Quiz: Mechanical Engineering
Each question carries 2 mark
Negative marking: 1/3 mark
Time: 10 Minutes
Q1. Two parallel forces 100 kN and 125 kN act on a body and have resultant of 25 kN. Then the two forces are
(a) Like parallel forces
(b) Unlike parallel forces
(c) Concurrent forces
(d) None of the above
Q2. The coefficient of rolling resistance is defined as
(a) Rolling resistance to lateral load
(b) Lateral load to rolling resistance
(c) Rolling resistance to normal load
(d) Normal load to rolling resistance
Q3. The moment of inertia of an area always least with respect to
(a) Bottom most axis
(b) Radius of gyration
(c) Central axis
(d) Centroidal axis
Q4. The vector P ̅=ai ̂+aj ̂+3k ̂ and Q ̅=ai ̂-2 ̂-k ̂ are perpendicular to each other. The positive value of a is
Q5. A vector A ̅ points vertically upward and B ̅ point towards north. The vector product A ̅×B ̅ is
(b) Along west
(c) Along east
(d) Vertically downward
Q6. When the bridges are extended over long routes or distance then _
(a) A rocker or a roller is used at the joints
(b) They are not extended to such a long distance
(c) The bridges are painted
(d) The roads are made narrow
Q7. Dry friction is also called _
(a) Column Friction
(b) Coulomb Friction
(c) Dry column friction
(d) Surface friction
Q8. The frictional force in the belts always acts __ to the surface of the application of the friction.
Q9. What is parallel axis theorem and to whom it is applied so that it can give the product of inertia of an area?
(a) Theorem used to add the two mutually perpendicular moment of inertias for areas
(b) Theorem used to add the two mutually perpendicular moment of inertias for volumes
(c) Theorem used to add the two mutually perpendicular moment of inertias for linear distances
(d) Theorem used to add the two mutually perpendicular moment of inertias for vectors
Q10. Whenever the distributed loading acts perpendicular to an area its intensity varies __
S1. Ans (b)
Sol. We know, according to parallelogram law,
R = 25 kN
P = 100 kN
Q = 125 kN
Then, from equation (1)
S2. Ans (c)
Sol. Coefficient of rolling resistance (μ_R) =(Rolling resistance(F_R))/(Normal load(N))
S3. Ans (d)
Sol. The first moments of area about any centroidal axis of the area is zero. Since, the centroid is located on the centroidal axis, the perpendicular distance from the centroid to the centroidal axis must be zero.
Sol. P ̅ is perpendicular to Q ̂ if P̂ . Q̂ = 0
⇒ (ai ̂+aj ̂+3k ̂ )(ai ̂-2j ̂-k ̂ )=0
⇒ a²-2a-3 = 0 ⇒ (a-3)(a+1) = 0
∵ a = 3 (positive value)
Sol. Keep fingers of your right hand towards sky and curl them towards north. Y our thumb will point towards west.
Sol. When the bridges are extended over long routes or distance then a rocker or a roller is used at the joints. This allows the bridge joints to move around. That is when the temperature is raised. The elongations and the contractions of the joints part are not much affected if the rollers and rockers are used.
Sol. The dry friction is acted upon the surfaces. And they are tangential to each other. As we know the friction is the phenomena that define that there is a resistance which is present there between the two surfaces. The dry friction is also termed as the Coulomb friction as it was given by C.A. Coulomb.
Sol. The friction is the phenomena that defines that there is a resistance which is present there between the two surfaces. This friction is applied tangentially to the surfaces in contact. Thus the main thing is that the forces on both of the surfaces act as a tangential to each other.
Sol. Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. That is the moment of inertia perpendicular to the surface in considerance.
Ans. The load intensity is varying linearly in the structures. Thus the intensity is not varying parabolically nor is it cubically. It cannot be a vector also. Thus the intensity is linearly varied.