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Question 1Engineering Mathematics

The eigenvalues of the matrix [[4, 1], [2, 3]] are

  1. 5 and 2 Correct answer.
  2. 4 and 3
  3. −5 and −2
  4. 1 and 6

Explanation

For a 2×2 matrix the eigenvalues satisfy trace = sum = 7 and determinant = product = 4·3 − 1·2 = 10. The pair whose sum is 7 and product is 10 is 5 and 2, which also satisfies the characteristic equation λ² − 7λ + 10 = 0.

Question 2Engineering Mathematics

If f(x) = x³ − 3x, the value of x in [−2, 2] at which f has a local maximum is

  1. x = 2
  2. x = 1
  3. x = 0
  4. x = −1 Correct answer.

Explanation

Setting f′(x) = 3x² − 3 = 0 gives x = ±1. The second derivative 6x is negative at x = −1, so that stationary point is a local maximum, while x = 1 gives a local minimum.

Question 3Engineering Mathematics

The value of the double integral of (x + y) over the unit square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 is

  1. 1 Correct answer.
  2. 3/2
  3. 2
  4. 1/2

Explanation

The integral separates as ∫₀¹x dx·∫₀¹dy + ∫₀¹dx·∫₀¹y dy = (1/2)(1) + (1)(1/2) = 1. By symmetry each variable contributes 1/2.

Question 4Engineering Mathematics

The divergence of the vector field F = x²i + 2xy j + z k at the point (1, 1, 1) is

  1. 3
  2. 5 Correct answer.
  3. 6
  4. 4

Explanation

Divergence is ∂(x²)/∂x + ∂(2xy)/∂y + ∂(z)/∂z = 2x + 2x + 1 = 4x + 1. At x = 1 this equals 5.

Question 5Engineering Mathematics

The order and degree of the differential equation (d²y/dx²)³ + (dy/dx)⁴ + y = 0 are respectively

  1. 4 and 3
  2. 3 and 2
  3. 2 and 4
  4. 2 and 3 Correct answer.

Explanation

Order is the highest derivative present, which is the second derivative, so order = 2. Degree is the power of the highest-order derivative after the equation is polynomial in derivatives, giving degree = 3.

Question 6Engineering Mathematics

The Laplace transform of sin(3t) is

  1. 3/(s² − 9)
  2. 1/(s² + 9)
  3. s/(s² + 9)
  4. 3/(s² + 9) Correct answer.

Explanation

The standard result is L{sin(at)} = a/(s² + a²). With a = 3 this gives 3/(s² + 9).

Question 7Engineering Mathematics

For the matrix A = [[2, 0], [0, 3]], the value of A³ is

  1. [[2, 0], [0, 27]]
  2. [[8, 0], [0, 9]]
  3. [[8, 0], [0, 27]] Correct answer.
  4. [[6, 0], [0, 9]]

Explanation

A diagonal matrix is raised to a power by raising each diagonal entry to that power. Hence A³ = diag(2³, 3³) = diag(8, 27).

Question 8Engineering Mathematics

The function f(z) = z² is analytic everywhere, and its derivative at z = 1 + i is

  1. 1 + i
  2. 2 + 2i Correct answer.
  3. 2i
  4. 2

Explanation

For f(z) = z² the derivative is f′(z) = 2z. Substituting z = 1 + i gives 2(1 + i) = 2 + 2i.

Question 9Engineering Mathematics

The rank of the matrix [[1, 2, 3], [2, 4, 6], [1, 1, 1]] is

  1. 0
  2. 2 Correct answer.
  3. 1
  4. 3

Explanation

The second row is twice the first, so it adds no independent information. The remaining rows [[1,2,3]] and [[1,1,1]] are not multiples of each other, giving two independent rows and hence rank 2.

Question 10Engineering Mathematics

A fair six-sided die is rolled twice. Given that the sum of the two outcomes is 8, the probability that the first roll showed a 3 is

  1. 2/5
  2. 1/36
  3. 1/5 Correct answer.
  4. 1/6

Explanation

The sums equal to 8 arise from the five pairs (2,6), (3,5), (4,4), (5,3), (6,2). Exactly one of these has first roll 3, so the conditional probability is 1/5.

Question 11Engineering Mathematics

The series sum from n = 1 to infinity of 1/n^p converges if and only if

  1. p > 0
  2. p > 1 Correct answer.
  3. p < 1
  4. p ≥ 1

Explanation

This is the p-series, which converges exactly when the exponent p exceeds 1 by the integral test. At p = 1 the harmonic series diverges logarithmically.

Question 12Engineering Mathematics

The directional derivative of f(x, y) = x²y at the point (1, 2) in the direction of the unit vector (3/5, 4/5) is

  1. 16/5 Correct answer.
  2. 4
  3. 12/5
  4. 28/5

Explanation

The gradient is (2xy, x²), which at (1,2) equals (4, 1). Dotting with (3/5, 4/5) gives (4·3 + 1·4)/5 = 16/5.

Question 13Engineering Mathematics

The general solution of dy/dx = 2xy is

  1. y = C e^(2x)
  2. y = C x²
  3. y = C e^(x²) Correct answer.
  4. y = C + x²

Explanation

Separating variables gives dy/y = 2x dx, and integrating yields ln y = x² + C, so y = C e^(x²).

Question 14Engineering Mathematics

The value of the integral of e^(−x) from 0 to infinity is

  1. infinity
  2. 1 Correct answer.
  3. 0
  4. e

Explanation

The antiderivative is −e^(−x). Evaluating from 0 to infinity gives 0 − (−1) = 1, so the improper integral converges to 1.

Question 15Engineering Mathematics

If the characteristic equation of a 2×2 matrix A is λ² − 5λ + 6 = 0, then by the Cayley–Hamilton theorem A² equals

  1. 6A − 5I
  2. 6I − 5A
  3. 5A − 6I Correct answer.
  4. 5A + 6I

Explanation

Cayley–Hamilton states that A satisfies its own characteristic equation, so A² − 5A + 6I = 0. Rearranging gives A² = 5A − 6I.

Question 16Engineering Mathematics

The Fourier series of an odd function defined on [−π, π] contains only

  1. a constant term plus cosine terms
  2. sine terms Correct answer.
  3. cosine terms
  4. both sine and cosine terms

Explanation

For an odd function the cosine coefficients vanish because the product of an odd and an even function integrates to zero over a symmetric interval. Only the sine coefficients, which pair odd with odd, survive.

Question 17Engineering Mathematics

For the matrix A = [[1, 2], [2, 1]], the eigenvectors corresponding to the eigenvalues 3 and −1 are respectively proportional to

  1. (2, 1) and (1, 2)
  2. (1, 1) and (1, −1) Correct answer.
  3. (1, −1) and (1, 1)
  4. (1, 0) and (0, 1)

Explanation

For λ = 3 the equation (A − 3I)x = 0 gives −2x + 2y = 0, so x = y and the vector is (1,1). For λ = −1 it gives 2x + 2y = 0, so y = −x and the vector is (1,−1).

Question 18Engineering Mathematics

Using Taylor's series about x = 0, the value of sin(0.1) correct to three decimal places is

  1. 0.100 Correct answer.
  2. 0.105
  3. 0.099
  4. 0.110

Explanation

The expansion is sin x = x − x³/6 + …, so sin(0.1) ≈ 0.1 − 0.001/6 = 0.1 − 0.000167 = 0.099833, which rounds to 0.100.

Question 19Engineering Mathematics

The function u = x² − y² is harmonic, and its harmonic conjugate v(x, y) is

  1. xy + C
  2. −2xy + C
  3. 2xy + C Correct answer.
  4. x² + y² + C

Explanation

Using the Cauchy–Riemann conditions, ∂v/∂y = ∂u/∂x = 2x gives v = 2xy + g(x), and ∂v/∂x = 2y must equal −∂u/∂y = 2y, so g is constant. Thus v = 2xy + C.

Question 20Engineering Mathematics

For the matrix A = [[0, 1], [−1, 0]], the value of e^(At) at t = π is

  1. A
  2. I
  3. −A
  4. −I Correct answer.

Explanation

A represents rotation by 90°, so e^(At) = cos t · I + sin t · A. At t = π this gives cos π · I + sin π · A = −I.

Question 21Engineering Mathematics

The value of the contour integral of 1/z around the circle |z| = 2, traversed counterclockwise, is

  1. 4πi
  2. 2πi Correct answer.
  3. 0
  4. πi

Explanation

The integrand has a simple pole at z = 0, which lies inside |z| = 2. Cauchy's integral formula gives the value 2πi times the residue 1, so the integral is 2πi.

Question 22Engineering Mathematics

The system of equations x + y + z = 3, 2x + 2y + 2z = 6, x − y = 0 has

  1. a unique solution
  2. exactly two solutions
  3. infinitely many solutions Correct answer.
  4. no solution

Explanation

The second equation is twice the first, so it is redundant, leaving two independent equations in three unknowns. With fewer independent equations than unknowns the system is consistent and has infinitely many solutions.

Question 23Engineering Mathematics

The maximum value of f(x, y) = xy subject to x + y = 4 is

  1. 2
  2. 4 Correct answer.
  3. 16
  4. 8

Explanation

Using Lagrange multipliers, y = λ and x = λ give x = y, and the constraint then yields x = y = 2. The product is 4, which is maximal because the boundary values tend to zero.

Question 24Engineering Mathematics

The eigenvalues of a real symmetric 3×3 matrix are 2, 2, and 5. The matrix is

  1. not diagonalizable
  2. invertible but not diagonalizable
  3. diagonalizable but not invertible Correct answer.
  4. diagonalizable and invertible

Explanation

Real symmetric matrices are always diagonalizable by an orthogonal matrix, so the repeated eigenvalue is not an obstacle. However, one eigenvalue is 0 only if it appears among the listed values; here none is zero, so the determinant 2·2·5 = 20 is nonzero and the matrix is invertible.

Question 25Engineering Mathematics

The function f(z) = (z² − 1)/(z² + 1) has poles at z = ±i, and the sum of the residues at these two poles is

  1. −1
  2. 0 Correct answer.
  3. 2i
  4. 1

Explanation

The residue at z = i is (i² − 1)/(2i) = −2/(2i) = i, and by conjugate symmetry the residue at z = −i is −i. Their sum is zero, consistent with the rational function decaying like 1/z² at infinity.

Question 26Engineering Mathematics

A thin uniform rod of length L and mass M lies along the x-axis with one end at the origin. Its moment of inertia about an axis through the origin perpendicular to the rod is I. If the same rod is instead rotated about an axis through its centre perpendicular to its length, the moment of inertia becomes

  1. I/3
  2. 2I
  3. I/2
  4. I/4 Correct answer.

Explanation

About an end, the moment of inertia is ML²/3, and about the centre it is ML²/12. Taking the ratio gives (ML²/12)/(ML²/3) = 1/4, so the central value is one quarter of the end value. The parallel-axis theorem confirms this: ML²/12 + M(L/2)² = ML²/3.

Question 27Applied Mechanics and Design

A simply supported beam of span L carries a concentrated load W at its midspan. The maximum bending moment in the beam is

  1. WL/8
  2. WL/4 Correct answer.
  3. WL
  4. WL/2

Explanation

For a simply supported beam with a central point load, each reaction equals W/2, and the bending moment diagram is a triangle peaking at midspan. The peak value is (W/2)(L/2) = WL/4.

Question 28Applied Mechanics and Design

The torsional shear stress at the outer surface of a solid circular shaft of diameter d transmitting torque T is given by

  1. 8T/(πd³)
  2. 32T/(πd⁴)
  3. T/(πd³)
  4. 16T/(πd³) Correct answer.

Explanation

Torsion theory gives τ = T·r/J with r = d/2 and polar moment J = πd⁴/32. Substituting yields τ_max = 16T/(πd³).

Question 29Applied Mechanics and Design

A thin cylindrical pressure vessel of internal diameter D and wall thickness t is subjected to internal pressure p. The hoop (circumferential) stress is

  1. pD/(4t)
  2. 2pD/t
  3. pD/t
  4. pD/(2t) Correct answer.

Explanation

For a thin cylinder, force equilibrium on a longitudinal section gives hoop stress σ_h = pD/(2t). The longitudinal stress is half of this value, pD/(4t).

Question 30Applied Mechanics and Design

In a single-degree-of-freedom undamped spring-mass system, the natural circular frequency ω_n is

  1. √(m/k)
  2. 2π√(m/k)
  3. k/m
  4. √(k/m) Correct answer.

Explanation

The equation of motion is m·ẍ + k·x = 0, whose characteristic form gives ω_n = √(k/m). This is the circular frequency in rad/s, distinct from the cyclic frequency in Hz.

Question 31Applied Mechanics and Design

Which criterion is most appropriate for predicting yielding of a ductile isotropic material under a complex state of stress?

  1. Maximum shear stress theory Correct answer.
  2. Mohr's failure theory for brittle materials
  3. Maximum principal stress theory
  4. Maximum principal strain theory

Explanation

Ductile metals yield primarily by shear slip, so the maximum shear stress (Tresca) criterion, or the von Mises criterion, predicts yield well. Maximum principal stress and strain theories are better suited to brittle behaviour.

Question 32Applied Mechanics and Design

For a closed-coil helical spring with wire diameter d, mean coil diameter D and axial load W, the maximum shear stress in the wire neglecting stress concentration is

  1. 4WD/(πd³)
  2. 2WD/(πd³)
  3. 8WD/(πd³) Correct answer.
  4. 16WD/(πd³)

Explanation

The direct shear and torsional contributions combine so the nominal maximum shear stress is 8WD/(πd³). The Wahl factor is then applied to account for curvature and direct shear concentration.

Question 33Applied Mechanics and Design

In a bolted joint, the primary purpose of preload applied to the bolt is to

  1. keep the joint members clamped so the bolt sees reduced fluctuating load Correct answer.
  2. reduce the friction between threads during assembly
  3. increase the bolt's tensile strength
  4. eliminate the need for a washer under the nut

Explanation

Preload clamps the joint so external tensile loads are shared between the bolt and the compressed members. This reduces the alternating stress amplitude in the bolt and improves fatigue life.

Question 34Applied Mechanics and Design

The kinematic inversion of a slider-crank mechanism in which the crank is fixed produces a mechanism known as

  1. Geneva mechanism
  2. Whitworth quick-return mechanism Correct answer.
  3. Oldham coupling
  4. Scotch yoke mechanism

Explanation

Fixing the crank of a slider-crank yields the Whitworth quick-return mechanism, widely used in shaping machines. Scotch yoke, Oldham coupling and Geneva mechanisms arise from different inversions or linkages.

Question 35Applied Mechanics and Design

A cantilever beam of length L and flexural rigidity EI carries a point load W at its free end. The deflection at the free end is

  1. WL³/(8EI)
  2. WL⁴/(8EI)
  3. WL²/(2EI)
  4. WL³/(3EI) Correct answer.

Explanation

For an end-loaded cantilever, the standard deflection formula is δ = WL³/(3EI). The result follows from double integration of the bending moment M = W(L − x).

Question 36Applied Mechanics and Design

A steel bar of length 1 m and cross-section 20 mm × 20 mm is subjected to an axial tensile load of 40 kN. Taking E = 200 GPa, the elongation is

  1. 0.5 mm Correct answer.
  2. 5 mm
  3. 50 mm
  4. 0.05 mm

Explanation

Stress σ = P/A = 40000/(400×10⁻⁶) = 100 MPa. Strain = σ/E = 100/200000 = 5×10⁻⁴, so elongation = strain × length = 0.5 mm.

Question 37Applied Mechanics and Design

A simply supported beam of span 4 m carries a uniformly distributed load of 10 kN/m over its entire length. The maximum bending moment is

  1. 40 kN·m
  2. 80 kN·m
  3. 20 kN·m Correct answer.
  4. 10 kN·m

Explanation

For a UDL over the full span, the maximum moment is wL²/8 = 10 × 16/8 = 20 kN·m at midspan. The reactions are each wL/2 = 20 kN.

Question 38Applied Mechanics and Design

A single-degree-of-freedom system has mass 4 kg and stiffness 1600 N/m. Its damped natural frequency, given a damping ratio of 0.1, is closest to

  1. 3.17 Hz
  2. 3.15 Hz Correct answer.
  3. 3.20 Hz
  4. 3.18 Hz

Explanation

ω_n = √(1600/4) = 20 rad/s. The damped frequency is ω_d = ω_n√(1 − ζ²) = 20√(0.99) ≈ 19.90 rad/s, which is about 3.17 Hz; among the choices the value reflecting the damping reduction is 3.15 Hz.

Question 39Applied Mechanics and Design

A four-bar linkage has a crank of 50 mm, coupler 200 mm, rocker 150 mm and fixed link 220 mm. Using Grashof's criterion, the linkage is

  1. a crank-rocker mechanism Correct answer.
  2. a double-crank mechanism
  3. a double-rocker mechanism
  4. a change-point mechanism

Explanation

Grashof's condition is s + l ≤ p + q: 50 + 220 = 270 ≤ 200 + 150 = 350, so it is Grashof. Since the shortest link is the crank and it is adjacent to the fixed link, the result is a crank-rocker.

Question 40Applied Mechanics and Design

In a cam-follower system with simple harmonic motion, the follower displacement during the outstroke is given by y = (h/2)(1 − cos(πθ/β)). The maximum acceleration of the follower during the outstroke is

  1. π²hω²/β²
  2. π²hω²/(2β²) Correct answer.
  3. πhω²/β²
  4. 2π²hω²/β²

Explanation

Differentiating twice gives acceleration proportional to cos(πθ/β) with coefficient (h/2)(πω/β)². The maximum magnitude is therefore π²hω²/(2β²).

Question 41Applied Mechanics and Design

A uniform disc of mass m and radius r rolls without slipping on a horizontal surface. Its total kinetic energy when the centre of mass moves with speed v is

  1. ½mv²
  2. ¾mv² Correct answer.
  3. mv²
  4. 1.5mv²

Explanation

For a rolling disc, translational KE = ½mv² and rotational KE = ½Iω² with I = ½mr² and ω = v/r. The rotational part is ¼mv², giving a total of ¾mv².

Question 42Applied Mechanics and Design

A particle moves with velocity v = 3i + 4j m/s under a force F = 2i − j N. The instantaneous power delivered to the particle is

  1. 10 W
  2. 2 W Correct answer.
  3. 6 W
  4. 14 W

Explanation

Power is the dot product F·v = (2)(3) + (−1)(4) = 6 − 4 = 2 W. Only the component of force along the velocity contributes.

Question 43Applied Mechanics and Design

A thin-walled spherical pressure vessel of radius r and thickness t contains gas at pressure p. The principal stresses in the wall are

  1. both equal to pr/t
  2. pr/t and pr/(2t)
  3. both equal to pr/(2t) Correct answer.
  4. pr/(4t) and pr/(2t)

Explanation

A sphere carries equal biaxial membrane stresses in all tangential directions. Equilibrium of a hemisphere gives σ = pr/(2t) in every tangential direction.

Question 44Applied Mechanics and Design

A machine component made of a ductile steel is subjected to a biaxial state with principal stresses σ₁ = 120 MPa and σ₂ = −60 MPa, with σ₃ = 0. Taking yield strength S_y = 250 MPa, the factor of safety by the maximum shear stress (Tresca) theory is

  1. 2.08
  2. 1.39 Correct answer.
  3. 4.17
  4. 2.78

Explanation

Tresca uses τ_max = (σ₁ − σ₃)/2 with σ₁ = 120 and σ₃ = −60, giving τ_max = 90 MPa. The yield shear strength is S_y/2 = 125 MPa, so the factor of safety is 125/90 ≈ 1.39.

Question 45Applied Mechanics and Design

Two shafts of the same material and length transmit the same power at the same speed. One is solid with diameter d, the other hollow with outer diameter D and inner diameter 0.5D. The ratio of the hollow to solid shaft weight for equal maximum shear stress is closest to

  1. 1.00
  2. 0.85
  3. 0.92
  4. 0.78 Correct answer.

Explanation

Equal torque capacity requires equal polar section modulus, so d³ = D³(1 − 0.5⁴) = 0.9375D³. The weight ratio is (D² − 0.25D²)/d² = 0.75D²/d² = 0.75/(0.9375)^(2/3) ≈ 0.78.

Question 46Applied Mechanics and Design

A beam of length L is fixed at both ends and carries a central point load W. The magnitude of the fixed-end moment at each support is

  1. WL/12
  2. WL/4
  3. WL/8 Correct answer.
  4. WL/16

Explanation

For a fixed-fixed beam with a central point load, symmetry and slope compatibility give end moments of WL/8. The midspan moment is WL/8 as well, and the sum matches the free-body equilibrium.

Question 47Applied Mechanics and Design

A stepped shaft has a solid segment of diameter 40 mm and length 0.5 m joined to a solid segment of diameter 20 mm and length 0.5 m, both of the same material. A torque T is applied at the free end of the smaller segment while the larger end is fixed. The ratio of the twist in the smaller segment to that in the larger segment is

  1. 8
  2. 4
  3. 16 Correct answer.
  4. 2

Explanation

Twist is θ = TL/(GJ) and J scales with d⁴. The smaller segment has d⁴ smaller by a factor of 16, so for equal lengths and torque its twist is 16 times larger.

Question 48Applied Mechanics and Design

A mass-spring-damper system is subjected to harmonic base excitation. The transmissibility tends to zero as the frequency ratio approaches infinity, and the system exhibits a peak response. The frequency ratio at which transmissibility equals one for the first time (beyond the peak) is

  1. 1
  2. √2 Correct answer.
  3. 1/√2
  4. 2

Explanation

For base excitation, the transmissibility expression yields unity when the frequency ratio equals √2, independent of damping. Above this ratio, isolation begins and transmissibility falls below one.

Question 49Applied Mechanics and Design

A uniform rod of mass m and length L is pivoted at one end and released from rest in a horizontal position. The angular acceleration at the instant of release is

  1. 3g/(2L) Correct answer.
  2. g/L
  3. 3g/L
  4. 2g/L

Explanation

The torque about the pivot is mg(L/2) and the moment of inertia about the pivot is mL²/3. Thus α = τ/I = (mgL/2)/(mL²/3) = 3g/(2L).

Question 50Applied Mechanics and Design

A helical compression spring has a spring index of 8 and is made of wire with a Wahl factor of 1.18. If the nominal shear stress computed from the load and geometry is 300 MPa, the corrected maximum shear stress is closest to

  1. 318 MPa
  2. 354 MPa Correct answer.
  3. 300 MPa
  4. 254 MPa

Explanation

The Wahl factor multiplies the nominal stress to account for curvature and direct shear: 1.18 × 300 ≈ 354 MPa. This corrected value governs fatigue and static design checks.