APSC AE Mechanical - Engineering Mechanics MCQs
1. A force of 200 N acts at an angle of 30° to the horizontal. What is the horizontal component of the force?
Answer: 173.2 N
Description: The horizontal component of a force is given by F_x = F cos(θ). Here, F = 200 N and θ = 30°. So, F_x = 200 × cos(30°) = 200 × 0.866 = 173.2 N.
Description: The horizontal component of a force is given by F_x = F cos(θ). Here, F = 200 N and θ = 30°. So, F_x = 200 × cos(30°) = 200 × 0.866 = 173.2 N.
2. The moment of a force about a point is zero when:
Answer: The line of action of the force passes through the point
Description: The moment of a force about a point is given by M = F × d, where d is the perpendicular distance from the point to the line of action of the force. If the force passes through the point, d = 0, so M = 0.
Description: The moment of a force about a point is given by M = F × d, where d is the perpendicular distance from the point to the line of action of the force. If the force passes through the point, d = 0, so M = 0.
3. A body of mass 10 kg is moving with an acceleration of 5 m/s². What is the force acting on it?
Answer: 50 N
Description: According to Newton’s second law, F = m × a. Here, m = 10 kg and a = 5 m/s². So, F = 10 × 5 = 50 N.
Description: According to Newton’s second law, F = m × a. Here, m = 10 kg and a = 5 m/s². So, F = 10 × 5 = 50 N.
4. Which of the following is a scalar quantity?
Answer: Work
Description: A scalar quantity has only magnitude and no direction. Work is defined as W = F × d × cos(θ), which results in a scalar value. Velocity, force, and momentum are vector quantities as they have both magnitude and direction.
Description: A scalar quantity has only magnitude and no direction. Work is defined as W = F × d × cos(θ), which results in a scalar value. Velocity, force, and momentum are vector quantities as they have both magnitude and direction.
5. A couple is produced by:
Answer: Two equal and opposite forces separated by a distance
Description: A couple consists of two equal and opposite forces that are not collinear, producing a pure rotational effect without any net translational motion.
Description: A couple consists of two equal and opposite forces that are not collinear, producing a pure rotational effect without any net translational motion.
6. A particle moves with a velocity of 10 m/s and is retarded at 2 m/s². How far will it travel before coming to rest?
Answer: 25 m
Description: Use the equation v² = u² + 2as, where v = 0 (final velocity), u = 10 m/s (initial velocity), and a = -2 m/s² (retardation). Solving for s: 0 = 10² + 2(-2)s => 0 = 100 - 4s => s = 100/4 = 25 m.
Description: Use the equation v² = u² + 2as, where v = 0 (final velocity), u = 10 m/s (initial velocity), and a = -2 m/s² (retardation). Solving for s: 0 = 10² + 2(-2)s => 0 = 100 - 4s => s = 100/4 = 25 m.
7. The condition for equilibrium of a body under coplanar forces is:
Answer: Both sum of forces and moments are zero
Description: For a body to be in equilibrium under coplanar forces, the resultant force (ΣF = 0) and the resultant moment (ΣM = 0) about any point must be zero, ensuring no translation or rotation.
Description: For a body to be in equilibrium under coplanar forces, the resultant force (ΣF = 0) and the resultant moment (ΣM = 0) about any point must be zero, ensuring no translation or rotation.
8. A 500 N force acts at a perpendicular distance of 0.4 m from a pivot. What is the moment about the pivot?
Answer: 200 Nm
Description: Moment is calculated as M = F × d. Here, F = 500 N and d = 0.4 m. So, M = 500 × 0.4 = 200 Nm.
Description: Moment is calculated as M = F × d. Here, F = 500 N and d = 0.4 m. So, M = 500 × 0.4 = 200 Nm.
9. Lami’s theorem is applicable to:
Answer: Three concurrent forces in equilibrium
Description: Lami’s theorem states that for three concurrent forces in equilibrium, each force is proportional to the sine of the angle between the other two forces.
Description: Lami’s theorem states that for three concurrent forces in equilibrium, each force is proportional to the sine of the angle between the other two forces.
10. The coefficient of friction between two surfaces is 0.2. If the normal force is 100 N, what is the frictional force?
Answer: 20 N
Description: Frictional force is given by F_f = μ × N, where μ is the coefficient of friction and N is the normal force. Here, μ = 0.2 and N = 100 N. So, F_f = 0.2 × 100 = 20 N.
Description: Frictional force is given by F_f = μ × N, where μ is the coefficient of friction and N is the normal force. Here, μ = 0.2 and N = 100 N. So, F_f = 0.2 × 100 = 20 N.
11. A truss is considered perfect if:
Answer: m = 2j - 3
Description: For a truss to be perfect (statically determinate), the number of members (m) and joints (j) must satisfy m = 2j - 3, ensuring the structure is stable without redundancy.
Description: For a truss to be perfect (statically determinate), the number of members (m) and joints (j) must satisfy m = 2j - 3, ensuring the structure is stable without redundancy.
12. The centroid of a triangle lies at:
Answer: The intersection of its medians
Description: The centroid of a triangle is the point where all three medians intersect, dividing each median in a 2:1 ratio.
Description: The centroid of a triangle is the point where all three medians intersect, dividing each median in a 2:1 ratio.
13. A car accelerates from rest at 4 m/s² for 5 seconds. What is its final velocity?
Answer: 20 m/s
Description: Using the equation v = u + at, where u = 0 (initial velocity), a = 4 m/s², and t = 5 s, we get v = 0 + 4 × 5 = 20 m/s.
Description: Using the equation v = u + at, where u = 0 (initial velocity), a = 4 m/s², and t = 5 s, we get v = 0 + 4 × 5 = 20 m/s.
14. The principle of virtual work is used to:
Answer: Analyze equilibrium of a system
Description: The principle of virtual work states that for a system in equilibrium, the virtual work done by all forces during a virtual displacement is zero, used to analyze static equilibrium.
Description: The principle of virtual work states that for a system in equilibrium, the virtual work done by all forces during a virtual displacement is zero, used to analyze static equilibrium.
15. A beam is supported at both ends with a point load at the center. The maximum bending moment occurs:
Answer: At the center
Description: For a simply supported beam with a point load at the center, the maximum bending moment occurs at the point of load application (center), calculated as M = (P × L)/4, where P is the load and L is the beam length.
Description: For a simply supported beam with a point load at the center, the maximum bending moment occurs at the point of load application (center), calculated as M = (P × L)/4, where P is the load and L is the beam length.
16. A block of weight 1000 N rests on an incline of 30°. What is the normal force?
Answer: 866 N
Description: The normal force on an incline is N = W × cos(θ), where W = 1000 N and θ = 30°. So, N = 1000 × cos(30°) = 1000 × 0.866 = 866 N.
Description: The normal force on an incline is N = W × cos(θ), where W = 1000 N and θ = 30°. So, N = 1000 × cos(30°) = 1000 × 0.866 = 866 N.
17. Varignon’s theorem is related to:
Answer: Moments of forces
Description: Varignon’s theorem states that the moment of a force about a point is equal to the sum of the moments of its components about the same point.
Description: Varignon’s theorem states that the moment of a force about a point is equal to the sum of the moments of its components about the same point.
18. A projectile is launched with an initial velocity of 20 m/s at 45°. What is the maximum height reached?
Answer: 10 m
Description: Maximum height is given by H = (u² sin²θ)/(2g), where u = 20 m/s, θ = 45°, and g = 9.81 m/s². So, H = (20² × sin²45°)/(2 × 9.81) = (400 × 0.5)/(19.62) ≈ 10 m.
Description: Maximum height is given by H = (u² sin²θ)/(2g), where u = 20 m/s, θ = 45°, and g = 9.81 m/s². So, H = (20² × sin²45°)/(2 × 9.81) = (400 × 0.5)/(19.62) ≈ 10 m.
19. The angle of friction is:
Answer: The angle between normal reaction and resultant force
Description: The angle of friction (φ) is the angle between the normal reaction and the resultant of the normal reaction and frictional force, given by tan(φ) = μ.
Description: The angle of friction (φ) is the angle between the normal reaction and the resultant of the normal reaction and frictional force, given by tan(φ) = μ.
20. A flywheel rotates with an angular acceleration of 2 rad/s². If it starts from rest, what is its angular velocity after 4 seconds?
Answer: 8 rad/s
Description: Angular velocity is given by ω = ω₀ + αt, where ω₀ = 0 (initial angular velocity), α = 2 rad/s², and t = 4 s. So, ω = 0 + 2 × 4 = 8 rad/s.
Description: Angular velocity is given by ω = ω₀ + αt, where ω₀ = 0 (initial angular velocity), α = 2 rad/s², and t = 4 s. So, ω = 0 + 2 × 4 = 8 rad/s.
21. The law of parallelogram of forces is used to:
Answer: Find the resultant of two forces
Description: The parallelogram law states that the resultant of two forces acting at a point is represented by the diagonal of a parallelogram formed by the two forces as adjacent sides.
Description: The parallelogram law states that the resultant of two forces acting at a point is represented by the diagonal of a parallelogram formed by the two forces as adjacent sides.
22. A 2 kg mass is dropped from a height of 5 m. What is its kinetic energy just before hitting the ground?
Answer: 100 J
Description: Potential energy at height h is mgh, which converts to kinetic energy just before impact. Here, m = 2 kg, g = 9.81 m/s², h = 5 m. So, KE = 2 × 9.81 × 5 ≈ 100 J.
Description: Potential energy at height h is mgh, which converts to kinetic energy just before impact. Here, m = 2 kg, g = 9.81 m/s², h = 5 m. So, KE = 2 × 9.81 × 5 ≈ 100 J.
23. The moment of inertia of a body depends on:
Answer: Mass and distribution of mass
Description: Moment of inertia (I) depends on the mass of the body and how that mass is distributed relative to the axis of rotation, given by I = Σmr².
Description: Moment of inertia (I) depends on the mass of the body and how that mass is distributed relative to the axis of rotation, given by I = Σmr².
24. A roller of weight 2000 N rests on a smooth surface. What is the normal reaction?
Answer: 2000 N
Description: On a smooth horizontal surface, the normal reaction equals the weight of the body. Here, W = 2000 N, so N = 2000 N.
Description: On a smooth horizontal surface, the normal reaction equals the weight of the body. Here, W = 2000 N, so N = 2000 N.
25. D’Alembert’s principle is used to:
Answer: Convert dynamic problems to static ones
Description: D’Alembert’s principle introduces an inertial force (ma) opposite to the acceleration, allowing dynamic systems to be analyzed as if in static equilibrium.
Description: D’Alembert’s principle introduces an inertial force (ma) opposite to the acceleration, allowing dynamic systems to be analyzed as if in static equilibrium.
26. A shaft rotates at 120 rpm. What is its angular velocity in rad/s?
Answer: 4π rad/s
Description: Angular velocity ω = (2π × N)/60, where N is rpm. Here, N = 120, so ω = (2π × 120)/60 = 4π rad/s.
Description: Angular velocity ω = (2π × N)/60, where N is rpm. Here, N = 120, so ω = (2π × 120)/60 = 4π rad/s.
27. The center of gravity of a body:
Answer: Depends on mass distribution
Description: The center of gravity is the point where the entire weight of a body acts, determined by the distribution of its mass, not necessarily at the geometric center.
Description: The center of gravity is the point where the entire weight of a body acts, determined by the distribution of its mass, not necessarily at the geometric center.
28. A force of 300 N is applied at 0.5 m from a pivot at 60° to the radius. What is the moment?
Answer: 129.9 Nm
Description: Moment is M = F × d × sin(θ), where F = 300 N, d = 0.5 m, θ = 60°. So, M = 300 × 0.5 × sin(60°) = 300 × 0.5 × 0.866 = 129.9 Nm.
Description: Moment is M = F × d × sin(θ), where F = 300 N, d = 0.5 m, θ = 60°. So, M = 300 × 0.5 × sin(60°) = 300 × 0.5 × 0.866 = 129.9 Nm.
29. The impulse-momentum principle states that:
Answer: Impulse equals change in momentum
Description: The impulse-momentum principle states that the impulse (F × t) applied to a body equals the change in its momentum (mΔv).
Description: The impulse-momentum principle states that the impulse (F × t) applied to a body equals the change in its momentum (mΔv).
30. A body is in dynamic equilibrium when:
Answer: It moves with constant velocity
Description: Dynamic equilibrium occurs when a body moves with constant velocity, implying no net force acts on it (ΣF = 0), as per Newton’s first law.
Description: Dynamic equilibrium occurs when a body moves with constant velocity, implying no net force acts on it (ΣF = 0), as per Newton’s first law.