Question 1

Projectile motion — definition check: time of flight of a projectile  The total time taken by a projectile to rise to its maximum height and then return to the ground is called the time of flight.  Choose True or False.

Accepted Answer

Concept The time of flight of a projectile is the total time from launch until it lands back at the original vertical level (ground). It includes the upward journey to maximum height and the downward return. Standard formula (level ground) Time of flight, T = &frac{2u \sin\alpha}{g}Time to reach maximum height, Tup = &frac{u \sin\alpha}{g}Time to fall back, Tdown = &frac{u \sin\alpha}{g}Thus, T = Tup + Tdown. Common pitfall Do not confuse time of flight with range (horizontal distance) or with time to reach maximum height (which is half of the time of flight on level ground). Final Answer True

Question 2

Compound pendulum — equivalent simple pendulum length  Find the length of an equivalent simple pendulum that would have the same frequency (same time period) as the given compound pendulum.  Choose the correct formula in terms of radius of gyration …

Accepted Answer

Given/Notation Compound pendulum: mass m, radius of gyration about C.G. = k Distance between pivot and C.G. = h Equivalent simple pendulum length = Le Concept/Approach Equate the time period of a compound pendulum to that of a simple pendulum. Time periods Compound: T = 2\pi \sqrt{\dfrac{I_A}{m g h}}, \; 	ext{where } I_A = I_G + m h^2 = m(k^2 + h^2)Simple: T = 2\pi \sqrt{\dfrac{L_e}{g}} Step-by-Step Set equal: \; 2\pi \sqrt{\dfrac{m(k^2 + h^2)}{m g h}} = 2\pi \sqrt{\dfrac{L_e}{g}} \Rightarrow \dfrac{k^2 + h^2}{h} = L_e \Rightarrow L_e = h + \dfrac{k^2}{h} Common pitfall Forgetting to use the parallel axis theorem: IA = IG + m h^2. Final Answer k^2/h + h

Question 3

Units and dimensions — SI unit of angular acceleration  Identify the correct SI unit for angular acceleration (rate of change of angular velocity).  Choose the correct option.

Accepted Answer

Concept Angular acceleration α = dω/dt. Angular velocity ω is in rad/s, so α is in rad/s2. Why the others are wrong N·m is torque (unit of moment). m/s is linear velocity. m/s2 is linear acceleration. Final Answer rad/s2

Question 4

Work and energy — when is work done equal to zero?  Assess the given conditions and determine under which situation(s) the mechanical work done is zero.  Choose the correct option.

Accepted Answer

Concept Work done W = 𝐹 ⃗ ⋅ 𝑠 ⃗ = 𝐹 𝑠 cos ⁡ 𝜃 F ⋅ s =Fscosθ. Work is zero if either the force is zero or the displacement is zero (or if the force is perpendicular to the displacement, θ = 90°). Assessment of options (a) s = 0 → W = 0. (b) F = 0 → W = 0. Therefore, either (a) or (b) implies zero work. Final Answer Either (a) or (b)

Question 5

Impact and collisions — elastic impact and relative velocity  For a perfectly elastic impact, relate the relative velocity after impact to the relative velocity before impact.  Choose the correct statement.

Accepted Answer

Concept Coefficient of restitution, e = 1 for a perfectly elastic impact. Key relation Relative velocity of separation = e × Relative velocity of approachFor e = 1, magnitude after impact equals magnitude before impact, but the direction reverses. Final Answer equal and opposite to

Question 6

Area moments — centroidal moment of inertia of a triangular section  Find the second moment of area (area M.I.) of a triangle of base b and height h about an axis through its C.G. and parallel to the base.  Choose the correct expression.

Accepted Answer

Concept The centroidal area moment of inertia of a triangle about a horizontal axis through the centroid (parallel to the base) is a standard result. Result IG, ∥base = $\dfrac{b h^3}{36}$ Why others are incorrect Values such as $bh^3/12$ or $bh^3/8$ correspond to different shapes/axes (e.g., rectangles) and are not applicable to a triangle about this specific centroidal axis. Final Answer bh3/36

Question 7

Projectile on an upward inclined plane — time of flight expression  Given: initial speed u, angle of projection α with the horizontal, and an upward plane inclination β with the horizontal.  Select the correct formula for the time of flight (until t…

Accepted Answer

Setup Inclined plane makes angle β with the horizontal. Projectile launched at speed u and angle α with the horizontal. Axes chosen: tangential to plane (‖) and normal to plane (⊥). Components Initial velocity along the plane: u∥ = u cos(α − β)Initial velocity normal to the plane: u⊥ = u sin(α − β)Acceleration along the plane: a∥ = g sin β (down the plane)Acceleration normal to the plane: a⊥ = g cos β (toward the plane) Step-by-Step (normal motion) Normal displacement y⊥(t) = u sin(α − β) t − &frac{1}{2} g cosβ \, t^2 Impact with the plane occurs when y⊥(t) returns to zero (excluding t = 0): u sin(α − β) t − &frac{1}{2} g cosβ \, t^2 = 0 \Rightarrow t = \dfrac{2u \sin(α − β)}{g \cosβ} Common pitfalls Using g instead of g cosβ for the normal component. Confusing (α − β) with (α + β); the effective launch angle relative to the plane is (α − β). Final Answer (2u sin(α − β)) / (g cos β)

Question 8

Projectile motion on inclined planes: Compare the range on a downward slope versus an upward slope for identical launch parameters.  Conditions: Same initial speed (u) and same launch angle (α) with respect to the horizontal; magnitude of plane incl…

Accepted Answer

Given data Same initial speed u and launch angle α for both cases. Same magnitude of plane inclination |β| (one case upward, the other downward). Concept/Approach For a projectile on an inclined plane, the range along the plane is R(β) = (2u2 cosα · sin(α − β)) / (g cos2β) where β is taken positive for an upward slope and negative for a downward slope. Step-by-step reasoning Upward plane: Rup ∝ sin(α − β). Downward plane: Rdown ∝ sin(α − (−β)) = sin(α + β). For 0 < β < α < 90°, we have sin(α + β) > sin(α − β). ⇒ Rdown > Rup. Common pitfall Do not compare horizontal ranges; the comparison is for range measured along the plane. Final Answer Greater than

Question 9

Vehicle turning on a circular path: Dependence of the banking/lean angle on mass.  Context: A vehicle moves along a horizontal circular path of radius R at speed v (steady turn).  Choose the correct statement regarding how the angle of inclination d…

Accepted Answer

Given data Vehicle negotiates a curve of radius R at speed v. Concept/Approach For a vehicle that leans (or a banked road), the required lean/bank angle θ satisfies tanθ = v2 / (gR) which does not include the vehicle mass m. Step-by-step reasoning Centripetal requirement: Resultant of weight and normal provides v2/R. From force balance, tanθ = (centripetal acceleration)/g = v2/(gR). No mass term appears ⇒ θ is independent of mass. Common pitfall Confusing tyre–road friction limits (which depend on normal reaction) with the geometric relation for the required angle, which is mass-independent. Final Answer Independent of mass

Question 10

Impending motion on an incline: Compare the plane’s inclination with the angle of friction for motion to start.  Condition: A block rests on a rough plane; angle of friction = φ; plane inclination = θ.  Choose when the body will begin to move down t…

Accepted Answer

Given data Angle of friction φ where tanφ = μ (μ = coefficient of friction). Plane inclination θ. Concept/Approach Limiting equilibrium down the plane occurs when tanθ = μ ⇒ θ = φ. Beyond this, friction cannot balance the component of weight down the plane and motion starts. Step-by-step reasoning If θ < φ ⇒ tanθ < μ ⇒ at rest. If θ = φ ⇒ limiting equilibrium (on the verge of motion). If θ > φ ⇒ sliding commences. Common pitfall Saying motion begins at θ = φ — that is only the brink. Actual motion requires θ > φ (assuming a nudge to overcome stiction). Final Answer When θ is greater than φ

Question 11

Momentum comparison at equal kinetic energy: Two masses m1 and m2 (m1 > m2).  Condition: Kinetic energies of both bodies are equal.  Choose the correct comparison of their momenta.

Accepted Answer

Given data m1 > m2 and K1 = K2. Concept/Approach For a body: K = p2 / (2m) ⇒ p = √(2mK). Step-by-step calculation p1 = √(2 m1 K), p2 = √(2 m2 K). Since m1 > m2 and K is common, √(m1) > √(m2). Therefore p1 > p2. Common pitfall Assuming equal KE means equal momentum — only true if masses are equal. Final Answer Momentum of m1 is greater than momentum of m2

Question 12

Basic definition check: Coefficient of friction equals limiting friction divided by normal reaction.  Assess the statement's correctness for dry (Coulomb) friction.

Accepted Answer

Concept/Approach The (static) coefficient of friction μs is defined at impending motion: μs = Flimiting / N. The kinetic coefficient μk is similarly defined with kinetic friction: μk = Fk / N. Conclusion The statement matches the standard definition of coefficient of friction (limiting friction/normal reaction for static). Hence it is True. Common pitfall Confusing the definition (a ratio) with a particular measured value; μ is material-pair dependent and dimensionless.

Question 13

Area properties: Polar moment of inertia of a circular section about an axis perpendicular to the plane through its centroid.  State the correct expression in terms of radius r or diameter d. Choose the correct option.

Accepted Answer

Given/Definitions For a circular area of radius r, the centroidal second moments are Ix = Iy = (π r4)/4. The polar moment about the perpendicular centroidal axis is J = Ix + Iy. Step-by-step calculation J = Ix + Iy = (π r4)/4 + (π r4)/4 = (π r4)/2. In terms of diameter d (r = d/2): J = (π (d/2)4)/2 = (π d4)/32. Verification Option (a) matches the r–form; the equivalent d–form is option (d). Final Answer J = (π r^4) / 2 (equivalently, (π d^4)/32).

Question 14

Collision fundamentals: Value of the coefficient of restitution for perfectly elastic bodies.  Interpret the statement: “The coefficient of restitution for elastic bodies is one.” Choose the correct truth value.

Accepted Answer

Concept/Approach Coefficient of restitution e is defined as the ratio of the relative speed of separation to the relative speed of approach along the line of impact. For a perfectly elastic impact, kinetic energy is conserved and e = 1. Common pitfall Assuming e depends on masses; it is a property of the collision pair and impact conditions. Final Answer True

Question 15

Resultant of concurrent forces: Direction angle θ of the resultant with the horizontal. Context: Several forces act at a point with components ΣF_x (horizontal) and ΣF_y (vertical). Choose the correct relation for the inclinati…

Accepted Answer

Given/Definitions Resultant vector R has components Rx = ΣFx, Ry = ΣFy. Derivation tanθ = (opposite)/(adjacent) = Ry / Rx = ΣFy / ΣFx. Verification/Notes When ΣFx = 0, the resultant is vertical (θ = 90°). Sign conventions determine quadrant but not the basic ratio. Final Answer tanθ = ΣFy / ΣFx

Question 16

Concept check: Definition of projectile range (horizontal range)  Statement to evaluate for truth: "The distance between the point of projection and the point where the projectile strikes the ground is known as the range of the projectile." Choose w…

Accepted Answer

Given data / statement The item defines the range of a projectile as the horizontal distance from launch point to the point where it lands. Concept In ideal projectile motion (uniform gravity, no air resistance, level landing), the range is the horizontal displacement between the point of projection and the point of impact. Clarification Standard notation: If initial speed is u and projection angle is θ, then on level ground the range is R = (u^2 sin 2θ)/g, which is the horizontal distance between start and landing. Common pitfall Do not confuse range with the maximum height (vertical displacement) or the time of flight (duration in the air). Final Answer Correct

Question 17

Dynamics: Tangential acceleration in uniform circular motion  Statement to evaluate for truth: "When a particle moves along a circular path with uniform velocity (constant speed), there will be no tangential acceleration." Choose whether the stateme…

Accepted Answer

Concept Acceleration in circular motion has two perpendicular components: tangential (changes speed) and normal/centripetal (changes direction). Reasoning For uniform circular motion, the speed is constant. Therefore the rate of change of speed is zero and the tangential acceleration component is zero. Only centripetal acceleration remains: an = v^2 / r = ω^2 r. Common pitfall Zero tangential acceleration does not imply zero total acceleration; direction still changes, so centripetal acceleration is nonzero. Final Answer Correct

Question 18

Projectile motion: Shape of trajectory under uniform gravity  Statement to evaluate: "The path (trajectory) of an ideal projectile is a parabola." Choose the correct truth value.

Accepted Answer

Concept With constant gravitational acceleration and no air resistance, horizontal motion has constant velocity while vertical motion has constant acceleration. Derivation sketch Let initial speed be u and angle θ. Then: x = (u cosθ) t y = (u sinθ) t − (1/2) g t^2 Eliminate t = x/(u cosθ) to get y = x tanθ − (g x^2)/(2 u^2 cos^2θ) which is a quadratic in x — i.e., a parabola. Final Answer True

Question 19

Mass moment of inertia of a uniform thin rod about its midpoint  Find the mass moment of inertia of a uniform thin rod of mass M and length l about an axis through its mid-point and perpendicular to its length. Choose the correct expression (use SI …

Accepted Answer

Given Uniform thin rod, length l, mass M, axis through centroid (midpoint) and perpendicular to the rod. Formula / Concept The standard mass moment of inertia of a slender rod about a centroidal axis normal to its length is: IG = (1/12) M l^2 Derivation sketch From the integral definition: I = ∫ r^2 dm Let the rod lie along x-axis from −l/2 to +l/2 with uniform linear density λ = M/l and dm = λ dx. Then I = ∫−l/2+l/2 x^2 λ dx = λ [x^3 / 3]−l/2+l/2 = (M/l) (2 (l/2)^3 / 3) = (1/12) M l^2. Units kg·m². Final Answer (1/12) M l^2

Question 20

Power units: Definition of the watt in joules per second  One watt (W) is equal to how many joules per second (J/s)? Choose the correct equivalence.

Accepted Answer

Concept Power is the rate of doing work or energy transfer per unit time. Definition 1 watt (W) = 1 joule per second (J/s): P = dW/dt. Final Answer 1 joule/s

Engineering Mechanics Questions