Mechanical Properties Of Solids

Modulus of rigidity is also called shear modulus represented by G which is the ratio of shearing stress (stress due to tangential force) to shearing strain.

Only solids exhibit shearing as they have a definite shape. Liquids cannot show shearing as they do not have definite shape and cannot be subjected to tangential force.

For the wire to break, we need to apply a minimum force called breaking force on the wire. This breaking force is dependent on the area of cross-section of the wire and the breaking stress of the wire which is characteristic of the material and does not depend on the dimensions (length and thickness). Thus, in this case, since only the length is being altered, no change will happen to the maximum load the wire can withstand.

The Young’s Modulus (Y) is defined as

∆l can be written as

Where l is the original length, α is the coefficient of heat expansion (a constant for a material) and ∆T is the temperature difference.

So,

As ∆T increases, Y decreases.

When the load is attached to the spring, the spring will get stretched. This stretching of the spring will change both its shape and length. The change in length will correspond to the longitudinal strain while that in shape shall be shear strain.

Since, all the wires need to have same tension and we know that Young’s modulus of the wires shall be the ratio of the tension they experience to their area of cross – section, we have

Similarly

Equating tensions, we get

Strain is defined as

Here,

Using Binomial expansion considering x << L, we have

Let (T) be the tension on the wires and (M) be the mass of the block, then

For this case, 2T = Mg

T = Mg/2

For this case, the sine components are cancelled.

2Tcosθ = Mg

T = Mg/2cosθ

The cosine of any angle will vary from 0 to 1 with 1 reaching at 0˚ which would then become the previous case. In this since the cosine will be less than one Tension in this case will be more than the previous case when the strings are vertical.

Thus, tension is least in this case.

Both rubber and steel rod will be subjected to same force, and will be elongated, but since, they have different elasticities, (steel being more elastic than rubber), steel rod doesn’t show any perceptible change in shape while the rubber rod elongates with the shape of the bottom edge tapered to a tip at the centre.

The linear limit is elastic limit and the ultimate tension strength is the breaking strength.

Since, the linear limit of material (ii) is more than that of (i), it is more elastic. Also, since the gap between the Ultimate Tension Strength and Fracture Point is less for material (ii) than for material (i), it is more brittle.

The tensile stress at the cross – section of an object is defined as the ratio of the force (perpendicular to the cross – section) to the area of cross – section.

Here, Tensile stress = F/A

Tension at any cross – section will be Tensile stress x Area of cross – section = F.

Thus, option (a) and (d) are correct.

For the wires to have equal stress

Where T_A and T_B is the tension in wires A and B while A_A and A_B are the areas cross – section of the wires.

Let the mass be placed at a distance of x from wire A. As the system is in equilibrium, we can take moments about the point where the mass is hanging.

T_A (x) = T_B (l – x)

Thus, mass (m) should be suspended close to wire B. Option (b) is correct.

Thus, for same strain

Let the mass be placed at a distance of y units from wire A. Taking moment at the point at which mass (m) is attached

Thus, strain will be equal in both wires when mass m is near the wire A.

An ideal liquid is non – compressible in nature which means its compressibility is zero. Now, bulk modulus is the reciprocal of compressibility thus, it will be infinity for an ideal liquid.

Shear modulus is the ratio of shear stress (which arises due to tangential force) and shear strain. A liquid cannot sustain tangential force; thus, shear modulus will be zero for it.

Since, stress depends on the deforming force and the area of the object, they both will experience same stress because they have the same diameter and are subject to same deforming force.

Since, both these wires are made up of different materials, thus, they will have different elongations, leading to different strain.

Young’s modulus (Y) is defined as the ratio of tensile or longitudinal stress to tensile or longitudinal strain.

According to question Y_steel > Y_rubber but both have same longitudinal strain.

No

Stress is defined as restoring force developed per unit area in a body. It is neither scalar nor a vector quantity, it is a tensor quantity.

Given that Identical Springs Implies that Area of cross-section and Length of both springs are equal. Also Equally Stretched implies that Force is Equal and combining both condition we get that stress are equal in both cases.

We know Work done in stretching is given by

Dividing both sides with will give us

And we know that

Therefore Diving equation RHS with Stress we get,

From Equation 2 we get a relation between work done and young

And we know that Y (steel)>Y (Copper)

Therefore W (Copper)>W (Steel)

For a perfect rigid body condition, no strain will be present in ideal condition.

Therefore,

We know that

And for A perfect rigid body , Therefore

Given that same material implies that young modulus is same for both cases.

For First wire we can write a relation between force and young modulus as,

Similarly we can write it for second fire, and let change in wire is k,

It is clear that both equations are same hence change in length is also equal.

Since heat is travelling in single direction, therefore it is a case of 1D Thermal Expansion or linear expansion.

We know for thermal expansion, change in length is given by

Re-arranging it will give,

Now, for calculating tension we will use a relation between young modulus and Force,

We know that Bulk Modulus is given by

Where,

Change in pressure (indirectly implies change in force around the solid which is under observation)

Change in volume of solid due to change in pressure.

Intial Volume of solid

Let assume that Pressure outside sea is and to a depth h the pressure is equal to

Therefore,

Where,

Change in pressure as we are going below sea level

Density of water

=height from sea level

Now it is given that volume decrease percentage which is equal to 0.1%.

And Bulk Modulus is 9.8×108 N m^-2 with sea water density equal to 10³ kg m^-3.Therefore putting values in equation 1, we get

Given in Questions:

Length=9.1 m

Radius=5*10^-3m

Tension=800N

Y=2*10¹¹N/m^-2

Putting the values in equation

Ivory will rise to a greater height after striking the floor as ivory is more elastic in nature compared to wet clay ball.

When we drop the balls, they have equal potential energy stored it as they are at same height. Now once they are dropped, their potential energy starts getting converted into kinetic energy due to which they gain speed. Now after touching the floor, their kinetic energy makes them deformed and this energy is stored in elastic energy. Now as Ivory has high elasticity compared to wet-clay ball, therefore it is able to re-gain its same more easily and at a faster rate compared to wet clay ball. While re-gaining its shape, it pushes itself toward force and due to this ivory ball is able to reach higher compared to wet-clay ball.

Now, in the given questions we are being asked about inclined plane at an angle θ. Considering only inclined plane part, we can resolve the forces parallel and perpendicular to it.

For calculating area, let us assume cross-section area of long bar is A. Therefore it as reference we can calculate cross-section of inclined plane as

Therefore tensile force is given by

Therefore, Tensile force is maximum when as

Shearing force is given by,

Therefore, Shearing force is maximum when as

(a)

Taking a cross-section of wire

At X=0, Tension=Mg and X=0, Tension= μgL + Mg. Integrating above equation will give

As force is a function of x, therefore relating force with young modulus we have to consider a small part in which force can be approximated to be constant .Let the length of that part is dx. Therefore, we can write

We have written the equation which is valid for only 1D,as it is given in question that lateral strains << longitudinal strains, therefore we can assume that only extension in length is happening only

Therefore change in radius is given by

Putting in the values from questions

(b) We have been given Yield Strength = 2.5×10⁸ N/m²

We know from earlier equations that Maximum Tensile force at the bottom will be equal to

We know that,

Putting in the values

Solving the above equation we get,

.

Force acting in this element due to rotating is

Where

df=force on that small element under consideration

dm=mass of element under consideration

ω=angular velocity (which will remain same for all particles)

x=distance from the axis of rotation

Now, Taking mass per unit length

We have taken 2L as we considering the length of half part to be L

Now we know a relation between young modulus and tension force, but since, force is a function of x hence we can only write the equation for a small element of length dx

Here we have represented extension in length as dr, since our original length is dx

Now we need to multiply it by 2 as we were considering only half length of rod,

Assuming that Triangle has taken any arbitrary shape,

Blue line is perpendicular to the base triangle of height h

Now since it forms two right angled triangles therefore, we can write

,

,

Simplifying further,

From 1 we can write

From triangle we can write

Substituting value of x

.

Re-arranging the terms we can write,

We have written an equation which is valid for triangles, therefore above equation will be valid for Equilateral (A=60)

Differentiating the above equation with its parameter,

Now we know that for equilateral angle

Change in lengths can be equated with thermal coeff,

,,

Putting in equation,

Re-arranging the terms,

Let us assume that Tree is uniform, and considering tree is just about to fall.

Let the distance from ground = h/2

Radius of curvature =R

Radius of Tree=r

When Tree will, one part of radius will remain but other will be different.

Using Properties of Triangle we can write,

And we know that d<<R, therefore its square will be less compared, hence neglecting d², we can write,

It is given in questions elastic torque is given by

It will be equal to,

It is clear from question that Ball will at least travel to natural length of spring and further it will stretch due its elastic nature of spring.

Let assume that distance travelled by stone of Mass = x

Now we know that it isn’t elastic collision, therefore Energy will conserve. By travelling a distance of Y it will gain Potential energy of spring which will store in spring whereas it will lose its potential energy.

Therefore loss in Potential energy =

And gain in Potential energy of spring =

Here we taking the difference as energy is gained by spring is for the extra distance it will travel from its natural length.

Equating both equations

Solving the quadratic equation we will get two solutions, the solution with positive will answer as we are dealing in positive domain of distance.

(b) Maximum will be attained when a particle isn’t experiencing any force. That is when a particle’s own gravitational force is equal to force due to elasticity of spring

Where p is distance from its natural length (No force is experienced before spring stretches out of its natural length)

Using Conservation of energy, we know that loss in potential energy will be gain in kinetic energy which will be gain by particle and gain in elastic energy of spring.

Equating p to,

And solving the quadratic equation and taking only positive sign for velocity as we dealing in positive domain, we get,

(c) Intuitively we know that particle will perform SHM, but we need to prove that. Therefore we need to reduce it to SHM equation.

We know that at particular distance y (from its initial position) we can write,

As the F_Net which will experienced by stone will due to gravitational force towards down and elastic force towards up. And we also know that acceleration is double derivative of distance travel therefore we can write,

To reduce more, we will assume that

Now double differentiating w w.r.t time we will get,

Substituting in above equation we get,

Therefore, its angular frequency is

For calculating its mean position we have to find a particular distance at which no force is acting,