1 Mark Question
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.
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.
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.
Answer : i) A machine having 100% efficiency is called an ideal machine.
ii) The relation between MA and VR is given by
MA = VR
Answer : i) It belongs to class II lever.
ii) Effort arm = FA + AB
= 40 + 60 = 100 cm
Load arm = FA = 40 cm
Answer : i) Inclined plane is a machine used to multiply force.
ii) Pulley is a machine which can change the direction of force applied.
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.
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 (VE) to the velocity of load (VL).
Answer : The below diagram shows the position of load (L), effort (E) and fulcrum (F).
Answer : The expression showing the relationship between mechanical advantage (MA), velocity ratio (VR) and efficiency (η), for a simple machine is given by
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.
Answer : Give, Load L = mg = 6 x 10 = 60 N [ g=10 m/s2]
Effort E = 70 N
Mechanical advantage, MA = ?
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.
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.
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.
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
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
Work output = Load x displacement of load
Answer : Given, Length l = 7.5 m; Height h = 2.5 m;
Load L = 1500 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.
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.
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?
Answer : Given, Velocity ration VR = 4
Efficiency η = 90%
i) Mechanical advantage,
ii) Effort E = ?, Load L = 300 N
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.
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
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
Answer : i) The labelled diagram of the system is shown below.
ii) Given VR = 4, Load L = 4T, Effort = T
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
Answer : i)The relation expressing mechanical advantage of a lever is given by
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.
Answer : i) The labelled diagram of a block and tackle system of pulleys is shown below.
ii) Given Load L = 4T, Effort E = T
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.
Thus, the pulleys in the lower block should be as light as possible.
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 = ?
7 Marks Question
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.