Q. A material has a Poisson's ratio of 0.3. If it is stretched in one direction, what is the expected lateral contraction ratio? (2019)
A.
0.3
B.
0.7
C.
0.5
D.
0.1
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Solution
Poisson's ratio (ν) = - (lateral strain) / (longitudinal strain), thus the lateral contraction ratio is 0.3.
Correct Answer: A — 0.3
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Q. A wire of length L and diameter d is stretched by a force F. If the diameter is halved while keeping the length constant, what happens to the stress in the wire? (2022)
A.
It doubles
B.
It quadruples
C.
It remains the same
D.
It halves
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Solution
Stress (σ) = Force (F) / Area (A). Halving the diameter increases the area by a factor of 4, thus stress quadruples.
Correct Answer: B — It quadruples
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Q. A wire of length L and diameter d is stretched by a force F. If the diameter is halved, what will be the new elongation for the same force? (2020)
A.
It will double
B.
It will quadruple
C.
It will remain the same
D.
It will halve
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Solution
Elongation is inversely proportional to the area. Halving the diameter reduces the area to a quarter, thus elongation quadruples.
Correct Answer: B — It will quadruple
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Q. A wire of length L and diameter d is stretched by a force F. If the diameter is halved while keeping the length constant, what happens to the stress? (2020)
A.
It doubles
B.
It quadruples
C.
It halves
D.
It remains the same
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Solution
Stress = Force / Area. Halving the diameter increases the area by a factor of 1/4, thus stress quadruples.
Correct Answer: B — It quadruples
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Q. A wire of length L and diameter d is stretched by a force F. If the diameter is doubled, what will be the new elongation if the same force is applied? (2019)
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Solution
Elongation is inversely proportional to the area. Doubling the diameter increases the area by a factor of 4, thus elongation becomes L/4.
Correct Answer: B — L/2
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Q. A wire of length L and diameter d is stretched by a force F. What is the expression for the elongation of the wire? (2020)
A.
(F * L) / (A * Y)
B.
(F * Y) / (A * L)
C.
(Y * A) / (F * L)
D.
(L * d) / (F * Y)
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Solution
Elongation (ΔL) = (F * L) / (A * Y), where A is the cross-sectional area and Y is Young's modulus.
Correct Answer: A — (F * L) / (A * Y)
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Q. If a material has a high shear modulus, what does it indicate about the material? (2023)
A.
It is very elastic
B.
It is very ductile
C.
It resists shear deformation
D.
It is very brittle
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Solution
A high shear modulus indicates that the material resists shear deformation.
Correct Answer: C — It resists shear deformation
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Q. If a material has a shear modulus of 80 GPa and a shear stress of 40 MPa is applied, what is the shear strain? (2023)
A.
0.0005
B.
0.005
C.
0.5
D.
0.05
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Solution
Shear strain (γ) = Shear stress (τ) / Shear modulus (G) = 40 MPa / 80 GPa = 0.0005.
Correct Answer: A — 0.0005
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Q. If the diameter of a wire is halved while keeping the length constant, what happens to its tensile strength? (2019)
A.
It doubles
B.
It halves
C.
It quadruples
D.
It remains the same
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Solution
Tensile strength is inversely proportional to the cross-sectional area. Halving the diameter reduces the area to a quarter, thus tensile strength quadruples.
Correct Answer: C — It quadruples
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Q. If the length of a wire is doubled while keeping the diameter constant, what happens to its resistance? (2019)
A.
It doubles
B.
It halves
C.
It quadruples
D.
It remains the same
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Solution
Resistance (R) is directly proportional to length (L). If length is doubled, resistance also doubles.
Correct Answer: A — It doubles
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Q. If the shear modulus of a material is 80 GPa, what is the ratio of shear stress to shear strain? (2021)
A.
80 GPa
B.
160 GPa
C.
40 GPa
D.
20 GPa
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Solution
Shear modulus (G) = Shear stress / Shear strain, thus it is 80 GPa.
Correct Answer: A — 80 GPa
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Q. What happens to the density of a material when it is compressed? (2019)
A.
It decreases
B.
It increases
C.
It remains the same
D.
It becomes zero
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Solution
Density increases when a material is compressed, as mass remains constant while volume decreases.
Correct Answer: B — It increases
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Q. What happens to the density of a solid when it is subjected to uniform compression? (2020)
A.
Increases
B.
Decreases
C.
Remains the same
D.
Depends on the material
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Solution
The density of a solid increases when it is subjected to uniform compression.
Correct Answer: A — Increases
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Q. What happens to the shear modulus if the shear stress is doubled while keeping the shear strain constant? (2019)
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
Shear modulus (G) is defined as shear stress divided by shear strain. If shear strain is constant, doubling shear stress does not change G.
Correct Answer: C — It remains the same
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Q. What happens to the volume of a material when it is subjected to uniform pressure? (2019)
A.
It increases
B.
It decreases
C.
It remains the same
D.
It depends on the material
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Solution
Uniform pressure typically causes a decrease in volume, known as compression.
Correct Answer: B — It decreases
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Q. What happens to the volume of a solid when it is subjected to uniform pressure? (2019)
A.
It increases
B.
It decreases
C.
It remains the same
D.
It depends on the material
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Solution
The volume of a solid generally decreases when subjected to uniform pressure, as per the bulk modulus.
Correct Answer: B — It decreases
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Q. What is the bulk modulus of a material if a pressure increase of 1000 kPa causes a volume decrease of 0.01 m³ in a 1 m³ sample? (2022)
A.
10000 kPa
B.
1000 kPa
C.
100 kPa
D.
100 kPa
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Solution
Bulk modulus (K) = -ΔP / (ΔV/V) = 1000 kPa / (0.01 m³ / 1 m³) = 100000 kPa = 10000 kPa.
Correct Answer: A — 10000 kPa
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Q. What is the bulk modulus of a material if a pressure increase of 1000 kPa results in a volume decrease of 0.01 m³ for a volume of 1 m³? (2022)
A.
10000 kPa
B.
1000 kPa
C.
100 kPa
D.
100000 kPa
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Solution
Bulk modulus (K) = -ΔP / (ΔV/V) = 1000 kPa / (0.01 m³ / 1 m³) = 100000 kPa.
Correct Answer: A — 10000 kPa
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Q. What is the bulk modulus of a material if a pressure increase of 1000 kPa results in a volume decrease of 0.5%? (2022)
A.
200000 kPa
B.
500000 kPa
C.
1000000 kPa
D.
250000 kPa
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Solution
Bulk modulus (K) = -ΔP / (ΔV/V) = 1000 kPa / 0.005 = 200000 kPa.
Correct Answer: B — 500000 kPa
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Q. What is the bulk modulus of a material if a pressure increase of 50 MPa results in a volume decrease of 0.02 m³ for a volume of 1 m³? (2022)
A.
2500 MPa
B.
500 MPa
C.
1000 MPa
D.
2000 MPa
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Solution
Bulk modulus (K) = -ΔP / (ΔV/V) = 50 MPa / (0.02 m³ / 1 m³) = 2500 MPa.
Correct Answer: A — 2500 MPa
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Q. What is the bulk modulus of a material if a pressure increase of 50 MPa results in a volume decrease of 0.002 m³ for a volume of 1 m³? (2022)
A.
25 GPa
B.
50 GPa
C.
100 GPa
D.
200 GPa
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Solution
Bulk modulus (K) = -ΔP / (ΔV/V) = -50 MPa / (-0.002 m³ / 1 m³) = 25000 MPa = 25 GPa.
Correct Answer: B — 50 GPa
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Q. What is the Poisson's ratio for a material if a longitudinal strain of 0.005 results in a lateral strain of 0.002? (2019)
A.
0.4
B.
0.5
C.
0.2
D.
0.1
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Solution
Poisson's ratio (ν) = Lateral strain / Longitudinal strain = 0.002 / 0.005 = 0.4.
Correct Answer: A — 0.4
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Q. What is the Poisson's ratio for a material if a longitudinal strain of 0.02 results in a lateral strain of 0.008? (2022)
A.
0.4
B.
0.5
C.
0.6
D.
0.7
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Solution
Poisson's ratio (ν) = -lateral strain / longitudinal strain = -0.008 / 0.02 = 0.4.
Correct Answer: A — 0.4
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Q. What is the Poisson's ratio for a material that experiences a lateral strain of 0.02 when subjected to a longitudinal strain of 0.01? (2023)
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Solution
Poisson's ratio (ν) = Lateral strain / Longitudinal strain = 0.02 / 0.01 = 2.
Correct Answer: A — 0.5
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Q. What is the Poisson's ratio for a material that experiences a lateral strain of 0.005 when subjected to a longitudinal strain of 0.01? (2021)
A.
0.5
B.
1
C.
0.25
D.
0.75
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Solution
Poisson's ratio (ν) = -lateral strain / longitudinal strain = -0.005 / 0.01 = 0.5.
Correct Answer: A — 0.5
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Q. What is the primary reason for the brittleness of certain materials? (2023)
A.
High ductility
B.
Low atomic bonding energy
C.
High atomic bonding energy
D.
High elasticity
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Solution
Brittleness is primarily due to high atomic bonding energy, which prevents the material from deforming plastically.
Correct Answer: C — High atomic bonding energy
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Q. What is the primary reason for the phenomenon of plastic deformation in materials? (2023)
A.
Elastic limit
B.
Yield strength
C.
Ultimate tensile strength
D.
Ductility
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Solution
Plastic deformation occurs when the material is subjected to stress beyond its yield strength.
Correct Answer: B — Yield strength
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Q. What is the shear modulus of a material if a shear stress of 100 MPa produces a shear strain of 0.02? (2022)
A.
5000 MPa
B.
2000 MPa
C.
1000 MPa
D.
2500 MPa
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Solution
Shear modulus (G) = Shear stress / Shear strain = 100 MPa / 0.02 = 5000 MPa.
Correct Answer: A — 5000 MPa
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Q. What is the shear modulus of a material if a shear stress of 100 MPa produces a shear strain of 0.05? (2021)
A.
2000 MPa
B.
500 MPa
C.
1000 MPa
D.
1500 MPa
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Solution
Shear modulus (G) = Shear stress / Shear strain = 100 MPa / 0.05 = 2000 MPa.
Correct Answer: A — 2000 MPa
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Q. What is the Young's modulus of a material if a stress of 200 MPa produces a strain of 0.01? (2021)
A.
2000 MPa
B.
20 MPa
C.
200 MPa
D.
2 MPa
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Solution
Young's modulus (E) = Stress / Strain = 200 MPa / 0.01 = 20000 MPa = 2000 MPa.
Correct Answer: A — 2000 MPa
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