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# Multiple choice question for engineering

## Set 1

1. A pantograph consists of

Answer: a [Reason:] In a pantograph, all the pairs are turning pairs. It has 4 links.

2. Which of these mechanisms gives an approximately straight line?
a) hart
b) watt
c) peaucellier
d) tchebicheff

Answer: b [Reason:] Watt’s mechanism is a crossed four bar chain mechanism and was used by Watt for his early steam engines to guide the piston rod in a cylinder to have an approximate straight line motion.

3. Which of these mechanism has six links?
a) tchebicheff
b) hart
c) peaucellier
d) watt

Answer: b [Reason:] Hart’s mechanism requires only six links as compared with the eight links required by the Peaucellier mechanism.

4. Which of these mechanisms use two identical mechanisms?
a) hart
b) watt
c) peaucellier
d) none of the mentioned

5. The Davis steering gear is not used because
a) it has turning pairs
b) it is costly
c) it has sliding pair
d) it does not fulfil the condition of correct gearing

Answer: b [Reason:] Though the gear is theoretically correct, but due to the presence of more sliding members, the wear will be increased which produces slackness between the sliding surfaces, thus eliminating the original accuracy. Hence Davis steering gear is not in common use.

6. The Davis steering gear fulfils the condition of correct steering at
a) two positions
b) three positions
c) all positions
d) one positions

Answer: c [Reason:] It can be used in all positions. Ackermann steering gear fulfils the condition of correct steering at only three positions.

7. The Ackermann steering gear fulfils the condition of correct steering at
a) no position
b) one position
c) three positions
d) all positions

Answer: c [Reason:] Ackermann steering gear fulfils the condition of correct steering at only three positions.

8. A Hooke’s joint is used to join two shafts which are
a) aligned
b) intersecting
c) parallel
d) none of the mentioned

Answer: b [Reason:] Hooke’s joint is used to connect two shafts, which are intersecting at a small angle.

9. The maximum velocity of the driven shaft of a Hooke’s joint is
a) ω1 cosα
b) ω1/cosα
c) ω1 sinα
d) ω1/sinα

Answer: b [Reason:] Maximum speed of the driven shaft, ω1(max) = ωcosα/ 1 – sin2α = ωcosα/cos2α = ω/cosα.

10. The maximum velocity of the driven shaft of a Hooke’s joint is at ϴ equal to
a) 00 and 1800
b) 900 and 2700
c) 900 and 1800
d) 1800 and 2700

Answer: a [Reason:] Maximum velocity is determined at 00 ,1800 and 3600.

## Set 2

1. The direction of Corioli’s component of acceleration is the direction
a) of relative velocity vector for the two coincident points rotated by 900 in the direction of the angular velocity of the rotation of the link
b) along the centripetal acceleration
c) along tangential acceleration
d) along perpendicular to angular velocity

Answer: a [Reason:] The direction of coriolis component of acceleration will not be changed in sign if both ω and v are reversed in direction. It is concluded that the direction of coriolis component of acceleration is obtained by rotating v, at 90°, about its origin in the same direction as that of ω.

2. In a shaper mechanism, the Corioli’s component of acceleration will
a) not exist
b) exist
c) depend on position of crank
d) none of the mentioned

3. The magnitude of tangential acceleration is equal to
a) velocity2 x crank radius
b) velocity2/ crank radius
c) (velocity/ crank radius)2
d) velocity x crank radius2

Answer: b [Reason:] The magnitude of tangential acceleration is equal to velocity2/ crank radius. The magnitude of the Corioli’s component of acceleration of a slider moving at velocity V on a link rotating at angular speed ω is 2Vω.

4. Tangential acceleration direction is
a) along the angular velocity
b) opposite to angular velocity
c) perpendicular to angular velocity
d) all of the mentioned

5. The magnitude of the Corioli’s component of acceleration of a slider moving at velocity V on a link rotating at angular speed ω is
a) Vω
b) 2Vω
c) Vω/2
d) 2V/ω

Answer: b [Reason:] The magnitude of tangential acceleration is equal to velocity2/ crank radius. The magnitude of the Corioli’s component of acceleration of a slider moving at velocity V on a link rotating at angular speed ω is 2Vω.

6. In a rotary engine the angular velocity of the cylinder center line is 25 rad/sec and the relative velocity of a point on the cylinder center line w.r.t. cylinder is 10 m/sec. Corioli’s acceleration will be
a) 500m/sec2
b) 250m/sec2
c) 1000m/sec2
d) 2000m/sec2

Answer: a [Reason:] Corioli’s component = 2Vω = 2 x 10 x 25 = 500500m/sec2.

7. Corioli’s component is encountered in
a) quick return mechanism of shaper
b) four bar chain mechanism
c) slider crank mechanism
d) all of the mentioned

Answer: a [Reason:] When a point on one link is sliding along another rotating link, such as in quick return motion mechanism, then the coriolis component of the acceleration must be calculated.

8. Klein’s construction gives a graphical construction for
a) slider-crank mechanism
b) velocity polygon
c) acceleration polygon
d) none of the mentioned

Answer: c [Reason:] Klein’s construction represents acceleration polygon.

9. The velocity of a slider with reference to a fixed point about which a bar is rotating and slider sliding on the bar will be
a) parallel to bar
b) perpendicular to bar
c) both of the mentioned
d) none of the mentioned

10. Klien’s construction can be used to determine acceleration of various parts when the crank is at
a) inner dead center
b) outer dead center
c) right angles to the link of the stroke
d) all of the mentioned

Answer: d [Reason:] Klien’s construction can be used to determine acceleration in all the mentioned position.

11. The number of dead centers in a crank driven slider crank mechanism are
a) 0
b) 2
c) 4
d) 6

12. Corioli’s component acts
a) perpendicular to sliding surfaces
b) along sliding surfaces
c) both of the mentioned
d) all of the mentioned

Answer: a [Reason:] The coriolis component of acceleration is always perpendicular to the link.

13. The sense of Coriol’s component is such that it
a) leads the sliding velocity vector by 900
b) lags the sliding velocity vector by 900
c) is along the sliding velocity vector by 900
d) leads the sliding velocity vector by 1800

Answer: a [Reason:] The direction of coriolis component of acceleration is obtained by rotating v, at 90°, about its origin in the same direction as that of ω.

14. Klien’s construction can be used when
a) crank has a uniform angular velocity
b) crank has non-uniform velocity
c) crank has uniform angular acceleration
d) crank has uniform angular velocity and angular acceleration

15. Klein’s construction is useful to determine
a) velocity of various parts
b) acceleration of various parts
c) displacement of various parts
d) angular acceleration of various parts

Answer: b [Reason:] Klien’s construction can be used to determine acceleration.

## Set 3

1. The total number of instantaneous centres for a mechanism consisting of n links are
a) n/2
b) n
c) n-1
d) n(n-1)/2

Answer: d [Reason:] The number of pairs of links or the number of instantaneous centres is the number of combinations of n links taken two at a time. Mathematically, number of instantaneous centres, n(n-1)/2.

2. According to Kennedy’s theorem, if three bodies move relatively to each other, their instantaneous centres will lie on
a) straight line
b) parabolic curve
c) triangle
d) rectangle

Answer: a [Reason:] The Aronhold Kennedy’s theorem states that if three bodies move relatively to each other, they have three instantaneous centres and lie on a straight line.

3. Which of the following property of the instantaneous center is correct?
a) A rigid link rotates instantaneously relative to another link at the instantaneous centre for the configuration of the mechanism considered.
b) The two rigid links have no linear velocity relative to each other at the instantaneous centre.
c) The velocity of the instantaneous centre relative to any third link is same whether the instantaneous centre is regarded as a point on the first link or on the second rigid link.
d) all of the mentioned

4. The magnitude of velocities of the points on a rigid link is
a) directly proportional to the distance from the points to the instantaneous centre and is parallel to the line joining the point to the instantaneous centre.
b) directly proportional to the distance from the points to the instantaneous centre and is perpendicular to the line joining the point to the instantaneous centre.
c) inversely proportional to the distance from the points to the instantaneous centre and is parallel to the line joining the point to the instantaneous centre.
d) inversely proportional to the distance from the points to the instantaneous centre and is perpendicular to the line joining the point to the instantaneous centre.

5. In a mechanism, the fixed instantaneous centres are those centres which
a) remain in the same place for all configurations of the mechanism
b) vary with the configuration of the mechanism
c) moves as the mechanism moves, but joints are of permanent nature
d) none of the mentioned

Answer: a [Reason:] Fixed instantaneous centres remain in the same place for all configurations of the mechanism.

6. The instantaneous centres, which moves as the mechanism moves but joints are of permanent nature, are called permanent instantaneous centres.
a) True
b) False

Answer: a [Reason:] The permanent instantaneous centres move when the mechanism moves, but the joints are of permanent nature. Fixed instantaneous centres remain in the same place for all configurations of the mechanism.

7. The instantaneous centres which vary with the configuration of mechanism, are called
a) permanent instantaneous centres
b) fixed instantaneous centres
c) neither fixed nor permanent instantaneous centres
d) none of the mentioned

Answer: c [Reason:] Neither fixed nor permanent instantaneous centres vary with the configuration of the mechanism.

8. When two links are connected by a pin joint, their instantaneous centre lies
a) on their point of contact
b) at the centre of curvature
c) at the centre of circle
d) at the pin joint

9. The two links are said to have a pure rolling contact, when their instantaneous centre __________ on their point of contact.
a) lies
b) does not lie

10. When a slider moves on a fixed link having ____________ their instantaneous center lies at infinity.
a) straight surface
b) curved surface
c) oval surface
d) none of the mentioned

Answer: a [Reason:] When a slider moves on a fixed link having curved surface, their instantaneous centre lies at the centre of curvature. When a slider moves on a fixed link having straight surface their instantaneous center lies at infinity.

11. When a slider moves on a fixed link having curved surface, their instantaneous centre lies
a) on their point of contact
b) at the centre of curvature
c) at the centre of circle
d) at the pin joint

Answer: b [Reason:] When a slider moves on a fixed link having curved surface, their instantaneous centre lies at the centre of curvature. When a slider moves on a fixed link having straight surface their instantaneous center lies at infinity.

12. A slider moving on a fixed link having constant radius of curvature will have its instantaneous centre at the center of the circle.
a) True
b) False

13. The instantaneous center of a rigid thin disc rolling on a plane rigid surface is located at
a) the centre of the disc
b) the point of contact
c) an infinite distance on the plane surface
d) the point on the circumference situated vertically opposite to the contact point

## Set 4

1. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. Calculate the acceleration.
a) 0.3 m/s2
b) 0.4 m/s2
c) 0.5 m/s2
d) 0.6 m/s2

Answer: b [Reason:] Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m First of all, let us consider the motion of the car from rest. Acceleration of the car Let a = Acceleration of the car. We know that v2 = u2 + 2as or, 202 = 0 + 2a x 500 = 1000a or, a = 202/1000 = 0.4 m/s2

2. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. Calculate the time taken to attain the speed.
a) 50 s
b) 60 s
c) 70 s
d) 80 s

Answer: a [Reason:] Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m First of all, let us consider the motion of the car from rest. Acceleration of the car Let a = Acceleration of the car. We know that v2 = u2 + 2as or, 202 = 0 + 2a x 500 = 1000a or, a = 202/1000 = 0.4 m/s2

Let t = Time taken by the car to attain the speed. We know that v = u + a.t ∴ 20 = 0 + 0.4 × t or t = 20/0.4 = 50 s

3. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. If a further acceleration raises the speed to 90 km. p.h. in 10 seconds, find this acceleration and the further distance moved.
a) 0.3 m/s2
b) 0.4 m/s2
c) 0.5 m/s2
d) 0.6 m/s2

Answer: c [Reason:] Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m First of all, let us consider the motion of the car from rest. Acceleration of the car Let a = Acceleration of the car. We know that v2 = u2 + 2as or, 202 = 0 + 2a x 500 = 1000a or, a = 202/1000 = 0.4 m/s2

Let t = Time taken by the car to attain the speed. We know that v = u + a.t ∴ 20 = 0 + 0.4 × t or t = 20/0.4 = 50 s

Now consider the motion of the car from 72 km.p.h. to 90 km.p.h. in 10 seconds. Given : Initial velocity, u = 72 km.p.h. = 20 m/s ; Final velocity, v = 96 km.p.h. = 25 m/s ; t = 10 s Let a = Acceleration of the car. We know that v = u + a.t 25 = 20 + a × 10 or a = (25 – 20)/10 = 0.5 m2

4. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. A further acceleration raises the speed to 90 km. p.h. in 10 seconds.The brakes are now applied to bring the car to rest under uniform retardation in 5 seconds. Find the distance travelled during braking.
a) 200 m
b) 300 m
c) 225 m
d) 335 m

Answer: c [Reason:] Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m First of all, let us consider the motion of the car from rest. Acceleration of the car Let a = Acceleration of the car. We know that v2 = u2 + 2as or, 202 = 0 + 2a x 500 = 1000a or, a = 202/1000 = 0.4 m/s2

Let t = Time taken by the car to attain the speed. We know that v = u + a.t ∴ 20 = 0 + 0.4 × t or t = 20/0.4 = 50 s

Now consider the motion of the car from 72 km.p.h. to 90 km.p.h. in 10 seconds. Given : Initial velocity, u = 72 km.p.h. = 20 m/s ; Final velocity, v = 96 km.p.h. = 25 m/s ; t = 10 s Let a = Acceleration of the car.

We know that v = u + a.t 25 = 20 + a × 10 or a = (25 – 20)/10 = 0.5 m2

We know that distance moved by the car, s = ut + 1/2 at2 = 20 x 10 + 1/2 0.5(10)2 = 225 m

5. A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds. What is its angular acceleration?

Answer: a [Reason:] Given : N0 = 0 or ω = 0 ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s ; t = 20s Angular acceleration Let α = Angular acceleration in rad/s2. We know that ω = ω0 + α.t or 209.5 = 0 + α × 20 ∴ α = 209.5 / 20 = 10.475 rad/s2

6. A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds.How many revolutions does the wheel make in attaining the speed of 2000 r.p.m.?
a) 333.4
b) 444.4
c) 555.4
d) 666.4

Answer: a [Reason:] Given : N0 = 0 or ω = 0 ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s ; t = 20s Angular acceleration Let α = Angular acceleration in rad/s2. We know that ω = ω0 + α.t or 209.5 = 0 + α × 20 ∴ α = 209.5 / 20 = 10.475 rad/s2

We know that the angular distance moved by the wheel during 2000 r.p.m. (i.e. when ω = 209.5 rad/s), θ = (ω0 + ω)t/2 = ( 0 + 209.5)20/2 = 2095 rad Since the angular distance moved by the wheel during one revolution is 2π radians, therefore number of revolutions made by the wheel, n = θ /2π = 2095/2π = 333.4

7. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the linear velocity at the beginning?
a) 288.6 m/s
b) 388.6 m/s
c) 488.6 m/s
d) 188.6 m/s

Answer: d [Reason:] Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ; N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s

Linear velocity at the beginning We know that linear velocity at the beginning, v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s

8. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the linear velocity at the end of the interval ?
a) 235.5 m/s
b) 335.5 m/s
c) 435.5 m/s
d) 535.5 m/s

Answer: a [Reason:] Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ; N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s

Linear velocity at the beginning We know that linear velocity at the beginning, v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s

Linear velocity at the end of 5 seconds We also know that linear velocity after 5 seconds, v5 = r . ω = 1.5 × 157 = 235.5 m/s

9. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the normal component of the acceleration of the mid-point of the bar after 5 seconds after the acceleration begins ?
a) 2.7 m/s2
b) 3.7 m/s2
c) 4.7 m/s2
d) 5.7 m/s2

Answer: c [Reason:] Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ; N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s

Linear velocity at the beginning We know that linear velocity at the beginning, v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s

Linear velocity at the end of 5 seconds We also know that linear velocity after 5 seconds, v5 = r . ω = 1.5 × 157 = 235.5 m/s

Let α = Constant angular acceleration. We know that ω = ω0+ α.t 157 = 125.7 + α × 5 or α = (157 – 125.7) /5 = 6.26 rad/s2

Radius corresponding to the middle point, r = 1.5 /2 = 0.75 m ∴ Tangential acceleration = α. r = 6.26 × 0.75 = 4.7 m/s2

10. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the tangential component of the acceleration of the mid-point of the bar after 5 seconds after the acceleration begins ?
a) 18287 m/s2
b) 18387 m/s2
c) 18487 m/s2
d) 18587 m/s2

Answer: c [Reason:] Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ; N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s

Linear velocity at the beginning We know that linear velocity at the beginning, v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s

Linear velocity at the end of 5 seconds We also know that linear velocity after 5 seconds, v5 = r . ω = 1.5 × 157 = 235.5 m/s

Let α = Constant angular acceleration. We know that ω = ω0+ α.t 157 = 125.7 + α × 5 or α = (157 – 125.7) /5 = 6.26 rad/s2

Radius corresponding to the middle point, r = 1.5 /2 = 0.75 m ∴ Tangential acceleration = α. r = 6.26 × 0.75 = 4.7 m/s2

Radial acceleration = ω2 . r = (157)2 0.75 = 18 487 m/s2

## Set 5

1. A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate mechanical advantage.
a) 6
b) 7
c) 8
d) 9

Answer: a [Reason:] Given, load raised, W = 360 N Effort applied, P = 60 N Distance moved by the effort, y = 1.8 m Distance moved by the load, x = 200 mm = 0.2 m Mechanical advantage, M.A. = W/P = 360/60 = 6.

2. A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate velocity ratio.
a) 6
b) 7
c) 8
d) 9

Answer: d [Reason:] Given, load raised, W = 360 N Effort applied, P = 60 N Distance moved by the effort, y = 1.8 m Distance moved by the load, x = 200 mm = 0.2 m Velocity ratio, V.R. = y/x = 1.8/0.2 = 9.

3. A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate efficiency at this load.
a) 44.44%
b) 55.55%
c) 66.66%
d) 77.77%

Answer: c [Reason:] Given, load raised, W = 360 N Effort applied, P = 60 N Distance moved by the effort, y = 1.8 m Distance moved by the load, x = 200 mm = 0.2 m Mechanical advantage, M.A. = W/P = 360/60 = 6 Velocity ratio, V.R. = y/x = 1.8/0.2 = 9 Efficiency at the load, ȵ = M.A./V.R. x 100 = 6/9 x 100 = 66.66%.

4. A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate effect of friction.
a) 10 N
b) 20 N
c) 30 N
d) 40 N

Answer: b [Reason:] Given, load raised, W = 360 N Effort applied, P = 60 N Distance moved by the effort, y = 1.8 m Distance moved by the load, x = 200 mm = 0.2 m Mechanical advantage, M.A. = W/P = 360/60 = 6 Velocity ratio, V.R. = y/x = 1.8/0.2 = 9 Effort lost in friction, FP = P – W/V.R. = 60 – 360/9 = 20 N.

5. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine law of the machine.
a) P = 1/10W +30
b) P = 1/20W +30
c) P = 1/30W +30
d) P = 1/40W +30

Answer: a [Reason:] Let the law of machine be P = mW + C where P = effort applied, W = load lifted and m and C being constants. when W = 200 N P = 50 N when W = 300 N P = 60 N Putting these values in the law of machine. 50 = 200m + C …………(i) 60 = 300m + C …………(ii)

Subtracting (i) and (ii), we get 1 = 10 m or, m = 1/10 Putting this value in equation (i), we get 50 = 200 x 1/10 + C C = 30

Hence, the machine follows the laws P = 1/10W +30.

6. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine efficiency to load of 200 N.
a) 10 %
b) 15 %
c) 20 %
d) 25 %

Answer: c [Reason:] When W = 200 N, P = 50 N M.A. = W/P = 200/50 = 4 V.R. = 20 Efficiency at this load ȵ = M.A./V.R. x 100 = 4/20 x 100 = 20 %.

7. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine efficiency to load of 300 N.
a) 10 %
b) 15 %
c) 20 %
d) 25 %

Answer: d [Reason:] When W = 300 N, P = 60 N M.A. = W/P = 300/60 = 5 V.R. = 20 Efficiency at this load ȵ = M.A./V.R. x 100 = 5/20 x 100 = 25 %.

8. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine effort loss in friction at 200 N.
a) 30 N
b) 35 N
c) 40 N
d) 45 N

Answer: c [Reason:] When W = 200 N, P = 50 N Effort lost in friction, FP = P – W/ V.R. = 50 – 200/20 = 40 N.

9. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine effort loss in friction at 200 N.
a) 30 N
b) 35 N
c) 40 N
d) 45 N

Answer: d [Reason:] When W = 300 N, P = 60 N Effort lost in friction, FP = P – W/ V.R. = 60 – 300/20 = 45 N.

10. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine maximum efficiency which can be expected from this machine.
a) 30 %
b) 40 %
c) 50 %
d) 60 %