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All questions of Mechanical Properties of Solids for NEET Exam
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 |
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
 | Madhuri Jadhav answered |
I think B
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 |
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.
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
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
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
 | Jyoti Kumar answered |
Elasticity is the property of a body, by virtue of which it tends to regain its original size and shape when the applied force is removed.
Explanation:
Elasticity is a property of materials that allows them to deform when a force is applied and then return to their original shape and size when the force is removed. This property is crucial in various fields such as engineering, physics, and materials science.
Key Points:
- Elasticity is the ability of a body to undergo deformation (change in shape or size) under the influence of an external force and return to its original shape and size when the force is removed.
- Elasticity is a fundamental property of materials and is related to the arrangement and behavior of atoms and molecules within the material.
- When a force is applied to an elastic material, it causes the atoms or molecules to move from their equilibrium positions, resulting in a temporary change in shape or size.
- The material stores the energy from the applied force in the form of potential energy, which enables it to return to its original shape and size once the force is removed.
- The restoring force that allows the material to regain its original shape and size is due to the intermolecular or atomic forces within the material.
- The magnitude of the deformation experienced by an elastic material is directly proportional to the applied force within the elastic limit. This relationship is described by Hooke's Law.
- If the applied force exceeds the elastic limit of the material, it may undergo permanent deformation or even fracture.
- Elastic materials exhibit a linear relationship between stress (force per unit area) and strain (deformation per unit length) within the elastic limit, which is represented by the elastic modulus or Young's modulus.
Conclusion:
In conclusion, elasticity is the property of a body that allows it to regain its original size and shape when the applied force is removed. It is a fundamental property of materials and is essential in various applications where materials need to withstand forces and deformations without undergoing permanent damage.
Explanation:
Elasticity is a property of materials that allows them to deform when a force is applied and then return to their original shape and size when the force is removed. This property is crucial in various fields such as engineering, physics, and materials science.
Key Points:
- Elasticity is the ability of a body to undergo deformation (change in shape or size) under the influence of an external force and return to its original shape and size when the force is removed.
- Elasticity is a fundamental property of materials and is related to the arrangement and behavior of atoms and molecules within the material.
- When a force is applied to an elastic material, it causes the atoms or molecules to move from their equilibrium positions, resulting in a temporary change in shape or size.
- The material stores the energy from the applied force in the form of potential energy, which enables it to return to its original shape and size once the force is removed.
- The restoring force that allows the material to regain its original shape and size is due to the intermolecular or atomic forces within the material.
- The magnitude of the deformation experienced by an elastic material is directly proportional to the applied force within the elastic limit. This relationship is described by Hooke's Law.
- If the applied force exceeds the elastic limit of the material, it may undergo permanent deformation or even fracture.
- Elastic materials exhibit a linear relationship between stress (force per unit area) and strain (deformation per unit length) within the elastic limit, which is represented by the elastic modulus or Young's modulus.
Conclusion:
In conclusion, elasticity is the property of a body that allows it to regain its original size and shape when the applied force is removed. It is a fundamental property of materials and is essential in various applications where materials need to withstand forces and deformations without undergoing permanent damage.
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.
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.
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 |
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
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
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
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
| | Ameya Unni answered |
Understanding Volume Strain
Volume strain is an important concept in mechanics and materials science that describes how a material deforms when subjected to external forces.
Definition of Volume Strain
- Volume strain is defined specifically as the ratio of the change in volume (ΔV) to the original volume (V0) of a material.
- Mathematically, it can be expressed as: Volume Strain = ΔV / V0.
Why Option B is Correct
- Change in Volume (ΔV): This represents the difference between the final volume after deformation and the initial volume before deformation.
- Original Volume (V0): This is the volume of the material before any external forces have been applied.
- Ratio Significance: By taking the ratio of the change in volume to the original volume, we obtain a dimensionless quantity that allows for comparison across different materials and conditions.
Other Options Explained
- Option A (Change in Volume V): This does not provide a comparative metric and lacks the necessary context of the original volume.
- Option C (Thrice the Original Volume): This is an arbitrary scaling that does not conform to the standard definition of volume strain.
- Option D (Twice the Original Volume): Similar to Option C, this does not reflect the true relationship defined in mechanics.
Conclusion
In conclusion, volume strain is fundamentally about understanding how a material's volume changes relative to its original volume, which is effectively captured by Option B. This definition is crucial for engineers and scientists to assess material behavior under stress.
Volume strain is an important concept in mechanics and materials science that describes how a material deforms when subjected to external forces.
Definition of Volume Strain
- Volume strain is defined specifically as the ratio of the change in volume (ΔV) to the original volume (V0) of a material.
- Mathematically, it can be expressed as: Volume Strain = ΔV / V0.
Why Option B is Correct
- Change in Volume (ΔV): This represents the difference between the final volume after deformation and the initial volume before deformation.
- Original Volume (V0): This is the volume of the material before any external forces have been applied.
- Ratio Significance: By taking the ratio of the change in volume to the original volume, we obtain a dimensionless quantity that allows for comparison across different materials and conditions.
Other Options Explained
- Option A (Change in Volume V): This does not provide a comparative metric and lacks the necessary context of the original volume.
- Option C (Thrice the Original Volume): This is an arbitrary scaling that does not conform to the standard definition of volume strain.
- Option D (Twice the Original Volume): Similar to Option C, this does not reflect the true relationship defined in mechanics.
Conclusion
In conclusion, volume strain is fundamentally about understanding how a material's volume changes relative to its original volume, which is effectively captured by Option B. This definition is crucial for engineers and scientists to assess material behavior under stress.
Elastomers are materials- a)which can be stretched without corresponding stress
- b)which cannot be stretched to cause large strains
- c)which cannot be stretched to beyond elastic limit
- d)which can be stretched to cause large strains
Correct answer is option 'D'. Can you explain this answer?
Elastomers are materials
a)
which can be stretched without corresponding stress
b)
which cannot be stretched to cause large strains
c)
which cannot be stretched to beyond elastic limit
d)
which can be stretched to cause large strains
| | Rajeev Saxena answered |
An elastomer is a polymer with viscoelasticity (i. e., both viscosity and elasticity) and very weak intermolecular forces, and generally low Young's modulus and high failure strain compared with other materials. Elastomer rubber compounds are made from five to ten ingredients, each ingredient playing a specific role. Polymer is the main component, and determines heat and chemical resistance, as well as low- temperature performance. Reinforcing filler is used, typically carbon black, for strength properties.
A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire? Given that the Young's modulus for steel is Young's modulus of the material is 2.00×1011 N/m2)- a)1.4 mm
- b)10 mm
- c)1.5 mm
- d)12.4 mm
Correct answer is option 'A'. Can you explain this answer?
A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire? Given that the Young's modulus for steel is Young's modulus of the material is 2.00×1011 N/m2)
a)
1.4 mm
b)
10 mm
c)
1.5 mm
d)
12.4 mm
 | EduRev NEET answered |
A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10−2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.- a)1.1 cm
- b)1.0 cm
- c)0.9 cm
- d)1.2 cm
Correct answer is option 'A'. Can you explain this answer?
A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10−2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.
a)
1.1 cm
b)
1.0 cm
c)
0.9 cm
d)
1.2 cm
 | Siddharth Mehra answered |
When a solid is deformed,- a)the atoms or molecules do not move from their equilibrium position
- b)only the atoms or molecules at some points move from their equilibrium position
- c)only the atoms or molecules of the surface move from their equilibrium position
- d)all the atoms or molecules are displaced from their equilibrium positions causing a change in inter atomic (or intermolecular) distances.
Correct answer is option 'D'. Can you explain this answer?
When a solid is deformed,
a)
the atoms or molecules do not move from their equilibrium position
b)
only the atoms or molecules at some points move from their equilibrium position
c)
only the atoms or molecules of the surface move from their equilibrium position
d)
all the atoms or molecules are displaced from their equilibrium positions causing a change in inter atomic (or intermolecular) distances.
 | Sravya Banerjee answered |
Explanation:External force permanently distubed the equilibrium position of the interatomic ( or intermolecular ) forces between the particles of solid bodies.
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 for a copper wire of the original length and diameter?- a)0.39 mm
- b)0.37 mm
- c)0.33 mm
- d)0.18 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 for a copper wire of the original length and diameter?
a)
0.39 mm
b)
0.37 mm
c)
0.33 mm
d)
0.18 mm
 | Darshan Nagesh answered |
A lead cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force F is applied to the upper face parallel to one of the edges. How large must F be to cause the cube to deform by 0.250 mm? (Shear modulus of lead = 0.6 × 1010Pa)- a)7.0 * 105 N
- b)6* 105 N
- c)6.6 * 105 N
- d)5.6 * 105 N
Correct answer is option 'C'. Can you explain this answer?
A lead cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force F is applied to the upper face parallel to one of the edges. How large must F be to cause the cube to deform by 0.250 mm? (Shear modulus of lead = 0.6 × 1010Pa)
a)
7.0 * 105 N
b)
6* 105 N
c)
6.6 * 105 N
d)
5.6 * 105 N
| | Sanchita Mukherjee answered |
Shear modulus or modulus of rigidity is- a)the ratio of shearing stress to the corresponding lateral strain
- b)the ratio of shearing stress to the corresponding shearing strain
- c)the ratio of longitudinal stress to the corresponding shearing strain
- d)the ratio of shearing strain to the corresponding shearing stress
Correct answer is option 'B'. Can you explain this answer?
Shear modulus or modulus of rigidity is
a)
the ratio of shearing stress to the corresponding lateral strain
b)
the ratio of shearing stress to the corresponding shearing strain
c)
the ratio of longitudinal stress to the corresponding shearing strain
d)
the ratio of shearing strain to the corresponding shearing stress
| | Ameya Choudhury answered |
Shear modulus or modulus of rigidity is the ratio of shearing stress to the corresponding shearing strain, by the definition of shear modulus.
In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 kg from a vertical steel wire 0.50 m long and 2.5 × 10−3 cm2in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 kg. Compute the elongation.- a)3.7 mm upper, 1.0 mm lower
- b)3.4 mm upper, 1.0 mm lower
- c)3.5 mm upper, 1.1 mm lower
- d)1.6 mm upper, 1.0 mm lower
Correct answer is option 'D'. Can you explain this answer?
In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 kg from a vertical steel wire 0.50 m long and 2.5 × 10−3 cm2in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 kg. Compute the elongation.
a)
3.7 mm upper, 1.0 mm lower
b)
3.4 mm upper, 1.0 mm lower
c)
3.5 mm upper, 1.1 mm lower
d)
1.6 mm upper, 1.0 mm lower
| | Muskaan Kumar answered |
Hence 1.6 mm upper, 1.0 mm lower is correct.
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? Take Young's modulus of copper as 42 × 109Pa- a)3.65 × 10-8
- b)3.65 × 10-3
- c)3.65 × 10-9
- d)3.65 × 10-2
Correct answer is option 'B'. Can you explain this answer?
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? Take Young's modulus of copper as 42 × 10
9
Paa)
3.65 × 10-8
b)
3.65 × 10-3
c)
3.65 × 10-9
d)
3.65 × 10-2
| | Anjana Sharma answered |
Given Data,
Length of the piece of copper = l = 19.1 mm = 19.1 × 10-3m
Breadth of the piece of copper = b = 15.2 mm = 15.2× 10-3m
Tension force applied on the piece of cooper, F = 44500N
Area of rectangular cross section of copper piece,
Area = l× b
⇒ Area = (19.1 × 10-3m) × (15.2× 10-3m)
⇒ Area = 2.9 × 10-4 m2
Modulus of elasticity of copper from standard list, η = 42× 109 N/m2
By definition, Modulus of elasticity, η = stress/strain
⇒ Strain = F/Aη
⇒ Strain = 3.65 × 10-3
Hence, the resulting strain is 3.65 × 10-3
Length of the piece of copper = l = 19.1 mm = 19.1 × 10-3m
Breadth of the piece of copper = b = 15.2 mm = 15.2× 10-3m
Tension force applied on the piece of cooper, F = 44500N
Area of rectangular cross section of copper piece,
Area = l× b
⇒ Area = (19.1 × 10-3m) × (15.2× 10-3m)
⇒ Area = 2.9 × 10-4 m2
Modulus of elasticity of copper from standard list, η = 42× 109 N/m2
By definition, Modulus of elasticity, η = stress/strain
⇒ Strain = F/Aη
⇒ Strain = 3.65 × 10-3
Hence, the resulting strain is 3.65 × 10-3
If the elastic limit of copper is 1.5 × 108 N/ m2, determine the minimum diameter a copper wire can have under a load of 10.0 kg if its elastic limit is not to be exceeded.- a)1.012 mm
- b)0.912 mm
- c)0.712 mm
- d)0.812 mm
Correct answer is option 'B'. Can you explain this answer?
If the elastic limit of copper is 1.5 × 108 N/ m2, determine the minimum diameter a copper wire can have under a load of 10.0 kg if its elastic limit is not to be exceeded.
a)
1.012 mm
b)
0.912 mm
c)
0.712 mm
d)
0.812 mm
 | Pooja Chauhan answered |
Understanding the Problem
To determine the minimum diameter of a copper wire under a load of 10.0 kg without exceeding its elastic limit, we will use the relationship between stress, force, and area.
Key Concepts
- Elastic Limit of Copper: 1.5 × 10^8 N/m²
- Load (Weight): 10 kg
- Force Calculation: Weight (Force) = mass × gravity
Step-by-Step Calculation
1. Calculate the Force (Weight):
- Weight = mass × gravity = 10 kg × 9.81 m/s² = 98.1 N
2. Determine the Required Area (A):
- Stress (σ) = Force (F) / Area (A)
- Rearranging gives: A = F / σ
- A = 98.1 N / (1.5 × 10^8 N/m²) = 6.54 × 10^-7 m²
3. Relating Area to Diameter:
- Area of a circle: A = π(d² / 4)
- Rearranging gives: d² = (4A) / π
- Thus, d = sqrt((4 × 6.54 × 10^-7) / π)
4. Calculate Diameter:
- d ≈ 0.912 mm
Conclusion
The minimum diameter of the copper wire that can safely support a load of 10 kg without exceeding its elastic limit is approximately 0.912 mm. Thus, the correct answer is option 'B'.
This calculation highlights the importance of material properties and dimensions in engineering applications to ensure safety and functionality.
To determine the minimum diameter of a copper wire under a load of 10.0 kg without exceeding its elastic limit, we will use the relationship between stress, force, and area.
Key Concepts
- Elastic Limit of Copper: 1.5 × 10^8 N/m²
- Load (Weight): 10 kg
- Force Calculation: Weight (Force) = mass × gravity
Step-by-Step Calculation
1. Calculate the Force (Weight):
- Weight = mass × gravity = 10 kg × 9.81 m/s² = 98.1 N
2. Determine the Required Area (A):
- Stress (σ) = Force (F) / Area (A)
- Rearranging gives: A = F / σ
- A = 98.1 N / (1.5 × 10^8 N/m²) = 6.54 × 10^-7 m²
3. Relating Area to Diameter:
- Area of a circle: A = π(d² / 4)
- Rearranging gives: d² = (4A) / π
- Thus, d = sqrt((4 × 6.54 × 10^-7) / π)
4. Calculate Diameter:
- d ≈ 0.912 mm
Conclusion
The minimum diameter of the copper wire that can safely support a load of 10 kg without exceeding its elastic limit is approximately 0.912 mm. Thus, the correct answer is option 'B'.
This calculation highlights the importance of material properties and dimensions in engineering applications to ensure safety and functionality.
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Dl on applying a force F, how much force is needed to stretch the second wire by the same amount?- a)9F
- b)6F
- c)4F
- d)F
Correct answer is option 'A'. Can you explain this answer?
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Dl on applying a force F, how much force is needed to stretch the second wire by the same amount?
a)
9F
b)
6F
c)
4F
d)
F
 | Ambition Institute answered |
Young’s modulus, 
Since initial volume of wires are same and their areas of cross sections are A and 3A so lengths are 3l and l respectively. For wire 1,
Since initial volume of wires are same and their areas of cross sections are A and 3A so lengths are 3l and l respectively. For wire 1,
For wire 1,
For wire 2, let F ′ force is applied
From eqns (i) and (ii),
⇒ F' = 9F
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