Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

# Multiple choice question for engineering

## Set 1

1. Rest and motion are relative terms. True or false?
a) True
b) False

Answer: a [Reason:] A passenger sitting in train is in rest with respect to his other passengers but is in motion with respect to the things outside the train. Thus, an object maybe at rest with respect to one object and, at the same time, in motion with respect to another object. Hence rest and motion are relative terms.

2. A train is under a journey of several hundred kilometres. How can it be regarded?
a) An object in motion
b) An object under rest
c) An object under absolute motion
d) A point object

Answer: d [Reason:] If the position of an object changes by distances much greater than its own size in a reasonable time, then the object maybe regarded as a point object.

3. A car is moving along a zigzag path on a level road. This is an example for which of the following?
a) Point object
b) Two dimensional motion
c) Three dimensional motion
d) One dimensional motion

Answer: b [Reason:] The motion of an object is said to be two dimensional if only two of the three coordinates specifying its position changes with time. Hence a car moving in a zigzag path on a level road is an example for two dimensional motions.

4. Which of the following can be regarded as an example for three dimensional motions?
a) Motion of planets around the sun
b) Motion of a train along a straight track
c) Motion of a free falling body
d) A kite flying on a windy day

Answer: d [Reason:] The motion of an object is said to be three dimensional if all the three coordinates specifying the position changes with respect time. Thus kite flying on a windy day is an example for three dimensional motions.

5. Displacement is a scalar quantity. True or false?
a) True
b) False

Answer: a [Reason:] Displacement is the change in potion of an object in a fixed direction. It has both magnitude and direction. Thus displacement is a vector quantity.

6. A body travels from A to B at 40m/s and from B to A at 60m/s. Calculate the average speed.
a) 0
b) 48m/s
c) 240m/s
d) 3.5m/s

Answer: b [Reason:] Total time taken by the body to travel from A to B and then from B to A, t1+t2 = AB/40 + AB/60 = AB/24 s Total distance covered = AB + BA = 2AB Average speed = 2AB/(t1+t2 ) = 48m/s

7. On a 60km track travels the first 30km with a uniform speed of 30km/h. How fast must the train travel the next 30km so as to average 40km.h for the entire trip?
a) 60km.h
b) 90km/h
c) 120km/h
d) 30km/h

Answer: a [Reason:] vav = (2v1 v2)/(v1+v2 ) 40 = (2×30×v2)/(30+v2 ) v2 = 60km/h

8. What is the acceleration of a bus approaching a bus stop?
a) Uniform acceleration
b) Instantaneous acceleration
c) Average acceleration
d) Negative acceleration

Answer: d [Reason:] If the velocity of an object decreases with time, its acceleration is negative. When a bus approaches its stop, its acceleration decreases, hence it has negative acceleration.

9. A jet plane starts from rest with an acceleration of 3m/s^2 and makes a run for 35s before taking off. What is the minimum length of the runway?
a) 105 m
b) 1837.5 m
c) 2451 m
d) 1204 m

Answer: b [Reason:] Minimum length of the driveway is given by s = ut + 1/2 at2 = 0 + 1/2 × 3 × 35 × 35 = 1837.5 m

10. A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving at a speed of 54km/h and the brakes cause a deceleration of 6.0m/s2, find the distance travelled by the car after he sees the need to put the brakes.
a) 18.75m
b) 225 m
c) 21.5 m
d) 12 m

Answer: c [Reason:] Distance covered in 0.20s = 15 × 0.20 = 3 m For motion with deceleration: U = 15m/s v = 0 a = -6 m/s2 As v2 – u2 = 2as s = 18.75 m Total distance travelled = 3.0 + 18.75 = 21.75 m

## Set 2

1. For motion in two or three dimensions, what is the angle between velocity and acceleration vectors?
a) 0°
b) 180°
c) Between 0° and 180°
d) 90°

Answer: c [Reason:] For motion in one dimension, the velocity and acceleration are always along the same line either in same direction or in opposite direction. For motion in two or three dimensions, the angle between velocity and acceleration vectors may have any value between 0° and 180°.

2. A cyclist moves along a circular path of radius 70m. If he completes one round in 11s, calculate total length of path.
a) 40m
b) 440m
c) 0m
d) 11m

Answer: b [Reason:] Radius of the circular path, r = 70m Time takes to complete one round, t = 11s Total length of the path, s = 2πr = 2×22/7×70 = 440m

3. An object thrown from an aeroplane is an example for
a) Projectile motion
b) Resolution of forces
c) Composition of vectors
d) Addition of vectors

Answer: a [Reason:] A projectile is the name given to any body which once thrown into space with some initial velocity, moves thereafter under the influence of gravity alone without being propelled by an engine or fuel. The path followed by a projectile is called its trajectory.

4. A ball is dropped downward from the roof of a building and simultaneously another ball is thrown in a horizontal direction, when will the balls reach the ground?
a) Same time and same place
b) The first ball will reach later than the second ball
c) The second ball will reach later than the second ball
d) Same time but different places

Answer: d [Reason:] Both the balls will reach the ground at the same time but at different places. The vertical motion is not affected by the horizontal motion. This is called principle of physical independence.

5. A body is projected horizontally from the top of a cliff with a velocity of 9.8m/s. What time elapses before horizontal and vertical velocities become equal? Take g = 9.8m/s2
a) 9.8s
b) 0s
c) 10s
d) 1s

Answer: d [Reason:] Horizontal velocity at any instant, vx = u = 9.8m/s Vertical velocity at any instant, vy = 0 + gt = 9.8t 9.8 = 9.8t t = 1s

6. Motion of the tip of second hand of the clock is an example for
a) Uniform circular motion
b) Projectile motion
c) Motion in a plane with uniform velocity
d) Motion in a plane with constant acceleration

Answer: a [Reason:] If a particle moves along a circular path with a constant speed then its motion is said to be a uniform circular motion. Hence, motion of the tip of the second hand of a clock is an example for uniform circular motion.

7. M0 L0 T(-1) is an example for
a) Angular displacement
b) Angular velocity
c) Frequency
d) Time period

Answer: b [Reason:] The rate of change of angular displacement is called angular velocity. It is measured in radian per second and its dimensional formula is M0 L0 T(-1)

8. A circular motion is an accelerated motion. True or false?
a) True
b) False

Answer: a [Reason:] In uniform circular motion, the direction of velocity vector which acts along the tangent to the path, remains constant. So circular motion is an accelerated motion.

9. In projectile motion, magnitude remains constant but the direction continuously changes. True or false?
a) True
b) False

Answer: b [Reason:] In projectile motion, both magnitude and direction of acceleration remains constant, while in uniform circular motion the magnitude remains constant but the direction continuously changes.

10. Calculate the angular speed of flywheel making 420 revolutions per minute.

Answer: d [Reason:] v = 420/60 revolution/ second ω = 2πv= 2 × 22/7 × 420/60 = 44rad/sec

11. Find the magnitude of the centripetal acceleration of a particle on the tip of a fan blade, 0.30 metre in diameter, rotation at 1200 rev/minute
a) 40 m/s2
b) 4737.6 m/s2
c) 245 m/s2
d) 20 m/s2

Answer: b [Reason:] v = 1200/60 = 20rps ω = 2πv = 2×π×20 = 40π rad/sec Magnitude of acceleration = rω2 = 0.30×(40π)2 = 4737.6 m/s2

12. Pressure is a vector. True or false?
a) True
b) False

Answer: b [Reason:] Pressure is always taken to be normal to the plane of the area on which it is acting. As this direction is unique, it does not need any specification. So pressure is not a vector.

13. What is the minimum number of coplanar vectors of different magnitudes which can give zero resultant?
a) One
b) Two
c) Three
d) Four

Answer: c [Reason:] If three vectors can be represented by the three sides of a triangle taken in the same order, then their resultant is a zero vector.

14. Two persons are pulling the ends of a string in such a way that the string is stretched horizontally. When a weight of 10kg is suspended in the middle of the string the string does not remain horizontal. Can the persons make it horizontal again by pulling it with a greater force?
a) Yes
b) No

Answer: b [Reason:] The persons cannot make it horizontal again by pulling it with a greater force because the vertical weight cannot be balanced by the horizontal force, however large the two forces may be.

## Set 3

1. In which of the following the atoms do not move from each other?
a) Shape memory alloys
b) Nano materials
c) Dielectrics
d) Static materials

Answer: b [Reason:] There exists a special type of material in which the atoms do not move away from each other and its size will be in the order of 1-100nanometers. These new materials are called nano-materials.

2. Which of the following uses radio frequency to produce nano-particles?
a) Plasma arching
b) Chemical vapour deposition
c) Sol-gel technique
d) Electro deposition

Answer: a [Reason:] Nano particles are produced by generating plasma using Radio frequency heating coils in plasma arching method. It consists of an evacuated chamber, wounded by high voltage RF coils.

3. Which of the following methods can be used to produce nano-powders of oxides?
a) Plasma arching
b) Sol-gel technique
c) Chemical vapour deposition
d) Mechanical crushing

Answer: c [Reason:] Chemical vapour deposition is used to prepare nano-powder. In this technique, initially the material is heated to form a gas i=and is allowed to deposit on a solid surface under vacuum condition, which forms nano-powders on the surface of the solid.

4. Which of the following is used to make both nano-particles and nano-powders?
a) Chemical vapour deposition
b) Sol-gel technique
c) Plasma arching
d) Electro deposition

Answer: b [Reason:] Sol-gel technique is based on hydrolysis of liquid precursors and formation of colloidal solutions. Only this method can be used for the preparation of nano-particles and nano-powders.

5. Which method can be used to prepare iron nitriles nano-crystals using ammonia gas?
a) Pulsed laser deposition
b) Sol-gel technique
c) Electro-deposition
d) Mechanical crushing

Answer: d [Reason:] Mechanical crushing is the method sued in the preparation of metal oxide nano crystals. In this method small balls are allowed to rotate inside a drum and are made to fall on a solid with high gravitational force which crushes the solid into nano-crystals. Thus iron nitriles nano-crystal can be made by using mechanical crushing.

6. Nano-particles exhibit super plastic behaviour. True or false?
a) True
b) False

Answer: a [Reason:] The hardness of nano-phase materials varies from material to material. This may be due to the phase transformation, stress relief, density and grain boundaries. Therefore nano-particles exhibit super plastic behaviour.

7. Which of the following is used to modify the optical properties of a material system?
a) Electricity
b) Magnetic field
c) Pressure
d) Light

Answer: d [Reason:] In nonlinear optics the modification of the optical properties of a material system is made by light. This is because according to electro optic effects, when the light is passed through a material it changes the properties of the medium.

8. Find the odd one out.
a) Frequency mixing
b) Second-harmonic generation
c) Optical mixing
d) Raman and Rayleigh scattering

Answer: a [Reason:] The rest of the mentioned effects exhibit Bire fingence. When light passes through a material the incident ray splits into two rays, one of the same wavelength and the other of different wavelength. This phenomenon is called double refraction. It is observed in NLO materials. Whereas frequency mixing is a passive material.

9. Which of the following is used in electro optic modulators?
a) Lithium tantalite
b) Barium sodium niobate
c) Lithium niobate
d) Lithium sodium niobate

Answer: c [Reason:] When an ordinary light is passed through the lithium niobate crystal, it is converted into polarised extraordinary light. Hence it is used in electro optic modulator.

## Set 4

1. Two plane harmonic sound waves are expressed by the equations:
y1 (x,t)=Acos(0.5πx-100πt)
y2 (x,t)=Acos(0.46πx-92πt)
All the parameters are in mks system. How many times does an observer hear maximum intensity in one second?
a) 4
b) 6
c) 8
d) 10

Answer: a [Reason:] For first wave, ω1=2πv_1=100π v1=50Hz Beat frequency=v1-v2=4Hz Hence the intensity of sound becomes maximum 4 times in one second.

2. A vibrating string of length l under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of sound in sir to 340m/s, the frequency n of the tuning fork in Hz is
a) 344
b) 336
c) 117.3
d) 109.3

Answer: a [Reason:] If v is the frequency of the string, then n-4=v Third harmonic of closed pipe, v=3v/4L Where L = length of the tube, and v = velocity of sound in air n=v+4=3v/4L+4=(3×340)/(4×0.75)+4=344Hz

3. A whistle giving out 450Hz approaches a stationary observer at a speed of 33m/s. The frequency heard by the observer in Hz is (speed of sound = 330m/s)
a) 409
b) 429
c) 517
d) 500

Answer: d [Reason:] v=v/(v-vs)×v v=330/(330-33)×450=500Hz

4. A train moves towards a stationary observer with speed 34m/s. The train sounds a whistle and its frequency registered by the observer is f1. If the train’s speed is reduced to 17m/s, the frequency registered is f2. If the speed of sound is 340m/s, then the ratio f1/f2 is
a) 18/19
b) ½
c) 2
d) 19/18

Answer: d [Reason:] For the stationary observer, f=v/(v-vs)×f f1=340/(340-34)×f f2=340/(340-17)×f f1/f2 =(340-17)/(340-34)=19/18

5. A siren placed at a railway platform is emitting sound of frequency 5kHz. A passenger sitting in a moving train ‘A’ records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B he records a frequency of 6kHz while approaching the same siren. The ratio of the velocity of train B to that train A is
a) 242/252
b) 2
c) 5/6
d) 11/6

Answer: b [Reason:] When the observer approaches stationary source, γ=(v+v0)/v×γ For train A,5.5=(v+vA)/v×5 or vA=0.1v For train B,6=(v+vB)/v×5 or vB=0.2v vB/vA =2

6. A transverse wave is described by the equation
y=y0 sin2π(ft-x/ʎ)
The maximum particles velocity is equal to four times the wave velocity if
a) ʎ=πy0/4
b) ʎ=πy0/2
c) ʎ=πy0
d) ʎ=2πy0

Answer: b [Reason:] y=y0 sin2π(ft-x/ʎ) u=dy/dt=2πfy0 cos2(ft-x/ʎ) umax=2πfy0 Wave velocity, v=fʎ 2πfy0=4×ʎf ʎ=(πy0)/2

7. The displacement of particles in a string stretched in the x-direction is represented by y. Among the following expressions for y, those describing wave motion are
a) coskxsinωt
b) k2 ω22 t2
c) cos2 (kx+ωt)
d) cos2 (k2 x22 t2)

Answer: a [Reason:] y= coskxsinωt is the only equation of standing wave. The conditions for a function of x and t to represent a wave is (∂2 y)/∂x2 =constant×(∂2 y)/∂t2 Only first expression satisfies this condition

8. A tube closed at one end and containing air produces, when excited, the fundamental note of frequency 512Hz. If the tube is open at both ends, the fundamental frequency that can be excited is (in Hz)
a) 1024
b) 512
c) 256
d) 128

Answer: a [Reason:] Fundamental frequency of a closed pipe, γ=v/4L=512Hz Fundamental frequency of open pipe γ=v/4L=2γ=2×512=1024Hz

9. An organ pipe P1 closed at one end of vibrating in its first harmonic and another pipe P2 open at both ends vibrating in its third harmonic are in resonance with a given tuning fork. The ratio of the length of P1 to that of P2 is
a) 8/3
b) 3/8
c) 1/6
d) 1/3

Answer: d [Reason:] First harmonic of closed pipe=Third harmonic of open pipe v/(4L1 )=3×v/(2L2) L1/L2 =1/6

10. A string of length 0.4m and mass 10(-2) kg is tightly clamped at its ends. The tension in the string is 1.6N. Identical wave pulses are produced at one end at equal intervals of time, ∆t. The minimum value of ∆t which allows constructive interference between successive pulses is
a) 0.05s
b) 0.10s
c) 0.20s
d) 0.40s

Answer: b [Reason:] Mass per unit length of the string, m=10-2/0.4=2.5×(10-2S)kg/m Velocity of the wave in the string, v=√(T/m)=√(1.6/(2.5×10))=8m/s For constructive interference between successive pulses, the minimum time interval is ∆tmin=2L/v=(2×0.4)/8=0.10s

## Set 5

1. A particle in simple harmonic motion is described by the displacement function x(t)=Acos⁡(ωt+θ). If the initial (t=0) position of the particle is 1cm and its initial velocity isπcm/s, what is its amplitude? The angular frequency is the particle is πrad/s
a) 1 cm
b) √2 cm
c) 2 cm
d) 2.5 cm

Answer: b [Reason:] v=ω√(A2-x2) π=π√(A2-1) A2-1=1 or A2=2 A=√2cm

2. A particle executes simple harmonic motion, its time period is 16s. If it passes through the centre of oscillation, then its velocity is 2 m/s at time 2s. The amplitude will be
a) 7.2m
b) 4cm
c) 6cm
d) 0.72m

Answer: a [Reason:] Here, t=2s,v=2m/s,T=16s v=Aωcosωt 2=A×2π/14×cos⁡(2π/16×2) A=(16√2)/π=7.2m

3. A body is executing the simple harmonic motion with an angular frequency of 2rad/sec. Velocity of the body at 20m displacement, when amplitude of motion is 60m, is
a) 90 m/s
b) 118 m/s
c) 113 m/s
d) 131 m/s

Answer: c [Reason:] v=ω√(A2-y2 )=2√(602-202 ) v=80√2 v=113m/s

4. A particle is executing simple harmonic motion of amplitude 10cm. Its time period of oscillation is π seconds. The velocity of the particle when it is 2 cm from extreme position is
a) 10 cm/s
b) 12 cm/s
c) 16√16 cm/s
d) 16 cm/s

Answer: b [Reason:] v=2π/T×√(A2-y2 ) v=2π/π √(102-82 ) =2×6=12cm/s

5. The magnitude of acceleration of particle executing simple harmonic motion at the position of maximum displacement is
a) Zero
b) Minimum
c) Maximum
d) Infinity

Answer: c [Reason:] Acceleration in simple harmonic motion is a=ω2 y At ymax=A,amax2 A Acceleration is maximum at the position of maximum displacement.

6. The maximum velocity and maximum acceleration of a body moving in a simple harmonic motion are 2m/s and 4m/s2 respectively. Then the angular velocity will be

Answer: c [Reason:] vmax=ωA,amax2 A ω=amax/vmax =4/2 ω=2rad/sec

7. A particle executing simple harmonic motion has amplitude 0.01 and frequency 60Hz. The maximum acceleration of the particle is
a) 144 π2 m/s2
b) 80 π2 m/s2
c) 120 π2 m/s2
d) 60 π2 m/s2

Answer: a [Reason:] amax2 A=4π2 v2 A =4π2×60×60×0.01=144 π2 m/s2

8. A particle having potential energy 1/3 of the maximum value at a distance of 4 cm from mean position. Amplitude of motion is
a) 4√3
b) 6/√2
c) 2/√6
d) 2√6

Answer: a [Reason:] Ep=1/3 E 1/2 ky2=1/3×1/2×kA2 A=√3 y=√3×4=4√3 cm

9. A particle executes simple harmonic motion of amplitude A. At what distance from the mean position is its kinetic energy equal to its potential energy?
a) 0.51A
b) 0.71A
c) 0.61A
d) 0.81A

Answer: b [Reason:] Ek=Ep 1/2 k(A2-y2 )=1/2×ky2 y=±A/√2 y=±0.71A

10. To show that a simple pendulum executes simple harmonic motion, it is necessary to assure that
a) Length of the pendulum is small
b) Amplitude of oscillation is small
c) Mass of the pendulum is small
d) Acceleration due to gravity is small

Answer: b [Reason:] Motion is simple harmonic only when amplitude of oscillation is small because only then f is proportional to x/.

11. Time period of a simple pendulum will be double, if we
a) Decrease the length 2 times
b) Decrease the length 4 times
c) Increase the length 2 times
d) Increase the length 4 times

Answer: d [Reason:] T=2π√(l/g) or T∝√l When length is increases four times, time period gets doubled

12. The time period of a simple pendulum is 2 sec. If its length is increased by 4 times, then its period becomes
a) 16 sec
b) 8 sec
c) 12 sec
d) 4 sec

Answer: d [Reason:] T∝√l,T∝√4l T/T=2 T=2T=2×2=4 sec

13. A simple pendulum is executing simple harmonic motion with a time period T. If the length of the pendulum is increased by 21%, the increase in the time period of the pendulum of increased length is
a) 10%
b) 30%
c) 21%
d) 50%

Answer: a [Reason:] As T is proportional to l, the percentage increase in time period on increasing the length by 21% =1/2×∆l/l×100=1/2×21=10.5%

14. A hollow spherical pendulum is filled with mercury has time period T. If mercury is thrown out completely, then the new time period
a) Increases
b) Decreases
c) Remains the same
d) First increases and then decreases

Answer: c [Reason:] Position of centre of gravity remains unaffected when mercury is thrown out. Hence effective length and time period remain same.

15. A simple pendulum is vibrating in an evacuated chamber. It will oscillate with
a) Constant amplitude
b) Increasing amplitude
c) Decreasing amplitude
d) First increasing amplitude and then decreasing amplitude