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# Multiple choice question for engineering

## Set 1

1. The unit of linear acceleration is
a) kg-m
b) m/s
c) m/s2

Answer: c [Reason:] Linear acceleration is defined as the rate of change of linear velocity of a body with respect to the time. i.e a = v/t and unit of velocity is m/s so, unit of linear acceleration becomes m/s2.

2. The angular velocity (in rad/s) of a body rotating at N r.p.m. is
a) π N/60
b) 2 π N/60
c) π N/120
d) π N/180

Answer: b [Reason:] Angular velocity is defined as the rate of change of angular displacement with respect to time. It is usually expressed by a Greek letter ω (omega). Mathematically, angular velocity, ω =dθ/dt If a body is rotating at the rate of N r.p.m. (revolutions per minute), then its angular velocity, ω = 2πΝ / 60 rad/s

3. The linear velocity of a body rotating at ω rad/s along a circular path of radius r is given by
a) ω.r
b) ω/r
c) ωs2.r
d) ωs2/r

Answer: a [Reason:] If the displacement is along a circular path, then the direction of linear velocity at any instant is along the tangent at that point. therefore, the linear velocity will be ω.r

4. When a particle moves along a straight path, then the particle has
a) tangential acceleration only
b) centripetal acceleration only
c) both tangential and centripetal acceleration
d) none of the mentioned

Answer: a [Reason:] The acceleration of a particle at any instant moving along a circular path in a direction tangential to that instant, is known as tangential component of acceleration or tangential acceleration.

5. When a particle moves with a uniform velocity along a circular path, then the particle has
a) tangential acceleration only
b) centripetal acceleration only
c) both tangential and centripetal acceleration
d) none of the mentioned

Answer: b [Reason:] The acceleration of a particle at any instant moving along a circular path in a direction normal to the tangent at that instant and directed towards the centre of the circular path is known as normal component of the acceleration or normal acceleration. It is also called radial or centripetal acceleration.

6. When the motion of a body is confined to only one plane, the motion is said to be
a) plane motion
b) rectilinear motion
c) curvilinear Motion
d) none of the mentioned

Answer: a [Reason:] When the motion of a body is confined to only one plane, the motion is said to be plane motion. The plane motion may be either rectilinear or curvilinear.

7. _______________ is the simplest type of motion and is along a straight line path.
a) plane motion
b) rectilinear motion
c) curvilinear Motion
d) none of the mentioned

Answer: b [Reason:] Rectilinear Motion is the simplest type of motion and is along a straight line path. Such a motion is also known as translatory motion.

8. _________________ is the motion along a curved path.
a) plane motion
b) rectilinear motion
c) curvilinear Motion
d) none of the mentioned

Answer: c [Reason:] Curvilinear Motion is the motion along a curved path. Such a motion, when confined to one plane, is called plane curvilinear motion.

9. Displacement of a body is a ___________ quantity.
a) scalar
b) vector
c) scalar and vector
d) none of the mentioned

Answer: b [Reason:] The displacement of a body is a vector quantity, as it has both magnitude and direction. Linear displacement may, therefore, be represented graphically by a straight line.

10. A train covers 60 miles between 2 p.m. and 4 p.m. How fast was it going at 3 p.m.?
a) 60 mph
b) 30 mph
c) 40 mph
d) 50 mph

Answer: b [Reason:] The speed is traveled distance (60 miles) divided by traveled time (4pm – 2pm = 2hours):

60 miles/ 2 hours = 30 mph

## Set 2

1. The force which acts along the radius of a circle and directed ____________ the centre of the circle is known as centripetal force.
a) away from
b) towards
c) at the
d) none of the mentioned

Answer: b [Reason:] Centripetal force acts radially inwards and is essential for circular motion.

2. The unit of mass moment of inertia in S.I. units is
a) m4
b) kgf-m-s2
c) kg-m2
d) N-m

Answer: c [Reason:] Moment of inertia is the distance, from a give reference, where the whole mass of body is assumed to be concentrated to give the same value of I. The unit of mass moment of inertia in S.I. units is kg-m2.

3. Joule is a unit of
a) force
b) work
c) power
d) none of the mentioned

Answer: b [Reason:] In S.I. system of units, the practical unit of work is N-m. It is the work done by a force of 1 newton, when it displaces a body through 1 metre. The work of 1 N-m is known as joule (briefly written as J ) such that 1 N-m = 1 J.

4. The energy possessed by a body, for doing work by virtue of its position, is called
a) potential energy
b) kinetic energy
c) electrical energy
d) chemical energy

Answer: a [Reason:] Potential energy is the energy possessed by a body for doing work, by virtue of its position. Kinetic energy is the energy possessed by a body, for doing work, by virtue of its mass and velocity of motion.

5. When a body of mass moment of inertia I (about a given axis) is rotated about that axis with an angular velocity, then the kinetic energy of rotation is
a) 0.5 I.ω
b) I.ω
c) 0.5 I.ω2
d) I.ω2

Answer: c [Reason:] When a body of mass moment of inertia I (about a given axis) is rotated about that axis, with an angular velocity ω, then it possesses some kinetic energy. In this case, Kinetic energy of rotation = 1/ 2I.ω2

When a body has both linear and angular motions e.g. in the locomotive driving wheels and wheels of a moving car, then the total kinetic energy of the body is equal to the sum of kinetic energies of translation and rotation. ∴ Total kinetic energy = 1/ 2mv2 +1/ 2I.ω2

6. The wheels of a moving car possess
a) potential energy only
b) kinetic energy of translation only
c) kinetic energy of rotation only
d) kinetic energy of translation and rotation both.

Answer: d [Reason:] in the locomotive driving wheels and wheels of a moving car, then the total kinetic energy of the body is equal to the sum of kinetic energies of translation and rotation.

7. The bodies which rebound after impact are called
a) inelastic bodies
b) elastic bodies
c) solid bodies
d) none of the mentioned

Answer: b [Reason:] The bodies, which rebound after impact are called elastic bodies and the bodies which does not rebound at all after its impact are called inelastic bodies.

8. The coefficient of restitution for inelastic bodies is
a) zero
b) between zero and one
c) one
d) more than one

Answer: a [Reason:] The process of regaining the original shape is called restitution. Inelastic bodies can not regain their original shapes. Therefore their coefficient of restitution is zero.

9. Which of the following statement is correct ?
a) The kinetic energy of a body during impact remains constant.
b) The kinetic energy of a body before impact is equal to the kinetic energy of a body after impact.
c) The kinetic energy of a body before impact is less than the kinetic energy of a body after impact.
d) The kinetic energy of a body before impact is more than the kinetic energy of a body after impact.

Answer: d [Reason:] Total kinetic energy of the system before impact, E1 = 1/2 m1 (u1)2 + 1/2 m2 (u2)2

When the two bodies move with the same velocity v after impact, then Kinetic energy of the system after impact,

E2= 1/2( m1 + m2) v2

∴ Loss of kinetic energy during impact, EL = E1 – E2

10. A body of mass m moving with a constant velocity v strikes another body of same mass m moving with same velocity but in opposite direction. The common velocity of both the bodies after collision is
a) v
b) 2 v
c) 4 v
d) 8 v

Answer: b [Reason:] If the body will move in opposite direction a negative sign would be there. We know that Common velocity = V12 Here both the velocities are same. Therefore Common velocity = V – (-V) = V + V = 2V

## Set 3

1. The two parallel and coplaner shafts are connected by gears having teeth parallel to the axis of the shaft. This arrangement is known as
a) spur gearing
b) helical gearing
c) bevel gearing
d) spiral gearing

Answer: a [Reason:] The two parallel and co-planar shafts connected by the gears. These gears are called spur gears and the arrangement is known as spur gearing.

2. The arrangement is called bevel gearing, when two __________ are connected by gears.
a) tension in the tight side of the belt
b) tension in the slack side of the belt
c) sum of the tensions on the tight side and slack side of the belt
d) average tension of the tight side and slack side of the belt

Answer: a [Reason:] The two non-parallel or intersecting, but coplanar shafts connected by gears. These gears are called bevel gears and the arrangement is known as bevel gearing.

3. When two non-intersecting and non-coplaner shafts are connected by gears,the arrangement is known as helical gearing.
a) True
b) False

Answer: b [Reason:] The two parallel and co-planar shafts connected by the gears. These gears are called spur gears and the arrangement is known as spur gearing. These gears have teeth parallel to the axis of the wheel. Another name given to the spur gearing is helical gearing.

4. The gears are termed as medium velocity gears, if their peripheral velocity is
a) 1-3 m/s
b) 3-15 m/s
c) 15-30 m/s
d) 30-50 m/s

Answer: b [Reason:] The gears having velocity less than 3 m/s are termed as low velocity gears and gears having velocity between 3 and 15 m/s are known as medium velocity gears.

5. An imaginary circle which by pure rolling action, gives the same motion as the actual gear, is called
b) dedendum circle
c) pitch circle
d) clearance circle

Answer: c [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.

6. The size of a gear is usually specified by
a) pressure angle
b) circular pitch
c) diametral pitch
d) pitch circle diameter

Answer: d [Reason:] Pitch circle diameter is the diameter of the pitch circle. The size of the gear is usually specified by the pitch circle diameter. It is also known as pitch diameter.

7. The radial distance of a tooth from the pitch circle to the bottom of the tooth is called
a) dedendum
c) clearance
d) working depth

Answer: a [Reason:] Dedendum is the radial distance of a tooth from the pitch circle to the bottom of the tooth. Addendum is the radial distance of a tooth from the pitch circle to the top of the tooth.

8. The addendum is the radial distance of tooth from the pitch circle to the top of the tooth.
a) True
b) False

Answer: a [Reason:] Dedendum is the radial distance of a tooth from the pitch circle to the bottom of the tooth. Addendum is the radial distance of a tooth from the pitch circle to the top of the tooth.

9. The working depth of a gear is radical distance from the
a) pitch circle to the bottom of a tooth
b) pitch circle to the top of a tooth
c) top of a tooth to the bottom of a tooth
d) addendum circle to the clearance circle

Answer: d [Reason:] Working depth is the radial distance from the addendum circle to the clearance circle. It is equal to the sum of the addendum of the two meshing gears.

10. The radial distance from the top of a tooth to the bottom of a tooth in a meshing gear, is called
a) dedendum
c) clearance
d) working depth

Answer: c [Reason:] Clearance is the radial distance from the top of the tooth to the bottom of the tooth, in a meshing gear. A circle passing through the top of the meshing gear is known as clearance circle.

## Set 4

1. A multiple disc clutch has five plates having four pairs of active friction surfaces. If the intensity of pressure is not to exceed 0.127 N/mm2, find the power transmitted at 500 r.p.m. The outer and inner radii of friction surfaces are 125 mm and 75 mm respectively. Assume uniform wear and take coefficient of friction = 0.3.
a) 17.8 kW
b) 18.8 kW
c) 19.8 kW
d) 20.8 kW

Answer: b [Reason:] n1 + n2 = 5 ; n = 4 ; p = 0.127 N/mm2 ; N = 500 r.p.m. or ω = 2π × 500/60 = 52.4 rad/s ; r1 = 125 mm ; r2 = 75 mm ; μ = 0.3 Since the intensity of pressure is maximum at the inner radius r2, therefore p.r2 = C or C = 0.127 × 75 = 9.525 N/mm We know that axial force required to engage the clutch, W = 2 π C (r1 – r2) = 2 π × 9.525 (125 – 75) = 2990 N and mean radius of the friction surfaces, R = r1 + r2/2 = 125 + 75/2 = 100 mm = 0.1 m We know that torque transmitted, T = n.μ.W.R = 4 × 0.3 × 2990 × 0.1 = 358.8 N-m ∴ Power transmitted, P = T.ω = 358.8 × 52.4 = 18 800 W = 18.8 kW.

2. A single plate clutch, with both sides effective, has outer and inner diameters 300 mm and 200 mm respectively. The maximum intensity of pressure at any point in the contact surface is not to exceed 0.1 N/mm2. If the coefficient of friction is 0.3, determine the power transmitted by a clutch at a speed 2500 r.p.m.
a) 61.693 kW
b) 71.693 kW
c) 81.693 kW
d) 91.693 kW

Answer: a [Reason:] Given : d1 = 300 mm or r1 = 150 mm ; d2 = 200 mm or r2 = 100 mm ; p = 0.1 N/mm2 ; μ = 0.3 ; N = 2500 r.p.m. or ω = 2π × 2500/60 = 261.8 rad/s Since the intensity of pressure ( p) is maximum at the inner radius (r2), therefore for uniform wear, p.r2 = C or C = 0.1 × 100 = 10 N/mm We know that the axial thrust, W = 2 π C (r1 – r2) = 2 π × 10 (150 – 100) = 3142 N and mean radius of the friction surfaces for uniform wear, R = r1 + r2/2 = 150 + 100/2 = 125 mm = 0.125m We know that torque transmitted, T = n.μ.W.R = 2 × 0.3 × 3142 × 0.125 = 235.65 N-m …( n = 2,for both sides of plate effective) ∴ Power transmitted by a clutch, P = T.ω = 235.65 × 261.8 = 61 693 W = 61.693 kW.

3. A 60 mm diameter shaft running in a bearing carries a load of 2000 N. If the coefficient of friction between the shaft and bearing is 0.03, find the power transmitted when it runs at 1440 r.p.m.
a) 171.4 W
b) 271.4 W
c) 371.4 W
d) 471.4 W

Answer: b [Reason:] Given : d = 60 mm or r = 30 mm = 0.03 m ; W = 2000 N ; μ = 0.03 ; N = 1440 r.p.m. or ω = 2π × 1440/60 = 150.8 rad/s We know that torque transmitted, T = μ.W.r = 0.03 × 2000 × 0.03 = 1.8 N-m ∴ Power transmitted, P = T.ω = 1.8 × 150.8 = 271.4 W.

4. The force of friction is inversely proportional to the normal load between the surfaces.
a) True
b) False

Answer: b [Reason:] The force of friction is directly proportional to the normal load between the surfaces.

5. The force of friction is dependent of the area of the contact surface for a given normal load.
a) True
b) False

Answer: b [Reason:] The force of friction is independent of the area of the contact surface for a given normal load.

6. The force of friction depends upon the material of which the contact surfaces are made.
a) True
b) False

Answer: a [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.

7. The force of friction is dependent of the velocity of sliding of one body relative to the other body.
a) True
b) False

Answer: b [Reason:] The force of friction is independent of the velocity of sliding of one body relative to the other body.

8. The force of friction is almost dependent of the load.
a) True
b) False

Answer: b [Reason:] The force of friction is almost independent of the load.

9. The force of friction is dependent of the substances of the bearing surfaces.
a) True
b) False

Answer: b [Reason:] The force of friction is independent of the substances of the bearing surfaces.

10. The force of friction is _____________ for different lubricants.
a) same
b) different
c) zero
d) none of the mentioned

Answer: a [Reason:] The force of friction is different for different lubricants.

## Set 5

1. The force of friction always acts in a direction, ___________ to that in which the body tends to move.
a) same
b) opposite
c) both of the mentioned
d) none of the mentioned

Answer: b [Reason:] The force of friction always acts in a direction, opposite to that in which the body tends to move.

2. The magnitude of the force of friction is ____________ to the force, which tends the body to move.
a) equal
b) different
c) both of the mentioned
d) none of the mentioned

Answer: a [Reason:] The magnitude of the force of friction is exactly equal to the force, which tends the body to move.

3. The magnitude of the limiting friction (F ) bears a constant ratio to the normal reaction (RN) between the two surfaces.
a) True
b) False

Answer: a [Reason:] The magnitude of the limiting friction (F ) bears a constant ratio to the normal reaction (RRN) between the two surfaces. Mathematically F/RRN = constant.

4. The force of friction is _____________ of the area of contact, between the two surfaces.
a) dependent
b) independent
c) both of the mentioned
d) none of the mentioned

Answer: b [Reason:] The force of friction is independent of the area of contact, between the two surfaces.

5. The force of friction does not depends upon the roughness of the surfaces.
a) True
b) False

Answer: b [Reason:] The force of friction depends upon the roughness of the surfaces.

6. The ratio of magnitude of the kinetic friction to the normal reaction between the two surfaces is_____________ than that in case of limiting friction.
a) greater
b) less
c) equal
d) none of the mentioned

Answer: b [Reason:] The magnitude of the kinetic friction bears a constant ratio to the normal reaction between the two surfaces. But this ratio is slightly less than that in case of limiting friction.

7. For moderate speeds, the force of friction
a) increases
b) decreases
c) remains constant
d) none of the mentioned

Answer: c [Reason:] For moderate speeds, the force of friction remains constant. But it decreases slightly with the increase of speed.

8. A body of mass 400 g slides on a rough horizontal surface. If the frictional force is 3.0 N, find the angle made by the contact force on the body with the vertical.
a) 350
b) 360
c) 370
d) 380

Answer: c [Reason:] Let the contact force on the block by the surface be F which makes an angle ϴ with the vertical. The component of F perpendicular to the contact surface is the normal force N and the component of F parallel to the surface is the friction f As the surface is horizontal, N is vertically upward. For vertical equilibrium, N = mg = (0.400 kg) (10 m/s2) = 4.0 N. The frictional force is f = 3.0 N. tan ϴ = f/N = 3/4 or, ϴ = tan-1 (3/4) = 370.

9. A body of mass 400 g slides on a rough horizontal surface. If the frictional force is 3.0 N, find the magnitude of the contact force.
a) 5 N
b) 10 N
c) 15 N
d) 20 N

Answer: a [Reason:] Let the contact force on the block by the surface be F which makes an angle ϴ with the vertical. The component of F perpendicular to the contact surface is the normal force N and the component of F parallel to the surface is the friction f As the surface is horizontal, N is vertically upward. For vertical equilibrium, N = mg = (0.400 kg) (10 m/s2) = 4.0 N. The frictional force is f = 3.0 N. F = √N2 = √42 + 32 = 5 N.

10. A heavy box of mass 20 kg is pulled on a horizontal surface by applying a horizontal force. If the coefficient of kinetic friction between the box and the horizontal surface is 0.25, find the force of friction exerted by the horizontal surface on the box.
a) 29 N
b) 39 N
c) 49 N
d) 59 N

Answer: c [Reason:] As the box slides on the horizontal surface, the surface exerts kinetic friction on the box. The magnitude of the kinetic friction is f = μN = μmg = 0.25 x (20 kg) x (9.8 m/s 2) = 49 N. This force acts in the direction opposite to the pull.

11. A boy (30 kg) sitting on his horse whips it. The horse speeds up at an average acceleration of 2.0 m/s 2.If the boy does not slide back, what is the force of friction exerted by the horse on the boy ?
a) 20 N
b) 30 N
c) 40 N
d) 60 N

Answer: d [Reason:] The forces acting on the boy are (i) the weight Mg. (ii) the normal contact force N and (iii) the static friction fs

As the boy does not slide back, its acceleration a is equal to the acceleration of the horse. As friction is the only horizontal force, it must act along the acceleration and its magnitude is given by Newton’s second law fs = Ma = (30 kg) (2.0 m/s2) = 60 N.

12. A boy (30 kg) sitting on his horse whips it. The horse speeds up at an average acceleration of 2.0 m/s 2.If the boy slides back during the acceleration, what can be said about the coefficient of static friction between the horse and the boy.
a) 0.10
b) 0.20
c) 0.30
d) 0.40

Answer: b [Reason:] If the boy slides back, the horse could not exert a friction of 60 N on the boy. The maximum force of static friction that the horse may exert on the boy is fs = μs = μsMg

μs(30 kg) (10m/s2) = μs 300 N where μs is the coefficient of static friction. Thus, μs(300 N) <60 N or, μs <60/300 = 0.20.

13. A wooden block is kept on a polished wooden plank and the inclination of the plank is gradually increased. It is found that the block starts slipping when the plank makes an angle of 18° with the horizontal. However, once started the block can continue with uniform speed if the inclination is reduced to 15°. Find the coefficient of static friction between the block and the plank.
a) tan 180
b) tan 150
c) tan 330
d) tan 30

Answer: a [Reason:] The coefficient of static friction is μs = tan 180.

14. A wooden block is kept on a polished wooden plank and the inclination of the plank is gradually increased. It is found that the block starts slipping when the plank makes an angle of 18° with the horizontal. However, once started the block can continue with uniform speed if the inclination is reduced to 15°. Find the coefficient of kinetic friction between the block and the plank.
a) tan 180
b) tan 150
c) tan 330
d) tan 30