Question 1

When the spring of a watch is wound, it will possess

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strain energy

Question 2

The angle of inclination with the vertical is __________ if the cyclist is running at a faster speed than that when he is running at a slower speed.

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more

Question 3

An irregular body may have more than one centre of gravity.

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No

Question 4

In a screw jack, the effort required to lift the load is given by(where W = Load lifted, α = Helix angle, and φ = Angle of friction.)

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P = W tan (α + φ)

Question 5

The resultant of the two forces P and Q is R. If Q is doubled, the new resultant is perpendicular to P. Then

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Q = R

Question 6

Whenever a force acts on a body and the body undergoes a displacement, then

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work is said to be done

Question 7

Moment of inertia of a circular section about its diameter (d) is

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πd4/64

Question 8

When a particle moves along a straight path, there will be centripetal acceleration as well as tangential acceleration.

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Incorrect

Question 9

Rotational kinematics — tangential (linear) acceleration:
A body is rotating along a circular path of radius r with an angular acceleration α (in rad/s^2). What is the expression for its tangential (linear) acceleration a along the path?

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Introduction / Context: Tangential (linear) acceleration connects rotational motion with linear motion at a radius. In mechanical and civil engineering problems involving pulleys, gears, flywheels, or curved vehicle paths, converting angular acceleration to linear acceleration is a routine step. Given Data / Assumptions: Circular motion of radius r (meters). Angular acceleration α (rad/s^2) is constant or instantaneous. Tangential acceleration a is the linear acceleration along the circle. Concept / Approach: By kinematics, linear and angular measures are related through the radius. The linear velocity is v = ω * r and the tangential acceleration is the time rate of change of v due to changing angular speed, so a_t depends on α and r directly. Step-by-Step Solution: Start from v = ω * r.Differentiate with respect to time: a_t = dv/dt = r * dω/dt.But dω/dt = α, hence a_t = r * α.Therefore, a = α * r. Verification / Alternative check: Units: α has units rad/s^2 (radian is dimensionless), multiplied by r (m) gives m/s^2, which are correct units for linear acceleration. Why Other Options Are Wrong: a = α / r: Inverts the dependence; dimensions become 1/(s^2·m), not m/s^2. a = r / α: Also dimensionally incorrect and physically implausible. none of these: Incorrect because a = α * r is the standard relation. Common Pitfalls: Mixing tangential acceleration a_t = α * r with normal (centripetal) acceleration a_n = v^2 / r = ω^2 * r; both exist simultaneously in non-uniform circular motion. Forgetting that radians are dimensionless, preserving unit consistency. Final Answer: a = α * r

Question 10

Plane pin-jointed frames — condition for imperfection:
A framed structure (simple, plane, pin-jointed) is considered imperfect if the number of members m is __________ (2j − 3), where j is the number of joints.

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Introduction / Context: For plane pin-jointed trusses, the simple test m = 2j − 3 helps classify whether a frame is statically determinate (perfect), unstable (deficient), or redundant (excess). This criterion is widely used in structural analysis as a quick screening before detailed calculations. Given Data / Assumptions: Plane, pin-jointed frame. No special geometric constraints (no internal mechanisms) beyond simple connections. m = number of members; j = number of joints. Concept / Approach: A simple (perfect) truss satisfies m = 2j − 3. If m < 2j − 3, the truss is deficient in members (mechanism/unstable). If m > 2j − 3, it is redundant and therefore imperfect from a purely statical-determinacy standpoint. Step-by-Step Solution: Perfect frame condition: m = 2j − 3.Deficient (unstable): m < 2j − 3.Redundant (statically indeterminate): m > 2j − 3.Imperfect = not perfect ⇒ m ≠ 2j − 3 ⇒ either less than or greater than 2j − 3. Verification / Alternative check: Classical examples: A triangular frame (j = 3) with m = 3 is perfect (3 = 2*3 − 3). Removing one member (m = 2) makes it a mechanism (deficient). Adding an extra member (m = 4) introduces redundancy (excess). Both non-equal cases are imperfect. Why Other Options Are Wrong: equal to: That is the perfect condition, not imperfect. less than: Only one type of imperfection (deficient) but misses redundant case. greater than: Only redundant case; misses deficient case. Common Pitfalls: Assuming m = 2j − 3 guarantees stability in all geometric layouts; certain special geometries can still be mechanisms. Confusing determinacy (solvability by statics alone) with structural safety or serviceability. Final Answer: either (b) or (c)

Question 11

Connected masses — effect of friction on string tension:
A 20 kg block on a rough horizontal plane is connected by a light string over a smooth pulley to a 5 kg hanging mass (free to move vertically). As the coefficient of friction between the bloc…

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Introduction / Context: This classic two-body system illustrates how friction alters internal force (tension) and acceleration. Understanding the direction and magnitude trends is crucial for conveyor, hoist, and brake problems in mechanical design. Given Data / Assumptions: Block on horizontal plane: m1 = 20 kg; hanging mass: m2 = 5 kg. Smooth pulley, light inextensible string. Kinetic/limiting friction modeled as F = μ * N = μ * m1 * g. Hanging mass tends to move downward; block tends to move toward the pulley. Concept / Approach: Apply Newton’s second law to each mass and eliminate acceleration to express tension T as a function of μ. Then inspect how T varies with μ. The algebra reveals monotonic behavior here. Step-by-Step Solution: For m2 (down positive): m2g − T = m2a.For m1 (toward pulley positive): T − μm1g = m1a.Eliminate a: add equations ⇒ m2g − μm1g = (m1 + m2)a ⇒ a = (m2g − μm1g)/(m1 + m2).Back-substitute into T = m1a + μm1g ⇒ T = m1m2g(1 + μ)/(m1 + m2).Hence, dT/dμ > 0 ⇒ tension increases with μ. Verification / Alternative check: At μ = 0 (frictionless plane), T = m1m2g/(m1 + m2). As μ grows, friction resists motion of m1 more, so the string must carry more load, raising T. Why Other Options Are Wrong: decrease: Opposite of derived trend. not be affected: Contradicted by the explicit dependence T ∝ (1 + μ). indeterminate: The model yields a clear analytic dependence on μ. Common Pitfalls: Assuming friction always reduces internal forces; here it reduces acceleration but increases tension. Sign mistakes when writing the equations of motion for each mass. Final Answer: increase

Question 12

Work and power — basic definition:
In mechanics, the rate at which work is done or energy is transferred per unit time is called:

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Introduction / Context: Power quantifies how quickly work is performed. In machines and structures, power ratings determine motor sizing, heat generation, and efficiency comparisons between processes performing the same total work in different times. Given Data / Assumptions: Work W measured in joules; time t in seconds. Average power P_avg = W / t; instantaneous power P = dW/dt. Mechanical context, no special constraints. Concept / Approach: Energy forms (potential, kinetic) are states; power is a rate. Therefore, the only term that represents rate of doing work is power. Step-by-Step Solution: Use definition: Power = work per unit time.Expressed as P = dW/dt or P = F * v when force and velocity are collinear.Hence, among the options, the correct term is power. Verification / Alternative check: Dimensional analysis: [Power] = joule/second = watt (W). Neither potential nor kinetic energy has time in the denominator; they are stored energy measures (J). Why Other Options Are Wrong: potential energy: Stored energy due to position/configuration. kinetic energy: Energy of motion, 0.5mv^2. none of these: Incorrect because power is explicitly listed and correct. Common Pitfalls: Confusing high energy with high power; a small energy released quickly can be high power. Final Answer: power

Question 13

Centroid of a right circular solid cone:
For a solid cone of height h, measured along the central vertical axis from its base (the circular face), the centre of gravity (centroid) lies at what distance from the base?

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Introduction / Context: Centroid locations are essential in calculating moments, hydrostatic forces, and stresses. For a homogeneous right circular solid cone, the centroid lies along the axis of symmetry between the base and the apex at a fixed ratio of the height. Given Data / Assumptions: Uniform density, solid cone. Height = h; base is the circular face. Distance measured upward from the base along the axis. Concept / Approach: By standard centroid formulas or integration (using similar discs), the y-coordinate of the centroid from the base is one-quarter of the height for a solid cone. This differs from a conical frustum or a hollow (conical shell), which have different centroid positions. Step-by-Step Solution (outline): Consider the cone as stacked thin disks of radius proportional to their distance from the apex.Compute ȳ = (∫ y dA or y dV) / (∫ dA or dV) using volume elements.Evaluation yields ȳ = h/4 from the base (or 3h/4 from the apex). Verification / Alternative check: Rule-of-thumb: for simple solids of revolution, the cone centroid is at 1/4 from the base, cylinder at 1/2, hemisphere (solid) at 3/8 from the base. The cone result is consistent with these benchmarks. Why Other Options Are Wrong: h/2: Mid-height; correct for uniform prism/cylinder, not a cone. h/3: Applies to triangular area centroid, not the solid cone volume. h/6: Too close to the base; not supported by integration. Common Pitfalls: Confusing area centroid of a triangular cross-section with volume centroid of the cone. Measuring from the apex instead of the base; from the apex the distance is 3h/4. Final Answer: h/4

Question 14

Equilibrium of three coplanar forces — Lami’s theorem check:
If three coplanar forces acting at a point keep a body in equilibrium, then each force is proportional to the sine of the angle between the other two forces. Is this statement correct?

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Introduction / Context: Lami’s theorem provides a direct relation between the magnitudes of three concurrent, coplanar forces in equilibrium and the angles between them. It is widely used for quick solutions without resolving forces into components. Given Data / Assumptions: Three forces are coplanar and concurrent at a point. The body is in equilibrium (net force zero). Angles considered are the included angles between the other two forces. Concept / Approach: Lami’s theorem states: F1/sin(α) = F2/sin(β) = F3/sin(γ), where α, β, γ are the angles between the other two forces. This follows from constructing the force triangle (closed polygon) and applying the sine rule. Step-by-Step Solution: Express equilibrium by a closed vector triangle.Apply the sine rule to the triangle relating side (force) to sine of opposite angle.Conclude proportionality: each force ∝ sine of angle between the other two forces. Verification / Alternative check: Resolving forces along two perpendicular axes gives two equations; eliminating one variable leads to the same sine relationships, confirming Lami’s theorem. Why Other Options Are Wrong: Incorrect: would contradict both geometric construction and component-resolution proof. Common Pitfalls: Mistaking the angle at a force tip for an external angle; use the included angle between the other two forces. Final Answer: Correct

Question 15

Units — work in the SI system:
Select the correct SI unit of work (and energy) used in physics and engineering.

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Introduction / Context: Clear unit identification prevents calculation errors and ensures proper interpretation of results. In SI, work and energy share the same unit, important for thermodynamics, mechanics, and electrical systems (via watt·second). Given Data / Assumptions: Standard SI (International System of Units). Work W = force * displacement in the direction of force. Concept / Approach: In SI, the unit of force is newton (N) and distance is meter (m). Therefore, work has unit N·m, which is defined as the joule (J). Step-by-Step Solution: Work W = F * s.Units: N * m = N·m = joule (J).Hence the correct SI unit is the joule. Verification / Alternative check: 1 joule = 1 newton-meter = 1 kg·m^2/s^2. This is consistent with kinetic energy 0.5mv^2 and potential energy mgh. Why Other Options Are Wrong: newton: Unit of force, not work. erg: CGS unit of work (1 erg = 1e−7 J). kg-m: Not an SI unit for work; dimensions do not match energy. Common Pitfalls: Using N·m as torque; while the unit matches joule dimensionally, torque is not energy. Context distinguishes them. Final Answer: joule

Question 16

Definition check — value of one joule:
One joule is numerically equal to which of the following?

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Introduction / Context: Knowing exact SI equivalences helps convert between descriptive and formula-based expressions. Engineers frequently toggle between J, N·m, and W·s when analyzing power and energy transfers. Given Data / Assumptions: SI definitions: joule (J), newton (N), meter (m). Work W = F * s. Concept / Approach: By definition, 1 joule is the work done when a force of 1 newton acts through a distance of 1 meter in the direction of the force. Hence 1 J = 1 N·m. Step-by-Step Solution: Apply definition: W = F*s.Set F = 1 N and s = 1 m in the same direction.Compute W = 1 * 1 = 1 N·m = 1 J. Verification / Alternative check: Dimensional consistency: 1 J = 1 kg·m^2/s^2, which equals the product of 1 N (kg·m/s^2) and 1 m, confirming 1 N·m. Why Other Options Are Wrong: 0.1 N·m, 10 N·m, 100 N·m: Incorrect scalings; only 1 N·m equals 1 J. Common Pitfalls: Confusing torque (also in N·m) with energy; context matters even though units match. Final Answer: 1 N·m

Question 17

Vibration of a mass–spring system (vertical helical spring):
One end of a helical spring is fixed and the other end carries a load W. The load executes simple harmonic motion vertically. If the static deflection under W is δ, what is the frequency …

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Introduction / Context: Small vertical vibrations of a mass hung from a spring are modeled as simple harmonic motion (SHM). Designers often use the static deflection δ under weight W to estimate natural frequency without directly computing spring stiffness and mass separately. Given Data / Assumptions: Load W causes a static deflection δ of the spring. Linear spring behavior; small oscillations about equilibrium. Gravity g acts downward; mass of spring neglected compared with load mass. Concept / Approach: Let spring stiffness be k. At static equilibrium, W = k * δ, so k = W / δ. With W = m * g, we have k = m * g / δ. The circular natural frequency is ω = sqrt(k / m). Frequency in hertz is f = ω / (2π). Step-by-Step Solution: From static deflection: k = W / δ = (m * g) / δ.Compute natural frequency: ω = sqrt(k / m) = sqrt((m * g / δ) / m) = sqrt(g / δ).Convert to hertz: f = ω / (2π) = (1 / 2π) * sqrt(g / δ). Verification / Alternative check: Unit check: g/δ has units (m/s^2)/m = 1/s^2, so sqrt(g/δ) is 1/s, matching angular frequency; dividing by 2π gives cycles per second (Hz). Why Other Options Are Wrong: (1/2π)*sqrt(δ/g): Inverted; would have units s, not 1/s. (1/2π)*sqrt(W/δ): Dimensionally inconsistent for frequency without dividing by sqrt(m). (1/2π)*sqrt(δ/W): Also dimensionally incorrect. Common Pitfalls: Forgetting to use static deflection to eliminate k and m cleanly. Confusing angular frequency ω with frequency f; ω = 2π f. Final Answer: f = (1 / 2π) * sqrt(g / δ)

Question 18

Varignon’s theorem of moments — correct statement:
For a system of coplanar forces acting on a particle or rigid body, Varignon’s theorem states that:

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Introduction / Context: Varignon’s theorem is a powerful shortcut in statics: the moment of a resultant force about any point equals the sum of moments of component forces about the same point. This underpins many beam, frame, and machine analyses. Given Data / Assumptions: Coplanar force system. Moments taken about the same reference point. Resultant exists (vector sum of forces). Concept / Approach: Instead of computing the moment of a large force directly, you can resolve the force into components, compute moments of components, and add them. Varignon guarantees equivalence to the moment of the original force (or the final resultant of several forces). Step-by-Step Solution: Let forces be F1, F2, ..., Fn.Define resultant R = ΣFi (vector sum).Varignon’s theorem: ΣM_O(Fi) = M_O(R), where M_O denotes moment about point O.Therefore, the correct statement is: the algebraic sum of moments of all forces about a point equals the moment of their resultant about the same point. Verification / Alternative check: Resolve each Fi into perpendicular components. Using perpendicular distances, moments add linearly. Geometric proof via the parallelogram of forces arrives at the same equality. Why Other Options Are Wrong: their algebraic sum is zero: That is a special equilibrium condition, not Varignon’s theorem. their lines of action are at equal distances: Not generally true. algebraic sum of moments is zero: Again, an equilibrium condition, not the theorem statement. Common Pitfalls: Applying the theorem but forgetting to preserve sign convention for clockwise/counter-clockwise moments. Final Answer: the algebraic sum of their moments about any point equals the moment of their resultant about the same point

Question 19

Area Moment of Inertia – Rectangle about a Centroidal Axis Parallel to depth d
For a rectangular section of width b and depth d, what is the second moment of area (area moment of inertia) about an axis that passes through the centroid and is parall…

Accepted Answer

Introduction / Context: In strength of materials and structural analysis, standard formulas for second moments of area (also called area moments of inertia) are used to evaluate bending stiffness about different centroidal axes. For a rectangle, the values differ depending on whether the reference axis is parallel to the width or the depth. Given Data / Assumptions: Plane figure: rectangle with width b (horizontal) and depth d (vertical). Axis: centroidal axis parallel to the depth d (vertical centroidal axis through the center). Homogeneous, thin area; standard definitions apply. Concept / Approach: The two principal centroidal area moments for a rectangle are: Ix (about the centroidal horizontal axis, parallel to width) Ix = (b * d^3) / 12 and Iy (about the centroidal vertical axis, parallel to depth) Iy = (d * b^3) / 12. The requested axis is parallel to d, so we need Iy. Step-by-Step Solution: Identify axis: centroidal vertical axis ⇒ use Iy. Write formula: Iy = (b^3 * d) / 12. Check dimensions: length^4 (b^3 * d) ⇒ correct for area moments. Verification / Alternative check: By symmetry, swapping b and d swaps the roles of Ix and Iy. If the axis were parallel to width, the answer would be (b * d^3) / 12 instead. Why Other Options Are Wrong: (b) corresponds to Ix, not the asked axis. (c) and (d) have the correct variables but wrong denominator (3 instead of 12). (e) has wrong power (dimensionally incorrect). Common Pitfalls: Confusing which axis (x or y) is parallel to width or depth. Final Answer: I = (b^3 * d) / 12

Engineering Mechanics Questions