Q.1
Work done is best given by___________
  • a) dU = Fdrcosθ
  • b) dU = Fdrsinθ
  • c) dU = Fdrcotθ
  • d) dU = Fdrdθ
Q.2
Calculate the Normal force developed between the body and the surface due to the work done by the force.
Questionengineering-mechanics-questions-answers2.jpg
  • a) 611N
  • b) 116N
  • c) 100N
  • d) 180N
Q.3
The supports in theare having more than three reaction forces. Because they are having three axis on which the components of the work needs to be zero.
  • a) The first part of the statement is false and other part is true
  • b) The first part of the statement is false and other part is false too
  • c) The first part of the statement is true and other part is false
  • d) The first part of the statement is true and other part is true too
Q.4
∑Fx=∑Fy=0 and ∑Fz=0 are vector equations for the three dimensions. They are satisfied when the body is achieved it state of equilibrium and the network done is zero.
  • a) True
  • b) False
Q.5
Calculate the frictional force developed between the body and the surface due to the work done by the force.
  • a) 160N
  • b) 16N
  • c) 10N
  • d) 180N
Q.6
Work done is best given by _____________
Questionengineering-mechanics-questions-answers6.jpg
  • a) dU = F.dr
  • b) dU = Fdrsinθ
  • c) dU = F.drcotθ
  • d) dU = Fxdrdθ
Q.7
We often determine the work done by the couples. A couple moment is developed when _______ of the attached member is prevented.
  • a) Translation
  • b) Rotation
  • c) Addition
  • d) Subtraction
Q.8
Work done by the couple is best given by___________
  • a) dU = Mdθ
  • b) dU = M.dθ
  • c) dU = Mxdθ
  • d) dU = Mθ
Q.9
What is not the condition for the equilibrium in three dimensional system of axis so as to calculate the unknown forces acting on the body?
  • a) ∑Fx=0
  • b) ∑Fy=0
  • c) ∑Fz=0
  • d) ∑F≠0
Q.10
Virtual Work done is best given by_______________
  • a) δU = Fδrcosθ
  • b) δU = Fδrsinθ
  • c) δU = Fδrcotθ
  • d) δU = Fδrδθ
Q.11
Virtual work done by the couple is best given by_____________
  • a) δU = Mδθ
  • b) δU = M.δθ
  • c) δU = Mxδθ
  • d) δU = Mθ
Q.12
Principle of virtual work done is having condition___________
  • a) δU = 0
  • b) δU = δF
  • c) δU = δM
  • d) δU = δD
Q.13
If solving the question incalculations is difficult, then use thesystem and then equate the total work done to zero.
  • a) True
  • b) False
Q.14
If the resolved force or the force which you get as the answer after solving the question is negative, then what does this implies?
  • a) The work done is in the reverse direction w.r.t the direction set in the free body diagram
  • b) The force is not in the reverse direction w.r.t the direction set in the free body diagram
  • c) The force component is not possible
  • d) The force is possible, but in the direction perpendicular to the resultant force
0 h : 0 m : 1 s