Machines Questions and Answers ICSE Class 10

1 Mark Question

Question 1 : What is the principle of an ideal machine?

Answer : An ideal machine is that in which there is no dissipation of energy in any manner. The work output is equal to the work input.

Question 2 : Draw a diagram to show how a single pulley can be used so as to have its ideal MA=2.

Answer :

Question 3 : A type of a single pulley is very often used as a machine even though it does not give any gain in mechanical advantage. i) Name the type of pulley used. ii) For what purpose is such a pulley used?

Answer : i) The pulley which is used in the arrangement is single fixed pulley. ii) Single fixed pulley is used only to change the direction of the force applied.

Question 4 : Under what condition will a set of gears produce? i) a gain in speed. ii) a gain in torque

Answer : i) The set of gears gain its speed when the number of teeth in driving gear is more than as in the driven gear. ii) The set of gears gain its torque when the number of teeth in driving gear is less than as in the driven gear.

Question 5 : i) What is meant by an ideal machine? ii) Write a relationship between the mechanical advantage (MA) and velocity ratio (VR) of an ideal machine?

Answer : i) A machine having 100% efficiency is called an ideal machine. ii) The relation between MA and VR is given by MA = VR

Question 6 : The diagram alongside shows a lever in use. i) To which class of lever does it belong? ii) If FA=40 cm, AB=60 cm, then find the mechanical advantage of the lever.

Answer : i) It belongs to class II lever. ii) Effort arm = FA + AB = 40 + 60 = 100 cm Load arm = FA = 40 cm Therefore,

M A = \frac{E f f o r t a r m}{L o a d a r m} = \frac{F B}{F A} = \frac{100}{40} = 2.5

Question 7 : Name a machine which can be used to i) multiply force ii) change the direction of the force applied.

Answer : i) Inclined plane is a machine used to multiply force. ii) Pulley is a machine which can change the direction of force applied.

Question 8 : i) Why is the mechanical advantage of a lever of the second order always greater than one? ii) Name the type of single pulley that has a mechanical advantage greater than one.

Answer : i) The effort arm is always greater than the load arm so the mechanical advantage of class II lever is always greater than one. ii) Single movable pulley is the type of the pulley whose mechanical advantage is greater than unity.

Question 9 : i) With reference to the terms mechanical advantage, velocity ratio and efficiency of a machine, name the term that will not change for a machine of a given design. ii) Define the term stated by you in pat(i).

Answer : i) Velocity ration will not change for a machine of a given design, because the distance moved by load in the same time is equal to the distance moved by the effort. ii) Velocity ratio is defined as the ratio of the velocity of effort (V_E) to the velocity of load (V_L).

V R = \frac{V_{E}}{V_{L}} = \frac{d_{E}}{d_{L}}

Question 10 : Copy the diagram of the forearm given below, indicate the positions of load, effort and fulcrum.

Answer : The below diagram shows the position of load (L), effort (E) and fulcrum (F).

Question 11 : Write an expression to show the relationship between mechanical advantage, velocity ratio and efficiency for a simple machine.

Answer : The expression showing the relationship between mechanical advantage (MA), velocity ratio (VR) and efficiency (η), for a simple machine is given by

η = \frac{M A}{V R} X 100

Question 12 : What class of levers has a mechanical advantage always greater than one? What change can be brought about in this lever to increase its mechanical advantage?

Answer : In class II lever effort arm is always greater than load arm. Therefore, mechanical advantage (MA = Effort arm/Load arm) is always greater than one. To increase its MA we need to decrease its load arm.

Question 13 : A woman draws water from a well using a fixed pulley. The mass of the bucket and water together is 6.0 kg. The force applied by the woman is 70 N. Calculate the mechanical advantage. [Take g=10 m/s²]

Answer : Give, Load L = mg = 6 x 10 = 60 N [ g=10 m/s²] Effort E = 70 N Mechanical advantage, MA = ?

M A = \frac{L}{E} = \frac{60}{70} = 0.857

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Question 14 : Give two reasons why the efficiency of a single movable pulley system is not 100%/.

Answer : In single movable pulley system, efficiency is not 100%, because i) the weight of the movable pulley with its frame is not negligible as compared to the effort. ii) the frictional force between the pulley wheel and the string used is not negligible. iii) the string is not inextensible and massless.

Question 15 : What is the relationship between the mechanical advantage and the velocity ration for i) ideal machine ii) practical machine

Answer : i) For an ideal machine, the work done by the machine is equal to the work done on the machine. Therefore, mechanical advantage is equal to velocity ratio. i.e., MA = VR. ii) For practical machine, mechanical advantage is less than velocity ratio as efficiency is less than 1.

Question 16 : To use a machine as a force multiplier, what types (class) of lever should preferably be used? Draw a sketch of such a lever.

Answer : Lever of II type (class) is used as a force multiplier in a machine. The sketch of such lever is as shown below. The effort arm is always greater than the load arm, so mechanical advantage of class II lever is always greater than 1.

Question 17 : Why is the mechanical advantage of a lever of third order is always less than 1? Give one example of this class of lever.

Answer : In the third order lever, the effort E is in between the fulcrum F and the load L. The effort arm is always smaller than the load arm. And, we know mechanical advantage is the ratio between the effort arm and load arm so its value is always less than 1. e.g., Sugar tongs, the forearm used for lifting the load.

3 Marks Questions

Question 18 : Derive a relationship between MA, VR and η of a machine.

Answer : Let a machine overcome a load L by the application of an effort E. In time t, the displacement of effort is dE and the displacement of load is dL. Work input = Effort x displacement of effort

= E x d_{E}

Work output = Load x displacement of load

= L x d_{L}

E f f i c i e n c y η = \frac{W o r k o u t p u t}{W o r k i n p u t} = \frac{L x d_{L}}{E x d_{E}} = \frac{L}{E} x \frac{d_{L}}{d_{E}} = \frac{L}{E} x \frac{1}{d_{E} / d_{L}} B u t \frac{L}{E} = M A a n d \frac{d_{E}}{d_{L}} = V R T h e r e f o r e, η = \frac{M A}{V R}

Question 19 : A code is pushing a box weighing 1500 N up an inclined plane 7.5 m long onto a platform, 2.5 m above the ground. i) Calculate the mechanical advantage of the inclined plane. ii) Calculate the effort applied by the coolie. iii) In actual practice, the coolie needs to apply more effort than what is calculated. Give one reason why you think the coolie needs to apply more effort.

Answer : Given, Length l = 7.5 m; Height h = 2.5 m; Load L = 1500 N

i) M A = \frac{l}{h} = \frac{7.5}{2.5} = 3 i i) E f f o r t = \frac{L o a d}{M A} = \frac{1500}{3} = 500 N

iii) In actual practice, due to some friction between the plane and the bottom face of the body which is pulled over it, the effort needed is certainly more than L sin θ, (θ is inclination of the plane) so that MA is always less then l/h , hence efficiency is less than one.

Question 20 : i) State the class of levers and the relative positions of load (L), effort (E) and fulcrum (F) in each of the following cases. a) A bottle opener b) Sugar tongs ii) Why is less effort needed to lift a load over an inclined plane as compared to lifting the load directly?

Answer : i) a) A bottle opener is a class II lever. In this case, the load is in between the fulcrum and the effort. b) Sugar tongs is a class III lever. In this case, the effort is in between the fulcrum and the load. ii) Less effort is needed to lift a load over an inclined plane as compared to lifting the load directly, because mechanical advantage increases.

Question 21 : A pulley system comprises two pulleys, one fixed and the other movable. i) Draw a labelled diagram of the arrangement and show clearly the directions of all the forces acting on it. ii) What change can be made in the movable pulley of this system to increase the mechanical advantage of the system?

Answer : i) The labelled diagram of the above arrangement is shown below. ii) To increase the mechanical advantage of the system we need to reduce the friction between the string and the movable pulley?

Question 22 : A pulley system has a velocity ratio of 4 and an efficiency of 90%. Calculate i) the mechanical advantage of the system. ii) the effort required to raise a load of 300 N by the system.

Answer : Given, Velocity ration VR = 4 Efficiency η = 90% i) Mechanical advantage,

M A = V R x \frac{η}{100} = 4 x \frac{90}{100} = 3.6 [A s η = \frac{M A}{V R}]

ii) Effort E = ?, Load L = 300 N

E = \frac{L}{M A} = \frac{300}{3.6} = 83.33 [M A = \frac{L}{E}]

Question 23 : Name the type of single pulley that can act as a multiplier. Draw a labelled diagram of the above named pulley.

Answer : The single pulley that can act as a force multiplier is single movable pulley. The labelled diagram of single movable pulley is shown below.

Question 24 : i) Define an inclined plane. ii) Draw a labelled sketch of a class II lever. Give one example of such a lever.

Answer : i) An inclined plane is a sloping surface kept inclined to the horizontal at an angle of θ. Its mechanical advantage is always greater than unity. ii) The labelled sketch of class II lever are

Question 25 : A cook uses a fire tong of length 29 cm to lift a piece of burning coal of mass 250 g. If he applies his effort at a distance of 7 cm from the fulcrum, what is the effort in SI unit?[Take g=10 m/s²] Answer : Given , mass = 250 g = 0.25 kg Load L = 0.25 x 10 N = 2.5 N $D i s \tan c e o f l o a d L_{d} = 28 c m = 0.28 m D i s \tan c e o f e f f o r t E_{d} = 7 c m = 0.07 m, E f f o r t E = ? A c c o r d i n g t o t h e p r i n c i p l e o f m o m e n t, w e h a v e E x E_{d} = L x L_{d} = > E = \frac{L x L_{d}}{E_{d}} = \frac{0.25 x 10 x 0.28}{0.07} = 10 N$

Question 26 : A pair of scissors and a pair of pliers belong to the same class of levers. i) Which one has mechanical advantage less than one? ii) State the usefulness of such a machine whose mechanical advantage is less than 1.

Answer : i) A pair of scissors has mechanical advantage less than one, because the blades are longer than its handles which is used to cut a piece of paper or cloth so that the blades move longer on the cloth or paper when the handles are moved a little. ii) A machine such as a single fixed pulley whose MA<1 is useful because it is easy and comfortable to apply effort (E) in the downward direction rather than upward direction. We can also use own body weight while applying effort in the downward direction.

4 Marks Questions

Question 27 : A block and tackle system of pulleys has a velocity ratio 4. i) Draw a labelled diagram of the system indicating clearly the points of application and directions of load and effort. ii) What is the value of the mechanical advantage of the given pulley system, if it is an ideal pulley system?

Answer : i) The labelled diagram of the system is shown below. ii) Given VR = 4, Load L = 4T, Effort = T

T h e r e f o r e, M A = \frac{L o a d}{E f f o r t} = \frac{4 T}{T} = 4

Question 28 : A uniform metre scale can be balanced at the 70.0 cm mark when a mass of 0.05 kg is hung from the 94.0 cm mark. i) Draw a diagram of the arrangement. ii) Find the mass of the metre scale.

Answer : i) The diagram of the above arrangement is shown below. ii) Let w be the weight of the scale, By principle of moments, we get w x (70-50) = 0.05 (94-70) => w x 20 = 0.05 x 24

= > w = \frac{0.05 x 24}{20} = 0.05 k g

Question 29 : i) Write a relation expressing the mechanical advantage of a lever. ii) Write an expression for the mechanical advantage of an inclined plane. iii) Give two reasons as to why the efficiency of a single movable pulley system is always less than 100%.

Answer : i)The relation expressing mechanical advantage of a lever is given by

M A = \frac{E f f o r t a r m}{L o a d a r m}

ii) The expression for the mechanical advantage of an inclined plane is given by where, l is the length of inclined plane, h is the height of the inclined plane.

Question 30 : i) Draw a labelled diagram of a block and tackle system of pulleys with two pulleys in each block. Indicate the directions of the load, effort and tension in the string. ii) Write down the relation between the load and the effort for the pulley system.

Answer : i) The labelled diagram of a block and tackle system of pulleys is shown below. ii) Given Load L = 4T, Effort E = T

T h e r e f o r e, \frac{L o a d}{E f f o r t} = \frac{4 T}{T} H e n c e, L = 4 E

Question 31 : A block and tackle pulley system has a velocity ratio 3. i) Draw a labelled diagram of this system. In your diagram, indicate clearly the points of application and the direction of the load and effort. ii) Why should the lower block of this pulley system be of negligible weight?

Answer : i) The labelled diagram of the system is shown below. ii) If the total weight (w) of pulley in the lower block is in the balanced position. Then,

E f f i c i e n c y η = \frac{M A}{V R} = \frac{n - \frac{w}{E}}{n} = 1 - \frac{w}{n E} F r o m t h e e x p r e s s i o n, i t i s c l e a r t h a t f o r g r e a t e r e f f i c i e c n y \frac{w}{n E} s h o u l d b e n e g l i g i b l e .

Thus, the pulleys in the lower block should be as light as possible.

Question 32 : The alongside figure shows the combination of a movable pulley P1 with a fixed pulley P2 used for lifting up a load w. i) State the function of the fixed pulley P2. ii) If the free end of the string moves through a distance x, find the distance by which the load w is raised. iii) Calculate the force to be applied at C to just raise the load w=20 kgf, neglecting the weight of the pulley P1, and friction.

Answer : i) The fixed pulley (P2) helps us to apply the effort in a convenient direction. ii) If the free end of the string moves through a distance x, then the load is raised by a distance of x/2. iii) Force, F = ?

L o a d, w = 20 k g f = 20 x 10 = 200 N (g = 10 m / s^{2}) T h e r e f o r e, F x x = w x \frac{x}{2} F = \frac{200}{2} = 100 N = 10 k g f

Question 33 : The radius of the driving wheel of a set of gears is 18 cm. It has 100 teeth and rotates at a speed of 30 rpm. The driven wheel rotates at a speed of 150 rpm. Calculate: i) the gear ratio ii) the number of teeth on the drive wheel iii) the radius of the driven wheel

Answer :

G i v e n, R a d i u s o f d r i v i n g w h e e l r_{A} = 18 c m = 0.18 m N u m b e r o f t e e t h o f d r i v i n g w h e e l, N_{A} = 100 L i n e a r s p e e d o f d r i v i n g w h e e l, ω_{A} = 30 r p m L i n e a r s p e e d o f d r i v e n w h e e l, ω_{B} = 150 r p m i) G e a r r a t i o = \frac{ω_{B}}{ω_{A}} = \frac{150}{30} = 5 i i) N u m b e r o f t e e t h o n t h e d r i v e n w h e e l, N_{B} = ? \frac{N_{A}}{N_{B}} = \frac{ω_{B}}{ω_{A}} = > N_{B} = N_{A} x \frac{1}{ω_{B} / ω_{A}} = 100 x \frac{1}{5} = 20 i i i) R a d i u s o f t h e d r i v e n w h e e l, r_{B} = ? \frac{r_{A}}{r_{B}} = \frac{ω_{B}}{ω_{A}} = > r_{B} = r_{A} x \frac{1}{ω_{B} / ω_{A}} = 0.18 x \frac{1}{5} = 0.036 m = 3.6 c m

7 Marks Question

Question 34 : In the diagram of a stationary wheel barrow, the centre of gravity is at A. The wheel and the leg are in contact with the ground. The horizontal distance between A and F is 50 cm and that between B and F is 150 cm. i) What is the direction of the force acting at A? Name the force. ii) What is the direction of the minimum force at B to keep the leg off the ground? What is this force called? iii) The weight of the wheel barrow is 15 kgf and it holds sand of weight 60 kgf. Calculate the minimum force required to keep the leg off the ground.

Answer : i) The direction of the force acting at A is vertically downwards. The force is named as load. ii) The direction of the minimum force at B to keep the leg off ground is acting vertically upwards. This force is called effort.

G i v e n, W e i g h t o f w h e e l w_{E} = 15 k g f W e i g h t o f s a n d w_{s} = 60 k g f M i n i m u m f o r c e E = ? T o t a l l o a d L = w_{E} + w_{s} = 15 + 60 = 75 k g f E f f o r t a r m E_{a} = 50 + 100 = 150 c m = 1.5 m L o a d a r m L_{a} = 50 c m = 0.5 m A p p l y i n g t h e p r i n c i p l e o f l e v e r, L o a d x l o a d a r m = E f f o r t x e f f o r t a r m = > L x L_{a} = E x E_{a} = > E = \frac{L x L_{a}}{E_{a}} = \frac{75 x 0.5}{1.5} = 25 k g f