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Multiple choice question for engineering

Set 1

1. The piston of a steam engine moves with simple harmonic motion. The crank rotates at 120 r.p.m. with a stroke of 2 metres. Find the velocity of the piston, when it is at a distance of 0.75 metre from the centre.
a) 8 m/s
b) 8.31 m/s
c) 9 m/s
d) none of the mentioned

View Answer

Answer: b [Reason:] Given : N = 120 r.p.m. or ω = 2π × 120/60 = 4π rad/s ; 2r = 2 m or r = 1 m; x = 0.75 m Velocity of the piston We know that velocity of the piston, v = ω√r2 – x2 = 4π√1 – (0.75)2 = 8.31 m/s.

2. The piston of a steam engine moves with simple harmonic motion. The crank rotates at 120 r.p.m. with a stroke of 2 metres. Find the acceleration of the piston, when it is at a distance of 0.75 metre from the centre.
a) 118.46 m/s2
b) 90 m/s2
c) 100 m/s2
d) none of the mentioned

View Answer

Answer: a [Reason:] Given : N = 120 r.p.m. or ω = 2π × 120/60 = 4π rad/s ; 2r = 2 m or r = 1 m; x = 0.75 m Velocity of the piston We know that velocity of the piston, v = ω√r2 – x2 = 4π√1 – (0.75)2 = 8.31 m/s

We also know that acceleration of the piston, a = ω2.x = (4π)2 0.75 = 118.46 m/s2.

3. Law of isochronism
a) states the time period (tp ) of a simple pendulum does not depend upon the mass of the body suspended at the free end of the string.
b) states the time period (tp ) of a simple pendulum does not depend upon its amplitude of vibration and remains the same, provided the angular amplitude (θ) does not exceed 4°.
c) states the time period (tp) of a simple pendulum is directly proportional to √L , where L is the length of the string.
d) none of the mentioned

View Answer

Answer: b [Reason:] It states the time period (tp ) of a simple pendulum does not depend upon its amplitude of vibration and remains the same, provided the angular amplitude (θ) does not exceed 4°.

4. Law of mass
a) states the time period (tp ) of a simple pendulum does not depend upon the mass of the body suspended at the free end of the string.
b) states the time period (tp ) of a simple pendulum does not depend upon its amplitude of vibration and remains the same, provided the angular amplitude (θ) does not exceed 4°.
c) states the time period (tp) of a simple pendulum is directly proportional to √L , where L is the length of the string.
d) none of the mentioned

View Answer

Answer: a [Reason:] It states the time period (tp ) of a simple pendulum does not depend upon the mass of the body suspended at the free end of the string.

5. Law of length
a) states the time period (tp ) of a simple pendulum does not depend upon the mass of the body suspended at the free end of the string.
b) states the time period (tp ) of a simple pendulum does not depend upon its amplitude of vibration and remains the same, provided the angular amplitude (θ) does not exceed 4°.
c) states the time period (tp) of a simple pendulum is directly proportional to √L , where L is the length of the string.
d) none of the mentioned

View Answer

Answer: c [Reason:] It states the time period (tp) of a simple pendulum is directly proportional to √L , where L is the length of the string.

6. Law of gravity
a) states the time period (tp ) of a simple pendulum does not depend upon the mass of the body suspended at the free end of the string.
b) states the time period (tp ) of a simple pendulum does not depend upon its amplitude of vibration and remains the same, provided the angular amplitude (θ) does not exceed 4°.
c) states the time period (tp) of a simple pendulum is directly proportional to √L , where L is the length of the string.
d) states the time period (tp ) of a simple pendulum is inversely proportional to √g , where g is the acceleration due to gravity.

View Answer

Answer: d [Reason:] It states the time period (tp ) of a simple pendulum is inversely proportional to √g , where g is the acceleration due to gravity.

7. A helical spring, of negligible mass, and which is found to extend 0.25 mm under a mass of 1.5 kg, is made to support a mass of 60 kg. The spring and the mass system is displaced vertically through 12.5 mm and released. Determine the frequency of natural vibration of the system.
a) 6 Hz
b) 4.98 Hz
c) 5.98 Hz
d) none of the mentioned

View Answer

Answer: b [Reason:] Given : m = 60 kg ; r = 12.5 mm = 0.0125 m ; x = 5 mm = 0.005 m Since a mass of 1.5 kg extends the spring by 0.25 mm, therefore a mass of 60 kg will extend the spring by an amount, δ = 0.25/1.5 x 60 = 10 mm = 0.01m We know that frequency of the system, n = 1/2π√g/δ = 1/2π√9.81/0.01 = 4.98 Hz.

8. A helical spring, of negligible mass, and which is found to extend 0.25 mm under a mass of 1.5 kg, is made to support a mass of 60 kg. The spring and the mass system is displaced vertically through 12.5 mm and released. Find the velocity of the mass, when it is 5 mm below its rest position.
a) 0.36 m/s
b) 0.46 m/s
c) 0.56 m/s
d) none of the mentioned

View Answer

Answer: a [Reason:] Given : m = 60 kg ; r = 12.5 mm = 0.0125 m ; x = 5 mm = 0.005 m Since a mass of 1.5 kg extends the spring by 0.25 mm, therefore a mass of 60 kg will extend the spring by an amount, δ = 0.25/1.5 x 60 = 10 mm = 0.01m We know that frequency of the system, n = 1/2π√g/δ = 1/2π√9.81/0.01 = 4.98 Hz

Let v = Linear velocity of the mass. We know that angular velocity, ω = √g/δ = √9.81/0.01 = 31.32 rad/s and v = ω√r2 – x2 = 31.32√(0.0125)2 − (0.005)2 = 0.36 m/s.

Set 2

1. In a wheel and differential axle, the diameter of the effort wheel is 400 mm. The radii of the load axles are 150mm and 100 mm respectively. The diameter of the rope is 10 mm. Find the load which can be lifted by an effort of 100 N, assuming an efficiency of the machine to be 75%.
a) 800 N
b) 725 N
c) 615 N
d) none of the mentioned

View Answer

Answer: c [Reason:] Diameter of effort wheel, D = 400 mm Diameter of longer axle, d1 = 2 x 150 = 300 mm Diameter of the smaller axle, d2 = 2 x 100 = 200 mm Diameter of the rope, dr = 10 mm

therefore, V.R. = 2(D + dr )/(d1 + dr ) – (d2 + dr ) = 2(400 + 10)/ (300 + 10) – (200 + 10) = 820/100 = 8.2 Effort, P = 100 N ȵ = 75% Let W = load which can be lifted by the machine

ȵ = M.A./V.R. 0.75 = W/P x 8.2 W = 0.75 x 100 x 8.2 = 615 N.

2. Four movable pulleys are arranged as in the first system. If the weight of each pulley is 5 N, calculate the effort which can lift a load of 10 kN.
a) 629.7 N
b) 615 N
c) 625 N
d) none of the mentioned

View Answer

Answer: a [Reason:] We know that M.A. = 2nW/W + w(2n – 1) where W = load to be lifted w = weight of each pulley n = no. of movable pulleys

therefore, M.A. = 24 x 10000/10000 + 5(24 – 1) = 10000/P P = 10000 + 5(24 – 1)/ 24 = 629.7 N.

3. A person weighing 600 N platform attached to the lower block of a system of 5 pulleys arranged in the second system. The platform and the lower block weigh 100N. The man himself supports by exerting a downward pull at the free end of the rope. Neglecting friction, the minimum pull of the man will be
a) 1000 N
b) 200 N
c) 116.7 N
d) none of the mentioned

View Answer

Answer: c [Reason:] n = total number of pulleys in the system = 5 W = 600 N Weight of the lower block and platform = 100 N Total weight = 600 + 100 = 700 N Let the pull exerted by the man be P newton. Due to this pull the effective load on the lower platform will reduce to (700 – P) nP = effective load = 700 – P therefore, 5P = 700 -P 5P = 700 P = 116.7 N.

4. Five pulleys are arranged in the second system of pulleys. When not loaded the effort required to raise the movable block is 35N. Further wastage in friction increases the pull at the rate of 3% of the load lifted. What is the effort required to raise a load of 2kN?
a) 500 N
b) 400 N
c) 495 N
d) none of the mentioned

View Answer

Answer: c [Reason:] n = no. of pulleys = 5 Frictional effort at zero loading = 35N Frictional effort at 2 kN loading = 35 + 2000 x 3/100 = 95 N When the system is considered frictionless nP = W 5P = 2000 P = 400 N Hence total effort = 400 + 95 = 495 N.

5. Five pulleys are arranged in the second system of pulleys. When not loadwd the effort required to raise the movable block is 35N. Further wastage in friction increases the pull at the rate of 3% of the load lifted. What is the efficiency of the system at 2kN?
a) 80%
b) 80.81%
c) 80.50%
d) none of the mentioned

View Answer

Answer: b [Reason:] n = no. of pulleys = 5 Frictional effort at zero loading = 35N Frictional effort at 2 kN loading = 35 + 2000 x 3/100 = 95 N When the system is considered frictionless nP = W 5P = 2000 P = 400 N Hence total effort = 400 + 95 = 495 N Efficiency at this load = effort without friction/effort with friction = 400/495 x 100 = 80.81%.

6. In a weston differential pulley block, the number of recesses in the smaller wheel is 9/10 of that of the larger wheel. If the efficiency of the machine is 50%, find the load lifted by an effort of 300N.
a) 2000N
b) 3000N
c) 4000N
d) none of the mentioned

View Answer

Answer: b [Reason:] Let the recesses in the larger wheel, n1 = 10 Recesses in the smaller wheel, n2 = 9/10 x 10 = 9

V.R. = 2n1 /n1 – n2 = 2×10/10 – 9 = 20 and mechanical advantage M.A. = W/P = W/300 efficiency = M.A./V.R. 0.5 = W/300×20 W = 3000N.

7. If the velocity ratio for an open belt drive is 8 and the speed of driving pulley is 800 r.p.m, then considering an elastic creep of 2% the speed of the driven pulley is
a) 104.04 r.p.m
b) 102.04 r.p.m
c) 100.04 r.p.m
d) 98.04 r.p.m

View Answer

Answer: d [Reason:] Velocity Ratio = Velocity of belt on driver/Velocity of belt on driven Velocity of belt on driven = 800/8 = 100 r.p.m Elastic creep = velocity of belt at driven pulley – Velocity of driven pulley 0.02 × Vp = [100-Vp] Vp = 100/1.02 = 98.04r.p.m.

8. If the angle of wrap on smaller pulley of diameter 250 mm is 1200 and diameter of larger pulley is twice the diameter of smaller pulley, then the centre distance between the pulleys for an open belt drive is
a) 1000 mm
b) 750 mm
c) 500 mm
d) 250 mm

View Answer

Answer: d [Reason:] sin α = (D -d)/2c Angle of wrap on smaller pulley = п – 2α 2п/3 = п – 2sin-1(D -d)/2c c = 250 mm.

9. In short open-belt drives, an idler pulley is used in order to decrease the angle of contact on the smaller pulley for higher power transmission.
a) True
b) False

View Answer

Answer: b [Reason:] In short open-belt drives, an idler pulley is used in order to increase the angle of contact on the smaller pulley for higher power transmission.

10. In design of arms of a pulley, in belt drive, the cross-section of the arm is elliptical with minor axis placed along the plane of rotation.
a) True
b) False

View Answer

Answer: b [Reason:] Arms of a pulley in belt drive are subjected to complete reversal of stresses and is designed for bending in the plane of rotation.

Set 3

1. In a squared and ground helical spring the effective number of turns is increased by
a) 1
b) 2
c) 1.5
d) 0

View Answer

Answer: b [Reason:] In a compression spring the ends can be plain, plain and ground, squared, squared and ground. The number of inactive turns in each case are 1.5,1,1 and 2 respectively.

2. Frequency of the fluctuating load on a helical compression spring should be
a) less than natural frequency of vibration
b) twenty times the natural frequency of vibration
c) slightly greater than the natural frequency
d) twenty times the natural frequency of vibration

View Answer

Answer: d [Reason:] If the frequency of application of load matches the natural frequency of vibration, very high stresses are induced in the spring material causing fatigue failure. It can be avoided by keeping the natural frequency at least 20 times greater.

3. Two concentric springs with stiffness equal to 100 N/mm and 80 N/mm respectively, when subjected to a load of 900 N will deflect by
a) 9 mm
b) 11.25 mm
c) 5 mm
d) 31.5 mm

View Answer

Answer: c [Reason:] In this case, total stiffness Kt is given by, 1/Kt = K1 + K2 = 100+ 80 = 180N/mm

Therefore, deflection = Force/stiffness = 900N/180N/mm = 5 mm.

4. Stiffness of the spring can be increased by
a) increasing the number of turns
b) increasing the free length
c) decreasing the number of turns
d) decreasing the spring wire diameter

View Answer

Answer: c [Reason:] Stiffness of the spring can be decreased by increasing the number of turns and increased by decreasing the number of turns.

5. The stress induced in an extra full length leaf in case of a prestresed leaf spring is
a) 1.5 times that in graduated leaves
b) equal to that in graduated leaves
c) dependent on the ratio of the number of extra full length and graduated leaves
d) none of the mentioned

View Answer

Answer: b [Reason:] The leaves which are cut from original triangular leaf, are termed as graduated leaves. In actual practice some extra leaves with the same length as that of top leaf are added to increase the stiffness of the spring.

6. Initial gap between two turns of a close coiled helical tension spring should be
a) 0.5 mm
b) based on the maximum deflection
c) 1 mm
d) 0

View Answer

Answer: d [Reason:] There is no gap between two turns of a close coiled helical tension spring.

7. Due to addition of extra full length leaves the deflection of a semi-elliptic spring
a) increases
b) decreases
c) does not change
d) is doubled

View Answer

Answer: b [Reason:] In actual practice some extra leaves with the same length as that of top leaf are added to increase the stiffness of the spring.

8. A connecting rod of mass 5.5 kg is placed on a horizontal platform whose mass is 1.5 kg. It is suspended by three equal wires, each 1.25 m long, from a rigid support. The wires are equally spaced round the circumference of a circle of 125 mm radius. When the c.g. of the connecting rod coincides with the axis of the circle, the platform makes 10 angular oscillations in 30 seconds. Determine the mass moment of inertia about an axis through its c.g.
a) 0.198 kg-m2
b) 1.198 kg-m2
c) 2.198 kg-m2
d) 3.198 kg-m2

View Answer

Answer: a [Reason:] Given : m1 = 5.5 kg ; m2 = 1.5 kg ; l = 1.25 m ; r = 125 mm = 0.125 m Since the platform makes 10 angular oscillations in 30 s, therefore frequency of oscillation, n = 10/30 = 1/3 Hz Let kG = Radius of gyration about an axis through the c.g. We know that frequency of oscillation (n)

1/3 = r/2πkG √g/l = 0.125/2πkG√9.81/1.25 = 0.056/kG

kG = 0.056 x 3 = 0.168 m and mass moment of inertia about an axis through its c.g., I = mk2G = (m1 + m2)k2G = (5.5 + 1.5) (0.168)2 kg-m2 = 0.198 kg-m2.

9. In order to find the radius of gyration of a car, it is suspended with its axis vertical from three parallel wires 2.5 metres long. The wires are attached to the rim at points spaced 120° apart and at equal distances 250 mm from the axis. It is found that the wheel makes 50 torsional oscillations of small amplitude about its axis in 170 seconds. Find the radius of gyration of the wheel.
a) 168 mm
b) 268 mm
c) 368 mm
d) 468 mm

View Answer

Answer: b [Reason:] Given : l = 2.5 m ; r = 250 mm = 0.25 m ; Since the wheel makes 50 torsional oscillations in 170 seconds, therefore frequency of oscillation, n = 50/170 = 5/17 Hz

Let kG = Radius of gyration of the wheel We know that frequency of oscillation (n),

5/17 = r/2πkG √g/l = 0.25/2πkG√9.81/2.5 = 0.079/kG

kG = 0.079 x 17/5 = 0.268 m = 268 mm.

10. A small connecting rod of mass 1.5 kg is suspended in a horizontal plane by two wires 1.25 m long. The wires are attached to the rod at points 120 mm on either side of the centre of gravity. If the rod makes 20 oscillations in 40 seconds, find the radius of gyration of the rod about a vertical axis through the centre of gravity.
a) 107 mm
b) 207 mm
c) 307 mm
d) 407 mm

View Answer

Answer: a [Reason:] Given : m = 1.5 kg ; l = 1.25 m ; x = y = 120 mm = 0.12 m Since the rod makes 20 oscillations in 40 s, therefore frequency of oscillation, n = 20/40 = 0.5 Hz

Let kG = Radius of gyration of the connecting rod. We know that frequency of oscillation (n),

0.5 = 1/2πkG √gxy/l = 1/2πkG √9.81 x 0.12 x 0.12/1.25 = 0.0535/k kG = 0.0535/0.5 = 0.107 m = 107 mm.

11. A small connecting rod of mass 1.5 kg is suspended in a horizontal plane by two wires 1.25 m long. The wires are attached to the rod at points 120 mm on either side of the centre of gravity. If the rod makes 20 oscillations in 40 seconds, find the mass moment of inertia of the rod about a vertical axis through the centre of gravity.
a) 0.014 kg-m2
b) 0.015 kg-m2
c) 0.016 kg-m2
d) 0.017 kg-m2

View Answer

Answer: d [Reason:] Given : m = 1.5 kg ; l = 1.25 m ; x = y = 120 mm = 0.12 m Since the rod makes 20 oscillations in 40 s, therefore frequency of oscillation, n = 20/40 = 0.5 Hz

Let kG = Radius of gyration of the connecting rod. We know that frequency of oscillation (n),

0.5 = 1/2πkG √gxy/l = 1/2πkG √9.81 x 0.12 x 0.12/1.25 = 0.0535/k kG = 0.0535/0.5 = 0.107 m = 107 mm We know that mass moment of inertia, I = mk2G = 1.5(0.107)2 = 0.017 kg-m2.

Set 4

1. A block whose mass is 650 gm is fastened to aspring constant K equals 65 N/m whose other end is fixed. The block is pulled a distance x = 11 cm from its equilibrium position at x = 0 on a smooth surface, and released from rest at t = 0. The maximum speed ‘S’ of the oscillating block is
a) 11 cm /sec
b) 11 m/sec
c) 11 mm/sec
d) 1.1 m/sec

View Answer

Answer: d [Reason:] Maximum speed of the spring will be at x = 0, when total spring energy will be converted into kinematic energy of mass. 1/2 kx2 = 1/2 mv2 1/2 x 65 x(0.11)2 = 1/2 x (0.650)V2 V = 1.1 m/sec.

2. Which of the following statements regarding laws governing the friction between dry surfaces are correct?
a) The friction force is directly proportional to the normal force.
b) The friction force is dependent on the materials of the contact surfaces.
c) The friction force is independent of the area of contact.
d) all of the mentioned

View Answer

Answer: d [Reason:] Following are the laws of solid friction : 1. The force of friction is directly proportional to the normal load between the surfaces. 2. The force of friction is independent of the area of the contact surface for a given normal load. 3. The force of friction depends upon the material of which the contact surfaces are made. 4. The force of friction is independent of the velocity of sliding of one body relative to the other body.

3. If two bodies one light and other heavy have equal kinetic energies, which one has a greater momentum
a) heavy body
b) light body
c) both have equal momentum
d) it depends on the actual velocities

View Answer

Answer: a [Reason:] 1/2m1V12 = 1/2m2V22 v2/V1 = √m1/m2 For momentum ratio, M1/M2 = √m1/m2.

4. A heavy block of mass m is slowly placed on a conveyer belt moving with speed v. If coefficient of friction between block and the belt is μ, the block will slide on the belt through distance
a) v/μg
b) v2/√μg
c) (v/μg)2
d) v2/2μg

View Answer

Answer: d [Reason:] Retardation due to friction force = μg V2 = 2.μg s s = v2/2μg.

5. A car moving with uniform acceleration cover 450 m in a 5 second interval, and covers 700 m in the next 5 second interval. The acceleration of the car is
a) 7 m/s2
b) 50 m/s2
c) 25 m/s2
d) 10 m/s2

View Answer

Answer: d [Reason:] s = ut + 1/2at2 at t = 5 sec, s = 450 450 = u(5) + 1/2a(25) at t = 10 sec, s = 450 + 700 = 1150 1150 = u(10) = 1/2a(100) a = 10 m/sec2.

6. A particle starts with a velocity 2 m/sec and moves on a straight-line track with retardation 0.1 m/s2. The time at which the particle is 15 m from the startin g point would be
a) 10 s
b) 20 s
c) 50 s
d) 40 s

View Answer

Answer: a [Reason:] u = 2m/sec a = 0.1 m/sec s = ut – 1/2at2 15 = 2t – 1/2(0.1)t2 t = 30, t = 10.

7. Two particles with masses in the ratio 1 : 4 are moving with equal kinetic energies. The magnitude of their linear momentums will conform to the ratio
a) 1 : 8
b) 1 : 2
c) √2 : 1
d) √2

View Answer

Answer: b [Reason:] 1/2m1V12 = 1/2m2V22 m1/m2 = (V2/V1)2 = 1/4 V2/V1 = 1/2 L1/L2 = m1V1/m2V2 = 1/2.

8. A stone is projected horizontally from a cliff at 10 m/sec and lands on the ground below at 20 m from the base of the cliff. Find the height of the cliff.
a) 18 m
b) 20 m
c) 22 m
d) 24 m

View Answer

Answer: b [Reason:] h = 1/2at2 h = 1/2 x 10 x 4 = 20m.

9. A car moving with speed u can be stopped in minimum distance x when brakes are applied. If the speed becomes n times, the minimum distance over which the car can be stopped would take the value
a) x/n
b) nx
c) x/n2
d) n2x

View Answer

Answer: d [Reason:] x = u2/2g x’ = (nu)2/2g x’ = n2x.

10. Ratio of the radii of planes P1 and P2 is k and ratio of the accelerations due to gravity on them is s. Ratio of escape velocities from them will be
a) ks
b) √ks
c) √k/s
d) √s/k

View Answer

Answer: d [Reason:] v1/v2 = √g1/g2 √R2/R1 v1/v2 = √s/√k = √s/k.

Set 5

1. The velocity of sliding _____________ the distance of the point of contact from the pitch point.
a) is directly proportional to
b) is inversaly proportional to
c) is equal to cosɸ multiplied by
d) does not depend upon

View Answer

Answer: a [Reason:] The velocity of sliding is the velocity of one tooth relative to its mating tooth along the common tangent at the point of contact.

2. In involute gears, the pressure angle is
a) dependent on the size of teeth
b) dependent on the size of gears
c) always constant
d) always variable

View Answer

Answer: c

3. In full depth 140 involute system, the smallest number of teeth in a pinion which meshes with rack without interference is
a) 12
b) 16
c) 25
d) 32

View Answer

Answer: d [Reason:] The minimum number of teeth on the pinion in order to avoid interference for 14.50 full depth involute are 32 and for 200 full depth involute teeth are 18.

4. The pressure angle for involute gears depends upon the size of teeth.
a) True
b) False

View Answer

Answer: b [Reason:] In a gear drive, the shape of the tooth depends upon the pressure angle.

5. The contact ratio is given by
a) Length of the path of approach/Circular pitch
b) Length of the path of recess/Circular pitch
c) Length of the arc of contact/Circular pitch
d) Length of the arc of approach/cosɸ

View Answer

Answer: c [Reason:] None

6. For an involute gear, the ratio of base circle radius and pitch circle radius is equal to
a) sinɸ
b) cosɸ
c) secɸ
d) cosecɸ

View Answer

Answer: b

7. Which of the following statement is correct for gears?
a) The addendum is less than the dedendum
b) The pitch circle diameter is the product of module and number of teeth
c) The contact ratio means the number of pairs of teeth in contact
d) All of the mentioned

View Answer

Answer: d

8. In a gear having involute teeth, the normal to the involute is a tangent to the
a) pitch circle
b) base circle
c) addendum circle
d) dedendum circle

View Answer

Answer: b [Reason:] Addendum circle is the circle drawn through the top of the teeth and is concentric with the pitch circle. Dedendum circle is the circle drawn through the bottom of the teeth. It is also called root circle. Pitch circle is an imaginary circle which by pure rolling action, would give the same motion as the actual gear.

9. The centre distance between two meshing involute gears is equal to
a) sum of base circle radii/cosɸ
b) difference of base circle radii/cosɸ
c) sum of pitch circle radii/cosɸ
d) difference of pitch circle radii/cosɸ

View Answer

Answer: a

10. When the tip of a tooth undercuts the root on its mating gear, it is known as interference.
a) True
b) False

View Answer

Answer: a