Buy BOOKS at Discounted Price

Mechanical Properties Of Solids

Class 11th Physics NCERT Exemplar Solution
Mcq I
  1. Modulus of rigidity of ideal liquids is
  2. The maximum load a wire can withstand without breaking, when its length is reduced to half…
  3. The temperature of a wire is doubled. The Young’s modulus of elasticity…
  4. A spring is stretched by applying a load to its free end. The strain produced in the…
  5. A rigid bar of mass M is supported symmetrically by three wires each of length l. Those at…
  6. A mild steel wire of length 2L and cross-sectional area A is stretched, well within…
  7. A rectangular frame is to be suspended symmetrically by two strings of equal length on two…
  8. Consider two cylindrical rods of identical dimensions, one of rubber and the other of…
Mcq Ii
  1. The stress-strain graphs for two materials are shown in Fig.9.3 (assume same scale).…
  2. A wire is suspended from the ceiling and stretched under the action of a weight F…
  3. A rod of length l and negligible mass is suspended at its two ends by two wires of steel…
  4. For an ideal liquid
  5. A copper and a steel wire of the same diameter are connected end to end. A deforming force…
Vsa
  1. The Young’s modulus for steel is much more than that for rubber. For the same longitudinal…
  2. Is stress a vector quantity?
  3. Identical springs of steel and copper are equally stretched. On which, more work will have…
  4. What is the Young’s modulus for a perfect rigid body?
  5. What is the Bulk modulus for a perfect rigid body?
Sa
  1. A wire of length L and radius r is clamped rigidly at one end. When the other end of the…
  2. A steel rod (Y= 2.0 × 1011 Nm-2; and α = 10-5 C-1) of length 1 m and area of cross-section…
  3. To what depth must a rubber ball be taken in deep sea so that its volume is decreased by…
  4. A truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and…
  5. Two identical solid balls, one of ivory and the other of wet-clay, are dropped from the…
La
  1. Consider a long steel bar under a tensile stress due to force’s F acting at the edges…
  2. (a) A steel wire of mass μ per unit length with a circular cross section has a radius of…
  3. A steel rod of length 2l, cross sectional area A and mass M is set rotating in a…
  4. An equilateral triangle ABC is formed by two Cu rods AB and BC and one Al rod. It is…
  5. In nature, the failure of structural members usually result from large torque because of…
  6. A stone of mass m is tied to an elastic string of negligible mass and spring constant k.…

Mcq I
Question 1.

Modulus of rigidity of ideal liquids is
A. infinity.

B. zero.

C. unity.

D. some finite small non-zero constant value.


Answer:

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.


Question 2.

The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will
A. be double.

B. be half.

C. be four times.

D. remain same.


Answer:

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.


Question 3.

The temperature of a wire is doubled. The Young’s modulus of elasticity
A. will also double.

B. will become four times.

C. will remain same.

D. will decrease.


Answer:

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.


Question 4.

A spring is stretched by applying a load to its free end. The strain produced in the spring is
A. volumetric.

B. shear.

C. longitudinal and shear.

D. longitudinal.


Answer:

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.


Question 5.

A rigid bar of mass M is supported symmetrically by three wires each of length l. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to
A. Y copper/ Yiron

B. √

C.

D.


Answer:

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




Question 6.

A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars (Fig. 9.1). A mass m is suspended from the midpoint of the wire. Strain in the wire is



A.

B.

C.

D.


Answer:


Strain is defined as



Here,



Using Binomial expansion considering x << L, we have




Question 7.

A rectangular frame is to be suspended symmetrically by two strings of equal length on two supports (Fig. 9.2). It can be done in one of the following three ways;

(a)

(b)

(c)

The tension in the strings will be

A. the same in all cases.

B. least in (a).

C. least in (b).

D. least in (c).


Answer:


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.


Question 8.

Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass M is attached to each of the free ends at the centre of the rods.
A. Both the rods will elongate but there shall be no perceptible change in shape.

B. The steel rod will elongate and change shape but the rubber rod will only elongate.

C. The steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse. (d) The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.


Answer:

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.



Mcq Ii
Question 1.

The stress-strain graphs for two materials are shown in Fig.9.3 (assume same scale).



A. Material (ii) is more elastic than material (i) and hence material (ii) is more brittle.

B. Material (i) and (ii) have the same elasticity and the same brittleness.

C. Material (ii) is elastic over a larger region of strain as compared to (i).

D. Material (ii) is more brittle than material (i).


Answer:

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.


Question 2.

A wire is suspended from the ceiling and stretched under the action of a weight F suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight.
A. Tensile stress at any cross-section A of the wire is F/A.

B. Tensile stress at any cross section is zero.

C. Tensile stress at any cross-section A of the wire is 2F/A.

D. Tension at any cross-section A of the wire is F.


Answer:


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.


Question 3.

A rod of length l and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths (Fig. 9.4). The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively

(YAI = 70 X 109 Nm-2 and Ysteel = 200 X 109 Nm-2)



A. Mass m should be suspended close to wire A to have equal stresses in both the wires.

B. Mass m should be suspended close to B to have equal stresses in both the wires.

C. Mass m should be suspended at the middle of the wires to have equal stresses in both the wires.

D. Mass m should be suspended close to wire A to have equal strain in both wires.


Answer:

For the wires to have equal stress


Where TA and TB is the tension in wires A and B while AA and AB 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.


TA (x) = TB (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.


Question 4.

For an ideal liquid
A. the bulk modulus is infinite.

B. the bulk modulus is zero.

C. the shear modulus is infinite.

D. the shear modulus is zero.


Answer:

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.


Question 5.

A copper and a steel wire of the same diameter are connected end to end. A deforming force F is applied to this composite wire which causes a total elongation of 1cm. The two wires will have
A. the same stress.

B. different stress.

C. the same strain.

D. different strain


Answer:

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.



Vsa
Question 1.

The Young’s modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?


Answer:

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


According to question Ysteel > Yrubber but both have same longitudinal strain.






Question 2.

Is stress a vector quantity?


Answer:

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.



Question 3.

Identical springs of steel and copper are equally stretched. On which, more work will have to be done?


Answer:

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)



Question 4.

What is the Young’s modulus for a perfect rigid body?


Answer:


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


Therefore,




Question 5.

What is the Bulk modulus for a perfect rigid body?


Answer:

We know that



And for A perfect rigid body , Therefore




Sa
Question 1.

A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by l. Another wire of the same material of length 2L and radius 2r, is pulled by a force 2f. Find the increase in length of this wire


Answer:

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.



Question 2.

A steel rod (Y= 2.0 × 1011 Nm-2; and α = 10-5 C-1) of length 1 m and area of cross-section 1 cm2 is heated from 0° C to 200° C, without being allowed to extend or bend. What is the tension produced in the rod?

Original Question has and α = 10-5


Answer:

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,







Question 3.

To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%. (The bulk modulus of rubber is 9.8×108 N m-2; and the density of sea water is 103 kg m-3.)

Water density is 103 kg m-3.


Answer:

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




Question 4.

A truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and has a radius of 5 mm. When the car just begins to move, the tension in the cable is 800 N. How much has the cable stretched? (Young’s modulus for steel is 2 × 1011 Nm-2.)


Answer:

Given in Questions:


Length=9.1 m


Radius=5*10-3m


Tension=800N


Y=2*1011N/m-2




Putting the values in equation




Question 5.

Two identical solid balls, one of ivory and the other of wet-clay, are dropped from the same height on the floor. Which one will rise to a greater height after striking the floor and why?


Answer:

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.




La
Question 1.

Consider a long steel bar under a tensile stress due to force’s F acting at the edges along the length of the bar (Fig. 9.5). Consider a plane making an angle θ with the length. What are the tensile and shearing stresses on this plane?



(a) For what angle is the tensile stress a maximum?

(b) For what angle is the shearing stress a maximum?


Answer:


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



Question 2.

(a) A steel wire of mass μ per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg m-3(Young’s modules Y=2×1011 Nm-2).

(b) If the yield strength of steel is 2.5×108 Nm-2, what is the maximum weight that can be hung at the lower end of the wire?


Answer:

(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,


.



Question 3.

A steel rod of length 2l, cross sectional area A and mass M is set rotating in a horizontal plane about an axis passing through the centre. If Y is the Young’s modulus for steel, find the extension in the length of the rod. (Assume the rod is uniform.)


Answer:


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,




Question 4.

An equilateral triangle ABC is formed by two Cu rods AB and BC and one Al rod. It is heated in such a way that temperature of each rod increases by ∆T. Find change in the angle ABC. [Coeff. of linear expansion for Cu is α1 ,Coeff. of linear expansion for Al is α2 ]


Answer:

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,




Question 5.

In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by . Y is the Young’s modulus, r is the radius of the trunk and R is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.


Answer:

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,






Question 6.

A stone of mass m is tied to an elastic string of negligible mass and spring constant k. The unstretched length of the string is L and has negligible mass. The other end of the string is fixed to a nail at a point P. Initially the stone is at the same level as the point P. The stone is dropped vertically from point P.

(a) Find the distance y from the top when the mass comes to rest for an instant, for the first time.

(b) What is the maximum velocity attained by the stone in this drop?

(c) What shall be the nature of the motion after the stone has reached its lowest point?


Answer:

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 FNet 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,