Select Page
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. If the diameter of the earth becomes half its present value but its average density remains unchanged then how would be the weight of an object on the surface of the earth affected?
a) The weight of the object remains unchanged
b) The weight is doubled
c) The weight will become one fourth of the present value
d) The weight is halved

Answer: d [Reason:] Acceleration due to gravity on the surface of the earth g=GM/R2 =G/R2 ×4/3 πR3 ρ=4/3 πGRρ When the diameter or radius becomes half its present value, g=4/3 πG R/2 ρ=g/2 Hence the weight of the object will be halved.

2. The centre of mass of a body is often called
a) Centre of attraction
b) Centre of repulsion
c) Centre of gravity
d) Centre of acceleration

Answer: d [Reason:] The torque due to gravity on a body acts as it its entire mass were concentrated at its centre of mass. That is why the centre of mass of a body is often called its centre of gravity.

3. The orbital speed of Jupiter is
a) Greater than the orbital speed of earth
b) Less than the orbital speed of earth
c) Equal to the orbital speed of the earth
d) Proportional to distance from the earth

Answer: b [Reason:] v0=√((GMsun)/r) As the Jupiter is at a larger distance from the sun than the earth, so the orbital speed of Jupiter is less than that of the earth.

4. Statement 1- The comets do not obey the Kepler’s law of planetary motion
Statement 2- The comets do not have elliptical orbits
a) Statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1
b) Statement 1 and statement 2 are correct but statement 2 is not the correct explanation of statement 1
c) Statement 1 is true but statement 2 is false
d) Both statement 1 and 2 are false

Answer: b [Reason:] Both the statements are true but the statement 2 is not the correct explanation of statement 1. Some of the comets are non- periodic and move along a hyperbolic or parabolic path. They do not obey Kepler’s laws of planetary motion.

5. The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2m on the surface of A. What is the height of jump by the same person on the planet B?
a) 2/9 m
b) 18 m
c) 6 m
d) 2/ m

Answer: b [Reason:] For the same gain of potential energy in both cases MgA hA=mgB hB 9×gB×2=gB×hB hB=18m

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

Answer: c [Reason:] As no external torque is being applied, by law of conservation of angular momentum, the ball will continue to move with the same angular velocity along the original orbit of the spacecraft.

7. If the radius of earth were to shrink by one percent (its mass remains the same), then the acceleration due to gravity on the earth’s surface
a) Would decrease
b) Would remain unchanged
c) Would increase
d) Cannot be predicted

Answer: c [Reason:] Acceleration due to gravity on the earth’s surface g=GM/R2 If R decreases, then g increases. Hence the acceleration due to gravity on the earth’s surface would increase if the radius of earth were to shrink by one percent.

8. If the distance between the earth and the sun were half its present value, the number of days in year would have been
a) 64.5
b) 129
c) 182.5
d) 730

Answer: b [Reason:] According to Kepler’s law of periods T2 αR3 (T2/T1 )2=(R2/R1 )3=((R1/2)/R1 )3=1/8 T1=1/(2√2)×T1=0.353×365days=129days

9. The earth revolves around the sun with an angular velocity of 2×10-7rad/s in a circular orbit of radius 1.5×108km. The force exerted by the sun on the earth, in newton, is
a) 30×1021
b) 27×1039
c) Zero
d) 18×1025

Answer: a [Reason:] Force exerted by the sun = Centripetal force F = Mrω2 F = 6×1024×1.5×1011×(2×10-7)2 = 36×1021 N

10. What is the weight of the body at the centre of the earth?
a) Infinity
b) Same as the weight of the body on the earth
c) Zero
d) Cannot be determined

Answer: c [Reason:] The weight of the body at the centre of the earth is zero. This is because the value of g is zero at the centre of the earth.

11. A clock fitted with a pendulum and another with a spring indicate correct time on earth. Which shows correct time on the moon?
a) Clock fitted with pendulum
b) Clock fitted with spring
c) Both the clocks
d) Neither of the clocks

Answer: c [Reason:] A clock fitted with spring will show the correct time on the moon, because its time period is not affected by the variation in g.

12. What is the effect of rotation of earth on the acceleration due to gravity?
a) The acceleration due to gravity decreases
b) The acceleration due to gravity increases
c) There is no effect
d) The acceleration due to gravity first increases and then decreases

Answer: a [Reason:] The acceleration due to gravity decreases due to rotation of the earth. This effect is zero at the poles and maximum at the equators.

13. What is the mass of a body that weighs 1N at a [place where g = 9.80m/s]?
a) 0kg
b) 1.02kg
c) 0.102kg
d) 102kg

Answer: c [Reason:] m = W/g m=1/9.80=0.102kg

14. Which of the following is an evidence to show that here must be force acting on earth and directed toward the sun?
a) Deviation of the falling bodies towards east
b) Revolution of the earth round the sun
c) Phenomenon of day and night
d) Apparent motion of sun around the earth

Answer: b [Reason:] The revolution of the earth around the sun is due to the centripetal force provided but the gravitational attraction of the sun on the earth.

## Set 2

1. In which of the following ways the accelerated motion of a body will occur under an external force?
a) Due to change in direction
b) Due to inability to change its direction
c) Due to inability to change its speed
d) Due to its ability to remain at rest

Answer: a [Reason:] When the force acts perpendicular to the direction of motion, it produces circular motion. Thus accelerated motion is always due to external force and it can occur due to change in the direction.

2. Internal forces have no role to play in acceleration. True or false?
a) True
b) False

Answer: a [Reason:] The cause of every accelerated motion is an external force. Internal forces have no role to play

3. A bullet of mass 0.04kg moving with a speed of 90m/s enters a wooden block and is stopped after a distance of 60cm. What is the average resistive force exerted by the block on the bullet?
a) 15N
b) 54N
c) 22.50N
d) 270N

Answer: d [Reason:] m = 0.04kg, u = 90m/s v = 0, s = 60cm = 0.60m v2-u2 = 2as a = -6750m/s2 Therefore, retardation force = 0.04×6750 = 270N

4. A scooterist moving with a speed of 36km/h sees a child standing in the middle of the road. He applies the brakes and brings the scooter to rest in 5s just in time to save the child. Calculate the average retarding force on the vehicle, if mass of the vehicle and the driver is 300kg.
a) 270N
b) 600N
c) 150N
d) 125N

Answer: b [Reason:] u = 36km/h = 10m/s, v = 0, t = 5s a = (v-u)/t = (0-10)/5 = -2m/s2 Retardation= 2m/s2 Average retarding force = 300×2 = 600N

5. Which of the following is an example for impulsive force?
a) A person in a moving bus
b) A horse suddenly stopping
c) Force exerted by a bat hitting a ball
d) Sharpening of knife

Answer: c [Reason:] A large force acting for a short time to produce a finite charge in momentum is called an impulsive force. Thus a force exerted by a ball hitting a bat is an example for impulsive force

6. A person is falling from a certain height. Which of the following is likely to happen?
a) If he falls on cement floor, he gets more injuries
b) If he falls on a heap of sand, he receives less injuries
c) He gets the same injuries in both the cases
d) He is unaffected

Answer: a [Reason:] In both the cases, the impulse is the same. On cemented floor, the person is topped abruptly. So the cemented floor exerts a large force of reaction causing him severe injuries. When he falls on a heap of sand, the sand yields under his weight. This decreases the force exerted by the floor on the person.

7. A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12m/s. If the mass of the ball is 0.15kg, determine the impulse imparted to the ball. (Assume linear motion of the ball)
a) -45Ns
b) -3.6Ns
c) 3.78Ns
d) 60Ns

Answer: b [Reason:] m = 0.15Kg, u = 12m/s, v = -12m/s Impulse = m(v-u) = 0.15(-12-12) = -3.6Ns The negative sign indicates that the direction of the impulse is from the batsman to the bowler.

8. While launching a rocket of mass 2×104 kg, a force of 5×105 N is applied for 20s. Calculate the velocity attained by the rockets at the end of 20s.
a) 3.6m/s
b) 300m/s
c) 120m/s
d) 500m/s

Answer: d [Reason:] m = 2×104kg F = 5×105N t = 20s u = 0 v = ? Impulse of force = F×t = m(v-u) 5×105×20 = 2×104(v-0) v = 500m/s

9. Swimming is an example for which of the following?
a) Newton’s first law of motion
b) Newton’s second law of motion
c) Newton’s third law of motion
d) Law of inertia

Answer: c [Reason:] While swimming, a man pushes water backward and in turn he is pushed forward. Thus for the backward action there is an equal and opposite forward reaction.

10. Single force can exist. True or false?
a) True
b) False

Answer: b [Reason:] The force always exists in pairs. The two forces act simultaneously. Any one of them may be called the action and the other will be called reaction.

## Set 3

1. A motion can be oscillatory but not simple harmonic. True or false?
a) True
b) False

Answer: a [Reason:] A motion can be oscillatory but not simple harmonic =. When a ball is dropped from a height on a perfectly elastic surface, the motion is oscillatory but not simple harmonic as restoring force is constant.

2. A particle executes simple harmonic motion between x = -A and x = +A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 to A is T2. Then
a) T1 < T2
b) T1 > T2
c) T1=T2
d) T1=2T2

Answer: a [Reason:] x=Asinωt For x=A/2, sinωT1=1/2 ωT1=π/6 or T1=π/2ω For x=A, sinω(T1+T2)=1 T1+T2=π/2ω T2=π/2ω-π/6ω=π/3ω=2T1

3. A particle free to move along the x-axis has potential energy given by U(x) = k [1-exp -x2] for -∞≤x≤+∞, where k is a positive constant of appropriate dimensions. Then
a) The particle is in unstable equilibrium at points away from the origin
b) There is a force directed away from the origin for any finite nonzero value of x
c) It has its minimum kinetic energy at the origin if its total mechanical energy is k/2
d) The motion is simple harmonic for small displacement from x=0

Answer: d [Reason:] U=k(1-e(-x2)) F=-dU/dx=-2kxe^(-x2)=-2kx(1-x2+⋯) For small x, F ≅-2kx This shows that the force is directed towards the origin and for smaller x, F∝x. Hence the motion is simple harmonic.

4. A simple pendulum has a time period T1, when on the earth’s surface; and T2, when taken to a height R above the earth’s surface (R is the radius of the earth). The value of T2/T1 is
a) 1
b) √2
c) 4
d) 1

Answer: d [Reason:] g/g=(R/(R+R))2=1/4 As T∝1/√g T2/T1 =√(g/g )=√(4/1)=2

5. A coin is placed on a horizontal platform, which undergoes vertical simple harmonic motion of angular frequency ω. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time
a) For an amplitude of g22
b) For an amplitude of g/ω2
c) At the highest position of the platform
d) At the mean position of the platform

Answer: b [Reason:] The coin will remain in contact with the platform if amax does not exceed g, that is amax is at the most equal to g. amax=g aω2=g a=g/ω2

6. A child swinging on a swing stands up. Then the time period of the swing will
a) Increase
b) Decrease
c) Remain the same
d) Increase, if the child is long and decreases if the child is short

Answer: b [Reason:] The child and the swing together constitute a pendulum of time period, T∝2π√(l/g) As the girl stands up, her centre of gravity is raised. The distance between the point of suspension and the centre of gravity decreases, that is the length l decreases. Hence the time period T decreases.

7. If a particle is oscillating on the same horizontal plane in the ground
a) It has only kinetic energy but no potential energy
b) It has only potential energy but no kinetic energy
c) It has both kinetic and potential energies
d) It has neither kinetic nor potential energies

Answer: c [Reason:] The particle has both kinetic and potential energies.

8. If the length of simple pendulum is tripled, what will be its new time period in terms of original period T?
a) 0.7T
b) 1.73T
c) T/2
d) T

Answer: b [Reason:] T∝√l T∝√3l T=√3l=1.732T

9. Statement: The time period of a pendulum on a satellite orbiting the earth is infinity [Reason:] The time period of a pendulum is inversely proportional to square to root of g
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

Answer: b [Reason:] Both the statement and reason are true but reason is not the correct explanation of the statement. In satellite, g=0 and hence the time period is infinity

10. A spring 40mm long is stretched by the acceleration of a force. If 10N force is required to stretch the spring through 1mm, then work done in stretching through 40mm is
a) 84J
b) 48J
c) 24J
d) 8J

Answer: d [Reason:] k=F/x=10N/1mm=10N/(10-3m)=(104)N/m Work done in stretching the spring through 40mm, W=1/2×kx2=1/2×104×0.04×0.04=8J

## Set 4

1. The equation of state for g of oxygen at a pressure P, and temperature T, when occupying a volume V, will be
a) PV = (5/32)RT
b) PV = 5RT
c) PV = (5/2)RT
d) PV = (5/16)RT

Answer: a [Reason:] Number of moles in 5g of oxygen=5/32 As PV = nRT Therefore, PV = (5/32) RT

2. Two gases are at absolute temperatures 300K and 350K respectively. What is the ratio of average kinetic energy of their molecules?
a) 7:6
b) 6:7
c) 36:49
d) 49:36

Answer: b [Reason:] E1/E2 =T1/T2 300/350=6/7 6:7

3. The gas having average speed four times as that of SO2 (molecular mass 64) is
a) He(molecular mass 64)
b) O2(molecular mass 4)
c) H2(molecular mass 32)
d) CH4(molecular mass 16)

Answer: a [Reason:] vrms=√(3RT/M) v1/v2 =√(M2/M1 ) v/4v=√(M2/64) M2=4 Hence the gas is He which has molecule mass 4

4. Assertion: Air pressure in a car tyre increases during driving [Reason:] Absolute zero temperature is not zero energy temperature
a) Both assertion and reason are true and reason is the correct explanation of the assertion
b) Both assertion and reason are true but reason is not correct explanation of the assertion
c) Assertion it true but reason is false
d) Both assertion and reason are false

Answer: b [Reason:] Both the assertion and reason are true but the reason is not a correct explanation of the assertion. Air pressure in a car tyre increases due to rise in temperature during driving.

5. R/Cv =0.67 for a gas. Then this gas is made up of which molecules?
a) Diatomic
b) Mixture of diatomic and polyatomic
c) Monoatomic
d) Polyatomic

Answer: c [Reason:] Cv=f/2×R=1/0.67×R Therefore, f=2/0.67=3 Hence the gas is monoatomic

6. If a given mass of gas occupies a volume of 10cc at 1 atmospheric pressure and temperature 100°C, what will be its volume at 4 atmospheric pressure, the temperature being the same?
a) 100cc
b) 400cc
c) 104cc
d) 2.5cc

Answer: d [Reason:] From Boyle’s law, P1 V1=P2 V2 1×10=4×V2 V2=10/4=2.5cc

7. If an ideal gas has a volume V at 27°C and it is heated at a constant pressure, so that its volume becomes 1.5V, then the value of final temperature is
a) 327K
b) 177°C
c) 873K
d) 600°C

Answer: b [Reason:] At constant pressure, V2/V1 =T2/T1 1.5V/V=T2/(273+27) T2=300×1.5=450K=177℃

8. Average velocity of a gas becomes 4 times. What will be the effect on rms velocity at the same temperature?
a) 1.4 times
b) 4 times
c) 3 times
d) 2 times

Answer: b [Reason:] v=0.92vrms vrms=1/0.92×v When v becomes 4 times, (vrms)=4/0.92×v=4vrms

9. The temperature of H2 at which the rms velocity of its molecules is seven times the rms velocity of the molecules of nitrogen at 300K is
a) 2100K
b) 1700K
c) 1350K
d) 1050K

Answer: d [Reason:] vrms=√(3RT/M) v1/v2 =√((T1 M2)/(T2 M1 )) V(H2)/V(N2) =√((T(H2)×28)/(300×2))=7 T(H2)=(49×600)/28=1050K

10. We have a jar filled with gas characterised by parameters P, V and T and another jar B filled with a gas with parameters 2P, V/4,2T; where the symbols have their usual meanings. The ratio of the number of molecules of jar A to those of jar B is
a) 1:1
b) 1:2
c) 2:1
d) 4:1

Answer: d [Reason:] n1 r=PV/T n2 r=(2P)(V/2)/2T=PV/4T=1/4 n1 r n1/n2 =4/1=4:1

## Set 5

1. If suddenly the gravitational force of attraction between earth and a satellite revolving around it becomes zero, then the satellite will
a) Continue to move in its orbit with the same velocity
b) Move tangentially to the original orbit with the same velocity
c) Become stationary in its orbit
d) Move towards the earth

Answer: b [Reason:] If the gravitational force suddenly becomes zero, the satellite will move tangentially to the original orbit with the velocity, it was revolving around the earth.

2. If the radius of the earth decreases by 1% and its mass remains the same, then the acceleration due to gravity
a) Increases by 1%
b) Decreases by 1%
c) Increases by 2%
d) Decreases by 2%

Answer: c [Reason:] g=GM/R2 For constant G and M ∆g/g×100=2∆R/R×100=-2(-1)%=+2% The value of g increases by 2%

3. Gravitational law is also called
a) Universal law
b) Newton’s law
c) Constant law
d) True law

Answer: a [Reason:] Gravitational law is also called universal law because of the fact that the law of gravitation holds good for any pair of bodies in the universe whether microscopically small or astronomically large in size.

4. If the diameter if the earth becomes twice its present value but its mass remains unchanged, then how would be the weight of an object on the surface of the earth affected?
a) 1/2 W
b) 1/3 W
c) 1/4 W
d) W

Answer: c [Reason:] Weight of the body, W=mg=GMm/R2 When the diameter or radius of the earth becomes double its present value, the weight of the body will be W’=GMm/((2R)2 )=1/4 W That is, the weight will become one-fourth of the present value.

5. A body weighs more at the poles than at the equator. True or false?
a) True
b) False

Answer: a [Reason:] At the polesgp>ge, Hence mgp>mge

6. At which part of the earth’s surface, the value of g is largest?
a) At the equator
b) Immediately below the equator
c) Immediately above the equator
d) At the poles

Answer: d [Reason:] Distance of poles from the centre of the earth is smaller than the distance of any other point on earth’s surface from its centre. At poles, no centrifugal force acts on the body. Hence the value of g is largest at the poles.

7. A body weighed 250N on the surface. Assuming the earth to be a sphere of uniform mass density, how much would it weigh half way down to the centre of earth?
a) 240N
b) 210N
c) 195N
d) 125N

Answer: d [Reason:] At depth, d=R/2 gd=g(1-d/R)=g(1-(R/2)/R)=g/2 New weight = mg/2=250/2=125N

8. There is no atmosphere on the moon, because
a) It is closer to the earth and also it has inactive inter gases in it
b) It is too far from the sun and also it has very low pressure on its outer surface
c) Escape velocity of gas molecules is greater than their root mean square velocity
d) Escape velocity of gas molecules is less than their root mean square velocity

Answer: d [Reason:] The required escape velocity of gas molecules is less than their root mean square velocity, therefore there is no atmosphere on the moon.

9. A missile is launched with a velocity less than escape velocity. The sum of its kinetic energy and potential energy is
a) Zero
b) Positive
c) Negative
d) First positive and then negative