option 2:\( \text{ML}^{-1}\text{T}^{-2}\) .

Concept:
The breaking strength of a rope is determined by its cross-sectional area, assuming the material and quality of the rope remain the same.

For a rope with a circular cross-section, the area A can be calculated using the formula:

A = πd² / 4

where d is the diameter of the rope.

Since the breaking strength is directly proportional to the cross-sectional area, when you change the diameter of the rope, the breaking strength changes in proportion to the square of the diameter.

F₂ / F₁ = A₂ / A₁ = d₂² / d₁²

Calculation:
Here.

Diameter of the first rope, d₁ = 3 cm = 0.03 m
Breaking strength of the first rope, F₁ = 1.5 × 10⁵ N
Diameter of the second rope, d₂ = 1.5 cm = 0.015 m
Cross-Sectional Area:

Area of the first rope, A₁,

A₁ = πd₁² / 4 = π(0.03)² / 4 = π × 0.0009 / 4 = 0.000225π m²

Area of the second rope, A₂,

A₂ = πd₂² / 4 = π(0.015)² / 4 = π × 0.000225 / 4 = 0.00005625π m²

The ratio of the breaking strengths is equal to the ratio of the cross-sectional areas,

F₂ / F₁ = A₂ / A₁ = 0.00005625π / 0.000225π = 1 / 4

F₂ = F₁ / 4 = 1.5 × 10⁵ N / 4 = 3.75 × 10⁴ N

∴ The correct option is 4

​Concept:

Young's Modulus (Y): Young's modulus is a measure of the stiffness of a material. It is defined as the ratio of stress to strain. The formula is:

Y = (F/A) / (ΔL/L)

Where F is the force applied, A is the cross-sectional area, ΔL is the change in length, and L is the original length of the wire.

Relation between ΔL and L: The increase in length (ΔL) of a wire due to stretching force can be expressed as:

ΔL = (F × L) / (A × Y)

Where A = πr² is the cross-sectional area of the wire.

For the same material, Young's modulus Y remains constant.

Calculation:

Here,

Initial wire: Length = L, Radius = r, Force = F.

Increase in length of the initial wire = 5 cm.

Second wire: Length = 4L, Radius = 4r, Force = 4F.​​​​

 ⇒ \(\frac{F / A}{Δ L / L}=Y\)

⇒ \(Δ L=\frac{F L}{A Y} \)

⇒ \(4 \times 4 \times \frac{1}{16}=1\)

⇒ ΔL2 = ΔL1 = 5 cm

 The increase in length of this wire is 5 cm.

Concept:

Young's modulus and the formula for the elongation of a wire under tensile stress.

Young's modulus Y is defined as:

Y= Stress/Strain
​The stress σ is given by:
σ=F/ A
where F is the force applied and A is the cross-sectional area of the wire.

The strain ϵ is given by:
ϵ= ΔL/L
where ΔL is the change in length (elongation) and L is the original length.

Combining these equations, we get:
Y= F/A⋅L/ΔL​

The cross-sectional area A of the wire with a radius 
r is: A=πr²
Substituting A into the elongation formula:
ΔL= FL/πr_²Y​

Calculation:

\(\mathrm{Y}=\frac{\text { stress }}{\text { strain }} \)

\(\mathrm{Y}=\frac{\frac{\mathrm{F}}{\mathrm{\pi r}^2}}{\frac{ℓ}{\mathrm{L}}}\)

\(\mathrm{F}=\mathrm{Y} \pi \mathrm{r}^2 \times \frac{ℓ}{\mathrm{L}}\)           ….(i) 

\(\mathrm{Y}=\frac{\frac{\mathrm{F} / 2}{\pi \mathrm{r}^2 / 4}}{\frac{Δ ℓ}{\mathrm{L}}} \)

\(\mathrm{F}=\mathrm{Y} \frac{Δ ℓ}{\mathrm{L}} \times 2 \times \frac{\pi \mathrm{r}^2}{4}\)

From (i) 

\(\mathrm{Y} \pi \mathrm{r}^2 \frac{ℓ}{\mathrm{L}}=\mathrm{Y} \frac{Δ ℓ}{\mathrm{L}} \frac{\pi \mathrm{r}^2}{2}\)

Δℓ = 2ℓ 

∴ The correct option is 4)

CONCEPT:

Stress

• When an external force is applied to a material it gets deformed.
• Due to this deformation, an internal resistance force develops in the material.
• This internal resistance force per unit cross-sectional area is called stress.

\(\Rightarrow σ=\frac{P}{A}\)

Where σ = stress, P = applied load and A = cross sectional-area

• Types of stress

Thermal expansion: Material expands while being heated and trying to acquire more space.

eg. When a rod is heated, its length increases and it is given by

                ∆l = α ∆T L

Where, α – coefficient of linear thermal expansion

                ∆T – Increase in temperature

                L – original length of the rod

• Here, the rod is placed between two fixed supports and it is heated, and as we know that metals expand on heating.
• Therefore, the rod will expand and will exert force on the supports.

Free body diagram

Here, F₁ = Is the force acted by support on rod and F₂ = Is the force aced by rod on supports.

• Support will restrict the rod to expand and the rod will act force on support and according to Newton’s 3^rd law support also act force on rod.
• Hence, a compressive force acts on the rod. Therefore option 1 is correct.

CONCEPT:

• Elasticity: The ability of materials to return to their original shape after a deforming is called elasticity.
• Plasticity: The property due to which the material didn't return to its initial position after deformation is called plasticity.
• Viscosity: The relative motion between different layers of liquid or gases is opposed by a force, which is known as the Viscous force and this property is known as Viscosity.

EXPLANATION:

• The property of a body, by virtue of which it tends to regain its original size and shape when the applied force is removed, is known as elasticity. So option 1 is correct.

CONCEPT:

• Elasticity is the ability of a body that resists the body to distort under any force and try to return to its original shape and size when that force is removed.
• Elasticity for different substances is calculated by different experiments using Stress and strain.

Values of Elasticity for some materials are:

EXPLANATION:

The highest value of elasticity from the given substances will be of Steel.

So the correct answer is option 4.

CONCEPT:

Equilibrium of a rigid body:

• A rigid body is said to be in mechanical equilibrium if both its linear momentum and angular momentum are not changing with time, or equivalently, the body has neither linear acceleration nor angular acceleration.
• Condition for the mechanical equilibrium:
  1. The total force, i.e. the vector sum of the forces, on the rigid body is zero.
  2. The total torque, i.e. the vector sum of the torques on the rigid body is zero.

\(⇒ \vec{F_1}+\vec{F_2}+...+\vec{F_n}=0\)

\(⇒ \vec{τ_1}+\vec{τ_2}+...+\vec{τ_n}=0\)

• If the forces on a rigid body are acting in the 3 dimensions, then six independent conditions to be satisfied for the mechanical equilibrium of a rigid body.
• If all the forces acting on the body are coplanar, then we need only three conditions to be satisfied for mechanical equilibrium.
• A body may be in partial equilibrium, i.e., it may be in translational equilibrium and not in rotational equilibrium, or it may be in rotational equilibrium and not in translational equilibrium.

EXPLANATION:

A body is rotating at its position with constant angular velocity:

• In this case, the body is not moving from its position so its velocity is zero.
• Therefore the linear acceleration of the body will be zero and the body is said to be in translational equilibrium.
• Since the angular velocity is constant so the angular acceleration of the body will also be zero and the body is said to be in rotational equilibrium.
• A rigid body is said to be in mechanical equilibrium if both its linear momentum and angular momentum are not changing with time, or equivalently, the body has neither linear acceleration nor angular acceleration. Hence, option 3 is correct.

CONCEPT:

• Torsion is the twisting of an object when a certain torque is applied.
• Torsion can also be defined as a moment of twist. where the moment is what we called torque.

EXPLANATION:

• The unit of torque is equivalent to the unit of torsion.
• Torque can be expressed as the product of force and perpendicular distance i.e.,

\(\tau = F \times {d_{perpendicular}}\)

• The unit of F is newtons and distance is meter.
• ∴ Torque = Nm
• Therefore the unit of torsion = torque = Nm. Therefore option 2 is correct.​

CONCEPT:

• Deforming force: The external force applied on a body that causes the same change in its configuration, i.e., Length, Volume, or Shape of an object, then such force is known as deforming force
• Perfectly elastic body: if a body completely regains its original shape and size after removal of an external deforming force, such a body is known as perfectly elastic.
• Elastic limit: It is a limit used to describe the maximum deforming force any material can withstand without getting permanently deformed (It will be used in Hooke’s law).

EXPLANATION:

• From the above example we can see that if a body regain its original position under an external, such object are termed as a perfectly elastic object
• Hence option 1 is correct among all.

CONCEPT:

Equilibrium of a rigid body:

• A rigid body is said to be in mechanical equilibrium if both its linear momentum and angular momentum are not changing with time, or equivalently, the body has neither linear acceleration nor angular acceleration.
• Condition for the mechanical equilibrium:
  1. The total force, i.e. the vector sum of the forces, on the rigid body is zero.
  2. The total torque, i.e. the vector sum of the torques on the rigid body is zero.

\(⇒ \vec{F_1}+\vec{F_2}+...+\vec{F_n}=0\)

\(⇒ \vec{τ_1}+\vec{τ_2}+...+\vec{τ_n}=0\)

• If the forces on a rigid body are acting in the 3 dimensions, then six independent conditions to be satisfied for the mechanical equilibrium of a rigid body.
• If all the forces acting on the body are coplanar, then we need only three conditions to be satisfied for mechanical equilibrium.
• A body may be in partial equilibrium, i.e., it may be in translational equilibrium and not in rotational equilibrium, or it may be in rotational equilibrium and not in translational equilibrium.

EXPLANATION:

Translational equilibrium:

• A rigid body is said to be in translational equilibrium if its linear momentum does not change with time, or equivalently, the linear acceleration of the body is zero.
• So if the body is at rest or it is moving with constant velocity then it is said to be in translational equilibrium. Hence, 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.
•  The standard unit of measure for stress is N/m².
• The stress is directly proportional to the load and here the load is in terms of bending.
• Bending: Bending is a process by which metal can be deformed by plastically deforming the material and changing its shape. The surface area of the material does not change much. Bending is deformation about one axis.

EXPLANATION:

• Since at the neutral axis there is no bending of the beam so no stress is developed at the neutral axis.
• Thus the bending stress at the neutral axis of the beam is zero. So option 3 is correct.

Concept:

• Malleability is the property by virtue of which a material may be hammered or rolled into thin sheets without rupture. This property generally increases with the increase in temperature.
• Malleability is the ability of a metal to exhibit large deformation or plastic response when being subjected to compressive force.
• Lead, soft steel, wrought iron, copper, and aluminum are some materials in order of diminishing malleability.
• Ductility is the property of the material that enables it to be drawn out or elongated to an appreciable extent before rupture occurs.
• The percentage elongation or percentage reduction in area before rupture of a test specimen is the measure of ductility. Normally if the percentage elongation exceeds 15% the material is ductile and if it is less than 5% the material is brittle.
• Lead, copper, aluminum, and mild steel are typical ductile materials.
• Brittleness is opposite to ductility. Brittle materials show little deformation before fracture and failure occurs suddenly without any warning i.e. it is the property of breaking without much permanent distortion. Normally if the elongation is less than 5% the material is brittle. E.g. cast iron, glass, ceramics are typical brittle materials.

Explanation:

From the above explanation we can see that ductility of metal is the property that enables it to be drawn out or elongated to an appreciable extent before rupture occurs.

Because of this property we can create long metal wires.

CONCEPT:

Equilibrium of a rigid body:

• A rigid body is said to be in mechanical equilibrium if both its linear momentum and angular momentum are not changing with time, or equivalently, the body has neither linear acceleration nor angular acceleration.
• Condition for the mechanical equilibrium:
  1. The total force, i.e. the vector sum of the forces, on the rigid body is zero.
  2. The total torque, i.e. the vector sum of the torques on the rigid body is zero.

\(⇒ \vec{F_1}+\vec{F_2}+...+\vec{F_n}=0\)

\(⇒ \vec{τ_1}+\vec{τ_2}+...+\vec{τ_n}=0\)

• If the forces on a rigid body are acting in the 3 dimensions, then six independent conditions to be satisfied for the mechanical equilibrium of a rigid body.
• If all the forces acting on the body are co-planar, then we need only three conditions to be satisfied for mechanical equilibrium.
• A body may be in partial equilibrium, i.e., it may be in translational equilibrium and not in rotational equilibrium, or it may be in rotational equilibrium and not in translational equilibrium.

Couple:

• A pair of forces of equal magnitude but acting in opposite directions with different lines of action is known as a couple or torque.
• A couple produces rotation without translation.
• Examples:
  1. When we open the lid of a bottle by turning it, our fingers are applying a couple to the lid.

EXPLANATION:

Given F₁ = F₂ = F

• If two equal and opposite forces are acting on a body then the resultant force on the body is given as,

⇒ F = F₁ - F₂

⇒ F = F - F

⇒ F = 0

• Therefore the resultant force on the body is zero, so the body will be in translational equilibrium.
• A pair of forces of equal magnitude but acting in opposite directions with different lines of action is known as a couple or torque.
• A couple produces rotation without translation. So if the line of action of the two forces is the same then the body will be in rotational equilibrium.
• But the body will not be in rotational equilibrium if the line of action is of the two forces are different.
• Therefore the body may or may not be in rotational equilibrium.

• Condition for the mechanical equilibrium:
  1. The total force, i.e. the vector sum of the forces, on the rigid body is zero.
  2. The total torque, i.e. the vector sum of the torques on the rigid body is zero.

• Hence, option 2 is correct.