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# Multiple choice question for engineering

## Set 1

1. The tooth profile most commonly used in gear drives for power transmission is
a) A cycloid
b) An involute
c) An ellipse
d) A parabola

Answer: b [Reason:] It is due to easy manufacturing.

2. There are six gears A, B, C, D, E, F in a compound train. The numbers of teeth in the gears are 20, 60, 30, 80, 25 and 75 respectively. The ratio of the angular speeds of the driven (F) to the driver (A) of the drive is
a) 1/24
b) 1/8
c) 4/15
d) 12

Answer: a [Reason:] The ratio of angular speeds of F to A = TA.TC.TE/TB.TD.TF = 20 x 30 x 25/60 x 80 x 75 = 1/24.

3. A fixed gear having 100 teeth meshes with another gear having 25 teeth, the centre lines of both the gears being joined by an arm so as to form an epicyclic gear train. The number of rotations made by the smaller gear for one rotation of the arm is
a) 3
b) 4
c) 5
d) 6

Answer: c [Reason:] Revolution of 25 teeth gear = 1 + T100/T25 = 1 + 100/25 =5.

4. Speed reduction in a gear box is achieved using a worm and worm wheel. The worm wheel has 30 teeth and a pitch diameter of 210 mm. If the pressure angle of the worm is 20o, what is the axial pitch of the worm?
a) 7 mm
b) 22 mm
c) 14 mm
d) 63 mm

Answer: b [Reason:] m = 210/30 = 7 P = πm = 22/7 x 7 = 22 mm Axial pitch = circular pitch of the worm wheel = πm.

5. A speed reducer unit consists of a double-threaded worm of pitch = 11 mm and a worm wheel of pitch diameter = 84 mm. The ratio of the output torque to the input to rque is
a) 7·6
b) 12
c) 24
d) 42

Answer: a [Reason:] Output torque/Input torque = pitch diameter of worm wheel/ pitch of worm = 84/11 = 7.6.

6. A pair of gears forms a rolling pair.
a) True
b) False

7. Spiral bevel gears designed to be used with an offset in their shafts are called ‘hypoid gears’
a) True
b) False

Answer: a [Reason:] The pitch surfaces of such gears are hyperboloids of revolution.

8. Gears with involute tooth profile transmit constant velocity ratios between shafts connected by them.
a) True
b) False

Answer: a [Reason:] For involute gears, the common normal at the point of contact between pairs of teeth always passes through the pictch point.

9. In the case of spur gears, the mating teeth execute pure rolling motion with respect to each other from the commencement of engagement to its termination.
a) True
b) False

Answer: a [Reason:] The involute profiles of the mating teeth are conjugate profiles which obey the law of gearing.

10. A pair of helical gears has fewer teeth in contact as compared to an equivalent pair of spur gears.
a) True
b) False

Answer: b [Reason:] In spur gears, the contact between meshing teeth occurs along the entire face width of the tooth, resulting in a sudden application of the load which, in turn, results in impact conditions and generates noise.

## Set 2

1. Which of the following screw thread is adopted for power transmission in either direction?

Answer: b [Reason:] A square thread, is adapted for the transmission of power in either direction. This thread results in maximum efficiency and minimum radial or bursting pressure on the nut.

2. Multiple threads are used to secure
a) low efficiency
b) high efficiency

Answer: b [Reason:] The power screws with multiple threads such as double, triple etc. are employed when it is desired to secure a large lead with fine threads or high efficiency. Such type of threads are usually found in high speed actuators.

3. Screws used for power transmission should have
a) low efficiency
b) high efficiency
d) strong teeth

4. If α denotes the lead angle and φ, the angle of friction, then the efficiency of the screw is written as
a) tan(α − φ)/tanα
b) tanα/tan (α − φ)
c) tan(α + φ)/tanα
d) tanα/tan (α + φ)

Answer: d [Reason:] Efficiency, η = Ideal effort/Actual effort = tanα/tan (α + φ).

5. A screw jack has square threads and the lead angle of the thread is α. The screw jack will be self locking when the coefficient of friction (μ) is
a) μ > tan α
b) μ = sin α
c) μ = cot α
d) μ = cosec α

Answer: a [Reason:] A screw will be self locking if the friction angle is greater than helix angle or coefficient of friction is greater than tangent of helix angle i.e. μ or tan φ > tan α.

6. To ensure self locking in a screw jack, it is essential that the helix angle is
a) larger than friction angle
b) smaller than friction angle
c) equal to friction angle
d) such as to give maximum efficiency in lifting

Answer: b [Reason:] A screw will be self locking if the friction angle is greater than helix angle or coefficient of friction is greater than tangent of helix angle i.e. μ or tan φ > tan α.

7. A screw is said to be self locking screw, if its efficiency is
a) less than 50%
b) more than 50%
c) equal to 50%
d) none of the mentioned

Answer: a [Reason:] Efficiency of self locking screws is less than 1/2 or 50%. If the efficiency is more than 50%, then the screw is said to be overhauling.

8. A screw is said to be over hauling screw, if its efficiency is
a) less than 50%
b) more than 50%
c) equal to 50%
d) none of the mentioned

Answer: b [Reason:] Efficiency of self locking screws is less than 1/2 or 50%. If the efficiency is more than 50%, then the screw is said to be overhauling.

9. While designing a screw in a screw jack against buckling failure, the end conditions for the screw are taken as
a) both ends fixed
b) both ends hinged
c) one end fixed and other end hinged
d) one end fixed and other end free.

Answer: d [Reason:] For buckling failure, The screw is considered to be a strut with lower end fixed and load end free. For one end fixed and the other end free, C = 0.25.

10. The load cup of a screw jack is made separate from the head of the spindle to
a) enhance the load carrying capacity of the jack
b) reduce the effort needed for lifting the working load
c) reduce the value of frictional torque required to be countered for lifting the load
d) prevent the rotation of load being lifted

Answer: d [Reason:] For the prevention of the rotation of load being lift, the load cup of a screw jack is made separate from the head of the spindle.

## Set 3

1. The condition of correct gearing is
a) pitch line velocities of teeth be same
b) radius of curvature of two profiles be same
c) common normal to the pitch surface cuts the line of centres at a fixed point
d) none of the mentioned

Answer: c [Reason:] The fundamental condition of correct gearing is the common normal at the point of contact between a pair of teeth must always pass through the pitch point.

2. Law of gearing is satisfied if
a) two surfaces slide smoothly
b) common normal at the point of contact passes through the pitch point on the line joining the centres of rotation
c) number of teeth = P.C.D. / module
d) addendum is greater than dedendum

Answer: b [Reason:] Law of gearing says that the common normal at the point of contact between a pair of teeth must always pass through the pitch point.

3. Involute profile is preferred to cyloidal because
a) the profile is easy to cut
b) only one curve is required to cut
c) the rack has straight line profile and hence can be cut accurately
d) none of the mentioned

Answer: b [Reason:] The face and flank of involute teeth are generated by a single curve where as in cycloidal gears, double curves (i.e. epi-cycloid and hypo-cycloid) are required for the face and flank respectively. Thus the involute teeth are easy to manufacture than cycloidal teeth. In involute system, the basic rack has straight teeth and the same can be cut with simple tools.

4. The contact ratio for gears is
a) zero
b) less than one
c) greater than one
d) none of the mentioned

Answer: c [Reason:] The theoretical minimum value for the contact ratio is one, that is there must always be at least one pair of teeth in contact for continuous action.

5. The maximum length of arc of contact for two mating gears, in order to avoid interference, is
a) (r + R) sin φ
b) (r + R) cos φ
c) (r + R) tan φ
d) none of the mentioned

Answer: c [Reason:] Interference may only be prevented, if the addendum circles of the two mating gears cut the common tangent to the base circles between the points of tangency. maximum length of arc of contact = (r + R) tan φ where r = Pitch circle radius of pinion, R = Pitch circle radius of driver, and φ = Pressure angle.

6. When the addenda on pinion and wheel is such that the path of approach and path of recess are half of their maximum possible values, then the length of the path of contact is given by
a) (r + R) sin φ/2
b) (r + R) cos φ/2
c) (r + R) tan φ/2
d) (r + R) cot φ/2

Answer: a [Reason:] In case the addenda on pinion and wheel is such that the path of approach and path of recess are half of their maximum possible values, then Path of approach, KP = 1/2 MP.

7. Interference can be avoided in involute gears with 20° pressure angle by
a) cutting involute correctly
b) using as small number of teeth as possible
c) using more than 20 teeth
d) using more than 8 teeth

8. The ratio of face width to transverse pitch of a helical gear with α as the helix angle is normally
a) more than 1.15/tan α
b) more than 1.05/tan α
c) more than 1/tan α
d) none of the mentioned

9. The maximum efficiency for spiral gears is
a) sin (θ + φ ) + 1/ cos (θ − φ ) + 1
b) cos (θ − φ) + 1/sin (θ + φ ) + 1
c) cos (θ + φ ) + 1/ cos (θ − φ ) + 1
d) cos (θ − φ) + 1/cos (θ + φ ) + 1

Answer: c [Reason:] ηmax = cos (θ + φ ) + 1/ cos (θ − φ ) + 1 where θ = Shaft angle, and φ = Friction angle.

10. For a speed ratio of 100, smallest gear box is obtained by using
a) a pair of spur gears
b) a pair of helical and a pair of spur gear compounded
c) a pair of bevel and a pair of spur gear compounded
d) a pair of helical and a pair of worm gear compounded

## Set 4

1. A body is said to vibrate with simple harmonic motion if its acceleration is proportional to the distance from the mean position.
a) True
b) False

Answer: a [Reason:] A body is said to move with simple harmonic motion, if it satisfies the following two conditions: a) Its acceleration is always directed towards the center, known as point of reference or mean position. b) Its acceleration is proportional to the distance from that point.

2. The maximum displacement of a body, from its mean position is called amplitude.
a) True
b) False

Answer: a [Reason:] The time taken for one complete revolution of the particle is called periodic time. The maximum displacement of a body from its mean position is called amplitude.

3. Frequency of vibrations is usually expressed in
a) number of cycles per hour
b) number of cycles per minute
c) number of cycles per second
d) none of the mentioned

Answer: c [Reason:] The number of cycles per second is called frequency. It is the reciprocal of periodic time.

4. The amplitude of vibrations is always ______________ the radius of the circle.
a) equal to
b) less than
c) greater than
d) none of the mentioned

5. The time taken by a particle for one complete oscillation is known as periodic time.
a) True
b) False

Answer: a [Reason:] The time taken for one complete revolution of the particle is called periodic time. The maximum displacement of a body from its mean position is called amplitude.

6. The periodic time is given by
a) ω/2п
b) 2п/ω
c) ω x 2п
d) п/ω

Answer: b [Reason:] Periodic time, tp = 2п/ω seconds where ω = Angular velocity of the particle in rad/s.

7. When a body moves with simple harmonic motion, the product of its periodic time and frequency is equal to
a) zero
b) one
c) п/2
d) п

Answer: b [Reason:] The number of cycles per second is called frequency. It is the reciprocal of periodic time. Hence, when it is multiplied it is equal to one.

8. The acceleration of the particle moving with simple harmonic motion is ____________ at the mean position.
a) zero
b) minimum
c) maximum
d) none of the mentioned

Answer: a [Reason:] The acceleration of a body is zero at the mean position and maximum when x = r.

9. The maximum velocity of a particle moving with simple harmonic motion is
a) ω
b) ωr
c) ω2r
d) ω/r

Answer: b [Reason:] The velocity of a moving body with simple harmonic motion at any instant is given by v = ω√r2 – x2 The velocity is maximum at the mean position i.e. when x = 0.

Hence, v = ωr.

10. When a particle moves round the circumference of a circle of radius r with ω rad/s, then its maximum acceleration is ω2r.
a) True
b) False

Answer: a [Reason:] The acceleration of a body moving with simple harmonic motion at any instant is given by a = ω2r.

11. If a simple pendulum oscillates with an amplitude 50 mm and time period 2s, then its maximum velocity is
a) 0.1 m/s
b) 0.15 m/s
c) 0.8 m/s
d) 0.16 m/s

Answer: b [Reason:] Maximum velocity vmax = ωA where ‘ω’ is the angular frequency and ‘A’ is the amplitude. Therefore vmax = (2π/T)A = (2π/2)×50×10-3 = 0.157 m/s.

12. A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
a) 1/ 2π√3
b) 2π√3
c) 2π/√3
d) √3/2π

Answer: b [Reason:] The magnitudes of the velocity and acceleration of the particle when its displacement is ‘y’ are ω√(A2 –y2) and ω2y respectively. Equating them, ω√(A2 –y2) = ω2y, from which ω = [√(A2 –y2)]/y = √(4 –1) = √3. Period T = 2π/ω = 2π/√3.

13. Suppose you place a sphere of mass ‘m’ and radius ‘r’ inside a smooth, heavy hemispherical bowl of radius of 37r placed on a horizontal table. If the sphere is given a small displacement, what is its period of oscillation?
a) 2π√(m/37rg)
b) 2π√(m/rg)
c) 12π√(r/g)
d) 2π√(r/g)

Answer: c [Reason:] The arrangement depicted in this question is similar to that of a simple pendulum. Instead of the usual string, you have a concave surface to confine the bob (sphere) to its path along the arc of a circle. The usual expression for the period, T = 2π√(L/g) holds here also, where the length L = 36r since the length of the pendulum is measured from the centre of gravity of the bob. The point of ‘suspension’ is evidently at the centre of the hemispherical bowl. The correct option is 12π√(r/g).

14. The instantaneous displacement of a simple harmonic oscillator is given by y = A cos(ωt + π/4). Its speed will be maximum at the time
a) 2π/ω
b) ω/2π
c) ω/π
d) π/4ω

Answer: d [Reason:] The velocity is the time derivative of displacement: v = dy/dt = -Aω(sin ωt + π/4). Its maximum magnitude equal to Aω is obtained when ωt = π/4, from which t = π/4ω.

15. A particle of mass 5 g is executing simple harmonic motion with an amplitude 0.3 m and time period π/5 s. The maximum value of the force acting on the particle is
a) 5 N
b) 4 N
c) 0.5 N
d) 0.15 N

Answer: d [Reason:] T = 2π√(m/k) where ‘k’ is the force constant, the solution becomes quite easy. From this, k = 4π2m/T2 = 4π2 ×5×10-3/(π/5)2 = 0.5. Since ‘k’ is the force for unit displacement, the maximum force is k times the maximum displacement (amplitude). Therefore maximum force = kA = 0.5×0.3 = 0.15N.

## Set 5

1. Coriolis component of acceleration is a component of translatory acceleration.
a) True
b) False

Answer: a [Reason:] Its unit is m/s2. Therefore translatory acceleration (at = 2ωV).

2. If the relative motion between two links of a mechanism is pure sliding, then the relative instantaneous center for these two links does not exist.
a) True
b) False

Answer: b [Reason:] It does exists at infinity distance. Kennedy theorem says number of instantaneous center N = n(n – 1)/2

3. A slider sliding at 10 cm/s on a link which is rotating at 60 r.p.m. is subjected to Coriolis acceleration of magnitude
a) 40п2 cm/s2
b) 0.4пcm/s2
c) 40пcm/s2
d) 4пcm/s2

Answer: c [Reason:] Coriolis acceleration = 2ωV = 2 x 2пN/60 x V = 2 x 2п x 60/60 x 10 = 40пcm/s2

4. A body in motion will be subjected to coriolis acceleration when that body is
a) in plane rotation with variable velocity
b) in plane translation with variable velocity
c) in plane motion which is a resultant of plane translation and rotation
d) restrained to rotate while sliding over another body

Answer: d [Reason:] When a point on one link is sliding along another rotating link, such as in quick return motion mechanism, then the coriolis component of the acceleration must be calculated.

5. Mechanism used to produce a diagram to an enlarged or reduced sacle
a) hart’s mechanism
b) pantograph
c) grasshopper mechanism
d) peaucellier’s mechanism

Answer: b [Reason:] Pantograph mechanism is used to produce a diagram to an enlarged or reduced sacle. Hart’s mechanism is used to produce exact straight line motion mechanism.

6. A straight line mechanism made up of turning pairs
a) hart’s mechanism
b) pantograph
c) grasshopper mechanism
d) none of the mentioned

Answer: a [Reason:] Exact straight line motion mechanisms made up of turning pairs are peaucellier’s mechanism and hart’s mechanism.

7. Approximate straight line motion consisting of one sliding pair
a) hart’s mechanism
b) pantograph
c) grasshopper mechanism
d) peaucellier’s mechanism

Answer: c [Reason:] Hart’s mechanism is used to produce exact straight line motion mechanism. Grasshopper mechanism gives approximate straight line motion consisting of one sliding pair.

8. Exact straight line motion mechanism
a) hart’s mechanism
b) pantograph
c) grasshopper mechanism
d) none of the mentioned

Answer: a [Reason:] Exact straight line motion mechanisms made up of turning pairs are peaucellier’s mechanism and hart’s mechanism.

9. The sense of Coriolis component 2ωV is the same as that of the relative velocity vector V rotated.
a) 450 in the direction of rotation of the link containing the path
b) 450 in the direction opposite to the rotation of the link containing the path
c) 900 in the direction of rotation of the link containing the path
d) 1800 in the direction opposite to the rotation of the link containing the path

Answer: c [Reason:] The sense of Coriolis component 2ωV is the same as that of the relative velocity vector V rotated 900 in the direction of rotation of the link containing the path. The direction of coriolis component is along a line rotated 900 from the sliding velocity in a direction same as that of the angular velocity of the slotted lever.

10. What is the direction of the coriolis component of acceleration in a slotted lever-crank mechanism?
a) Along the sliding velocity vector
b) Along the direction of the crank
c) Along a line rotated 900 from the sliding velocity vector in a direction opposite to the angular velocity of the slotted lever
d) Along a line rotated 900 from the sliding velocity in a direction same as that of the angular velocity of the slotted lever.