Self Studies

Mechanical Properties of Solids Test 2

Result Self Studies

Mechanical Properties of Solids Test 2
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0

    A perfectly rigid body is one

    Solution

    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.

  • Question 2
    1 / -0

    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

    Solution

    Stress = F⟂/A = (Mass x Acceleration⟂)/Cross-sectional Area

    = (550 kg x 9.8 m/s2)/ (0.30×10−4 m2)

    = 1.8×108 Pa

    Young's ModulusSteel (Y) = 20×1010 Pa = Stress/Strain

    => Strain = Stress/Y = 1.8×108 Pa / 20×1010 Pa

    = 9.0×10−4 

  • Question 3
    1 / -0

    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 11 × 1010Pa

    Solution

    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

  • Question 4
    1 / -0

    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

    Solution

    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

  • Question 5
    1 / -0

    Elasticity is the property of a body, by virtue of which

    Solution

    Explanation:When external force is applied on the solid bodies, the solid bodies get deformed. The atoms or molecules are displaced from their equilibrium positions causing a change in the interatomic ( or intermolicular ) distances. When the deforming foce is removed, the interatomic forces tend to drive them back to their original postions. Thus the body regains its original shape and size.

  • Question 6
    1 / -0

    Elastomers are materials

    Solution

    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.

  • Question 7
    1 / -0

    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

    Solution

    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

  • Question 8
    1 / -0

    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?

    Solution

  • Question 9
    1 / -0

    The importance of the elastic behavior of materials is

    Solution

    Explanation:More the elastic a material is , more it has the property to regain its original position which is required in construction works.

  • Question 10
    1 / -0

    Shear modulus or modulus of rigidity is

    Solution

    Shear modulus or modulus of rigidity is the ratio of shearing stress to the corresponding shearing strain, by the definition of shear modulus.

  • Question 11
    1 / -0

    Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm. Bulk modulus of glass 37 GPa.

  • Question 12
    1 / -0

    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.

    Solution

  • Question 13
    1 / -0

    When a solid is deformed,

    Solution

    Explanation:External force permanently distubed the equilibrium position of the interatomic ( or intermolecular ) forces between the particles of solid bodies.

  • Question 14
    1 / -0

    Rectangular section is rarely used in beams because

    Solution

    I section is generally used as a beam because of its high section modulus as its most of the area is situated away from its 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.

  • Question 15
    1 / -0

    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?

    Solution

  • Question 16
    1 / -0

    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.

    Solution

    https://cdn3.edurev.in/ApplicationImages/Temp/0_c95efa72-9c79-4220-8b1e-2e75a06fe6fd_lg.png
    https://cdn3.edurev.in/ApplicationImages/Temp/0_3ef72341-6651-46da-98eb-e29fd272a7c6_lg.png
    https://cdn3.edurev.in/ApplicationImages/Temp/0_27789e82-45ef-4705-80d1-3c4d214c5cc7_lg.png

  • Question 17
    1 / -0

    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?

    Solution

    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

  • Question 18
    1 / -0

    Stress is

  • Question 19
    1 / -0

    Columns are loaded in

    Solution

    Correct Answer :- d

    Explanation : The load applied to a column would place the column in compression; conversely, a load hanging from a rod would place the rod in tension. Strain is the deformation of a structural member because of stress within the member.

  • Question 20
    1 / -0

    The S.I unit of stress is

    Solution

    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.

  • Question 21
    1 / -0

    Material is said to be ductile if

    Solution

    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.

  • Question 22
    1 / -0

    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.

    Solution

    Hence 1.6 mm upper, 1.0 mm lower is correct.

  • Question 23
    1 / -0

    volume strain is defined

  • Question 24
    1 / -0

    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?

    Solution

  • Question 25
    1 / -0

    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)

    Solution

  • Question 26
    1 / -0

    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.

    Solution

  • Question 27
    1 / -0

    According to Hooke’s law

    Solution

    Hook’s Law states that for small deformations the stress and strain are proportional to each other.

  • Question 28
    1 / -0

    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?

  • Question 29
    1 / -0

    With reference to figure the elastic zone is

    Solution

    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.

  • Question 30
    1 / -0

    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?

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now