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All questions of Mechanical Properties of Solids for NEET Exam
The materials which have the same elastic properties in all directions are called __________.- a)Isotropic
- b)Brittle
- c)Homogeneous
- d)Hard
Correct answer is option 'A'. Can you explain this answer?
The materials which have the same elastic properties in all directions are called __________.
a)
Isotropic
b)
Brittle
c)
Homogeneous
d)
Hard
|
Pooja Shah answered |
Same elastic properties in all direction is called the homogenity of a material.
Material is said to be ductile if- a)a large amount of plastic deformation takes place between the elastic limit and the fracture point
- b)fracture occurs soon after the elastic limit is passed
- c)material cross section is not significantly reduced at failure
- d)material breaks suddenly at little elongation
Correct answer is option 'A'. Can you explain this answer?
Material is said to be ductile if
a)
a large amount of plastic deformation takes place between the elastic limit and the fracture point
b)
fracture occurs soon after the elastic limit is passed
c)
material cross section is not significantly reduced at failure
d)
material breaks suddenly at little elongation
|
Nandini Iyer answered |
A ductile material is one that can withstand a large amount of plastic deformation between the elastic limit and the fracture point.
A material that breaks suddenly when elongated or fracture occurs in it soon after the elastic limit is crossed is called a brittle material.
A ductile material that exhibits extra elongation or deformation and does not fracture is also referred as superplastic material.
A material that breaks suddenly when elongated or fracture occurs in it soon after the elastic limit is crossed is called a brittle material.
A ductile material that exhibits extra elongation or deformation and does not fracture is also referred as superplastic material.
A 200-kg load is hung on a wire with a length of 4.00 m, a cross-sectional area of 0.200 × 10−4 m2, and a Young’s modulus of 8.00 × 1010N/ m2. What is its increase in length?- a)4.80 mm
- b)4.90 mm
- c)4.80 mm2.4.90 mm
- d)80 mm2.4.90 mm3.
Correct answer is option 'C'. Can you explain this answer?
A 200-kg load is hung on a wire with a length of 4.00 m, a cross-sectional area of 0.200 × 10−4 m2, and a Young’s modulus of 8.00 × 1010N/ m2. What is its increase in length?
a)
4.80 mm
b)
4.90 mm
c)
4.80 mm2.4.90 mm
d)
80 mm2.4.90 mm3.
|
Jyoti Sengupta answered |
Putty or mud is an example of- a)semi elastic
- b)plastic material
- c)elastomers
- d)elastic material
Correct answer is option 'B'. Can you explain this answer?
Putty or mud is an example of
a)
semi elastic
b)
plastic material
c)
elastomers
d)
elastic material
|
New Words answered |
Since mud or putty have no gross tendency to regain their previous shape & they get permanently deformed, they are close to ideal plastics.
- Perfectly plastic bodies are those that retain their distorted shape or size after external pressures applied to them have been released.
- Partially elastic bodies are those that partially return to their former shape or size when external forces are eliminated.
- The shape or size of mud or putty is now altered when force is applied, and the body does not return to its original state when the force is removed. Only perfectly plastic bodies, as already established, exhibit this feature.
- Additionally, many elasticity moduli types are used to measure elasticity.
- The ratio of stress to strain given to a body under various variations in dimension is what makes up an elastic modulus.
- Young's Modulus, Shear Modulus, and Bulk Modulus are the three forms of elastic moduli. Since Young's modulus or longitudinal stress by strain for mud or putty is almost zero, they are theoretically proven to be almost ideal plastic bodies.
Hence, putty or mud is an example of a perfectly plastic body.
Which of the following represents Hooke’s Law?- a)Stress = k x strain
- b)
- c)Stress = k x (strain)2
- d)Stress2 = k x strain
Correct answer is option 'A'. Can you explain this answer?
Which of the following represents Hooke’s Law?
a)
Stress = k x strain
b)
c)
Stress = k x (strain)2
d)
Stress2 = k x strain
|
Gaurav Kumar answered |
The law states that the strain in a solid is proportional to the applied stress within the elastic limit of that solid body i.e. stress = k x strain.
You hang a flood lamp from the end of a vertical steel wire. The flood lamp stretches the wire 0.18 mm and the stress is proportional to the strain. How much would it have stretched if the wire had the same length but twice the diameter?- a)0.065 mm
- b)0.055 mm
- c)0.045 mm
- d)0.075 mm
Correct answer is option 'C'. Can you explain this answer?
You hang a flood lamp from the end of a vertical steel wire. The flood lamp stretches the wire 0.18 mm and the stress is proportional to the strain. How much would it have stretched if the wire had the same length but twice the diameter?
a)
0.065 mm
b)
0.055 mm
c)
0.045 mm
d)
0.075 mm
|
Pritam Kapoor answered |
Given:
- The flood lamp stretches the wire by 0.18 mm
- The stress is proportional to the strain
To find:
- How much would it have stretched if the wire had the same length but twice the diameter
Let's begin by understanding the given information.
Stress is defined as the force per unit area and is denoted by the symbol σ (sigma). Mathematically, stress is given by:
σ = F / A
where F is the force applied and A is the area over which the force is applied.
Strain is defined as the change in length per unit length and is denoted by the symbol ε (epsilon). Mathematically, strain is given by:
ε = ΔL / L
where ΔL is the change in length and L is the original length.
From the given information, we know that the stress is proportional to the strain. This can be expressed mathematically as:
σ ∝ ε
or
σ = kε
where k is a constant of proportionality.
Now, let's apply this information to the problem at hand.
When the flood lamp is hung from the wire, it exerts a force on the wire which causes it to stretch. Let's assume that the original diameter of the wire is d and the original length is L.
From the given information, we know that the stress is proportional to the strain. Therefore, we can write:
σ = kε
where σ is the stress, k is a constant of proportionality, and ε is the strain.
The stress can be calculated using the formula:
σ = F / A
where F is the force applied and A is the cross-sectional area of the wire.
The force applied is the weight of the flood lamp, which can be calculated using the formula:
F = mg
where m is the mass of the flood lamp and g is the acceleration due to gravity.
The cross-sectional area of the wire can be calculated using the formula:
A = πd^2 / 4
where d is the diameter of the wire.
Therefore, we can write:
σ = (mg) / (πd^2 / 4)
The strain can be calculated using the formula:
ε = ΔL / L
where ΔL is the change in length and L is the original length.
From the given information, we know that the flood lamp stretches the wire by 0.18 mm. Therefore, we can write:
ε = 0.18 / L
Now, let's combine the equations for stress and strain:
σ = kε
σ = (mg) / (πd^2 / 4)
ε = 0.18 / L
Substituting the values of σ and ε, we get:
(mg) / (πd^2 / 4) = k (0.18 / L)
Simplifying, we get:
k = (mgL) / (0.18πd^2)
Now, let's use this value of k to calculate the change in length when the diameter of the wire is doubled.
When the diameter of the wire is doubled, the cross-sectional area of the wire becomes 4 times the original area. Therefore, the new diameter is 2d and the new cross-sectional area is:
A' = π(2d)^2 / 4 = 4πd^2
Using the same formula for stress,
- The flood lamp stretches the wire by 0.18 mm
- The stress is proportional to the strain
To find:
- How much would it have stretched if the wire had the same length but twice the diameter
Let's begin by understanding the given information.
Stress is defined as the force per unit area and is denoted by the symbol σ (sigma). Mathematically, stress is given by:
σ = F / A
where F is the force applied and A is the area over which the force is applied.
Strain is defined as the change in length per unit length and is denoted by the symbol ε (epsilon). Mathematically, strain is given by:
ε = ΔL / L
where ΔL is the change in length and L is the original length.
From the given information, we know that the stress is proportional to the strain. This can be expressed mathematically as:
σ ∝ ε
or
σ = kε
where k is a constant of proportionality.
Now, let's apply this information to the problem at hand.
When the flood lamp is hung from the wire, it exerts a force on the wire which causes it to stretch. Let's assume that the original diameter of the wire is d and the original length is L.
From the given information, we know that the stress is proportional to the strain. Therefore, we can write:
σ = kε
where σ is the stress, k is a constant of proportionality, and ε is the strain.
The stress can be calculated using the formula:
σ = F / A
where F is the force applied and A is the cross-sectional area of the wire.
The force applied is the weight of the flood lamp, which can be calculated using the formula:
F = mg
where m is the mass of the flood lamp and g is the acceleration due to gravity.
The cross-sectional area of the wire can be calculated using the formula:
A = πd^2 / 4
where d is the diameter of the wire.
Therefore, we can write:
σ = (mg) / (πd^2 / 4)
The strain can be calculated using the formula:
ε = ΔL / L
where ΔL is the change in length and L is the original length.
From the given information, we know that the flood lamp stretches the wire by 0.18 mm. Therefore, we can write:
ε = 0.18 / L
Now, let's combine the equations for stress and strain:
σ = kε
σ = (mg) / (πd^2 / 4)
ε = 0.18 / L
Substituting the values of σ and ε, we get:
(mg) / (πd^2 / 4) = k (0.18 / L)
Simplifying, we get:
k = (mgL) / (0.18πd^2)
Now, let's use this value of k to calculate the change in length when the diameter of the wire is doubled.
When the diameter of the wire is doubled, the cross-sectional area of the wire becomes 4 times the original area. Therefore, the new diameter is 2d and the new cross-sectional area is:
A' = π(2d)^2 / 4 = 4πd^2
Using the same formula for stress,
If we compress or elongate the solids, what happens to their potential energy?- a)P.E. of solid decreases whether it is compressed or elongated
- b)P.E. of solid increases if it is elongated
- c)P.E. of solid increases whether it is compressed or elongated
- d)P.E. of solid increases if it is compressed
Correct answer is option 'C'. Can you explain this answer?
If we compress or elongate the solids, what happens to their potential energy?
a)
P.E. of solid decreases whether it is compressed or elongated
b)
P.E. of solid increases if it is elongated
c)
P.E. of solid increases whether it is compressed or elongated
d)
P.E. of solid increases if it is compressed
|
Anjali Iyer answered |
When a soild is compressed or elongated, it's P.E increases. Potential energy increases because work has to be done against force of repulsion during compression and against force of attraction during elongation.
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10−5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10−5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?- a)1.2
- b)1.6
- c)1.8
- d)2.0
Correct answer is option 'C'. Can you explain this answer?
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10−5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10−5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
a)
1.2
b)
1.6
c)
1.8
d)
2.0
|
Krishna Iyer answered |
What diameter should a 10-m-long steel wire have if we do not want it to stretch more than 0.5 cm under a tension of 940 N? Take Young's modulus of steel as 20 × 1010 Pa- a)3.2 mm
- b)3.0 mm
- c)3.4 mm
- d)3.6 mm
Correct answer is option 'C'. Can you explain this answer?
What diameter should a 10-m-long steel wire have if we do not want it to stretch more than 0.5 cm under a tension of 940 N? Take Young's modulus of steel as 20 × 1010 Pa
a)
3.2 mm
b)
3.0 mm
c)
3.4 mm
d)
3.6 mm
|
|
Rajesh Gupta answered |
Y=F x l/A x Δ l
Δ l=0.5cm=0.5x10-2m, l=10M, F=940N
Y=20x1010pa
20x1010=940x10/πr2x0.5x10-10
πr2=94x100/5x10-3x2x1011=94x102/10x108
r2=94/π x 10-7 =2.99 x 10-6
r2 ≅3x10-6
r=1.13x10-10 m
diameter=2r=3.6mm
Δ l=0.5cm=0.5x10-2m, l=10M, F=940N
Y=20x1010pa
20x1010=940x10/πr2x0.5x10-10
πr2=94x100/5x10-3x2x1011=94x102/10x108
r2=94/π x 10-7 =2.99 x 10-6
r2 ≅3x10-6
r=1.13x10-10 m
diameter=2r=3.6mm
With reference to figure the elastic zone is
- a)BC
- b)CD
- c)AB
- d)OA
Correct answer is option 'D'. Can you explain this answer?
With reference to figure the elastic zone is
a)
BC
b)
CD
c)
AB
d)
OA
|
Hansa Sharma answered |
Hooke’s law: a law stating that the strain in a solid is proportional to the applied stress within the elastic limit of that solid.
In the OA line Hooke’s law is valid because stress is directly proportional to strain.
In the OA line Hooke’s law is valid because stress is directly proportional to strain.
The modulus of elasticity of steel is greater than that of rubber because under the same stress- a)the strain in steel is less than rubber
- b)the strain in steel is more than rubber
- c)elongation in steel is more than in rubber
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The modulus of elasticity of steel is greater than that of rubber because under the same stress
a)
the strain in steel is less than rubber
b)
the strain in steel is more than rubber
c)
elongation in steel is more than in rubber
d)
none of these
|
Om Desai answered |
Modulus of elasticity= stress/strain =(F/A)/(ΔL/L) So, for same stress Modulus of elasticity ∝( L/ΔL)and ΔL for rubber is more as compared to steel so Modulus of elasticity for rubber will be less as they are inversely proportional and also ΔL/L is less for steel
You hang a flood lamp from the end of a vertical steel wire. The flood lamp stretches the wire 0.18 mm and the stress is proportional to the strain. How much would it have stretched if the wire were twice as long?- a)0.36 mm
- b)0.34 mm
- c)0.38 mm
- d)0.40 mm
Correct answer is option 'A'. Can you explain this answer?
You hang a flood lamp from the end of a vertical steel wire. The flood lamp stretches the wire 0.18 mm and the stress is proportional to the strain. How much would it have stretched if the wire were twice as long?
a)
0.36 mm
b)
0.34 mm
c)
0.38 mm
d)
0.40 mm
|
|
Rajesh Gupta answered |
F/A = e/L
F/A = 0.18/l ( let the length be l)
F/A = x/2l
0.18/L = x/2l
0.18×2l = xl
0.36l = xl
x = 0.36
F/A = 0.18/l ( let the length be l)
F/A = x/2l
0.18/L = x/2l
0.18×2l = xl
0.36l = xl
x = 0.36
The S.I unit of stress is- a)Watt
- b)Joule
- c)Pascal
- d)Newton
Correct answer is option 'C'. Can you explain this answer?
The S.I unit of stress is
a)
Watt
b)
Joule
c)
Pascal
d)
Newton
|
|
Krishna Iyer answered |
Stress has its own SI unit called the Pascal. 1 Pascal (Pa) is equal to 1 N/m2. In imperial units stress is measured in pound force per square inch which is often shortened to "psi". The dimension of stress is same as that of pressure.
A solid cylinder can be subjected to- a)tensile stress
- b)compressive stress
- c)shearing stress
- d)all the above
Correct answer is option 'D'. Can you explain this answer?
A solid cylinder can be subjected to
a)
tensile stress
b)
compressive stress
c)
shearing stress
d)
all the above
|
Infinity Academy answered |
To answer this question you should know the definitions of tensile, compressive and shear stress.
Tensile stress causes change (increases the length of cylinder) in the length of the object, compressive strength changes the volume of the object (it can be applied from all sides of the object), shear stress is applied parallel to the surface of an object (in case of the cylinder shear stress will be parallel to circular surface) we can apply all the three stresses in case of cylinder hence option D is correct.
Tensile stress causes change (increases the length of cylinder) in the length of the object, compressive strength changes the volume of the object (it can be applied from all sides of the object), shear stress is applied parallel to the surface of an object (in case of the cylinder shear stress will be parallel to circular surface) we can apply all the three stresses in case of cylinder hence option D is correct.
Tissue of aorta blood vessel is an example of- a)state of permanant deformation
- b)perfectly plastic body
- c)perfectly elastic body
- d)elastomer
Correct answer is option 'D'. Can you explain this answer?
Tissue of aorta blood vessel is an example of
a)
state of permanant deformation
b)
perfectly plastic body
c)
perfectly elastic body
d)
elastomer
|
|
Gaurav Kumar answered |
Elastomers are popular in vascular engineering applications, as they offer the ability to design implants that match the compliance of native tissue.
Substances like tissue of aorta can be stretched to cause large strain.
Substances like tissue of aorta can be stretched to cause large strain.
Yield point may be defined as the point where- a)working stress ends and breaking stress begins
- b)breaking stress ends and working stress begins
- c)elastic behavior ends and plastic behavior begins
- d)plastic behavior ends and elastic behavior begins
Correct answer is option 'C'. Can you explain this answer?
Yield point may be defined as the point where
a)
working stress ends and breaking stress begins
b)
breaking stress ends and working stress begins
c)
elastic behavior ends and plastic behavior begins
d)
plastic behavior ends and elastic behavior begins
|
|
Neha Joshi answered |
At the yield point, stress is not proportional to strain i.e: hooke's law is not obeyed. Hence the elastic behaviour ends and plastic behaviour begins.
If proportional limit is not exceeded, energy per unit volume in stretched wire is- a)1/2 x stress x strain
- b)stress x strain
- c)stress x strain 2
- d)1/2 stress x strain 2
Correct answer is 'A'. Can you explain this answer?
If proportional limit is not exceeded, energy per unit volume in stretched wire is
a)
1/2 x stress x strain
b)
stress x strain
c)
stress x strain 2
d)
1/2 stress x strain 2
|
Suresh Reddy answered |
strain energy=1/2×stress × strain Work done by a force on a wire
W =2LAy(ΔL)2/2L
=1/2(yALΔ/L)ΔL
=1/2(yΔL/L)(ΔL/L)(AL)
=1/2(Stress)(Strain)(Volume)
(Work)/(volume)=1/2(stress)(strain)
W =2LAy(ΔL)2/2L
=1/2(yALΔ/L)ΔL
=1/2(yΔL/L)(ΔL/L)(AL)
=1/2(Stress)(Strain)(Volume)
(Work)/(volume)=1/2(stress)(strain)
When a 13.2 kg mass is placed on top of a vertical spring, the spring compresses by 5.93 cm. Find the force constant of the spring.- a)2281 N/m
- b)3181 N/m
- c)2175 N/m
- d)3085 N/m
Correct answer is option 'C'. Can you explain this answer?
When a 13.2 kg mass is placed on top of a vertical spring, the spring compresses by 5.93 cm. Find the force constant of the spring.
a)
2281 N/m
b)
3181 N/m
c)
2175 N/m
d)
3085 N/m
|
|
Preeti Iyer answered |
From Hooke's Law: F =kx
The force on the spring is the weight of the object , 13.2*9.8 = 129 N
x = 5.93 cm =0.0593 m
129 = (0.0593) x
x = 2175 N/m
When an elastic material with Young’s modulus Y is subjected to stretching stress S, elastic energy stored per unit volume of the material is- a)YS/2
- b)S2Y/2 [1989]
- c)S2/2Y
- d)S/2Y
Correct answer is option 'C'. Can you explain this answer?
When an elastic material with Young’s modulus Y is subjected to stretching stress S, elastic energy stored per unit volume of the material is
a)
YS/2
b)
S2Y/2 [1989]
c)
S2/2Y
d)
S/2Y
|
Aniket Chawla answered |
Energy stored per unit volume
stress (stress / Young' s modulus) (stress)2 /( Young ' s modulus)
Elasticity is the property of a body, by virtue of which- a)it remains in original size and shape when the force is applied
- b)it changes size and shape when the force is applied and stays in that shape when applied force is removed
- c)it tends to regain its original size and shape when the applied force is removed
- d)it is distorted or stretches without the application of force
Correct answer is option 'C'. Can you explain this answer?
Elasticity is the property of a body, by virtue of which
a)
it remains in original size and shape when the force is applied
b)
it changes size and shape when the force is applied and stays in that shape when applied force is removed
c)
it tends to regain its original size and shape when the applied force is removed
d)
it is distorted or stretches without the application of force
|
Suraj Raj answered |
A
A body is said to be perfectly plastic if- a)it does not recover its original shape / size when the deforming force is removed
- b)it has the property of stretching indefinitely
- c)it expands without breaking, on subjection of large strain
- d)it is not effected by external force
Correct answer is option 'A'. Can you explain this answer?
A body is said to be perfectly plastic if
a)
it does not recover its original shape / size when the deforming force is removed
b)
it has the property of stretching indefinitely
c)
it expands without breaking, on subjection of large strain
d)
it is not effected by external force
|
|
Neha Joshi answered |
A body is said to be plastic when its coefficient of restitution or reformation is zero that means that whatever hits it loses its all kinetic energy once the body gets deformed does not reform.
An iron bar of length l cm and cross section A cm2 is pulled by a force F dynes from the ends so as to produce an elongation l cm. Which of the following statements is correct?- a)Elongation is directly proportional to area of cross-section A.
- b)Elongation is inversely proportional to length
- c)Elongation is inversely proportional to area of cross-section A.
- d)Elongation is directly proportional to Young’s modulus.
Correct answer is option 'C'. Can you explain this answer?
An iron bar of length l cm and cross section A cm2 is pulled by a force F dynes from the ends so as to produce an elongation l cm. Which of the following statements is correct?
a)
Elongation is directly proportional to area of cross-section A.
b)
Elongation is inversely proportional to length
c)
Elongation is inversely proportional to area of cross-section A.
d)
Elongation is directly proportional to Young’s modulus.
|
|
Gaurav Kumar answered |
F =AYΔl/L
Δl = FL/AY
Δl = FL/AY
The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied ? [NEET 2013]- a)Length = 100 cm, diameter = 1 mm
- b)Length = 200 cm, diameter = 2 mm
- c)Length = 300 cm, diameter = 3 mm
- d)Length = 50 cm, diameter = 0.5 mm
Correct answer is option 'D'. Can you explain this answer?
The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied ? [NEET 2013]
a)
Length = 100 cm, diameter = 1 mm
b)
Length = 200 cm, diameter = 2 mm
c)
Length = 300 cm, diameter = 3 mm
d)
Length = 50 cm, diameter = 0.5 mm
|
Akshat Chavan answered |
So, extension, 
[∵ F and Y are constant]
The ratio of 
is maximum for case (d).
Hence, option (d) is correct.
is maximum for case (d).Hence, option (d) is correct.
Which of the following represents Hooke’s Law?- a)Stress = Estrain2
- b)Stress = Estrain
- c)Stress2 = Estrain
- d)
Correct answer is option 'B'. Can you explain this answer?
Which of the following represents Hooke’s Law?
a)
Stress = Estrain2
b)
Stress = Estrain
c)
Stress2 = Estrain
d)
|
|
Hansa Sharma answered |
E=stress/strain
Strain x E= Stress
Strain x E= Stress
Young’s modulus is defined as the ratio of longitudinal stress to longitudinal strain. Which of the following is the unit of Young’s modulus?- a)dyne per square cm.
- b)dyne cm2
- c)dyne/cm
- d)dyne cm
Correct answer is option 'A'. Can you explain this answer?
Young’s modulus is defined as the ratio of longitudinal stress to longitudinal strain. Which of the following is the unit of Young’s modulus?
a)
dyne per square cm.
b)
dyne cm2
c)
dyne/cm
d)
dyne cm
|
|
Riya Banerjee answered |
Stress = Force per area and strain is unit less.
Therefore in CGS units dyne per cm square.
Hence A is correct.
Therefore in CGS units dyne per cm square.
Hence A is correct.
According to Hooke’s law- a)For small deformations the stress and strain are inversely proportional to each other
- b)For small deformations the stress is proportional to square of strain
- c)For large deformations the stress and strain are proportional to each other
- d)For small deformations the stress and strain are proportional to each other
Correct answer is option 'D'. Can you explain this answer?
According to Hooke’s law
a)
For small deformations the stress and strain are inversely proportional to each other
b)
For small deformations the stress is proportional to square of strain
c)
For large deformations the stress and strain are proportional to each other
d)
For small deformations the stress and strain are proportional to each other
|
|
Krishna Iyer answered |
Hook’s Law states that for small deformations the stress and strain are proportional to each other.
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 107Pa? Assume that each rivet is to carry one quarter of the load.- a)749 kN
- b)856 kN
- c)652 N
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 107Pa? Assume that each rivet is to carry one quarter of the load.
a)
749 kN
b)
856 kN
c)
652 N
d)
None of these
|
|
Sanchita Mukherjee answered |
In Hooke’s law, the constant of proportionality signifies- a)modulus of elasticity
- b)modulus of strain
- c)elasticity of wire
- d)modulus of stress
Correct answer is option 'A'. Can you explain this answer?
In Hooke’s law, the constant of proportionality signifies
a)
modulus of elasticity
b)
modulus of strain
c)
elasticity of wire
d)
modulus of stress
|
|
Nayanika Chakraborty answered |
Introduction:
Hooke's Law is a principle in physics that relates the force applied to a spring or elastic material to the resulting deformation or change in length of the material. It states that the force applied to a spring is directly proportional to the displacement or change in length of the spring.
Explanation:
The constant of proportionality in Hooke's Law is known as the modulus of elasticity or Young's modulus. It is represented by the symbol 'E' and is a measure of the stiffness or rigidity of a material. The modulus of elasticity signifies how much a material will deform when a force is applied to it.
Modulus of Elasticity:
The modulus of elasticity is a material property that describes how it responds to stress. It is defined as the ratio of stress to strain within the elastic limit of the material. In other words, it measures how much stress a material can withstand before it starts to deform permanently.
Modulus of Strain:
The modulus of strain is not a property used in Hooke's Law. Strain is the measure of deformation or change in length of a material, and the modulus of strain is not directly related to the constant of proportionality in Hooke's Law.
Elasticity of Wire:
The elasticity of a wire refers to its ability to return to its original shape after being stretched or deformed. It is related to Hooke's Law as the law describes the linear relationship between the force applied to a wire and the resulting deformation or change in length of the wire.
Modulus of Stress:
The modulus of stress is not a term used in Hooke's Law. Stress is defined as the force applied per unit area of a material, and the modulus of stress is not directly related to the constant of proportionality in Hooke's Law.
Conclusion:
In conclusion, the constant of proportionality in Hooke's Law signifies the modulus of elasticity. It is a measure of the stiffness or rigidity of a material and describes how much a material will deform when a force is applied to it. The modulus of elasticity is a fundamental property used to understand the behavior of elastic materials and is essential in various fields such as engineering and materials science.
Hooke's Law is a principle in physics that relates the force applied to a spring or elastic material to the resulting deformation or change in length of the material. It states that the force applied to a spring is directly proportional to the displacement or change in length of the spring.
Explanation:
The constant of proportionality in Hooke's Law is known as the modulus of elasticity or Young's modulus. It is represented by the symbol 'E' and is a measure of the stiffness or rigidity of a material. The modulus of elasticity signifies how much a material will deform when a force is applied to it.
Modulus of Elasticity:
The modulus of elasticity is a material property that describes how it responds to stress. It is defined as the ratio of stress to strain within the elastic limit of the material. In other words, it measures how much stress a material can withstand before it starts to deform permanently.
Modulus of Strain:
The modulus of strain is not a property used in Hooke's Law. Strain is the measure of deformation or change in length of a material, and the modulus of strain is not directly related to the constant of proportionality in Hooke's Law.
Elasticity of Wire:
The elasticity of a wire refers to its ability to return to its original shape after being stretched or deformed. It is related to Hooke's Law as the law describes the linear relationship between the force applied to a wire and the resulting deformation or change in length of the wire.
Modulus of Stress:
The modulus of stress is not a term used in Hooke's Law. Stress is defined as the force applied per unit area of a material, and the modulus of stress is not directly related to the constant of proportionality in Hooke's Law.
Conclusion:
In conclusion, the constant of proportionality in Hooke's Law signifies the modulus of elasticity. It is a measure of the stiffness or rigidity of a material and describes how much a material will deform when a force is applied to it. The modulus of elasticity is a fundamental property used to understand the behavior of elastic materials and is essential in various fields such as engineering and materials science.
A body is said to be perfectly elastic if- a)it can move freely
- b)it is not effected by external force
- c)it recovers its original shape / size when the deforming force is removed
- d)its surface is perfectly smooth
Correct answer is option 'C'. Can you explain this answer?
A body is said to be perfectly elastic if
a)
it can move freely
b)
it is not effected by external force
c)
it recovers its original shape / size when the deforming force is removed
d)
its surface is perfectly smooth
|
|
Hansa Sharma answered |
A body is said to be perfectly elastic when its coefficient of restitution is 1 or we observe total reformation in the body. That means that whatever hits it doesn't lose any of its kinetic energy or once the body gets totally reformed after a collision.
The volume of a spherical body is decreased by 10-3% when it is subjected to pressure of 40 atmospheres. Find the bulk modulus of body.
(1 atm = 1.01 x 105 N/m2).
- a)4.04 x 1011 N/m2
- b)4.04 x 106N/m2
- c)4.04 x 103 N/m2
- d)4.04 x 102 N/m2
Correct answer is option 'A'. Can you explain this answer?
The volume of a spherical body is decreased by 10-3% when it is subjected to pressure of 40 atmospheres. Find the bulk modulus of body.
(1 atm = 1.01 x 105 N/m2).
(1 atm = 1.01 x 105 N/m2).
a)
4.04 x 1011 N/m2
b)
4.04 x 106N/m2
c)
4.04 x 103 N/m2
d)
4.04 x 102 N/m2
|
|
Geetika Shah answered |
We know that magnitude of bulk modulus
K=[P/(dV/V)]
Now, percentage change in volume is. 10-3 %
Therefore, (dV/V)x100=10-3
So, dV/V=10-5
Hence, k is,
K=40atm/10-5=40x1.01x105 /10-5 =4.04x1011 N/m2
K=[P/(dV/V)]
Now, percentage change in volume is. 10-3 %
Therefore, (dV/V)x100=10-3
So, dV/V=10-5
Hence, k is,
K=40atm/10-5=40x1.01x105 /10-5 =4.04x1011 N/m2
Two wires A and B are of the same material. Their lengths are in the ratio 1 : 2 and the diameter are in the ratio 2 : 1. If they are pulled by the same force, then increase in length will be in the ratio- a)2 : 1
- b)1 : 4 [1988]
- c)1 : 8
- d)8 : 1
Correct answer is option 'C'. Can you explain this answer?
Two wires A and B are of the same material. Their lengths are in the ratio 1 : 2 and the diameter are in the ratio 2 : 1. If they are pulled by the same force, then increase in length will be in the ratio
a)
2 : 1
b)
1 : 4 [1988]
c)
1 : 8
d)
8 : 1
|
Nilotpal Gupta answered |
We know that Young's modulus
Since Y, F are same for both the wires, we have,
So, ℓ1 : ℓ2 = 1 : 8
How does the modulus of elasticity change with temperature?- a)It does not depend on temperature
- b)It decreases with increase in temperature
- c)It increases with increase in temperature
- d)It sometimes increases, sometimes decreases with increase in temperature
Correct answer is option 'B'. Can you explain this answer?
How does the modulus of elasticity change with temperature?
a)
It does not depend on temperature
b)
It decreases with increase in temperature
c)
It increases with increase in temperature
d)
It sometimes increases, sometimes decreases with increase in temperature
|
|
Pooja Shah answered |
As the temperature increases, the inter-atomic distance also increases. So, it results in an increase in the area (stress = force /area). As the area increases stress decreases, this results in a decrease in Young's modulus.
The stress which is set up in the body due to increase in its dimensions is called- a)volumetric stress
- b)Tangential stress
- c)Normal stress
- d)Tensile stress
Correct answer is option 'D'. Can you explain this answer?
The stress which is set up in the body due to increase in its dimensions is called
a)
volumetric stress
b)
Tangential stress
c)
Normal stress
d)
Tensile stress
|
|
Rahul Bansal answered |
Tensile stress (or tension) is the stress state leading to expansion; that is, the length of a material tends to increase in the tensile direction. The volume of the material stays constant. When equal and opposite forces are applied on a body, then the stress due to this force is called tensile stress.
The area occupied below the stress-strain graph and above strain axis gives the value of
- a)work done in producing extension
- b)energy stored in the material
- c)Restoring force.
- d)energy density of the material
Correct answer is option 'D'. Can you explain this answer?
The area occupied below the stress-strain graph and above strain axis gives the value of
a)
work done in producing extension
b)
energy stored in the material
c)
Restoring force.
d)
energy density of the material
|
|
Jayant Mishra answered |
The area under the stress-strain curve represents the mechanical energy per unit volume consumed by the material. This is true in the elastic range of the graph where the energy is reversibly sorted within the material. Area under the stress strain curve depicts the energy absorbed by the material prior to failure.
A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2. It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the elongation. Take Young's modulus of steel as 20 × 1010 Pa- a)2.0 mm
- b)1.6 mm
- c)2.2 mm
- d)1.8 mm
Correct answer is option 'D'. Can you explain this answer?
A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2. It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the elongation. Take Young's modulus of steel as 20 × 1010 Pa
a)
2.0 mm
b)
1.6 mm
c)
2.2 mm
d)
1.8 mm
|
|
Gaurav Kumar answered |
σ=Stress and ε=strain
σ=F/A= (550kg) × (9.81m/s2)3×10-5m2/=0.18GPA
ε=Δl/l0=σ/Υ=0.18×109/200×109=9×10-4
Δl=εl0= (9×10-4) (2m) = 0.0018m=1.8mm
σ=F/A= (550kg) × (9.81m/s2)3×10-5m2/=0.18GPA
ε=Δl/l0=σ/Υ=0.18×109/200×109=9×10-4
Δl=εl0= (9×10-4) (2m) = 0.0018m=1.8mm
What type of stress is produced in a body when the deforming force produces sheer strain?
- a)shear stress
- b)Longitudinal stress
- c)Normal stress
- d)Tangential stress
Correct answer is option 'A'. Can you explain this answer?
What type of stress is produced in a body when the deforming force produces sheer strain?
a)
shear stress
b)
Longitudinal stress
c)
Normal stress
d)
Tangential stress
|
|
Pooja Shah answered |
Explanation:
When a deforming force is applied to a body in a direction parallel to its surface, it produces a shear strain. Shear strain is the deformation that occurs when one layer of a material slides past another layer. This type of deformation produces shear stress, which is the force that is acting parallel to the surface of the material. Shear stress is calculated as the force per unit area and is expressed in units of pressure, such as pascals (Pa) or pounds per square inch (psi).
volume strain is defined- a)as the change in volume ΔV
- b)as the ratio of change in volume (ΔV) to the original volume V
- c)as the ratio of change in volume (ΔV) to thrice the original volume V
- d)as the ratio of change in volume (ΔV) to twice the original volume V
Correct answer is option 'B'. Can you explain this answer?
volume strain is defined
a)
as the change in volume ΔV
b)
as the ratio of change in volume (ΔV) to the original volume V
c)
as the ratio of change in volume (ΔV) to thrice the original volume V
d)
as the ratio of change in volume (ΔV) to twice the original volume V
|
Imk Pathsala answered |
Volume Strain Definition
- Definition: Volume strain is defined as the ratio of change in volume (ΔV) to the original volume V.
Detailed Explanation
- Volume strain is a measure of how much a material changes in volume under stress.
- It is calculated by taking the change in volume and dividing it by the original volume of the material.
- Volume strain is usually expressed as a decimal or a percentage.
- It is an important concept in physics and engineering, especially in the study of materials and structures.
- Understanding volume strain is crucial for predicting how materials will behave under different conditions and loads.
Rectangular section is rarely used in beams because- a)stresses are uniform in a rectangular section
- b)more material in kilos is required
- c)stresses are always plastic in rectangular section
- d)less material in kilos is required
Correct answer is option 'B'. Can you explain this answer?
Rectangular section is rarely used in beams because
a)
stresses are uniform in a rectangular section
b)
more material in kilos is required
c)
stresses are always plastic in rectangular section
d)
less material in kilos is required
|
|
Anjali Iyer answered |
I section is generally used as a beam because of its high section modulus as it's most of the area is situated away from it's neutral axis hence it has high moment of inertia i.e high section modulus i.e high moment carrying capacity which is the major requirement for a good beam section.
A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2 . It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the resulting strain. Young's Modulus of Steel (Y) = 20×1010 Pa
- a)8.0 × 10−4
- b)10.0 × 10−4
- c)7.0 × 10−4
- d)9.0 × 10−4
Correct answer is option 'D'. Can you explain this answer?
A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2 . It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the resulting strain. Young's Modulus of Steel (Y) = 20×1010 Pa
a)
8.0 × 10−4
b)
10.0 × 10−4
c)
7.0 × 10−4
d)
9.0 × 10−4
|
|
Naina Bansal answered |
from the Hooke’s law :
ε = σ/Y = (1.8 x 10^8 Pa)/(20 x 10^10 Pa)
= 9.0 x 10^−4
Two wires P and Q of same length and material but radii in the ratio 2 : 1 are suspended from a rigid support. Find the ratio of strain produced in the wires when both are under same force.- a)1:2
- b)4:1
- c)1:4
- d)2:1
Correct answer is option 'C'. Can you explain this answer?
Two wires P and Q of same length and material but radii in the ratio 2 : 1 are suspended from a rigid support. Find the ratio of strain produced in the wires when both are under same force.
a)
1:2
b)
4:1
c)
1:4
d)
2:1
|
|
Upasana Bose answered |
Using Hooke ‘s Law we get
Stress directly proportional to stress = Load/Area=F/pie*r*r
And rp:rq=2:1
When both the wires are under the same stress,strain produced will be the same.
When both the wires are under the same stress,strain produced will be the same.
2.when both the wires are loaded by same weight then
Strain p/strain q=(rq)2/(rp)2=¼
The length of the wire is increased by 1 mm on the application of a given load. In a wire of the same material but of length and radius twice that of the first, on application of the same force, extension produced is- a)0.25 mm
- b)2 mm
- c)4 mm
- d)0.5 mm
Correct answer is option 'D'. Can you explain this answer?
The length of the wire is increased by 1 mm on the application of a given load. In a wire of the same material but of length and radius twice that of the first, on application of the same force, extension produced is
a)
0.25 mm
b)
2 mm
c)
4 mm
d)
0.5 mm
|
|
Shreya Gupta answered |
Let F be the load applied on the wire of length L, area of cross-section a; such that the produced extension is I ( = 1 mm )
F=Y a l/L.... ( i )
Now, when same force F is applied on the wire of same material with length 2L and radius 2 R , let the extension produced be r .
Then, l' = F(2L) Y ( 4 a )
Or, substituting the value of F from ( i ) , we get
I' = 1 / 2
or, 1' = 1mm/ 2 = 0.5 mm
An Indian rubber cube of side 10 cm has one side fixed while a tangential force of 1800 N is applied to the opposite side. Find the shear strain produced. Take η = 2 x 106 N/ m2.- a)0.9
- b)0.009
- c)0.09
- d)9
Correct answer is option 'C'. Can you explain this answer?
An Indian rubber cube of side 10 cm has one side fixed while a tangential force of 1800 N is applied to the opposite side. Find the shear strain produced. Take η = 2 x 106 N/ m2.
a)
0.9
b)
0.009
c)
0.09
d)
9
|
Mohit Rajpoot answered |
To calculate the shear strain, we use the formula:
shear strain = shear stress / shear modulus.
Shear stress is defined as force/area.
Here, the area is 10 cm × 10 cm = 0.1 m × 0.1 m = 0.01 m².
Therefore, the shear stress is 1800 N / 0.01 m² = 180,000 N/m².
Given the shear modulus (η) is 2 × 10⁶ N/m²,
the shear strain = 180,000 N/m² / 2 × 10⁶ N/m² = 0.09.
Hence, the correct answer is option c) 0.09.
Columns are loaded in- a)shear
- b)hydraulic stress
- c)tension
- d)compression
Correct answer is option 'D'. Can you explain this answer?
Columns are loaded in
a)
shear
b)
hydraulic stress
c)
tension
d)
compression
|
Madhavan Patel answered |
Alternation of generations (also known as metagenesis) is the type of life cycle that occurs in those plants and algae in the Archaeplastida and the Heterokontophyta that have distinct sexual haploid and asexual diploid stages.
A perfectly rigid body is one- a)whose shape and size change on application of force
- b)which does not move on application of force
- c)which starts flowing like water on application of force
- d)whose shape and size do not change on application of force
Correct answer is option 'D'. Can you explain this answer?
A perfectly rigid body is one
a)
whose shape and size change on application of force
b)
which does not move on application of force
c)
which starts flowing like water on application of force
d)
whose shape and size do not change on application of force
|
|
Arka Bose answered |
Explanation:A perfectly rigid body is hypothetical in nature but for some phenomena ( in rotational bodies ) we assume bodies are perfectly rigid i.e. the intermolecular forces are always in equilibrium irrespective of the external forces due to which their shape and size are constant.
When too many people stand on a bridge it collapses, why?- a)Due to increase in stress
- b)Due to overweight
- c)Due to improper construction
- d)Due to friction
Correct answer is option 'A'. Can you explain this answer?
When too many people stand on a bridge it collapses, why?
a)
Due to increase in stress
b)
Due to overweight
c)
Due to improper construction
d)
Due to friction
|
Sai Mishra answered |
Stress is the force per unit area experienced by the body and strength is the ability to withstand the stress. When stress becomes greater than strength, accidents happen.
After prolonged use, springs deform permanently because of- a)its rigidity
- b)elastic fatigue
- c)elastic after effect
- d)plastic fatigue
Correct answer is option 'B'. Can you explain this answer?
After prolonged use, springs deform permanently because of
a)
its rigidity
b)
elastic fatigue
c)
elastic after effect
d)
plastic fatigue
|
|
Rohan Singh answered |
Springs deform permanently because of elastic fatigueness. The elasticity of the material of spring is lost and it deforms permanently.
Materials that show very small plastic range beyond elastic limit are called- a)Brittle materials
- b)Elastomers
- c)Elastic materials
- d)Ductile materials
Correct answer is option 'A'. Can you explain this answer?
Materials that show very small plastic range beyond elastic limit are called
a)
Brittle materials
b)
Elastomers
c)
Elastic materials
d)
Ductile materials
|
|
Anjali Iyer answered |
Brittle materials have shorter plastic region, i.e., the proportional and fracture points are close.
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