Question Bank Machines - Testlabz

Class-X Questions Bank 1

Question Bank

Machines

1. (a) What is a simple machine?

(b) State three functions of a machine.

or

What is the purpose of a machine?

Ans.(a) Machine : It is a device, by which we can overcome resistance, or gain

speed by applying comparatively small force, at a convenient point, and in

a desired direction.

(b) (i) A machine multiplies effort, i.e., it lifts more load for small effort.

(ii) A machine multiplies speed, i.e., effort acts through smaller distance and

load moves through larger distance.

(iii) A machine changes the direction of the applied effort.

2. (a) Name six simple machines.

(b) Give one practical example of each machine named in (a).

Ans. (a) (i) Lever, (ii) Pulley, (iii) Wheel and axle,

(iv) Inclined plane, (v) Wedge, (vi) Screw

(b) (i) Lever : The handle of a waterpump helps in multiplying effort.

(ii) Pulley : A single movable pulley is used for changing the direction of

the effort.


(iii) Wheel and axle : A screwdriver is the example.

(iv) Inclined plane : A staircase is an example.

(v) Wedge : A knife is an example.

(vi) Screw : A jackscrew is an example.

3. What do you understand by the term ideal machine?

Ans. A machine whose parts are weightless and frictionless is called an ideal

machine.

4. Explain the term mechanical advantage and state its unit.

Ans.The ratio between the useful load (resistance over come) moved by a machine

to the effort applied on it is called mechanical advantage. It is a pure number

and has no unit.

5. Define the term velocity ratio and state its unit.

Ans. It is the ratio of velocity at which effort is applied to the velocity at which the

load moves. It is a pure number and has no unit.

6. Define the term ideal mechanical advantage and state its unit?

Ans. The ratio between the total load (useful load + load of moving parts of

machine) moved by a machine to the effort applied on it is called ideal

mechanical advantage. It is a pure number and has no unit.

7. Define the term efficiency of a machine and state its unit?

Ans. The ratio between the mechanical advantage and velocity ratio is called its

efficiency. It is a pure number and has no units.


8. Prove that efficiency of a machine is the ratio between actual mechanical

advantage and velocity ratio.

Or

State the relationship between mechanical advantage velocity ratio and

efficiency.

Ans. Let ‘l’ be the useful load, which moves through a distance ‘d’, when an effort

‘E’ acts through a distance ‘D’.

Input = E × D; Output = l × d

Now, Efficiency of machine

(η ) = OutputInput

= × D M.A

= ÷ =E × D E d V.Rl d l

( ) = ηM.AV.R

9. Give two reasons, why a machine cannot be 100% efficient.

Ans. (i) A part of the effort is wasted in overcoming friction.

(ii) A part of the effort is wasted in moving the movable parts of the machine.

Thus, a machine cannot be 100% efficient.

10. What is a lever?

Ans. A lever is a straight or bent rigid bar capable of turning around a fixed point

called fulcrum.


11. What is the principle of lever?

Ans.When a lever is in equilibrium, the clockwise movements of the load about the

fulcrum is equal to the anticlockwise moment of the effort about the fulcrum,

provided the lever is weightless and offers no friction.

12. Name and define three classes of levers and give two examples for each kind.

Ans. (i) Lever of 1st order.

A lever in which the fulcrum acts in the middle, the load on one side and the

effort on the other side is called lever of the 1st order.

Scissors and crow-bar are the examples of lever of 1st order.

(ii) Lever of 2nd order

A lever in which the load acts in the middle, the effort on one side and

fulcurm on the other side is called lever of 2nd order.

A nut cracker and a wheel barrow are the examples of lever of 2nd order.

(iii) Lever of 3rd order

A lever in which effort acts in the middle, the load on one side and fulcrum

on the other side is called lever of 3rd order.

A fire tong and fishing rod are examples of lever of 3rd order.

13. How will you determine the order (kind) of a lever?

Ans. Locate, which amongst the effort, the fulcrum or the load is in the middle. If the

fulcrum is in the middle, the lever is of the first order. If the load is in the

middle, the lever is of the second order. If the effort is in the middle, the lever is

of the third order.


14. (a) Why the lever of the second order has mechanical advantage more than one?

(b) Why the lever of the third order has mechanical advantage less than one?

Give one example of this class of lever.

Ans. The mechanical advantage of lever is given by the expression, effort arm ÷ load

arm.

(a) In case of lever of second order, the effort arm is always longer than the load

arm and hence, its mechanical advantage is more than 1.

(b) In case of lever of the third order, the effort arm is always smaller than the

load arm and hence, its mechanical advantage is less than 1. For example, a

fire tong is a lever of third order.

15. To which order do the following levers belong and why?

(i) Railway signal, (ii) a man cutting bread with knife, (iii) a boy writing on a

piece of paper, (iv) nut-cracker, (v) handle of water pump, (vi) see-saw,

(vii) forceps, (viii) a man rowing a boat, (ix) lock and key,

(x) opening of a soda water bottle, (xi) closing a door, (xii) motor car foot brake,

(xiii) nail cutter, (xiv) a fishing rod, (xv) a lemon squeezer.

Ans. Levers of Ist order

(i) Railway signal

(ii) Handle of water pump

(iii) See-saw

(iv) Motor car foot brake

(v) Nail cutter


Levers of IInd order

(i) Nut-cracker

(ii) A man rowing a boat

(iii) Lock and key

(iv) Opening of soda-water bottle

(v) A lemon squeezer.

(vi) Closing a door

Levers of IIIrd order

(i) A man cutting bread with knife

(ii) A boy writing on a piece of paper

(iii) Forceps

(iv) A fishing rod

16. Explain why the cutting edges of scissors are made longer as compared to the

cutting edges of a metal cutter.

Ans. Effort gets multiplied only, if effort arm is longer than load arm. In case of

scissors, as effort need not be multiplied, therefore the effort arm is smaller

than load arm.

However, in case of a metal cutter, the resistance due to metal is very large.

Thus, effort needs multiplication. It is achieved by making cutting edge (load

arm) smaller than the effort arm.

17. (a) The diagram shows a wheelbarrow. In the diagram mark fulcrum. Also draw

arrows to show the direction of load and effort.


(b) What class of lever is wheelbarrow?

(c) Give one more example of same class of lever.

Ans. (a) Shown in the diagram.

(b) Lever of the second order.

(c) Nut-cracker.

18. By stating clearly, the position of load (L); effort (E) and fulcrum (F), state the

class of levers to which the following belong :

(i) Pliers, (ii) Sugar tongs, (iii) Scissors, (iv) Nut-cracker.

Ans. (i) Plier is a lever of the first order, as the fulcrum is in the middle, the load at

one end and the effort at the other end.

(ii) Sugar tongs is a lever of the third order, as the effort is in the middle, the

load at one end and the fulcrum at other end.

(iii) Scissors is a lever of the first order, as the fulcrum is in the middle, the load

at one end and the effort at the other end.

(iv) Nut cracker is a lever of the second order, as the load is in the middle, the

fulcrum at one end and the effort at the other.


19. Diagram shows a weightless lever in equilibrium. Neglect friction at the fulcrum F.

(i) State the principle of moments as applied to above lever.

(ii) Define mechanical advantage and calculate its value for given lever.

(iii)Name the type of lever, which has mechanical advantage greater than one.

Ans. (i) Principle of moments states that, when a lever is in equilibrium, then the

effort multiplied by the effort arm is equal to the load multiplied by the load

arm.

(ii) The ratio between the useful load lifted to the effort applied is called

mechanical advantage.

In the diagram L × BF = E × AF

L AF AF

= or M.A. =E BF BF

(iii) Lever of the second order has mechanical advantage greater than one.


20. Give one example each of class I lever, where mechanical advantage is :

(a) more than one, (b) less than one.

What is the use of lever, if its mechanical advantage is less than 1.

Ans. (a) The mechanical advantage of the handle of a hand pump is more than one.

(b) The mechanical advantage of the scissors is less than one.

If mechanical advantage is less than 1, the load moves through a large

distance, when the effort acts through a small distance.

21. A pair of scissors and a pair of pliers both belong to the same class of levers.

Name the class of lever. Which one has mechanical advantage less than one.

Ans. Pair of scissors as well as pair of pliers belong to lever of first class. The pair of

scissors have mechanical advantage less than 1.

22. Give an example of a class I lever, which has mechanical advantage equal to 1.

Ans. In case of physical balance (class I lever), the mechanical advantage is 1.

23. Draw a diagram of lever which is always used as force multiplier.

Ans. Lever of second order is always force multiplier.

24. What type of lever is formed by the human arm while raising a load from a table

placed in front. Where is the position of falcrum.

Ans. Human arm acts as a lever of 3rd class.

The fulcrum is the elbow of the human arm.


25. Class III levers have mechanical advantage less than one. Why are they used?

Ans. The class III levers move the load through large distance when effort is applied

through small distance. These levers are used as speed multipliers.

26. (a) Define pulley. By drawing diagram calculate : (1) Mechanical advantage,

(2) Velocity ratio of single fixed pulley.

(b) Why is single fixed pulley commonly used, in spite of the fact that its

mechanical advantage is less than one?

Ans. (a) Pulley is a flat circular disc, having a groove in its edge and is capable of

turning around a fixed point, passing through its centre and commonly

called axle.

Let L be the load lifted by an effort E, such that T is the tension in the rope.

∴ L = T ......... (i)

E = T .........(ii)

Comparing (i) and (ii) L = E; LE

= 1

or M.A = 1

Let ‘d’ be the distance through which effort acts, as well as load lifted.

∴ V.R = Distance through which effort moves ÷ Distance through which

load moves = d ÷ d = 1.

(b) It helps in changing the direction of effort applied. As it is far easier, to

apply effort in downward direction, therefore single fixed pulley is widely

used.


27. (a) Draw a neat diagram of a “single movable pulley system.”

(b) Why is the fixed pulley used in the above system?

(c) Why is single movable pulley system superior to single fixed pulley system?

Ans. (a) The diagram is given below.

(b) Single fixed pulley changes the direction of effort applied.

(c) It is because, its mechanical advantage is more than 1.


28. (i)A pulley system has a velocity ratio 3. Draw a labelled diagram of the pulley

system.

(ii) What is the mechanical advantage of above system.

Ans. (i) Shown in figure alongside.

(ii) M.A. = No. of pulleys in system

= 3.

29. In a single fixed pulley, if the effort moves by the distance x downward by what

height is load raised upward.

Ans.Velocity ratio of single fixed pulley is 1. Thus, if the effort acts through a

distance x in downward direction, the load will move through a distance x in

upward direction.


30. What is a single movable pulley system? What is mechanical advantage in ideal

condition?

Ans. A system of two single pulleys, such that one pulley is fixed at higher platform

(commonly called fixed pulley) and the second pulley is connected to load

(commonly called movable pulley) is called single movable pulley system.

When a rope or string is passed around the movable and fixed pulley and effort

is applied on the free end, the load moves upward along with movable pulley

which also rotates about its axis at the same time.

The ideal mechanical advantage of single movable pulley is 2.

31. In which direction the force need to be applied, when a single pulley is used with

a mechanical advantage greater than 1? Draw the diagram for the arrangement.

Ans.The force need to be applied in the upward direction as shown in the diagram

given below


32. Show how a single pulley can be used to reduce effort required to overcome a

given load. Draw the diagram of the system. Why is it generally more

convenient to use two pulleys for this?

Ans.For diagram, refer to figure in Q 6.

It is generally convenient to apply effort in the downward direction rather than

upward direction. Thus, to change the direction of effort applied another single

pulley fixed at higher platform is used.

33. Give reasons for the following :

(i) In a single fixed pulley, the velocity ratio is always more than the mechanical

advantage.

(ii) The efficiency of a pulley is always less than 100%.

(iii) In case of block and tackle arrangement, the mechanical advantage

increases with the increase in number of pulleys.

Ans. (i) In a single fixed pulley, the load moves through the same distance as the

distance through which effort is applied. Thus, velocity ratio being the ratio

of distance through which effort moves to the distance through which load

moves is 1.

However, a part of effort is wasted in moving the single fixed pulley around

its axis as well as friction between the rope and pulley. Thus, effort is able to

lift less load than actual effort applied and hence mechanical advantage is

less than 1.


(ii) In any pulley system, the mechanical advantage is always less than the

velocity ratio as a part of effort is wasted in overcoming friction and weight

of movable block.

Now, the efficiency is the ratio between mechanical advantage and velocity

ratio, therefore, it is always less than 1 or 100%.

(iii) The mechanical advantage of a pulley system is given by the expression.

MA = (number of pulleys in pulley system – Ex

)

where x is the weight of movable block.

Now, as the number of pulleys increases, the factor Ex

remains the same.

Therefore, on the whole mechanical advantage increases.

34. Name a machine which is used as : (i) force multiplier, (ii) speed multiplier,

(iii) change in the direction of force applied.

Ans. (i) Lever of 2nd class is based as force multiplier as is in case of nut cracker.

(ii) Lever of 3rd class is used as speed multiplier as is in the case of spade or

fishing rod.

(iii) Single fixed pulley is a machine which changes the direction of force

applied.

35. (a) Define inclined plane.

(b) How does inclined plane acts as a machine?

(c) Name four examples of inclined plane in daily life.


Ans. (a) A smooth, rigid and flat surface, inclined at some angle to the horizontal

surface is called inclined plane.

(b) As the smaller effort applied along the inclined plane, which is made

available in the vertically upward direction, therefore, inclined plane acts as

a machine.

(c) (i) Roads constructed on hills.

(ii) Staircases in homes

(iii) A gangway on a sailing ship.

(iv) A flat plank used for rolling barrels into a truck.

36. (a )State the expression for mechanical advantage of an inclined plane.

(b) What do you understand by the term gradient of inclined plane? Support

your answer with an example.

Ans. (a) Mech. advantage = Length of inclined planeheight of inclined plane

lh

(b) The ratio between the vertical distance moved by a body to the horizontal

distance travelled along inclined plane is called gradient of inclined plane.

For example, if a body rises vertically upward by 1 m, while travelling

100 m along the hill road, then gradient of hill road (inclined plane)

is 1

100 = 0.01.


37. The force needed to push a load up an inclined plane is less than the force

needed to lift it directly. Give a reason.

Ans. The force needed to push the load along the inclined plane acts through a larger

distance as compared to distance through which load rises up vertically. This in

turn increases the velocity ratio and hence mechanical advantage. Thus, a lesser

force is required.

38. Steeper the inclined plane, more is the effort needed to push a load up the plane.

Explain it.

Ans. Mechanical advantage of an inclined plane is given by the expression :

1MA =

sin θ

As the inclined plane gets steeper, the value of sin θ increases and hence the

mechanical advantage decrease. Thus, a greater effort is required to push a load

up on a steeper inclined plane.

39. State whether the mechanical advantage of an inclined plane is equal to 1, less

than 1 or greater than 1.

Ans. Mechanical advantage of an inclined plane is always greater than 1.


Numerical

1. In operating a water pump, a resistance of 480 N is overcome by an effort of

72N. If the distance of fulcrum from the point where resistance acts is

0.2 m, find the distance of fulcrum from where the effort acts.

Ans. Load = 480 N;

Effort = 72 N;

Distance of load from fulcrum = 0.2 m.

Effort × Distance of effort from fulcrum

= Load × Distance of load from fulcrum

72 N × D = 480 N × 0.25 m

∴ D = 480 N × 0.25 m

72 N

= 1.67 m.

2. A uniform plank of sea-saw is 5 m long and supported at its centre. A boy

weighing 40 kg, sits at a distance of 1.5 m from one end of see-saw. Where must a

girl weighing 25 kg sit on the other end of see-saw, so as to balance the weight of

boy.

Ans. Mass of girl × distance of girl from fulcrum = Mass of boy × distance of boy from fulcrum ∴ 25 kg × distance of girl from fulcrum = 40 kg × 1.5 m. ∴ Distance of girl from fulcrum

40 kg ×1.5 m

=25 kg

= 2.4 m..


3. A crow bar of length 120 cm has its fulcrum situated at a distance of 20 cm

from load. Calculate mechanical advantage of crow bar.

Ans. Distance of load from fulcrum (d) = 20 cm

∴ Distance of effort from fulcrum (D) = (120 – 20) = 100 cm.

Mechanical advantage of crow bar

D 100 cm

= 520 cmd

.

4. A handle of a nut cracker is 16 cm long and a nut is placed 2 cm from its hinge.

If a force of 4 kgf is applied at the end of handle to crack it, what weight, if

simply placed on the nut, will crack it?

Ans. Distance of nut from fulcrum (d) = 2 cm.

Distance of effort from fulcrum (D) = 16 cm.

Load × Load arm = Effort × Effort arm

Load × 2 cm = 4 kgf × 16 cm

∴ Load = 64 kgf

2 = 32 kgf.

Thus, if a load of 32 kgf is placed on nut, it will crack.

5. An effort of 50 kgf is applied at the end of lever of second order, which

supports a load of 750 kgf, such that load is at a distance of 0.1 m from hinge.

Find the length of lever.

Ans. E = 50 kgf; L = 750 kg;

Load arm (d) = 0.1 m; Effort arm (D) = ?


Effort × Effort arm = Load × Load arm.

50 kgf × D = 750 kgf × 0.1 m.

∴ D = 750 × 0.1 m

50 = 1.5 m.

∴ Length of lever = 1.5 m.

6. A machine displaces a load of 125 kgf through a distance of 0.30 m, when an

effort of 12.5 kgf acts through a distance of 4.0 m. Calculate : (i) velocity ratio,

(ii) mechanical advantage, (iii) % age efficiency of machine.

Ans. L = 125 kgf; d = 0.30 m; E = 12.5 kgf;

D = 4.0 m

(i) Velocity ratio = D ÷ d = 4.0 m ÷ 0.30 m

= 13.33.

(ii) Mechanical advantage = 125 kgf12.5 kgf

LE

= 10.

(iii) %age efficiency of machine

M.A 10

= ×100 = ×100 =V.R 13.33

75%

7. Calculate : (i) velocity ratio, (ii) mechanical advantage, (iii) %age efficiency of a

machine, which overcomes a resistance of 800 N through a distance of 0.12 m,

when an effort of 160 N acts through a distance of 0.72 m.

Ans. L = 800 N;

d = 0.12 m;


E = 160 N;

D = 0.72 m.

(i) Velocity ratio = D 0.72 m

=0.12 md

= 6

(ii) Mechanical advantage = EL

800 N160 N

= 5.

(iii) %age efficiency M.A

100V.R

5

1006

= 83.33%.

8. An effort of 500 N is applied through a distance of 0.50 m on a machine, whose

efficiency is 90%, such that resistance is overcome through a distance of

0.04 m. Calculate : (i) V.R, (ii) M.A, (iii) resistance overcome by machine.

Ans. E = 500 N;

D = 0.50 m;

η = 90%;

d = 0.04 m.


(i) V.R. = D 0.50 m

0.04 md

= 12.5.

(ii) M.A. = η× V.R = 90

100 × 12.5 = 11.25.

(iii) M.A. = LE

∴ L = E × M.A.

= 500 N × 11.25

= 5625 N.

9. A crow bar of length 2.0 m is used as a machine, to lift a box of 100 kgf by

placing a fulcrum at a distance of 0.1 m from the box. Calculate : (i) velocity

ratio, (ii) mechanical advantage, (iii) effort required. What assumption has been

made, in solving this problem?

Ans. Length of crow bar = 2.0 m.

Length of load arm (d) = 0.1 m.

∴ Length of effort arm (D) = (2.0 – 0.1) = 1.9 m.

∴ (i) Velocity ratio = D 1.9 md 0.1 m

= 19.

(ii) Mechanical advantage = LE

= V.R = 19

(iii) M.A. = LE


∴ E = L ÷ M.A.

= 100 kgf ÷ 19

= 5.26 kgf.

It is assumed (i) The crow bar is weightless.

(ii) There is no friction at the fulcrum.

10. A person is carrying a load of 25 kgf, suspended from a wooden staff, such that

load projects 8 cm from shoulder. Where must the person apply an effort of 7.5

kgf, so as to balance the staff?

Ans. L = 25 kgf ; d = 8 cm ; E = 7.5 kgf ; D = ?

Load × Load arm = Effort × Effort arm.

25 kgf × 8 cm = 7.5 kgf × D

∴ Effort arm, D = 25× 8cm

7.5 = 26.67 cm.

11. Figure shows a wheelbarrow with C as centre of gravity, such that its leg is in

contact with ground.

(i) What is the direction of force acting at C? Name the force.

(ii)What is the direction of minimum force acting at A to keep the leg off the

ground? What is the force called?

(iii) The weight of wheel borrow is 15 kgf and it holds 60 kgf of sand. Calculate

the minimum force to keep the leg off the ground?


Ans. (i) The force is acting vertically downward. The force is called LOAD.

(ii) The minimum force acts vertically upward at A. The force is called

EFFORT.

(iii) Weight of wheel barrow and sand

= (15 + 60) kgf

= 75 kgf

Load arm = 50 cm

Effort arm = (100 + 50) cm

= 150 cm.

By the principle of lever :

Effort × Effort arm = Load × Load arm

E × 150 cm = 75 kgf × 50 cm

∴ E = 75 × 50

150 kgf

= 25 kgf.

12. A 4 m long rod of negligible weight is to be balanced about a point 125 cm from

one end. A load of 18 kgf is suspended at a point 60 cm from the support on the

shorter arm.

(i) A weight W is placed 250 cm from the support on the longer arm. Find W.


(ii) If W = 5 kgf, where it must be kept to balance the rod?

(iii) To which class of lever does it belong?

Ans. (i) Load (L) = 18 kgf

Load arm (d) = 60 cm

Effort (W) = ?

Effort arm (D) = 250 cm

By the principle of lever


W × 250 cm = 18 kgf × 60 cm

W = 18 60 kgf cm

250 cm = 4.32 kgf

(ii) Load (L) = 18 kgf

Load arm (d) = 60 cm

Effort (W) = 5 kgf

Effort arm (D) = ?

By the principle of lever


5 kgf × Effort arm = 18 kgf × 60 cm

Effort arm = 18 kgf × 60 cm

5 kgf

= 216 cm (iii) It is 1st class lever.


13. A pair of scissors has its blades 15 cm long, while its handles 7.5 cm long. What is its mechanical advantage?

Ans. Mechanical advantage = Effort arm 7.5 cm

= =Load arm 15 cm

0.5

14. A force of 5 kgf is required to cut a metal sheet. A shears used for cutting the

metal sheet has its blade 5 cm long, while the handle is 10 cm long. What effort

is needed to cut the sheet?

Ans. Load (resistance overcome) = 5 kgf

Load arm = 5 cm

Effort required = ?

Effort arm = 10 cm

By the principle of lever :


Effort × 10 cm = 5 kgf × 5 cm

Effort = 5 kgf × 5 cm

10 cm = 2.5 kgf

15. Diagram below shows a lever in use.

(i) To which class of lever does it belong?

(ii) If AB = 1 m, AF = 0.4 m, find its mechanical advantage.

(iii) Calculate the value of E.

Ans. (i) It is class I lever.

(ii) Mechanical advantage


Effort arm=

Load arm(1 m – 0.4 cm)

=0.4 m

0.6 m= =

0.4 m1.5

(iii) Mechanical advantage = LoadEffort

1.5 = 15 kgfEffort

Effort = 15 kgf

1.5 = 10 kgf

16. Figure below shows a uniform metre scale kept in equilibrium, when supported

at 60 cm mark and mass M is suspended from the 90 cm mark. State with reason,

whether the weight of scale is greater, less than or equal to the mass M.

Ans. The weight of the scale is greater than M. It is because arm on the side of M is

30 cm and on side of weight of scale is 10 cm. So, to balance the scale, weight

of scale should be more than M.


17. A pulley system has velocity ratio 3 and an efficiency of 80%. Calculate :

(i) Mechanical advantage of system (ii) Value of effort required to raise a load

of 300 N.

Ans.V.R = 3 ; η = 80% ; M.A. = ? ;

E = ? ; L = 300 N.

(i) M.A. = η × V.R = 80

100 × 3 = 2.4.

(ii) LE

= M.A.

∴ E = L 300 N

M.A 2.4 = 125 N.

18. A pulley system has five pulleys in all and is 90% efficient. Calculate : (i)

mechanical advantage, (ii) effort required to lift a load of 1000 N,

(ii) resistance due to movable parts of machine and friction.

Ans.Number of pulleys in system = 5;

Load = 1000 N. Efficiency = 90%.

Velocity ratio = Number of pulleys in system = 5

(i) M.A. = η × V.R. = 90

100 × 5 = 4.5.

(ii) E = L 1000 N

M.A 4.5 = 222.2 N.

(iii) Wt. of movable parts and friction x =

E (V.R – M.A) = 222.2 N (5 – 4.5) = 111.1 N.


19. A pulley system has four pulleys in all and is 80% efficient. Calculate : (i)

mechanical advantage, (ii) load lifted by an effort of 1400 N, (iii) resistance of

movable parts of machine.

Ans.Velocity ratio = No. of pulleys in the system = 4

(i) M.A. = η × V.R = 80

100 × 4 = 3.2.

(ii) Load (L) = M.A. × E = 3.2 × 1400 = 4480 N.

(iii) Resistance of movable parts of machine

x = E (V.R – M.A.) = 1400(4 – 3.2)

= 1400 × 0.8 = 1120 N.

20. Diagram shows a pulley arrangement.

(i) Copy the diagram, and mark direction of force due to tension acting on the

movable pulley.

(ii) What is the purpose of fixed pulley?

(iii) If tension is T newtons, deduce the relation between T and E.

(iv) Calculate the velocity ratio of the arrangement.

(v) Assuming the efficiency to be 100%, what is the mechanical advantage?

(vi) Calculate effort E.

(vii) State two factors that will reduce efficiency of arrangement.

Ans. (i) Shown in the diagram.

(ii) It changes the direction of effort from downward to upward direction.

(iii) T = E.


(iv) Let ‘d’ be the distance through which rope is pulled by the effort. In order to

maintain equilibrium, the distance through which each segment of the rope

shortens is d/2.

∴ V.R. = d ÷ = 2.

(v) M.A.

η =V.R.

100 M.A.100 2

M.A.= 2.

(vi) M.A. = LE

L 100 N

E =M.A 2

= 50 N.

(vii) (1) The friction of movable parts reduces efficiency.

(2) The weight of movable pulley reduces efficiency.


21. A pulley system can lift a load of 1200 N by an effort of 250 N. If the resistance

due to weight of movable parts and friction is 300 N, calculate : (i) Mechanical

advantage, (ii) Velocity ratio, (iii) Total number of pulleys in system, (iv)

Efficiency of system.

Ans. L = 1200 N ; E = 250 N ; x = 300 N

(i) M.A. = L 1200

= = 4E 250

.8

(ii) x = E (V.R. – M.A.)

∴ 300 = 250 (V.R. – 4.8)

∴ 250 × V.R. = 300 + 1200

V.R. = 1500250

= 6.

(iii) Total number of pulleys in system = velocity ratio = 6.

(iv) M.A. 4.8

η = ×100 = ×100 = .V.R. 6

80%

22. An effort of 240 N overcomes a useful load of 1000 N, when applied on the block

and tackle system of pulleys, such that weight of movable blocks and friction etc.

is 200 N. Calculate : (i) mechanical advantage, (ii) velocity ratio, (iii) total

number of pulleys in system, (iv) efficiency of system.

Ans. E = 240 N ; l = 1000 N ; x = 200 N

(i) 1000 N

M.A. 4.167E 240 Nl

(ii) x = E(V.R. – M.A.)


200 = 240 (V.R. – 4.167)

240 × V.R. = 240 × 4.167 + 200

240 × V.R. = 1000 + 200

V.R. = 1200240

= 5.

(iii) Total number of pulleys in the system

= velocity ratio = 5.

(iv) M.A. 4.167

η = 100 100V.R. 5

= 83.34%.

23. A woman draws water from a well using fixed pulley. The mass of bucket and

water together is 6 kg. The force applied by the woman is 70 N. Calculate

mechanical advantage.[Take g = 10 ms–2]

Ans.Mass of water and bucket = 6 kg

Load lifted by the woman = mg

= 6 kg × 10 ms–2

= 60 N

Effort applied by the woman = 70 N

MA Load 6007

= = = 0.857Effort 7007

s

24. A fixed pulley is driven by 100 kg mass falling at a rate of 8.0 m in 4 s. It lifts a

load of 500 kgf. Calculate the power input to the pulley taking force of gravity on


1 kg = 10 N. If the efficiency of pulley is 75%, find the height to which load is

raised in 4.0 s.

Ans. (i) Force of 100 kg of falling mass

F = 100 × 10 N = 1000 N

Speed of fall (v) = 8.0 m 2 m

4 s s

Power input = F × v 2m

= 100 N ×s

= 2000 J

(ii) Efficiency = Power outputPower input

75 Power output

100 2000 J

Power output = 75 2000 J

100

= 1500 J

Now, power output = Force × Velocity

1500 J = 500 kgf × Velocity

1500 J = 5000 N × Velocity

Velocity = 1500 J5000 N

= 0.3 ms–1

Distance through which load moves in 4 s


= 0.3 ms–1 × 4 s = 1.2 m

25. In a block and tackle system consisting of 3 pulleys, a load of 75 kgf is raised

by an effort of 25 kgf. Find the mechanical advantage, velocity ratio and

efficiency.

Ans. Mechanical advantage = L 75 kgf

=E 25 kgf

= 3


%age mechanical advantage

= M.A 3

100 100 100%V.R 3

26. A block and tackle system has 5 pulleys. If an effort of 1000 N is needed to raise

a load of 4500 N. Calculate : (i) mechanical advantage, (ii) velocity ratio and

(iii) efficiency of the system.

Ans. Mechanical advantage L 4500 N

4.5E 1000 N


%age mechanical advantage

= MA 4.5

×100 = 100 90%VR 5

27. A block and tackle system has velocity ratio 3. A man can exert a pull of 200

kgf. What is the maximum load he can raise with this pulley system, if its

efficiency is 60%?


Ans. Efficiency M.A

(η) =V.R

∴ 60 MA

=100 3

∴ MA = 60 × 3

1.8100

Now, MA = El

1.8 = 200 kgf

l

load (l) = 1.8 × 200 kgf = 360 kgf.

28. A boy has to lift a load of mass 50 kg to a height of 1 m. What effort is required to

lift it vertically upward?However, boy can exert a maximum force of 250 N, so he

uses an inclined plane to lift the load up. What should be the minimum length of the

plank used by him? [Take g = 10 ms–2]

Ans. (i) Effort required to lift the load directly

= mg = 50 kg × 10ms–2 = 500 N (ii) Actual effort the boy can apply (E) = 250 N

Load to be lift (l) = 500 N

Vertical height (h) = 1 m

Length of inclined plane (L) = ?

For the inclined plane :

L

El

h


500 N ×1 m

L 2E 250 N

l h m

29. A coolie uses sloping wooden plank of length 2.0 m to push up a drum of mass

100 kg into a truck at a height of 1.0 m.

(i) What is the mechanical advantage of sloping plank?

(ii) How much effort is needed to push the drum up into the truck?

(iii) What assumption have you made in arriving at the answer in part (ii) above?

Ans. (i) Mechanical advantage = L 2.0 m

21.0 mh

(ii) Mechanical advantage = El

2 = 100 kgf

E

100 kgf

E2

50 kgf

(iii) Assumption : It is assumed that there is no friction between the drum and

the surface of the inclined plane.


30. An effort of 20 kgf is applied on a barrel of weight 200 kgf in order to roll it up

an inclined plane through a vertical height of 1.5 m. Calculate the length of the

inclined plane.

Ans. E = 20 kgf; L = 200 kgf ; h = 1.5 m; l = ? For inclined plane,

LE

lh

200 kgf

=20 kgf 1.5 m

l

l = 10 × 1.5 m = 15 m. 31. The gradient of a hill road is 1 : 50. Calculate the weight of a truck which is

pushed up this hill road, when its engine generates a force of 250 kgf.

Ans. Gradient = 150

hl

501

lh

Also, for inclined plane

LE

lh

L 5250 kgf 1

0

∴ L = 50 × 250 kgf


= 12500 kgf.

32. The engine of a car develops a force of 800 N. If this car is moving on a hill

road covers a vertical distance of 100 m, calculate the length of road along

which it moves. The load of car is 1000 kgf. [Take g = 10 ms–2]

Ans. E = 800 N ; L = 1000 kg × 10 ms–2 = 10,000 N ;

h = 100 m ; l = ?

For inclined plane, LE

lh

10,000 N

800 N 100 ml

10000 100

800l

= 1250 m.

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