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Multiple choice question for engineering

Set 1

1. What is the principle of fibre optical communication?
a) Frequency modulation
b) Population inversion
c) Total internal reflection
d) Doppler Effect

View Answer

Answer: c [Reason:] In optical fibres, the light entering the fibre does not encounter any new surfaces, but repeatedly they hit the same surface. The reason for confining the light beam inside the fibres is the total internal reflection.

2. What is the other name for maximum external incident angle?
a) Optical angle
b) Total internal reflection angle
c) Refraction angle
d) Wave guide acceptance angle

View Answer

Answer: d [Reason:] Only this rays which pass within the acceptance angle will be totally reflected. Therefore, light incident on the core within the maximum external incident angle can be coupled into the fibre to propagate. This angle is called as wave guide acceptance angle.

3. A single mode fibre has low intermodal dispersion than multimode. True or false?
a) True
b) False

View Answer

Answer: a [Reason:] In both single and multimode fibres the refractive indices will be in step by step. Since a single mode has less dispersion than multimode, the single mode step index fibre also has low intermodal dispersion compared to multimode step index fibre.

4. How does the refractive index vary in Graded Index fibre?
a) Tangentially
b) Radially
c) Longitudinally
d) Transversely

View Answer

Answer: b [Reason:] The refractive index of the core is maximum along the fibre axis and it gradually decreases. Here the refractive index varies radially from the axis of the fibre. Hence it is called graded index fibre.

5. Which of the following has more distortion?
a) Single step-index fibre
b) Graded index fibre
c) Multimode step-index fibre
d) Glass fibre

View Answer

Answer: c [Reason:] When rays travel through longer distances there will be some difference in reflected angles. Hence high angle rays arrive later than low angle rays. Therefore the signal pulses are broadened thereby results in a distorted output.

6. In which of the following there is no distortion?
a) Graded index fibre
b) Multimode step-index fibre
c) Single step-index fibre
d) Glass fibre

View Answer

Answer: a [Reason:] The light travels with different speeds in different paths because of the variation in their refractive indices. At the outer edge it travels faster than near the centre But almost all the rays reach the exit end at same time due to helical path. Thus, there is no dispersion in the pulses and hence the output is not a distorted output.

7. Which of the following loss occurs inside the fibre?
a) Radiative loss
b) Scattering
c) Absorption
d) Attenuation

View Answer

Answer: b [Reason:] Scattering is a wavelength dependent loss. Since the glass used in fabrication of fibres, the disordered structure of glass will make some vibrations in the refractive index inside the fibre. This causes Rayleigh scattering.

8. What causes microscopic bend?
a) Uniform pressure
b) Non-uniform volume
c) Uniform volume
d) Non-uniform pressure

View Answer

Answer: d [Reason:] Micro-bends losses are caused due to non-uniformities inside the fibre. This micro-bends in fibre appears due to non-uniform pressures created during the cabling of fibre.

9. When more than one mode is propagating, how is it dispersed?
a) Dispersion
b) Inter-modal dispersion
c) Material dispersion
d) Waveguide dispersion

View Answer

Answer: b [Reason:] When more than one mode is propagating through a fibre, then inter modal dispersion will occur. Since many modes are propagating, they will have different wavelengths and will take different time to propagate through the fibre.

10. A fibre optic telephone transmission can handle more than thousands of voice channels. True or false?
a) True
b) False

View Answer

Answer: a [Reason:] Optical fibre has larger bandwidth hence it can handle large number of channels for communication.

11. Which of the following is known as fibre optic back bone?
a) Telecommunication
b) Cable television
c) Delay lines
d) Bus topology

View Answer

Answer: d [Reason:] Each computer on the network is connected to the rest of the computers by the optical wiring scheme called bus topology, which is an application known as fibre optic back bone.

12. Calculate the numerical aperture of an optical fibre whose core and cladding are made of materials of refractive index 1.6 and 1.5 respectively.
a) 0.55677
b) 55.77
c) 0.2458
d) 0.647852

View Answer

Answer: a [Reason:] Numerical aperture = √(n2-n22 ) Numerical aperture = 0.55677

13. A step-index fibre has a numerical aperture of 0.26, a core refractive index of 1.5 and a core diameter of 100micrometer. Calculate the acceptance angle.
a) 1.47°
b) 15.07°
c) 2.18°
d) 24.15°

View Answer

Answer: b [Reason:] sin i = (Numerical aperture)/n sin i = 15.07°

Set 2

1. The mass of the moon is 1% of mass of the earth. The ratio of gravitational pull on moon and that of moon in earth will be
a) 1:1
b) 1:10
c) 1:100
d) 2:1

View Answer

Answer: a [Reason:] Gravitational forces are mutually equal and opposite

2. If the earth loses its gravity, then for a body
a) Weight becomes zero
b) Mass becomes zero
c) Neither mass nor weight is zero
d) Both mass and weight are zero

View Answer

Answer: a [Reason:] Weight = mg As g becomes zero, weight is zero. Mass is the quantity of matter which is independent of g

3. A ball is dropped from a spacecraft revolving around the earth at a height of 120km. What will happen to the ball?
a) It will go very far in the space
b) It will move with the same speed tangentially to the spacecraft
c) It will fall down to the earth gradually
d) It will continue to move with the same speed along the original orbit of the spacecraft

View Answer

Answer: d [Reason:] In the absence of external torque the ball will continue to move with the same speed along the original orbit of spacecraft.

4. The velocity with which a projectile must be fired so that it escapes the earth’s gravitational field doesn’t depend on
a) Mass of the earth
b) Mass of the projectile
c) Radius of the orbit
d) Universal gravitational constant

View Answer

Answer: a [Reason:] Escape velocity does not depend on the mass of the projectile

5. The escape velocity of a projectile from the earth is approximately
a) 7km/sec
b) 11.2km.sec
c) 112km/sec
d) 1.1km/sec

View Answer

Answer: b [Reason:] From the earth’s surface, ve=√2gR=11.2km/sec

6. The escape velocity of a body projected vertically upwards from the surface of the earth is 11km/sec. If the body is projected at an angle of 45° with the vertically, the escape velocity will be
a) 22km/sec
b) 11km/sec
c) 11√2km/sec
d) 11/√2km/sec

View Answer

Answer: b [Reason:] Escape velocity does not depend on the angle of projection of the projectile. Hence the escape velocity remains the same as the earlier condition.

7. The escape velocity from the earth is 11.2km/sec. Another plant is having a mass 1000 times and radius 10 times that of the earth, then escape velocity at that planet will be
a) 11.2km/sec
b) 112km/sec
c) 1.12km/sec
d) 1120km/sec

View Answer

Answer: b [Reason:] ve=√(2GM/R)=11.2km/sec (ve)=√((2G×100M)/(R/4))=2/9 √(2GM/R) (ve)=2/9×11.2=2.5km/sec

8. Black hole is
a) Super surface of atmosphere
b) Ozone layer
c) Super dense planetary material
d) A hole on the surface of earth

View Answer

Answer: c [Reason:] Black hole is a super dense planetary material formed due tot he continued compression the core of a star during supernova explosion.

9. If the orbital radius of the earth is made 4 times, then find the duration of the year if earlier it was T
a) T
b) 2T
c) 4T
d) 8T

View Answer

Answer: d [Reason:] T2/T1 =(4r/r )(3/2)=(4)(3/2)=8 T2=8T1=8T

10. If radius of the earth is reduced, then
a) Time duration is reduced
b) Earth rotates slower
c) Time period of earth decreases
d) Duration of day increases

View Answer

Answer: c [Reason:] I=2/5 MR2 When R decreases, the moment of inertia decreases. But Iω=consatnt ω=2π/T increases Hence the period of rotation T decreases.

11. A point mass m is placed inside a spherical shell of radius R and mass M at a distance R/2 from the centre of the shell. The gravitational force exerted by the shell on the point mass is
a) GMm/R2
b) GMm/R
c) Zero
d) 4GMm/R2

View Answer

Answer: c [Reason:] At inner position of the shell, the gravitational field intensity (E) is zero. Therefore, F=mE=0

12. A man is standing on an international space station which is orbiting earth at an altitude 520km with a constant speed 7.6km/s. If the man’s weight is 50kg, his acceleration is
a) 7.6 km/s2
b) 7.6 m/s2
c) 8.4 m/s2
d) 10 m/s2

View Answer

Answer: c [Reason:] gh=R2/(R+h)2 g gh=(6400×103)2/[(6400+520)×103]2 ×10=8.4m/s2

Set 3

1. The circular motion of a particle whose speed is constant is
a) Periodic but not simple harmonic
b) Simple harmonic but not periodic
c) Periodic and simple harmonic
d) Neither periodic not simple harmonic

View Answer

Answer: a [Reason:] Uniform circular motion is a periodic motion but not simple harmonic

2. Which of the following is simple harmonic motion?
a) Particle moving in a circle with uniform speed
b) Wave moving through a string fixed at both ends
c) Earth spinning about its axis
d) Ball bouncing between two vertical walls

View Answer

Answer: b [Reason:] Wave moving through a string fixed at both ends has simple harmonic nature

3. A particle executes simple harmonic motion along x-axis. The force acting on it is given by
a) Acos(kx)
b) Ae(-kx)
c) Akx
d) –Akx

View Answer

Answer: d [Reason:] F=-Akx implies that the force is proportional or displacement and acts in its opposite direction. So it represents simple harmonic motion.

4. Which one of the following represents simple harmonic motion?
a) Acceleration = kx
b) Acceleration = k0 x+k1 x2
c) Acceleration = -k(x+a)
d) Acceleration = k(x+a)

View Answer

Answer: c [Reason:] Acceleration = -kX, X = x+a Thus the acceleration is proportional to displacement and acts in its opposite direction. Hence, acceleration = -k(x+a) represents simple harmonic motion.

5. A particle executing simple harmonic motion of amplitude 5cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is
a) 4Hz
b) 3Hz
c) 2Hz
d) 1Hz

View Answer

Answer: d [Reason:] vmax=2πvA 31.4=2×3.14v×5 v=1Hz

6. A particle executes simple harmonic oscillation. Its amplitude is a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
a) T/8
b) T/12
c) T/2
d) T/4

View Answer

Answer: b [Reason:] y=asinωt a/2=asin⁡(2πt/T) sin⁡(2πt/T)=sin⁡(π/6) 2πt/T=π/6 T=T/12

7. A simple harmonic oscillator has an amplitude A and time period T. The time require by it to travel from x = A to x = A/2 is
a) T/6
b) T/4
c) T/3
d) T/2

View Answer

Answer: a [Reason:] As the oscillator starts from x=A, we can take x=acosωt a/2=acos⁡(2πt/T) cos⁡(2πt/T)=1/2=cos⁡(π/6) 2πt/T=π/6 or t=T/6

8. If a simple harmonic oscillator has got a displacement of 0.02m and acceleration equal to 2m/s2 at any time, the angular frequency of the oscillator is equal to
a) 10 rad/s
b) 0.1 rad/s
c) 100 rad/s
d) 1 rad/s

View Answer

Answer: a [Reason:] a=-ω2 y ω2=a/y=2/0.02=100 ω=10rad/s

9. The phase difference between acceleration of a particle executing simple harmonic motion and the instantaneous velocity is
a) π
b) 0.707π
c) Zero
d) 0.5π

View Answer

Answer: c [Reason:] The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is π/2

10. Which one of the following statements is true for the speed v and the acceleration of a particle executing simple harmonic motion?
a) When c is maximum, a is maximum
b) Value of a is zero, whatever maybe the vale of v
c) When v is zero, a is zero
d) When v is maximum, a is zero

View Answer

Answer: d [Reason:] In simple harmonic motion, acceleration is ahead of velocity in phase by π/2 rad. So when velocity is maximum, acceleration is zero and vice versa

Set 4

1. The motion of the earth about its axis is periodic and simple harmonic. True or false?
a) True
b) False

View Answer

Answer: b [Reason:] The earth takes 24 hours to complete its rotation about its axis, but the concept of to and fro motion is absent, and hence the rotation of the earth is periodic and not simple harmonic.

2. An object of mass 0.2kg executes simple harmonic motion along the x-axis with a frequency of (25/π)Hz. At the position x = 0.04, the object has kinetic energy of 0.5J and potential energy 0.4J. The amplitude of oscillation is
a) 6cm
b) 4cm
c) 8cm
d) 2cm

View Answer

Answer: a [Reason:] Total energy, E=2π2 mv2 A2 0.5+0.4=2π2×0.2×(25/π)2 A2 A2=0.9/(0.4×252) A=3/(2×25)=3/50 m=6cm

3. A spring of force constant 800N/m has an extension of 5cm. The work done in extending it from 5cm to 15cm is
a) 8J
b) 16J
c) 24J
d) 32J

View Answer

Answer: a [Reason:] At x1 = 5 cm, U1=1/2×k(x1)2=1/2×800×0.052=1J At x2=15cm, U2=1/2×k(x2)2=1/2×800×0.152=9J W=U2-U1=9-1=8J

4. A simple pendulum is attached to the roof of a lift. If time period of oscillation, when the lift is stationary is T, then frequency of oscillation when the lift falls freely, will be
a) Zero
b) T
c) 1/T
d) ∞

View Answer

Answer: a [Reason:] In a freely falling lift, g=0 v=1/2π×√(g/l)=1/2π×√(0/l)=0

5. There is a simple pendulum hanging from the ceiling of a lift. When the lift is standstill, the time period of the pendulum is T. If the resultant acceleration becomes g/4, then the new time period of the pendulum is
a) 0.8T
b) 0.25T
c) 2T
d) 4T

View Answer

Answer: c [Reason:] T=2×√(l/g) T=2π×√((l/g)/4) T=2T

6. A lightly damped oscillator with a frequency v is set in motion by a harmonic driving force of frequency v’. When v’ is lesser than v, then response of the oscillator is controlled by
a) Spring constant
b) Inertia of the mass
c) Oscillator frequency
d) Damping coefficient

View Answer

Answer: a [Reason:] Frequency of driving force is lesser than frequency v of damped oscillator. The vibrations are nearly in phase with the driving force and response of the oscillator is controlled by spring constant.

7. Statement: In simple harmonic motion, the velocity is maximum, when the acceleration is minimum [Reason:] Displacement and velocity in simple harmonic motion is differ in phase by π/2
a) Both statement and reason are true and reason is the correct explanation of the statement
b) Both statement and reason are true but reason is not the correct explanation of the statement
c) Statement is true, but reason is false
d) Statement and reason are false

View Answer

Answer: b [Reason:] Both statement and reason are true but reason is not the correct explanation of the statement, In fact, the phase difference between velocity and acceleration is π/2

8. What is time period of a pendulum hanged in a satellite? (T is the time period on earth)
a) Zero
b) T
c) Infinite
d) T/√6

View Answer

Answer: c [Reason:] In a satellite, g= 0 T=2π√(l/g)=2π√(l/0)=∞

9. Which of the following functions represents a simple harmonic oscillation?
a) sinωt-cosωt
b) sinωt+sin2ωt
c) sinωt-sin2ωt
d) sin2 ωt

View Answer

Answer: a [Reason:] y=sinωt-cosωt dy/dt=ωcosωt+ωsinωt (d2 y)/dt2 =-ω2 sinωt+ω2 cosωt =-ω2 (sinωt-cosωt) a=-ω2 y that is a∝y This satisfies the condition of simple harmonic motion

10. The displacement of a simple harmonic motion doing oscillation when kinetic energy = potential energy (amplitude = 4cm) is
a) 2√2cm
b) 2cm
c) 1/√2 cm
d) √2 cm

View Answer

Answer: a [Reason:] When kinetic energy = potential energy, y=a/√2=4/√2=2√2 cm

Set 5

1. The length of a simple pendulum executing simple harmonic motion is increased by 21%. The percentage increase in the time period of the pendulum of increases length is
a) 50%
b) 21%
c) 30%
d) 10.5%

View Answer

Answer: d [Reason:] Time period, T=2π√(l/g) The percentage increase in time period is given by, ∆T/T×100=1/2×∆l/l×100 ∆T/T×100=1/2×21%=10.5%

2. A spring of spring constant 5×103N/m is stretched initially by 5cm from the unstretched position. Then the work done to stretch is further by another 5cmis
a) 6.25Nm
b) 12.50Nm
c) 18.75Nm
d) 25Nm

View Answer

Answer: c [Reason:] W=1/2×k((x2)2-(x1)2) W=1/2×5×103×(0.102-0.052) W=18.75J

3. A particle is executing simple harmonic motion at midpoint of mean position and extremely. What is the potential energy in terms of total energy (E)?
a) E/4
b) E/16
c) E/2
d) E/8

View Answer

Answer: a [Reason:] At y = A/2, Potential energy = 1/2×kx2 Potential energy=1/2×k×A2/22 =1/4×1/2×k×A2=1/4×E

4. A mass m is suspended from a spring. Its frequency of oscillation is f. The spring is cut into two halves and the same mass is suspended from one of the two pieces of the spring. The frequency of oscillation of mass will be
a) √2 f
b) f/2
c) f
d) 2f

View Answer

Answer: a [Reason:] f=1/2π×√(k/m) When spring is cut into two halves, spring constant of each half is 2k. f=1/2π×√(2k/m)=√2 f

5. A particle executes simple harmonic motion with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
a) T/12
b) T/8
c) T/4
d) T/2

View Answer

Answer: a [Reason:] y=asinωt a/2=asin⁡(2π/T)t sin⁡(2π/T)t=sin⁡(π/6) 2π/T×t=π/6 or t=T/12

6. A particle executes simple harmonic motion with an angular velocity of 3.5 rad/sec and maximum velocity acceleration 7.5 m/s2 respectively. The amplitude of oscillations is
a) 0.28m
b) 0.36m
c) 0.707m
d) Zero

View Answer

Answer: d [Reason:] amax2 A A=amax2 =7.5/(3.5×3.5)=0.61m

7. The time period of a simple pendulum on a satellite, orbiting around the earth, is
a) Infinite
b) Zero
c) 84.6 min
d) 24 hours

View Answer

Answer: a [Reason:] In a satellite, g=0 T=2π√(l/g)=2π√(l/0)=∞

8. A simple pendulum has a time period T. The pendulum is completely immersed in a non-viscous liquid, whose density is 1/10th of that of the material of the bob. The time period of the pendulum is immersed in the liquid is
a) T
b) T/10
c) √(9/10) T
d) √(10/9) T

View Answer

Answer: d [Reason:] In air, T=2π√(l/g) Let ρ be the density of the bob material. When the bob is immersed in a non-viscous liquid of density ρ0=ρ/10, time period becomes T=2π√(l/(1-(ρ0/ρ) )g)=2π√(l/(9/10)g) T=√(10/9) T

9. Two bodies M and N of equal masses are suspended from separate massless spring of spring constants k1 and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of variation of M to that of N is
a) k1/k2
b) √(k1/k2)
c) k2/k1
d) √(k2/k1 )

View Answer

Answer: d [Reason:] vmax (A)=vmax (B) ω1 A12 A2 √(k1/m)×A1=√(k2/m)×A2 A1/A2 =√(k2/k1)

10. In forced oscillation of a particle, the amplitude is maximum for a frequency ω1 of the force, while the energy is maximum for a frequency ω2 of the force. Then
a) ω12
b) ω1 is lessed than ω2
c) ω1 is lesser than ω2, when damping is small and ω1 is greater than ω2, when damping is large
d) ω1 is lesser than ω2

View Answer

Answer: a [Reason:] Only in case of resonance, both the amplitude and energy of oscillation are maximum. Hence, ω12

11. If a simple pendulum oscillates with amplitude of 50mm and time period of 2s, then its maximum velocity is
a) 0.10m/s
b) 0.16m/s
c) 0.24m/s
d) 0.32m/s

View Answer

Answer: b [Reason:] vmax=ωA=2π/T×A vmax=(2×3.14×0.05)/2=0.16m/s

12. If the period of oscillation of mass m suspended from a spring is 2s, then the period of mass 4m will be
a) 1s
b) 4s
c) 8s
d) 16s

View Answer

Answer: b [Reason:] T2/T1 =√(m2/m1)=√(4m/m)=2 T2=2T1=2×2=4s

13. Statement: Resonance is a special case of forced vibration in which the nature and frequency of vibration of the body is the same as the impressed frequency and the amplitude of forced vibration, is maximum [Reason:] The amplitude of forced vibrations of a body increases with an increase in the frequency of the externally impressed periodic force
a) Both statement and reason are true and reason is the correct explanation of the statement
b) Both statement and reason are true but reason is not the correct explanation of the statement
c) Statement is true, but reason is false
d) Statement and reason are false

View Answer

Answer: c [Reason:] The statement is true but the reason is false. The amplitude of forced vibrations increases when the frequency of the impressed force approaches the natural frequency of the driven body.

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