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# Multiple choice question for engineering

## Set 1

1. When the length of the vibrating segment of a sonometer wire is increased by 1%, the percentage change in its frequency is
a) 100/101
b) 99/100
c) 1
d) 2

Answer: c [Reason:] v=1/2L×√(T/m) For constant T and m, ∆v/v×100=∆L/L×100=1% Frequency will decrease by 1%

2. In an experiment with sonometer a tuning fork of frequency 256Hz resonates with a length of 25cm and another tuning fork resonates with a length of 16xcm. Tension of the spring remaining constant the frequency of the second tuning fork is
a) 163.84Hz
b) 400Hz
c) 320Hz
d) 20.4Hz

Answer: b [Reason:] v∝1/L v2/v1 =L1/L2 v2=L1/L2 ×v1=25/16×256=400Hz

3. An open pipe resonates with a tuning fork of frequency 500Hz. It is observed that two successive nodes are formed at distances 16 and 46cm from the open end. The speed of sound in air in the pipe is
a) 230 m/s
b) 300 m/s
c) 320 m/s
d) 360 m/s

Answer: b [Reason:] v=2γ(l2-l1) v=2×500(46-16)cm/s =30000cm/s=300m/s

4. The velocity of sound in open ended tube in 330m/s, the frequency of wave is 1.1 kHz and length of tube is 30cm, then number of harmonics that it will emit is
a) 2
b) 3
c) 4
d) 5

Answer: a [Reason:] For an open tube, vn=nv/2L 1.1×103=(n×330)/(2×0.30) n=2

5. An organ pipe, open at both ends produce 5 beats per second when vibrated with a source of frequency 200Hz. The second harmonic of the same pipe produces 10 beats per second with a source of frequency 420 Hz. The frequency of source is
a) 195Hz
b) 205Hz
c) 190Hz
d) 210Hz

Answer: b [Reason:] Fundamental frequency of open pipe, f=200±5=195Hz or 205Hz Second harmonics of open pipe, 2f=420±10=410Hz or 430Hz f=205Hz or 215Hz

6. Following two wave trains are approaching each other
y1=asin200πt, y2=asin208πt.
The number of beats heard per second is
a) 8
b) 4
c) 1
d) Zero

Answer: c [Reason:] ω1=2πv1=200π or v1=100Hz ω2=2πv2=208π or v2=105Hz Beat frequency=v2-v1=4Hz

7. Two waves of wavelength 99cm and 100cm both travelling with velocity 396m/s are made to interfere. The number of beats produced by them per second is
a) 1
b) 2
c) 4
d) 8

Answer: c [Reason:] Beat frequency=v[(1/ʎ1)-(1/ʎ2) ] =396[(1/0.99)-(1/1) ]=4Hz

8. Two waves are propagating with same amplitude and nearly same frequency in opposite direction, they result in
a) Beats
b) Stationary wave
c) Resonance
d) Wave packet

Answer: b [Reason:] Stationary waves are formed when two waves of same frequency travelling in opposite directions are superimposed

9. A tuning fork A produces 4beats/s with another tuning fork B of frequency 320Hz. On filing one of the prongs of A, 4beats/s is again heard when sounded with the same fork B. Then, the frequency of the fork A before filing is
a) 328Hz
b) 316Hz
C) 324Hz
d) 320Hz

Answer: b [Reason:] Frequency of A=320±4=324 or 316Hz. As frequency increases on filing, so frequency of A=316Hz (lower value).

10. A source emits a sound of frequency of 400Hz, but the listener hears it’s to be 390Hz. Then
a) The listener is moving towards the source
b) The source is moving towards the listener
c) The listener is moving away from the source
d) The listener has a defective ear

Answer: c [Reason:] As apparent frequency is lesser than actual frequency, the listener is moving away from the source.

## Set 2

1. Which of the following is an inferior planet to earth?
a) Mercury
b) Saturn
c) Pluto
d) Neptune

Answer: a [Reason:] The planets which are closer to the sun than earth are called inferior planets. Therefore Mercury and Venus are the inferior planets. The other planets are superior planets.

2. What is the method for determining the distance of an inferior planet?
a) Kepler’s third law of planetary motion
b) Triangulation method
c) Parallax method
d) Copernicus method

Answer: d [Reason:] Copernicus assumed circular orbits for the planets. The angle formed at the earth between the earth-planet direction and earth-sun direction is called the planet’s elongation.

3. What is the method for determining the distance of a superior planet?
a) Copernicus method
b) Kepler’s third law of planetary motion
c) Parallax method
d) Triangulation method

Answer: b [Reason:] The distance of a superior planet can be found using Kepler’s third law of plantary motion. This law states the square of the period of revolution of a planet around the sun is proportional to the cube of the semi-major axis of the orbit.

4. Which of the following is used to find the depth of the sea-bed?
a) Laser method
b) Sonar method
d) Reflection or echo method

Answer: b [Reason:] The word sonar stands for sound navigation and ranging. On a sonar ultrasonic waves are transmitted through the ocean. They are reflected by the submerged rocks and received by the receiver. By measuring the time delay of the receipt, the distance can be determined.

5. The shadow of a tower standing on a level plane is found to be 50m longer when the sun’s altitude is 30° that when it is 60°. Find the height if the tower.
a) 1.732 m
b) 43.3 m
c) 25 m
d) 25√3 m

Answer: b [Reason:] h = d/(cotθ2-cotθ1 ) Here, d = 50m θ1 = 60° θ2 = 30° h = 43.3 m

6. The moon is observed from two diametrically opposite points A and B on the earth. The angle θ subtended at the moon by the two directions of observation is 1°54’. Given the diameter of the earth to be 1.276×107 m, compute the distance of the moon from the earth.
a) 3.84×108 m
b) 1.276×107 m
c) 3.84m
d) 1.27m

Answer: a [Reason:] Here parallactic angle, θ = 1°54’ = 114’ = (114×60)’’ = 114×60×4.85×10-6rad = 3.32×10-2 rad Basic, b = AB = 1.276×107m The distance of the moon from the earth, S=b/θ = 3.84×108m

7. The angular diameter of the sun is 1920’’. If the distance of the sun from the earth is 1.5×1011 m, what is the linear diameter of the sun?
a) 4.85×10-6 m
b) 1.4×109m
c) 2.35×108 m
d) 1.4×108m

Answer: b [Reason:] Distance of the sun from the earth = 1.5×1011m Angular diameter of the sun = 1920×4.85×10-6 rad Linear diameter of the sun = 1.5×1011×1920×4.85×10-6 = 1.4×109 m.

8. In a submarine fitted with SONAR, the time interval between the generation of an ultrasonic wave and the receipt of its echo is 200s. What is the distance of the enemy submarine? The speed of the sound in water is 1.450km/s.
a) 2.811 km
b) 112.5 km
c) 145 km
d) 100 km

Answer: c [Reason:] The distance of the enemy submarine = (speed ×time)/2 = 145 km

9. A 35mm wide slide with a 24mm×36mm picture is projected on a screen placed 12cm from the slide. The image of the slide picture on the screen measures 1m×1.5m. What is the linear magnification of the projector-screen arrangement?
a) 150.6
b) 1736
c) 1524
d) 41.67

Answer: d [Reason:] Areal magnification = (1×1.5)/(24×10-3)×36×10-3 ) = 1736 Linear magnification = √1736 = 41.67

10. If the size of an atom ( ≅1A ) were enlarged to the tip of a sharp pin ( ≅10-5 ), how large would the height of Mount Everest be?
a) 107m
b) 108m
c) 1010m
d) 109m

Answer: d [Reason:] Magnification = 105 Apparent height of Mount Everest = Actual height × magnification = 104 × 105 = 109m

## Set 3

1. A body of mass 5kg hangs from a spring and oscillates with a time period of 2π seconds. If the body is removed, the length of the spring will decrease by
a) g/k meters
b) k/g meters
c) 2π meters
d) g meters

Answer: d [Reason:] Here T=2π√(m/k)=2π or m/k=1 In equilibrium, mg=kd d=mg/k=1×g=g meters

2. The time period of mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be
a) T/4
b) T
c) T/2
d) 2T

Answer: c [Reason:] T=2π√(m/k),k=F/x When the spring is cut into four equal parts, k=F/(x/4)=4(F/x)=4k T=2π√(m/k )=2π√(m/4k)=T/2

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

Answer: d [Reason:] T/T=√(l/l)=√(4l/l)=2 T=2T=2×2=4sec

4. If the length of a simple pendulum is increased by 2%, then the time period
a) Increases by 1%
b) Decreases by 1%
c) Increases by 2%
d) Decreases by 2%

Answer: a [Reason:] T∝√l Percentage increase in time period, ∆T/T×100=1/2×∆l/l×100 =1/2×2%=1%

5. A second’s pendulum is mounted in a rocket. Its period of oscillation will decrease when rocket is
a) Moving down with uniform acceleration
b) Moving around the earth in geostationary orbit
c) Moving up with uniform velocity
d) Moving up with uniform acceleration

Answer: a [Reason:] When at rest, T=2π√(l/g) When the rocket moves up with uniform acceleration, T=2π√(l/(g+a)) Clearly, T&ltT

6. A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by T=2π√(l/g), where g is equal to
a) g
b) g-a
c) g+a
d) √(g2+a2 )

Answer: d [Reason:] As g and a are acting along perpendicular directions, the effective value of acceleration due to gravity is g=√(g2+a2 )

7. In case of a forced vibration, the resonance peak becomes very sharp when the
a) Damping force is small
b) Restoring force is small
c) Applied periodic force is small
d) Quality factor is small

Answer: a [Reason:] When the damping force is small, the resonance peak is high and narrow.

8. A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force Fsinωt. If the amplitude of the particle is maximum for ω=ω1 and the energy of the particle maximum for ω=ω2, then
a) ω1≠ω0 and ω20
b) ω10 and ω20
c) ω1≠ω0 and ω2≠ω0
d) ω1≠ω0 and ω2≠ω0

Answer: a [Reason:] In a driven harmonic oscillator, the energy is maximum at ω20 and amplitude is maximum at frequency, ω1 is lesser than ω0 in the presence of a damping of force. Therefore, ω1≠ω0 and ω20

9. Two simple pendulums of lengths 5m and 20m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed
a) 2 oscillations
b) 1 oscillation
c) 5 oscillations
d) 3 oscillations

Answer: b [Reason:] T1/T2 =√(l1/l2 )=√(5/20)=1/2 T1=2T2 When the longer pendulum completes 2 oscillations, the shorter pendulum completes one oscillation and both are again in same phase.

10. The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of π results in the displacement of the particle along
a) Circle
b) Figure of eight
c) Straight line
d) Ellipse

Answer: c [Reason:] Let x = asinωt y=bsin(ωt+π)=-bsinωt x/a=y/b y=-b/a×x This is the equation of a straight line.

## Set 4

1. When a pebble is dropped into a pond of still water, what happens?
a) Particles move
b) Waves move
c) The pebble moves
d) Water moves

Answer: b [Reason:] When a pebble is thrown in still water, a circular pattern of alternate crests spread out. The kinetic energy makes the particles to oscillate which comes in contact with it. The energy gets transferred to the particles of next layer which also begins to oscillate. Thus it is the disturbance or waves that move forward and not the particles of the medium.

2. Mechanical waves are called elastic waves. True or false?
a) True
b) False

Answer: a [Reason:] Waves which require a medium for their propagation are called mechanical waves. They are also called elastic waves because they depend on the elastic properties of medium.

3. What are the essential properties a medium must possess for the propagation of mechanical waves?
a) Stable pressure
b) Maximum friction
c) Constant temperature
d) Minimum friction

Answer: d [Reason:] The friction force amongst the particles of the medium should be negligibly small so that they continue oscillating for a sufficiently long time and the wave travels a sufficiently long distance through the medium

4. Transverse waves can be formed in fluids. True or false?
a) True
b) False

Answer: b [Reason:] Transverse waves travel in the form of crests and troughs. They involve changes in shape of the medium. So they can be transmitted through media which have rigidity. As fluids do not sustain shearing stress, transverse waves cannot be formed in them.

5. Which of the following waves can be transmitted through solids, liquids and gases?
a) Transverse waves
b) Electromagnetic waves
c) Mechanical waves
d) Longitudinal waves

Answer: d [Reason:] Longitudinal waves involve changes in volume and density of the medium. Since all media can sustain compressive stress, longitudinal waves can be transmitted through all the three types of media.

6. For an aluminium the modulus of rigidity is 2.1×1010 N/m2 and density is 2.7×103 kg/m3. Find the speed of transverse waves in the medium.
a) 27.9×103 m/s
b) 2.79×103 m/s
c) 25.14×103 m/s
d) 24.1×103 m/s

Answer: b [Reason:] Speed = √(Ƞ/ƿ) Speed = 2.79×103 m/s

7. Sound travels through a gas under which of the following condition?
a) Isothermal condition
b) Non-isothermal condition
d) Transverse condition

Answer: c [Reason:] The compressions and rarefactions are formed so rapidly that the heat generated in the regions of compressions does not get time to pass into the regions of rarefactions so as to equalise the temperature. So when sound travels through gas, the temperature remains constant. Therefore, it is adiabatic.

8. What kind of wave is formed in organ pipes?
a) Transverse stationary waves
b) Electromagnetic waves
c) Mechanical waves
d) Longitudinal stationary waves

Answer: d [Reason:] When two identical longitudinal waves travelling in opposite directions overlap, a longitudinal stationary wave is formed. Thus, the waves produced in organ pipes are longitudinal stationary waves.

9. A wave transmits momentum. Can’t it transfer angular momentum?
a) Yes
b) No

Answer: b [Reason:] A wave transmitting momentum cannot transmit angular momentum because transfer of angular momentum means the action of a torque which causes rotator motion.

10. What is the most fundamental property of wave?
a) Temperature
b) Pressure
c) Frequency
d) Wavelength

Answer: c [Reason:] When a wave travels from one medium to other, its wavelength as well as velocity may change. This is the reason that frequency is the fundamental property of wave.

11. Which of the following is also known as pressure waves?
a) Transverse waves
b) Longitudinal waves
c) Mechanical waves
d) Stationary waves

Answer: b [Reason:] Longitudinal waves travel in a medium as series of alternate compressions and rarefactions and hence are called pressure waves.

12. In which medium sound travels faster?
a) Solid
b) Liquid
c) Gas
d) Water vapour

Answer: a [Reason:] Sound travels in solid with highest speed because the coefficient of elasticity of solids is much greater than the coefficient of elasticity of liquids and gases.

## Set 5

1. The total energy of the universe is constant. True or false?
a) True
b) False

Answer: a [Reason:] As the entire universe may be regarded as an isolated system, the total energy of the universe is constant. If one part of the universe loses energy, another part must gain an equal amount of energy.

2. How much mass is converted into energy per day in Tarapur nuclear power plant operated at 107 kW?
a) 10g
b) 9g
c) 9.6g
d) 2g

Answer: c [Reason:] Power, P = 107 kW = 1010J/s Time, t=1day=24×60×60s Energy produced per day, E = Pt = 864×1012J E = mc2 m = E/c2 = 9.6g

3. A machine gun fires 60 bullets per minute, with a velocity of 700m/s. If each bullet has a mass of 50g, find the power developed by the gun.
a) 1225W
b) 12250W
c) 122.5W
d) 122W

Answer: b [Reason:] Mass of the bullets = 60×50 = 3000g = 3kg v = 700m/s t = 1min=60s Power = W/t = (Kinetic energy)/t = 12250W

4. For a collision to occur, the actual physical contact is necessary. True or false?
a) True
b) False

Answer: b [Reason:] A collision is said to occur between two bodies, either if they physically collide against each other or if the path of one is affected by the force exerted by the other. For a collision to take place, the actual physical contact is not necessary.

5. Which of the following is an example for inelastic collision?
a) Collision between two vehicles
b) Collision between glass balls
c) A bullet fired into a wooden block
d) Collision between two railway compartments

Answer: a [Reason:] If there is a loss of kinetic energy during a collision, it is called an inelastic collision. Collision between two vehicles is an example for inelastic collision.

6. Mud thrown on a wall and sticking to it is an example for
a) Inelastic collision
b) Elastic collision
c) Super elastic collision
d) Perfectly inelastic collision

Answer: d [Reason:] If two bodies stick together after the collision and move as a single body with a common velocity, then the collision is said to be perfectly inelastic collision. A mud thrown on a wall sticks to the wall, hence it is an example for perfectly inelastic collision.

7. Collision between two carom coins is an example for
a) Oblique collision
b) Perfectly inelastic collision
c) Inelastic collision
d) Elastic collision

Answer: a [Reason:] If two bodies do not move along the same straight line path but lie in the same plane before and after the collision, the collision is said to be oblique or two dimensional collisions.

8. When a light body collides with a massive body at rest
a) The light body rebounds after collision
b) The light body Remains at rest
c) The massive body rebounds after collision
d) No reaction happens

Answer: a [Reason:] When a light body collides against a massive body at rest, the light body rebounds after collision with an equal and opposite velocity while the massive body practically remains at rest. A light ball striking a wall rebounds almost with the same speed and the wall remains at rest.

9. When a massive body collides against a light body at rest
a) The light body starts moving
b) The light body rebounds
c) The velocity of the bodies get exchanged
d) The massive body comes to rest