Question 1

Units and conversion — express one newton (N) in dynes (CGS system).

Accepted Answer

Introduction / Context: Engineering often requires converting between SI and CGS units. Force units include the newton (N) in SI and the dyne in CGS. Knowing the relationship enables quick dimensional checks and conversions in legacy literature and laboratory measurements. Given Data / Assumptions: 1 N = 1 kg * m/s^2. 1 dyne = 1 g * cm/s^2. Conversion factors: 1 kg = 10^3 g; 1 m = 10^2 cm. Concept / Approach: Express the SI unit in CGS base units and collect powers of 10. This direct dimensional conversion avoids errors from intermediate approximations. Step-by-Step Solution: Write 1 N = 1 kg * m / s^2.Convert kg to g: 1 kg = 10^3 g.Convert m to cm: 1 m = 10^2 cm.Thus 1 N = (10^3 g) * (10^2 cm) / s^2 = 10^5 (g * cm / s^2) = 10^5 dynes. Verification / Alternative check: Reverse check: 1 dyne = 10^-5 N, consistent with standard references. Why Other Options Are Wrong: 10^3, 10^4, 10^6, and 10^7 dynes do not satisfy the base unit conversion factors. Common Pitfalls: Forgetting to convert both mass and length simultaneously; mixing MKS, SI, and CGS frameworks. Final Answer: 10^5 dynes

Question 2

Centroid of a composite lamina — a square hole is punched out of a circular plate; the square's diagonal equals the circle's radius r. The distance of the centroid of the remainder from the corner of the square that lies on the circle's circumferenc…

Accepted Answer

cannot be determined from the given data

Question 3

Kinematics — if a body's velocity increases by equal amounts in equal intervals of time, the motion is described as:

Accepted Answer

Introduction / Context: Describing motion accurately requires distinguishing among forms of acceleration. When the velocity change per unit time is constant, many equations of motion simplify and are widely used in mechanics of materials and dynamics. Given Data / Assumptions: Velocity increases by equal amounts in equal time intervals. Motion is along a straight line for simplicity, but the definition holds vectorially. Concept / Approach: Acceleration is the rate of change of velocity with respect to time. If delta v / delta t is constant, acceleration is uniform (constant). This is the fundamental assumption behind the constant-acceleration kinematic equations. Step-by-Step Solution: Define a = dv/dt.Given equal increments of v in equal t, dv/dt is constant ⇒ a is uniform.Hence the motion exhibits uniform acceleration (not retardation). Verification / Alternative check: Plot v–t: a straight line slope indicates constant a; integrating gives x–t quadratic, consistent with s = ut + 0.5a*t^2. Why Other Options Are Wrong: Variable acceleration implies non-constant dv/dt. Retardation refers to negative acceleration (decreasing speed), which contradicts “increases”. Common Pitfalls: Confusing constant speed with constant acceleration; assuming acceleration constant only if force is constant without considering mass variations. Final Answer: a uniform acceleration

Question 4

Impulse–momentum (recoil) — a 0.2 kg bullet is fired horizontally at 25 m/s from a gun mounted on a carriage; the gun plus carriage mass is 100 kg. Find the recoil velocity of the gun–carriage system.

Accepted Answer

Introduction / Context: Gun recoil problems illustrate conservation of linear momentum. Before firing, the system is at rest; after firing, the forward momentum of the projectile is balanced by an equal and opposite momentum of the gun–carriage assembly (neglecting external impulses and resistances over the short firing interval). Given Data / Assumptions: Bullet mass mb = 0.2 kg; bullet speed vb = 25 m/s (horizontal). Gun plus carriage mass Mg = 100 kg; recoil speed vg = ? System initially at rest; external horizontal impulse ≈ 0 over firing instant. Concept / Approach: Apply conservation of linear momentum along the line of fire: total momentum before equals total momentum after. The bullet acquires forward momentum; the gun recoils with equal magnitude opposite in direction. Step-by-Step Solution: Initial momentum = 0.Final momentum = mb * vb − Mg * vg = 0 (opposite signs).Solve for vg: Mg * vg = mb * vb ⇒ vg = (mb * vb) / Mg.Compute: vg = (0.2 * 25) / 100 = 5 / 100 = 0.05 m/s (opposite to bullet). Verification / Alternative check (if short method exists): Unit check: kg*m/s divided by kg gives m/s; magnitude is small due to large gun mass, as expected. Why Other Options Are Wrong: 1.00 m/s and 1.5 m/s would require much lighter gun mass. 0.01 m/s or 0.005 m/s underestimate momentum balance. Common Pitfalls (misconceptions, mistakes): Using 200 g as 200 kg by unit error; adding rather than balancing momenta; ignoring sign convention (recoil direction is opposite). Final Answer: 0.05 m/s

Question 5

System of forces — forces whose lines of action intersect at a single point but lie in different planes are termed:

Accepted Answer

Introduction / Context: Classifying force systems helps select appropriate equilibrium equations. In three dimensions, concurrency (meeting at a point) and coplanarity (lying in one plane) are independent attributes. Recognizing a non-coplanar concurrent system allows simplification to moment sums about the intersection point along with vector force balance. Given Data / Assumptions: All forces intersect at a common point. Their lines of action are not confined to a single plane. Concept / Approach: Concurrent implies a single intersection point; non-coplanar implies three-dimensional arrangement. Such systems are solvable using ΣF = 0 in vector form without needing separate couple-moment equations about arbitrary origins (since moments about the concurrency point from each force are zero). Step-by-Step Solution: Identify concurrency: lines of action intersect at one point.Identify non-coplanarity: cannot fit all lines of action in one plane.Therefore the correct classification is non-coplanar concurrent forces.Apply 3D equilibrium: ΣFx = ΣFy = ΣFz = 0. Verification / Alternative check: Drawing skew lines converging at a point confirms three-dimensional geometry with concurrency. Why Other Options Are Wrong: Coplanar non-concurrent: either coplanar or non-concurrent is violated. Non-coplanar non-concurrent do not meet at one point. “Intersecting forces (same plane)” contradict non-coplanarity. Common Pitfalls: Confusing skew lines with parallel forces; assuming concurrency implies coplanarity. Final Answer: non-coplanar concurrent forces

Question 6

Friction fundamentals — select the true relationship(s) among angle of friction, angle of repose, and coefficient of friction.

Accepted Answer

Introduction / Context: Static friction at impending motion is characterized by an angle whose tangent is the coefficient of friction. For a body on a rough inclined plane, the critical inclination at which sliding starts—angle of repose—is numerically equal to the angle of friction. These relations are foundational in machine design and geotechnical slope stability analogies. Given Data / Assumptions: Coulomb friction model with coefficient μ. Impending motion (limiting equilibrium). Concept / Approach: Define the angle of friction φ by tan φ = μ, arising from resultant reaction making angle φ with the normal at limiting friction. For a block on an incline, impending motion occurs when tan θr = μ, so θr (angle of repose) equals φ. Therefore, tan θr = μ and θr = φ hold together. Step-by-Step Solution: At limiting equilibrium: F = μN; resultant reaction R makes angle φ with normal where tan φ = F/N = μ.On an incline at angle θ: equilibrium gives tan θr = μ at impending motion.Hence θr = φ and tan θr = μ, confirming (a), (b), and (c).Therefore, the inclusive option (d) is correct. Verification / Alternative check: Graphical friction circle or wedge analysis reproduces the same identities at limiting conditions. Why Other Options Are Wrong: “None” contradicts standard Coulomb relations; choosing any single statement would omit other equally true relations. Common Pitfalls: Confusing kinetic and static coefficients; applying relations away from the limiting state; mixing radians and degrees in computations. Final Answer: All the above.

Question 7

Relative velocity of two ships — at an instant, ship A travels 6 km/h due east and ship B travels 8 km/h due north. Find the magnitude of the velocity of B relative to A (in km/h).

Accepted Answer

Introduction / Context: Relative velocity is essential in navigation, collision avoidance, and kinematics. It is found by vector subtraction of velocities. For perpendicular velocity directions (east and north), the relative speed forms the hypotenuse of a right triangle with component differences as legs. Given Data / Assumptions: v_A = 6 km/h east (positive x-axis). v_B = 8 km/h north (positive y-axis). Flat-earth approximation over short duration; instantaneous velocities. Concept / Approach: The velocity of B relative to A is v_B/A = v_B − v_A. Since the velocities are orthogonal, the magnitude is found using the Pythagorean theorem from the component differences. Step-by-Step Solution: Write vectors: v_A = (6, 0), v_B = (0, 8) in km/h.Compute v_B/A = (0 − 6, 8 − 0) = (−6, 8).Find magnitude: |v_B/A| = sqrt( (−6)^2 + 8^2 ) = sqrt(36 + 64) = sqrt(100 ) = 10 km/h.Direction is arctan(8/6) ≈ 53.13° north of west (not required but useful). Verification / Alternative check (if short method exists): Since components are 6 and 8, recall the 6–8–10 Pythagorean triple, confirming 10 km/h immediately. Why Other Options Are Wrong: 7 and 2 km/h arise from incorrect addition or subtraction of magnitudes; 1 km/h is unrealistic; 14 km/h is simple addition ignoring vector nature. Common Pitfalls (misconceptions, mistakes): Adding speeds as scalars; forgetting that relative velocity is a vector difference; mixing units (e.g., km/h with m/s) without conversion. Final Answer: 10 km/h

Question 8

Weight variation with location: If g1 and g2 are the local gravitational accelerations at two mountain locations A and B respectively, by what factor will the weight of a body change when transported from A to B (assume mass remains constant)?

Accepted Answer

Introduction / Context: In engineering mechanics and physics, the weight of a body is the gravitational force acting on its mass. Although the mass of a body is invariant with location, its weight can change slightly from place to place because the local acceleration due to gravity varies with altitude, latitude, and underlying geology. Given Data / Assumptions: Two locations: mountain A with gravitational acceleration g1 and mountain B with gravitational acceleration g2. The same body (constant mass m) is moved from A to B. Air buoyancy and other minor effects are neglected. Concept / Approach: Weight W is defined as W = m * g. For the same mass m at different locations with acceleration g, weight scales directly with g. Therefore, the ratio of the new weight to the old weight equals the ratio of the corresponding gravitational accelerations. Step-by-Step Solution: Let W_A = m * g1 at A.Let W_B = m * g2 at B.The multiplication factor when moving A → B is W_B / W_A = (m * g2) / (m * g1) = g2 / g1. Verification / Alternative check: If g2 = g1 there is no change and g2 / g1 = 1, which is consistent. If g2 is smaller (higher altitude), the factor is less than 1, indicating reduced weight—again consistent with experience. Why Other Options Are Wrong: g1 / g2 reverses the change direction. (g1 + g2) / 2 is an average, not a scaling factor for weight change. (g2 - g1) / g1 gives a fractional change, not the required multiplier. 'Independent of location' contradicts the definition W = m * g. Common Pitfalls: Confusing mass (constant) with weight (varies with g). Final Answer: g2 / g1

Question 9

Statically determinate beam with missing figure: The beam is supported by a hinge at A and a roller at B. What is the reaction R_A at the hinged support A (units: t)?

Accepted Answer

10.6 t

Question 10

Projectile in a low-height horizontal tunnel: A particle is fired inside a tunnel of clear height 554 cm with initial speed 60 m/s. To achieve maximum horizontal range without striking the roof, what should be the angle of projection (degrees from h…

Accepted Answer

Introduction / Context: This problem blends projectile kinematics with a geometric constraint: the projectile must not exceed the tunnel roof height. The usual 45° angle that maximizes range in open space is impossible here because the peak height would be far above the tunnel ceiling. Therefore, we optimize the range subject to a maximum-height constraint. Given Data / Assumptions: Tunnel height H = 554 cm = 5.54 m. Initial speed u = 60 m/s. Standard gravity g = 9.81 m/s^2. No air resistance; ground and roof are horizontal. Concept / Approach: The trajectory peak height for a projectile at angle theta is H_max = u^2 * sin^2(theta) / (2 * g). To avoid striking the roof, we require H_max <= H. For maximum allowed range under this constraint, the optimal angle is the largest theta that still satisfies the height limit (otherwise we could slightly increase theta and gain more range without hitting the roof). Step-by-Step Solution: Apply height constraint: u^2 * sin^2(theta) / (2 * g) = H (at the limiting optimal angle).Solve for sin(theta): sin(theta) = sqrt(2 * g * H) / u.Compute: 2 * g * H = 2 * 9.81 * 5.54 = 108.7 (approx).Take the square root: sqrt(108.7) ≈ 10.43.Compute sin(theta) = 10.43 / 60 ≈ 0.174 ⇒ theta ≈ 10.0°. Verification / Alternative check: Open-terrain maximum range at 45° would have H_max = u^2 / (4 * g) = 3600 / 39.24 ≈ 91.8 m >> 5.54 m, confirming the roof is the active constraint. With theta ≈ 10°, the projectile just clears the roof at the apex and thus gives the largest possible range within the tunnel. Why Other Options Are Wrong: 8°–9° are feasible but give a lower range than the constraint-limiting angle. 11°–12° would exceed the allowed peak height and strike the roof. Common Pitfalls: Maximizing range using 45° without checking height restrictions. Using degrees vs. radians inconsistently in calculator steps. Final Answer: 10°

Question 11

Kinematics on a line: For a particle with position x(t) = t^2 * (t + 1) = t^3 + t^2, what is the instantaneous acceleration a(t)?

Accepted Answer

Introduction / Context: This is a direct application of single-variable calculus to particle motion. Position x(t) is given; velocity is the first derivative dx/dt, and acceleration is the time derivative of velocity, d^2x/dt^2. Such questions are common in basic dynamics and control of motion along a line. Given Data / Assumptions: Position: x(t) = t^3 + t^2 (units consistent but unspecified). Differentiation rules apply; motion is along a straight line. Concept / Approach: Velocity v(t) = dx/dt; Acceleration a(t) = dv/dt. Differentiate term-by-term using power rules. Step-by-Step Solution: Compute velocity: v(t) = d/dt(t^3 + t^2) = 3t^2 + 2t.Compute acceleration: a(t) = d/dt(3t^2 + 2t) = 6t + 2.Therefore, a(t) = 6t + 2. Verification / Alternative check: As a quick check, units are consistent: if x is in m and t in s, then v is m/s and a is m/s^2; nothing contradictory arises from the differentiation. Why Other Options Are Wrong: 3t^2 + 2t is the velocity, not the acceleration. 6t - 2 and 3t - 2 alter constant terms incorrectly. 3t^3 - 2t is unrelated to correct differentiation steps. Common Pitfalls: Differentiating once and mistaking v(t) for a(t). Final Answer: 6t + 2

Question 12

Newton’s law of impact: Two bodies move in the same direction with approach velocities u1 and u2; after a 1D elastic collision their separation velocities are v1 and v2. What is the coefficient of restitution e?

Accepted Answer

Introduction / Context: In one-dimensional collisions along the line of impact, Newton’s law of restitution connects the relative speeds before and after impact. The coefficient of restitution e ranges from 0 (perfectly plastic) to 1 (perfectly elastic) and quantifies how ‘‘bouncy’’ the collision is. Given Data / Assumptions: Motion is collinear (same straight line). u1 and u2 are pre-impact velocities of approach (same direction so u1 > u2 for approach). v1 and v2 are post-impact velocities of separation (v2 > v1 if they separate). Concept / Approach: Definition: e = (relative speed of separation) / (relative speed of approach). With our sign convention (same direction along a line), the relative approach speed is u1 - u2 and the relative separation speed is v2 - v1. Step-by-Step Solution: Write the definition: e = (relative separation) / (relative approach).For same-direction motion: relative approach = u1 - u2.Relative separation = v2 - v1.Therefore, e = (v2 - v1) / (u1 - u2). Verification / Alternative check: When e = 1, v2 - v1 = u1 - u2 (perfectly elastic). When e = 0, v2 = v1 (no rebound), matching physical intuition. Why Other Options Are Wrong: Swapping sums for differences or reversing numerator/denominator breaks the standard definition. Common Pitfalls: Sign confusion: always use relative speeds (magnitudes along the line), not algebraic velocities with arbitrary signs for this simple formula. Final Answer: e = (v2 - v1) / (u1 - u2)

Question 13

Uniform rod with a suspended load—equilibrium moment (figure unspecified): A 9 m uniform rod (weight 40 kg) is pivoted at a point 2 m from one end; a 120 kg load is suspended as described. What perpendicular force at the far end keeps the rod in equ…

Accepted Answer

10 kg

Question 14

Centroid of a thin hollow cone: For a thin conical surface of height h, the center of gravity measured along the axis above the base lies at what distance?

Accepted Answer

Introduction / Context: Locating centroids is essential for computing moments, stability, and support reactions. A thin hollow cone (a conical shell) has mass concentrated on its surface, not its volume, so its centroid differs from that of a solid cone. Given Data / Assumptions: Geometric shape: thin conical surface of height h and negligible thickness. Uniform surface density. Axis of symmetry is vertical; base is a circle. Concept / Approach: For a solid right circular cone, the centroid lies at h/4 from the base along the axis. For a thin conical shell, the centroid shifts upward because the mass is distributed near the surface farther from the axis compared to volume distribution. The standard result for a thin hollow cone is h/3 above the base along the axis. Step-by-Step Solution: Model the shell as a stack of small circular rings along the slant height.Use symmetry to place the centroid on the axis.Integrate ring contributions (proportional to radius) along height to obtain the centroid position y = h / 3 above the base. Verification / Alternative check: The value h/3 lies above h/4 (solid cone), consistent with intuition that a shell’s mass is farther from the base centroid than a solid’s mass average. Why Other Options Are Wrong: h/4 is for a solid cone, not a shell. h/2, 2h/3, 3h/4 place the centroid unrealistically high for this geometry. Common Pitfalls: Confusing solid and hollow cone centroid formulas. Final Answer: h / 3

Question 15

Linear momentum: The product of mass and velocity of a moving body is termed what in mechanics?

Accepted Answer

Introduction / Context: Basic mechanics distinguishes between several related but different quantities: momentum, impulse, work, power, and moment. Correct terminology is crucial for applying conservation laws and solving collision and dynamics problems. Given Data / Assumptions: Mass m and velocity v describe translational motion. Concept / Approach: Linear momentum p is defined as p = m * v. It is a vector quantity pointing in the direction of motion. Momentum conservation is a cornerstone principle for analyzing isolated systems and impacts. Step-by-Step Solution: Identify the definition: momentum = mass * velocity.Therefore, the correct term for m * v is momentum. Verification / Alternative check: Impulse is the time integral of force and equals the change in momentum; moment relates to rotational effects (force * perpendicular distance); power is the rate of doing work. None of these equal m * v. Why Other Options Are Wrong: Moment pertains to rotation, not translation. Impulse is change of momentum, not momentum itself. Power is work rate; energy is capacity to do work. Common Pitfalls: Using 'moment' when you really mean 'momentum' in translational problems. Final Answer: momentum

Question 16

Pendulum clock at altitude (figure inconsistent): A beat-seconds pendulum that keeps correct time at sea level loses 27 s per day at the top of a mountain. Using the given Earth radius (stated as 600 km), estimate the mountain height above sea level.

Accepted Answer

2000 metres

Question 17

Screw jack performance: What is the maximum theoretical efficiency of a screw jack expressed in terms of the angle of friction phi (for a square-thread jack in raising mode)?

Accepted Answer

Introduction / Context: Screw jacks convert torque into axial lifting force via a power screw. Efficiency depends on thread lead angle and friction. There exists a condition that maximizes efficiency for a given angle of friction, leading to a compact expression for the maximum value achievable (ignoring collar friction and other losses). Given Data / Assumptions: Square-thread jack, raising a load slowly (quasi-static). Angle of friction = phi, with mu = tan phi. Collar friction and other parasitic losses neglected for the formula. Concept / Approach: For raising, efficiency eta can be written in terms of thread lead angle alpha and friction as eta = tan alpha / (tan alpha + mu * sec alpha). Maximizing eta with respect to alpha for a given phi yields a closed-form maximum depending only on phi. Step-by-Step Solution: Start from eta(alpha) = tan alpha / (tan alpha + mu * sec alpha), with mu = tan phi.Differentiate eta with respect to alpha and set derivative to zero to find optimal alpha.Solve the resulting condition to obtain the compact result: eta_max = (1 - sin phi) / (1 + sin phi). Verification / Alternative check: Check limits: as phi → 0 (no friction), sin phi → 0, so eta_max → 1 (100%). As phi increases, eta_max decreases monotonically, which matches physical intuition. Why Other Options Are Wrong: tan- and cos-based forms listed do not emerge from maximizing the screw jack efficiency expression. cos^2 phi does not match known limiting behavior as phi → 0. 1 - tan phi can be negative for moderate phi and is dimensionally inconsistent as an efficiency formula. Common Pitfalls: Confusing the general efficiency expression (depends on alpha) with the maximum efficiency (depends only on phi). Final Answer: eta_max = (1 - sin phi) / (1 + sin phi)

Question 18

Satellite motion fundamentals: Which of the following statements about gravity and satellite launch/flight is correct in the context of basic orbital mechanics and gravitational intensity with height?

Accepted Answer

Introduction / Context: This conceptual question checks your understanding of how gravitational acceleration varies with altitude and what that implies for satellites. In orbital mechanics, satellites remain in free-fall around Earth; the magnitude of gravitational acceleration g decreases with altitude according to an inverse-square law. Recognizing correct physical statements and filtering out incorrect units or exaggerations is essential for basic space engineering literacy. Given Data / Assumptions: We are discussing Earth's gravitational field and satellites at altitudes above the surface. Standard physics: g decreases with distance from Earth's center following g = GM / r^2. No drag or thrust is considered in the simple statements. Concept / Approach: Gravitational intensity is not constant; it decreases with height. Therefore, any statement that says "g declines with height" is correct. By contrast, citing a fixed decrease in satellite speed of "32 km per second" is dimensionally and physically incorrect, because 32 km/s is an enormous speed (orbital speeds are typically around 7–11 km/s). We must identify the single precise statement consistent with physics. Step-by-Step Solution: Recall the law: g ∝ 1 / r^2, where r is the distance from Earth's center.As altitude increases (r increases), g decreases—this is the definition of declining gravitational intensity.Select the option that explicitly states this: gravitational intensity declines with height. Verification / Alternative check: At low Earth orbit (a few hundred kilometres), g is still close to 9.8 m/s^2 but measurably lower; weightlessness is due to free-fall, not zero gravity. This verifies the direction of change stated in the correct option. Why Other Options Are Wrong: “Nature plays an important role …” is vague and not a technical statement. “Earth's gravity reduces speed by 32 km/s” is nonsensical in units and magnitude. “Gravitational force relents as the satellite climbs higher” is informal wording; the precise and correct technical statement is that g declines with height. “All the above” fails due to incorrect items included. Common Pitfalls: Confusing weightlessness with zero gravity; g remains significant in orbit. Accepting statements with incorrect units or unrealistic magnitudes. Final Answer: The gravitational intensity (value of g) declines with increasing height above the Earth

Question 19

Centroid of a solid right circular cone: Along its axis of symmetry, at what height above the base does the centre of gravity (C.G.) of a solid cone of height h lie?

Accepted Answer

Introduction / Context: Locating the centre of gravity (centroid) of common 3D bodies is a staple in engineering mechanics because it directly affects support reactions, stability, and dynamic response. A right circular cone is a standard test case that distinguishes between solid bodies and thin shells. Given Data / Assumptions: Solid right circular cone of height h and uniform density. We measure the centroid position along the axis of symmetry from the base plane. No cavities or truncations; this is a full cone, not a frustum. Concept / Approach: For a solid cone, the centroid lies at a quarter of the height from the base along the axis. This result can be derived by integrating contributions of thin circular disks stacked along the height, whose areas scale with the square of radius (and hence linearly with height for a right cone). Step-by-Step Solution: Model the cone as differential disks at height y with radius proportional to y.Set up the centroid integral y_c = (∫y dA) / (∫dA) for the volume elements (or use standard solid formula).Perform the integration to obtain y_c = h/4 above the base. Verification / Alternative check: Contrast with a thin hollow cone (conical shell): its centroid lies at h/3 above the base, which is higher because mass is concentrated near the surface rather than the interior. This cross-check prevents mixing formulas. Why Other Options Are Wrong: h/2 places the centroid at mid-height—incorrect for a cone narrowing toward the apex. h/3 is for a hollow (thin) conical shell, not a solid cone. h/5 and h/6 are too close to the base and do not match the derived result. Common Pitfalls: Confusing solid vs. hollow cone centroid locations (h/4 vs. h/3). Measuring from the wrong reference (apex vs. base). Final Answer: h/4

Question 20

Units in mechanics: The joule (J) is the SI unit of which physical quantity in engineering mechanics?

Accepted Answer

Introduction / Context: Clear understanding of SI units prevents dimensional mistakes in design and analysis. The joule (J) frequently appears in energy, work, and heat calculations across mechanical, civil, and electrical engineering applications. Given Data / Assumptions: We use SI (International System of Units). Typical base units: newton (N) for force, second (s) for time, metre (m) for length. Concept / Approach: Work is defined as the line integral of force along a displacement, with units newton-metre (N·m). In SI, 1 joule is exactly 1 N·m. While torque also has units N·m, in SI the joule is reserved for work/energy, and torque is expressed as N·m without being called a joule to avoid confusion. Step-by-Step Solution: Recall definitions: work W = F * s for constant force parallel to motion.Unit of force is newton, unit of distance is metre → unit of work is N·m = joule.Therefore, the joule corresponds to work (and energy). Verification / Alternative check: Energy forms (kinetic, potential, electrical) are all measured in joules, reinforcing the identification. Why Other Options Are Wrong: Force uses the newton (N), not joule. Power uses the watt (W) = J/s. Torque is quoted in N·m, but not named as joule in SI usage. “None of these” is incorrect because joule clearly denotes work/energy. Common Pitfalls: Calling torque “joules”; keep the naming distinction to avoid ambiguity. Final Answer: work

Applied Mechanics Questions