Theory of Machines - ISRO Mechanical & Aerospace Subject-Wise PYQ Practice

Practice topic-wise Theory of Machines previous year questions for ISRO Mechanical & Aerospace with instant step-by-step solutions and formula breakdowns.

All ISRO Mechanical & Aerospace papers

37 questions.

Question 1

Year: 2008 | Topic: Theory of Machines

Angle of dwell of cam is defined as the angle

  • A. during which the follower returns to its initial position
  • B. of rotation of the cam for definite displacement of the follower
  • C. through which the cam rotates during the period in which the follower remains in the highest position
  • D. moved by the cam from the instant the follower begins to rise, till it reaches its highest position

Answer: C. through which the cam rotates during the period in which the follower remains in the highest position

Explanation: Angle of dwell of cam is defined as the angle through which the cam rotates during the period in which the follower remains in the highest position.

Question 2

Year: 2008 | Topic: Theory of Machines

The supply of working fluid to the engine to suit the load condition is controlled by

  • A. Meyer's expansion valve
  • B. D-slide valve
  • C. Governor
  • D. Throttle valve.

Answer: C. Governor

Explanation: The supply of working fluid to the engine to suit the load condition is controlled by a governor.

Question 3

Year: 2008 | Topic: Theory of Machines

A pantograph is a device that reproduces a displacement

  • A. On the same scale
  • B. On any scale
  • C. On reduced scale
  • D. On enlarged scale

Answer: B. On any scale

Explanation: A Pantograph is a mechanical linkage connected in a manner based on parallelograms so that the movement of one pin produces identical movements in a second pin. It can trace an identical, enlarged, or miniaturized copy, meaning it reproduces on any scale.

Question 4

Year: 2008 | Topic: Theory of Machines

In the case of gyroscopic effect, the planes of spin, gyroscopic couple and precession are

  • A. In same plane
  • B. In different planes
  • C. Any two in perpendicular planes and third in different plane
  • D. Mutually perpendicular

Answer: D. Mutually perpendicular

Explanation: To achieve the effective gyroscopic effect, the planes of spin, gyroscopic couple and axis of precession are mutually perpendicular to each other.

Question 5

Year: 2008 | Topic: Theory of Machines

Which of the following is a lower pair?

  • A. Piston and cylinder
  • B. Cam and follower
  • C. Belt drive
  • D. Gear

Answer: A. Piston and cylinder

Explanation: When two elements of a pair are joined together with surface contact between them, the joint is called a lower pair. Prismatic pair (piston and cylinder), spherical pair (Ball and socket), screw pair etc. are examples of lower pairs.

Question 6

Year: 2014 | Topic: Theory of Machines

A body A of weight 10 kg with its C.G. 25 cm from the axis of rotation is to be balanced by another body B of weight 5 kg. The radius at which the C.G. of weight B should be placed is

  • A. 2 cm
  • B. 12.5 cm
  • C. 50 cm
  • D. None of these

Answer: C. 50 cm

Explanation: To statically balance a rotating system, the moments of the masses about the axis of rotation must be equal. Therefore, m1 * r1 = m2 * r2. Given body A has a mass m1 = 10 kg at radius r1 = 25 cm, and body B has mass m2 = 5 kg. Substituting these values yields 10 * 25 = 5 * r2. Solving for r2 gives r2 = (10 * 25) / 5 = 50 cm.

Question 7

Year: 2014 | Topic: Theory of Machines

The fundamental natural frequency of the cantilever beam with point load P acting at the free end, in rad/sec is

  • A. $\sqrt{\frac{3EIg}{PL^{3}}}$
  • B. $\sqrt{\frac{PL^{3}}{3EIg}}$
  • C. $\frac{1}{2\pi}\sqrt{\frac{3EIg}{PL^{3}}}$
  • D. $\frac{1}{2\pi}\sqrt{\frac{PL^{3}}{3EIg}}$

Answer: A. $\sqrt{\frac{3EIg}{PL^{3}}}$

Explanation: The static deflection of a cantilever beam with a point load P at the free end is $\Delta = \frac{PL^{3}}{3EI}$. The natural circular frequency is given by $\omega_{n} = \sqrt{\frac{g}{\Delta}}$. Substituting the expression for deflection gives $\omega_{n} = \sqrt{\frac{g}{PL^{3}/3EI}} = \sqrt{\frac{3EIg}{PL^{3}}}$ rad/sec.

Question 8

Year: 2014 | Topic: Theory of Machines

Determine the number of degrees of freedom used in the analysis of the mechanical system of below mentioned figure.

  • A. 3
  • B. 7
  • C. 5
  • D. 4

Answer: D. 4

Explanation: In the mechanism described by the problem context, there are independent motions that define the system's configuration. Specifically, there are two independent rotations of the pulleys and two independent vertical motions of the masses B and C. This results in a total of 4 degrees of freedom.

Question 9

Year: 2014 | Topic: Theory of Machines

In the figure shown below, the relative velocity of link 1 with respect to link 2 is 15 m/s. The link 2 rotates at a constant speed of 150 rpm. The magnitude of Coriolis component of acceleration of link 1 is

  • A. $201~m/s^{2}$
  • B. $239~m/s^{2}$
  • C. $418~m/s^{2}$
  • D. $471~m/s^{2}$

Answer: D. $471~m/s^{2}$

Explanation: The Coriolis component of acceleration is calculated using the formula $a_c = 2v\omega$, where $v$ is the sliding velocity and $\omega$ is the angular velocity of the rotating link. Given $v = 15$ m/s and $N = 150$ rpm. The angular velocity $\omega = \frac{2\pi N}{60} = \frac{2\pi \times 150}{60} = 5\pi$ rad/s. Substituting the values gives $a_c = 2 \times 15 \times 5\pi = 150\pi \approx 471.2$ m/s², which is approximately 471 m/s².

Question 10

Year: 2014 | Topic: Theory of Machines

Determine $m_{eq}$ and $k_{eq}$ for the system in below Fig., when X, the downward displacement of a block, measured from the system's equilibrium position, is used as the generalized coordinate. Take I as the M.I. of the wheel.

  • A. $K_{eq}=3k, m_{eq}=m+\frac{I}{r^{2}}$
  • B. $K_{eq}=3k, m_{eq}=mr^{2}+I$
  • C. $K_{eq}=3k, m_{eq}=m-\frac{I}{r^{2}}$
  • D. $K_{eq}=k/3, m_{eq}=m+\frac{I}{r^{2}}$

Answer: A. $K_{eq}=3k, m_{eq}=m+\frac{I}{r^{2}}$

Explanation: The springs with stiffness $k$ and $2k$ are connected in parallel, so the equivalent stiffness is $k_{eq} = k + 2k = 3k$. The total kinetic energy of the system is the sum of the translational kinetic energy of the mass $m$ and the rotational kinetic energy of the wheel: $E = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}I\omega²$. Since $x = r\omega$, we can substitute $\omega = \dot{x}/r$ into the equation: $E = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}I(\frac{\dot{x}}{r})^2 = \frac{1}{2}\dot{x}^2(m + \frac{I}{r²})$. Therefore, the equivalent mass is $m_{eq} = m + \frac{I}{r²}$.

Question 11

Year: 2014 | Topic: Theory of Machines

The system described by the following governing equation of motion, will have how many natural frequencies?
$\begin{bmatrix}m&0\\ 0&m\end{bmatrix}\begin{Bmatrix}\ddot{x}_{1}\\ \ddot{x}_{2}\end{Bmatrix}+\begin{bmatrix}2k&-k\\ -k&3k\end{bmatrix}\begin{Bmatrix}x_{1}\\ x_{2}\end{Bmatrix}=\begin{Bmatrix}0\\ 0\end{Bmatrix}$

  • A. 4
  • B. 3
  • C. 2
  • D. 1

Answer: C. 2

Explanation: The number of natural frequencies of a vibratory system is equal to the number of its degrees of freedom, which directly corresponds to the number of eigenvalues of its stiffness and mass matrices. Since the given matrices are of the order $2 \times 2$, there will be exactly two eigenvalues, and hence, two natural frequencies.

Question 12

Year: 2014 | Topic: Theory of Machines

Find out the correct freebody diagram from the middle mass for the given spring-mass system in below figure

  • A. $2k(X_{1}-X_{2})$ [left], $k(X_{3}-X_{2})$ [right]
  • B. $2k(X_{2}-X_{1})$ [left], $k(X_{3}-X_{2})$ [right]
  • C. $2k(X_{2}-X_{1})$ [left], $k(X_{2}-X_{3})$ [right]
  • D. $2k(X_{1}-X_{2})$ [left], $k(X_{2}-X_{3})$ [right]

Answer: B. $2k(X_{2}-X_{1})$ [left], $k(X_{3}-X_{2})$ [right]

Explanation: When the middle mass ($2m$) is displaced by $X_{2}$, the restoring force from the left spring (stiffness $2k$) depends on its relative elongation $(X_{2}-X_{1})$ and acts to oppose this motion. Similarly, the right spring (stiffness $k$) exerts a force proportional to $(X_{3}-X_{2})$. The correct free-body representation shows these internal restoring forces acting on the middle mass correctly oriented against relative displacements.

Question 13

Year: 2014 | Topic: Theory of Machines

For what value of m (mass) will resonance occur for the spring mass system given below with the given spring constants?
Excitation: $200\sin(50t)$
Springs: $2\times 10^{5}\text{ N/m}$ and $1\times 10^{5}\text{ N/m}$

  • A. 85kg
  • B. 150kg
  • C. 100kg
  • D. 120kg

Answer: D. 120kg

Explanation: The two springs are connected in parallel, so their equivalent stiffness is $k_{eq} = k_{1} + k_{2} = 2\times 10^{5} + 1\times 10^{5} = 3\times 10^{5}\text{ N/m}$. From the excitation force $200\sin(50t)$, the excitation frequency is $\omega = 50\text{ rad/sec}$. Resonance occurs when the natural frequency $\omega_{n}$ equals the excitation frequency $\omega$, i.e., $\omega = \sqrt{\frac{k_{eq}}{m}}$. Rearranging for mass: $m = \frac{k_{eq}}{\omega²} = \frac{3\times 10^{5}}{50^{2}} = \frac{300000}{2500} = 120\text{ kg}$.

Question 14

Year: 2014 | Topic: Theory of Machines

Which type of Gear meshing should be employed to transmit torque to a linear force?

  • A. Epicyclical Gear
  • B. Helical Gear
  • C. Bevel Gear
  • D. Rack and pinion

Answer: D. Rack and pinion

Explanation: A rack and pinion gear arrangement is specifically designed to convert rotational motion (torque) into linear motion (linear force). Therefore, it is the correct type of gear meshing to employ for this purpose.

Question 15

Year: 2015 | Topic: Theory of Machines

As shown in the figure a mass of 100 kg is held between two springs. The natural frequency of vibration of the system in cycle/s is (given both springs have stiffness 20 kN/m)

  • A. $1/2\pi$
  • B. $5/\pi$
  • C. $10/\pi$
  • D. $20/\pi$

Answer: C. $10/\pi$

Explanation: The mass is held between two springs which are effectively in a parallel configuration. The equivalent stiffness $k_{eq} = k_1 + k_2 = 20 + 20 = 40$ kN/m = $40000$ N/m. The natural circular frequency is $\omega_n = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{40000}{100}} = 20$ rad/sec. Since $\omega_n = 2\pi f_n$, the natural frequency $f_n = \frac{20}{2\pi} = \frac{10}{\pi}$ cycles/s (Hz).

Question 16

Year: 2016 | Topic: Theory of Machines

The piston of a steam engine moves with simple harmonic motion. The speed of rotation of crank is 120 rpm with a stroke of 2 m. What is the velocity of piston when it is 0.5 m from the centre?

  • A. 4 * pi * $\sqrt{3}$
  • B. pi * $\sqrt{3}$
  • C. 2 * pi * $\sqrt{3}$
  • D. 3 * pi * $\sqrt{3}$

Answer: C. 2 * pi * $\sqrt{3}$

Explanation: The crank radius (r) is half the stroke, so r = 2 / 2 = 1 m. The position from the start of the stroke is x = 1.5 m (which is 0.5 m from the center). Using the displacement equation x = r(1 - cos(theta)), we have 1.5 = 1(1 - cos(theta)), yielding cos(theta) = -0.5, thus theta = 120 degrees. The velocity of the piston is given by v = r * omega * sin(theta). With omega = (2 * pi * 120 / 60), v = 1 * (4 * pi) * sin(120 degrees) = 4 * pi * ($\sqrt{3}$ / 2) = 2 * pi * $\sqrt{3}$ m/s.

Question 17

Year: 2016 | Topic: Theory of Machines

The reading of a spring balance is from 0 to 200 N and is 10 cm long. A body suspended from the spring balance is observed to oscillate vertically at 2 Hz. The mass of the body is nearly

  • A. 22.5 kg
  • B. 12.5 kg
  • C. 37 kg
  • D. 45 kg

Answer: B. 12.5 kg

Explanation: The stiffness (k) of the spring is calculated as Force divided by deflection: k = 200 N / 0.1 m = 2000 N/m. The natural frequency of the oscillation is given by f = (1 / (2 * pi)) * $\sqrt{k / m}$. Substituting the known frequency f = 2 Hz, we get 2 = (1 / (2 * pi)) * $\sqrt{2000 / m}$, which simplifies to 4 * pi = $\sqrt{2000 / m}$. Squaring both sides yields 16 * pi² = 2000 / m, resulting in m = 2000 / (16 * pi²). This evaluates to approximately 12.66 kg, which closely matches 12.5 kg.

Question 18

Year: 2017 | Topic: Vibrations

A cantilever beam of length 'L' has natural frequency ω₁. If the beam is broken into halves, the natural frequency of the remaining cantilever beam ω₂ will be such that

  • A. ω₂ < ω₁
  • B. ω₂ > ω₁
  • C. ω₂ = ω₁
  • D. Cannot be obtained from the given data

Answer: B. ω₂ > ω₁

Explanation: Natural frequency ω is proportional to √(k/m). Stiffness k of a cantilever is proportional to EI/L³. If L becomes L/2, stiffness k becomes 8 times the original. Frequency is proportional to √k, so it increases. ω is inversely proportional to L², so halving the length increases the frequency by a factor of 4. The provided solution notes ω ∝ 1/L^(3/2), which also leads to an increase.

Question 19

Year: 2017 | Topic: Theory of Machines

Which of the following is FALSE?

  • A. A reverted gear train is one in which the first and last gears are on the same axis.
  • B. Train value for a gear is the ratio of product of driven tooth numbers to product of driving tooth numbers.
  • C. A planetary gear train is one in which the axes of some of the gears may have a motion.
  • D. A compound gear train is one which has two or more gears on each axis.

Answer: B. Train value for a gear is the ratio of product of driven tooth numbers to product of driving tooth numbers.

Explanation: Train value is the reciprocal of the speed ratio. It is defined as the ratio of the product of driving tooth numbers to the product of driven tooth numbers.

Question 20

Year: 2017 | Topic: Vibrations

Which of the following is TRUE for vibration of a mechanical system

  • A. Damping ratio is the ratio of the critical damping to the actual damping
  • B. The damping can be obtained using logarithmic decrement method
  • C. For damped systems, amplitude becomes infinity at resonance
  • D. Amplitude ratio is unity when the exciting force frequency is √2 times the natural frequency

Answer: D. Amplitude ratio is unity when the exciting force frequency is √2 times the natural frequency

Explanation: (d) is correct. For a transmissibility calculation, when the frequency ratio (ω/ωn) is √2, the transmissibility T.R. becomes 1, regardless of the damping ratio.

Question 21

Year: 2018 | Topic: Theory of Machines

For the spring system given in figure, the equivalent stiffness is

  • A. 5K/2
  • B. 2K/5
  • C. K
  • D. 3K

Answer: C. K

Explanation: Upper two springs are in parallel, so equivalent stiffness K1 = K + K = 2K. This equivalent spring is in series with the bottom 2K spring. 1/K_eq = 1/2K + 1/2K = 2/2K = 1/K. Thus, K_eq = K.

Question 22

Year: 2018 | Topic: Theory of Machines

If the natural frequency of a spring mass system on earth is f, its natural frequency on moon where acceleration due to gravity is one-sixth of that on earth will be

  • A. f/6
  • B. 6f
  • C. f/$\sqrt{6}$
  • D. f

Answer: D. f

Explanation: Natural frequency f = (1 / 2*pi) * $\sqrt{k/m}$. It does not depend on gravity, so it remains the same on Earth and the moon.

Question 23

Year: 2018 | Topic: Theory of Machines

A circular solid disc of uniform thickness 20mm, radius 400 mm and mass 10 kg, is used as a flywheel. If it rotates at 300 rpm, the kinetic energy of the flywheel, in Joules is

  • A. 395
  • B. 790
  • C. 1580
  • D. 3160

Answer: A. 395

Explanation: KE = (1/2)*I*omega². I = (m*r²)/2 = (10 * 0.4^2) / 2 = 0.8 kg-m². omega = 2*pi*300/60 = 10*pi rad/s. KE = 0.5 * 0.8 * (10*pi)^2 = 40 * pi² = 394.78 J, approximately 395 J.

Question 24

Year: 2018 | Topic: Theory of Machines

If the length of the cantilever beam is reduced to 1/4th, the natural frequency of the mass M at the end of this cantilever beam of negligible mass will be increased by a factor of

  • A. 2
  • B. $\sqrt{8}$
  • C. 4
  • D. 8

Answer: D. 8

Explanation: Deflection delta = (Mg*L³)/(3EI). Natural frequency f is proportional to 1/$\sqrt{delta}$, which means f is proportional to 1/L^(3/2). f2/f1 = (L1/L2)^(3/2) = (L / (L/4))^(3/2) = 4^(3/2) = 8.

Question 25

Year: 2018 | Topic: Theory of Machines

A perfectly balanced rotor shaft supported between short bearings at both ends shows a deflection of 2 mm at middle. Then the critical speed of the shaft in rpm will be nearly,

  • A. 335
  • B. 670
  • C. 2010
  • D. 1340

Answer: B. 670

Explanation: Omega = $\sqrt{g / delta}$ = $\sqrt{9.81 / 0.002}$ = 70.035 rad/s. Critical speed N_c = (60 * Omega) / (2 * pi) = (60 * 70.035) / (2 * 3.14) = 669.12 rpm, which is approximately 670 rpm.

Question 26

Year: 2024 | Topic: Theory of Machines

A slider moving outward on a rod OR with a velocity of 3 m/s and acceleration of 9 m/s². The rod has an angular velocity of 7 rad/s counter clockwise about O and an angular acceleration of 12 rad/s² clockwise. The magnitude of absolute acceleration of the slider at location A in m/s² is (where OA = 1m)

  • A. 20
  • B. 50
  • C. 67
  • D. 79

Answer: B. 50

Explanation: Radial acceleration ar = d^2r/dt² - r*omega² = 9 - 1*(7^2) = 9 - 49 = -40 m/s². Tangential acceleration at = r*alpha + 2*(dr/dt)*omega = 1*(-12) + 2*3*7 = -12 + 42 = 30 m/s². Absolute acceleration = $\sqrt{ar² + at²}$ = $\sqrt{(-40}$^2 + 30^2) = 50 m/s².

Question 27

Year: 2024 | Topic: Theory of Machines

The configuration of a planar four bar mechanism with frictionless joints is shown in the figure below. The length of the links are L1 = 40 mm, L2 = 15 mm, L3 = 35 mm and L4 = 30 mm. The suffixes 1, 2, 3 and 4 represents the fixed, input, coupler and output link respectively. If Ti and To are the torque to the input and output link, what is the mechanical advantage (To/Ti) of the mechanism?

  • A. 2
  • B. 0.5
  • C. Infinity
  • D. 0

Answer: C. Infinity

Explanation: In the given configuration, links 2 and 3 are collinear (180 degrees). This is a toggle position where a finite input torque creates an infinite output torque (theoretically) because the mechanical advantage becomes infinite. The ratio of output torque to input torque (To/Ti) approaches infinity.

Question 28

Year: 2024 | Topic: Theory of Machines

A mass m is attached to two identical springs having stiffness k as shown in figure below. The natural frequency of the vibrating system is

  • A. (1/2pi) * $\sqrt{k/m}$
  • B. (1/2pi) * $\sqrt{2k/m}$
  • C. (1/2pi) * $\sqrt{k/2m}$
  • D. (1/2pi) * $\sqrt{4k/m}$

Answer: B. (1/2pi) * $\sqrt{2k/m}$

Explanation: The two springs are attached to fixed supports on either side of the mass. When the mass is displaced by x, both springs exert a restoring force of kx in the same direction. Thus, they act in parallel. Equivalent stiffness K_eq = k + k = 2k. Natural frequency = (1/2pi) * $\sqrt{2k/m}$.

Question 29

Year: 2024 | Topic: Theory of Machines

A damped spring-mass system is forced to oscillate at t=0 by the application of an impulse force. Which of the following are true?

  • A. If the damping ratio is between 0 and 1, the system will show no oscillations
  • B. If the damping ratio is between 0 and 1, the system will eventually come to rest
  • C. If the damping ratio is 0, the system will eventually come to rest
  • D. The system will always oscillate

Answer: B. If the damping ratio is between 0 and 1, the system will eventually come to rest

Explanation: For an underdamped system (damping ratio between 0 and 1), it will oscillate with decaying amplitude and eventually come to rest due to energy dissipation. A damping ratio of 0 would mean it oscillates forever.

Question 30

Year: 2024 | Topic: Robotics

What is the revolving joint of the robot called?

  • A. V-Joint
  • B. L-Joint
  • C. O-Joint
  • D. T-Joint

Answer: A. V-Joint

Explanation: In robotics, a revolving joint where the input link is parallel to the axis of rotation and the output link is perpendicular to it is standardly referred to as a V-joint (reVolving joint).

Question 31

Year: 2024 | Topic: Theory of Machines

A planar four bar mechanism with all revolute joints has link lengths AB=30 mm, BC=120 mm, CD=100 mm and AD=80 mm. The link to be fixed to obtain a double crank mechanism is

  • A. AB
  • B. BC
  • C. CD
  • D. AD

Answer: A. AB

Explanation: By Grashof's law, Shortest + Longest (S+L) = 30 + 120 = 150 mm. The sum of the other two links (P+Q) = 100 + 80 = 180 mm. Since S+L < P+Q, it is a Grashof linkage. To obtain a double crank mechanism, the shortest link (AB) must be fixed.

Question 32

Year: 2024 | Topic: Theory of Machines

A cantilever beam of cross section A, moment of inertia I and length L is having natural frequency $omega_1$. If the beam is accidentally broken into two halves, the natural frequency of the remaining cantilever beam $omega_2$ will be such that it

  • A. Increases ($omega_2 > omega_1$)
  • B. Decreases ($omega_2 < omega_1$)
  • C. Remains same $omega_2 = omega_1$
  • D. Cannot be obtained from the given data

Answer: A. Increases ($omega_2 > omega_1$)

Explanation: For a continuous cantilever beam, the natural frequency $omega_n$ is proportional to $1/L²$. If the beam breaks and the length is halved to $L/2$, the new natural frequency will be proportional to $1/(L/2)^2 = 4/L²$, making it significantly higher. Therefore, $omega_2 > omega_1$.

Question 33

Year: 2024 | Topic: Theory of Machines

The equation of motion of a spring-mass-damper system is given by x'' + 4x' + 16x = 12 sin(5t). What is the damping factor of the system?

  • A. 1
  • B. 0.5
  • C. 0.25
  • D. 0.75

Answer: B. 0.5

Explanation: The standard differential equation of motion is x'' + 2*zeta*omega_n*x' + omega_n^2*x = F(t). Comparing this with the given equation: omega_n^2 = 16, so omega_n = 4 rad/s. Also, 2*zeta*omega_n = 4. Substituting omega_n gives 2*zeta*(4) = 4, so 8*zeta = 4, which means zeta = 0.5.

Question 34

Year: 2024 | Topic: Theory of Machines

A gear train comprises of 4 spur gears - the driven gear G1, two idlers G2, G3 and the driving gear G4. The radii of G1, G2, G3 and G4 are r, r/2, r/3 and 2r, respectively. What is the ratio of angular velocities of G1 to G4 and their direction of rotation?

  • A. Ratio 1, opposite direction
  • B. Ratio = 1/6, same direction
  • C. Ratio = 2, Opposite direction
  • D. Ratio = 1/2, same direction

Answer: C. Ratio = 2, Opposite direction

Explanation: In a simple gear train with idlers, the speed ratio only depends on the driving and driven gears. Ratio = omega_1 / omega_4 = Radius_4 / Radius_1 = 2r / r = 2. With 4 gears (1 driver, 2 idlers, 1 driven), there are 3 meshes. Each mesh reverses rotation. Therefore, G4 and G1 will rotate in opposite directions.

Question 35

Year: 2024 | Topic: Theory of Machines

A cam and oscillating follower mechanism is shown in the figure below. The Mobility of the mechanism is

  • A. -1
  • B. 0
  • C. 1
  • D. 2

Answer: C. 1

Explanation: The mechanism has 4 links: Ground (1), Cam (2), Follower (3), and Roller (4). It has 3 lower pairs (ground-cam, ground-follower, follower-roller) and 1 higher pair (cam-roller). Using Kutzbach criterion: F = 3(L-1) - 2J1 - J2 = 3(4-1) - 2(3) - 1 = 2. However, the rotation of the roller is a redundant degree of freedom that does not affect the input-output kinematic relationship. Thus, the effective mobility (functional degree of freedom) is 2 - 1 = 1.

Question 36

Year: 2024 | Topic: Theory of Machines

What is true in respect of the self locking screw jack with square thread? (I) The friction angle is more than the helix angle of the thread. (II) The efficiency of the screw jack should not be more than 50%

  • A. Only I
  • B. Only II
  • C. Both I and II
  • D. None of the above

Answer: C. Both I and II

Explanation: For a screw to be self-locking, the friction angle must be greater than or equal to the helix angle (lead angle) of the thread. Consequently, the mathematical maximum efficiency of a self-locking square-threaded screw jack is 50%. Therefore, both statements are true.

Question 37

Year: 2025 | Topic: Robotics

Degrees of freedom (DOF) of a robot manipulator describe:

  • A. Number of independent motions it can perform
  • B. Payload capacity
  • C. Power consumption
  • D. Workspace volume

Answer: A. Number of independent motions it can perform

Explanation: DOF indicates independent coordinates required to specify the manipulator's pose.