Q: Question 11Ramesh went for fishing with his fishing rod. Identify the class of lever for a fishing rod. If a fish exerts a load of 10 N and the distance between fulcrum and fish is 1.5 m, and the effort applied is 0.5 m from the fulcrum, what is the force required by the fisher to pull the fish?

A fishing rod is an example of a Class III lever.Here,Fulcrum : at the handle (where the rod pivots)Effort : applied by the fisher's hand along the rodLoad : the fish at the end of the rodIn Class III levers : Fulcrum — Effort — LoadThe principle of moments states that —Moment of the load about the fulcrum = Moment of the effort about the fulcrum∴ Effort × Effort Arm = Load × Load Arm .......... (i)Given :Load (L) = 10 NLoad arm (distance between fulcrum and fish) = 1.5 mEffort arm (distance between fulcrum and applied effort) = 0.5 mFrom (i)Effort x 0.5 = 10 x 1.5⇒ Effort = 150.5\dfrac{15}{0.5}0.515 = 30 N

Q: Question 14The diagram below shows a rod lifting a stone.(a) Mark position of fulcrum F and draw arrows to show the directions of load L and effort E.(b) What class of lever is the rod?(c) Give one more example of the same class of lever stated in part (b).

(a) Below diagram shows the position of fulcrum F and the directions of load L and effort E:(b) The rod is a Class II lever as the load is between fulcrum and effort.(c) A nut cracker is also an example of a Class II lever.

Q: Question 15State the kind of lever which always has the mechanical advantage less than 1. Draw a labelled diagram of such a lever.

The Class III levers have mechanical advantage always less than 1.

Q: Question 16Explain why mechanical mechanical advantage of class III lever is always less than 1.

The fulcrum F and the load L are at the two ends of the lever and the effort E is somewhere in between the fulcrum F and the load L.The effort and the load are on the same side of fulcrum but in opposite directions and the effort arm is always smaller than the load arm. Therefore, the mechanical advantage of class III lever is always less than 1.

Q: Question 17Classes III levers have mechanical advantage less than 1. Why are they then used?

Class III levers have mechanical advantage less than 1 so as effort arm is always less than the load arm, so we do not get gain in force, but we get gain in speed, i.e. a larger displacement of load is obtained by a smaller displacement of effort.

Q: Question 18State the class of levers and the relative positions of load (L), effort (E) and fulcrum (F) in(a) a bottle opener, and(b) sugar tongs.

(a) A bottle opener is an example of Class II lever, as the fulcrum F and the effort E are at the two ends of the lever and the load is in between the fulcrum F and the effort E.(b) Sugar tongs is an example of Class III lever, as the fulcrum F and the load L are at the two ends of the lever and the effort E is in between the fulcrum F and the load L.

Question 1

Question 1Derive a relationship between mechanical advantage, velocity ratio and efficiency of a machine.

Accepted Answer

Suppose a machine overcomes a load L by the application of effort E.Let,the displacement of effort = dE andthe displacement of the load = dL in time t.Work Input=effort×displacement of effort=E×dEWork Output=load×displacement of load=L×dLEfficiency η=work outputwork inputη=L×dLE×dEη=LE×dLdE=LE×1dEdL\text{Work Input} = effort \times \text{displacement of effort} \[0.5em] = E \times d_E \[0.5em] \text{Work Output} = load \times \text{displacement of load} \[0.5em] = L \times d_L \[0.5em] \text{Efficiency η} = \dfrac{\text{work output}}{\text{work input}} \[0.5em] η = \dfrac{L \times d_L}{E \times d_E} \[0.5em] η = \dfrac{L}{E} \times \dfrac{d_L}{d_E} \[0.5em] = \dfrac{L}{E} \times \dfrac{1}{\dfrac{d_E}{d_L}} \[0.5em]Work Input=effort×displacement of effort=E×dE​Work Output=load×displacement of load=L×dL​Efficiency η=work inputwork output​η=E×dE​L×dL​​η=EL​×dE​dL​​=EL​×dL​dE​​1​But we know that,LE=M.A.\dfrac{L}{E} = M.A. \[0.5em]EL​=M.A.anddEdL=V.R.\dfrac{d_E}{d_L} = V.R. \[0.5em]dL​dE​​=V.R.Substituting the above two values in the formula for efficiency we getη=M.A.V.R.⇒M.A.=V.R.×ηη = \dfrac{M.A.}{V.R.} \[0.5em] \Rightarrow M.A. = V.R. \times η \[0.5em]η=V.R.M.A.​⇒M.A.=V.R.×ηHence, the mechanical advantage of a machine is equal to the product of its efficiency and velocity ratio.

Question 2

Question 5A man uses a crowbar of length 1.5 m to raise a load of 75 kgf by putting a sharp edge below the bar at a distance 1 m from his hand.(a) Draw a diagram of the arrangement showing the fulcrum (F), load (L) and effort (E) with their directions.(b) State the kind of lever.(c) Calculate — (i) load arm, (ii) effort arm, (iii) mechanical advantage, and (iv) the effort needed.

Accepted Answer

(a) Below is the diagram of the crowbar showing the fulcrum (F), load (L) and effort (E) with their directions:(b) Crowbar belongs to class I lever.(c)Given,(i) Total length = 1.5 mEffort arm = 1 mLoad arm=Total length−Effort armLoad arm=1.5m−1m⇒Load arm=0.5m\text{Load arm} = \text{Total length} - \text{Effort arm} \[0.5em] \text{Load arm} = 1.5m - 1m \[0.5em] \Rightarrow \text{Load arm} = 0.5 m \[0.5em]Load arm=Total length−Effort armLoad arm=1.5m−1m⇒Load arm=0.5m(ii) Effort arm = 1 m(iii)Mechanical advantage M.A.=Effort armLoad armM.A.=10.5⇒M.A.=2\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \[0.5em] M.A. = \dfrac{1}{0.5} \[0.5em] \Rightarrow M.A. = 2 \[0.5em]Mechanical advantage M.A.=Load armEffort arm​M.A.=0.51​⇒M.A.=2(iv) Given,Load = 75 kgfWe know that,Effort=LoadM.A.\text{Effort} =\dfrac{\text{Load}}{\text{M.A.}} \[0.5em]Effort=M.A.Load​Substituting the values in formula we get,Effort=752⇒Effort=37.5 kgf\text{Effort} =\dfrac{75}{2} \[0.5em] \Rightarrow \text{Effort} = 37.5 \text{ kgf} \[0.5em]Effort=275​⇒Effort=37.5 kgf

Question 3

Question 6A pair of scissors is used to cut a piece of cloth by keeping it at a distance of 8.0 cm from its rivet and applying an effort of 10 kgf by fingers at a distance 2.0 cm from the rivet.(a) Find: (i) the mechanical advantage of scissors and (ii) the load offered by the cloth.(b) How does the pair of scissors act - as a force multiplier or as a speed multiplier?

Accepted Answer

(a)(i) We know that,Mechanical advantage M.A.=Effort armLoad arm\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \[0.5em]Mechanical advantage M.A.=Load armEffort arm​Given,Load arm = 8.0 cmEffort arm = 2 cmSubstituting the values in the formula we get,M.A.=28⇒M.A.=0.25M.A. = \dfrac{2}{8} \[0.5em] \Rightarrow M.A. = 0.25 \[0.5em]M.A.=82​⇒M.A.=0.25(ii) We know,Load=M.A.×EffortLoad = M.A. \times Effort \[0.5em]Load=M.A.×EffortGiven,Effort = 10 kgfSubstituting the values in the formula we get,Load=0.25×10⇒Load=2.5kgfLoad = 0.25 \times 10 \[0.5em] \Rightarrow Load = 2.5 kgf \[0.5em]Load=0.25×10⇒Load=2.5kgf(b) As the mechanical advantage is less than 1 i.e. M.A. < 1. Hence, the pair of scissors acts as a speed multiplier as there is larger displacement of load than displacement of effort.

Question 4

Question 11Ramesh went for fishing with his fishing rod. Identify the class of lever for a fishing rod. If a fish exerts a load of 10 N and the distance between fulcrum and fish is 1.5 m, and the effort applied is 0.5 m from the fulcrum, what is the force required by the fisher to pull the fish?

Accepted Answer

A fishing rod is an example of a Class III lever.Here,Fulcrum : at the handle (where the rod pivots)Effort : applied by the fisher's hand along the rodLoad : the fish at the end of the rodIn Class III levers : Fulcrum — Effort — LoadThe principle of moments states that —Moment of the load about the fulcrum = Moment of the effort about the fulcrum∴ Effort × Effort Arm = Load × Load Arm .......... (i)Given :Load (L) = 10 NLoad arm (distance between fulcrum and fish) = 1.5 mEffort arm (distance between fulcrum and applied effort) = 0.5 mFrom (i)Effort x 0.5 = 10 x 1.5⇒ Effort = 150.5\dfrac{15}{0.5}0.515​ = 30 N

Question 5

Question 13A fire tongs has its arms 20 cm long. It is used to lift a coal of weight 1.5 kgf by applying an effort at a distance 15 cm from the fulcrum.Find:(i) the mechanical advantage of fire tongs and(ii) the effort needed.

Accepted Answer

(i) We know that,Mechanical advantage M.A.=Effort armLoad arm\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \[0.5em]Mechanical advantage M.A.=Load armEffort arm​Given,Load arm = 20 cmEffort arm = 15 cmSubstituting the values in the formula we get,M.A.=1520⇒M.A.=0.75M.A. = \dfrac{15}{20} \[0.5em] \Rightarrow M.A. = 0.75 \[0.5em]M.A.=2015​⇒M.A.=0.75(ii) We know that,Effort=LoadM.A.\text{Effort} = \dfrac{\text{Load}}{\text{M.A.}} \[0.5em]Effort=M.A.Load​Given,Load = 1.5 kgfSubstituting the values in the formula we get,Effort=1.50.75⇒Effort=2 kgf\text{Effort} = \dfrac{1.5}{0.75} \[0.5em] \Rightarrow \text{Effort} = 2 \text{ kgf} \[0.5em]Effort=0.751.5​⇒Effort=2 kgf

Question 6

Question 14The diagram below shows a rod lifting a stone.(a) Mark position of fulcrum F and draw arrows to show the directions of load L and effort E.(b) What class of lever is the rod?(c) Give one more example of the same class of lever stated in part (b).

Accepted Answer

(a) Below diagram shows the position of fulcrum F and the directions of load L and effort E:(b) The rod is a Class II lever as the load is between fulcrum and effort.(c) A nut cracker is also an example of a Class II lever.

Question 7

Question 15State the kind of lever which always has the mechanical advantage less than 1. Draw a labelled diagram of such a lever.

Accepted Answer

The Class III levers have mechanical advantage always less than 1.

Question 8

Question 16Explain why mechanical mechanical advantage of class III lever is always less than 1.

Accepted Answer

The fulcrum F and the load L are at the two ends of the lever and the effort E is somewhere in between the fulcrum F and the load L.The effort and the load are on the same side of fulcrum but in opposite directions and the effort arm is always smaller than the load arm. Therefore, the mechanical advantage of class III lever is always less than 1.

Question 9

Question 17Classes III levers have mechanical advantage less than 1. Why are they then used?

Accepted Answer

Class III levers have mechanical advantage less than 1 so as effort arm is always less than the load arm, so we do not get gain in force, but we get gain in speed, i.e. a larger displacement of load is obtained by a smaller displacement of effort.

Question 10

Question 18State the class of levers and the relative positions of load (L), effort (E) and fulcrum (F) in(a) a bottle opener, and(b) sugar tongs.

Accepted Answer

(a) A bottle opener is an example of Class II lever, as the fulcrum F and the effort E are at the two ends of the lever and the load is in between the fulcrum F and the effort E.(b) Sugar tongs is an example of Class III lever, as the fulcrum F and the load L are at the two ends of the lever and the effort E is in between the fulcrum F and the load L.

Question 11

Question 19Indicate the positions of load L, effort E and fulcrum F in the forearm shown below. Name the class of lever.

Accepted Answer

The positions of load L, effort E and fulcrum F in the forearm are shown below:It is a Class III lever.

Question 12

Question 7Draw a diagram of combination of three movable pulleys and one fixed pulley to lift up a load. In the diagram, show the directions of load, effort and tension in each strand. Find:(i) the mechanical advantage,(ii) velocity ratio and(iii) the efficiency of the combination in the ideal situation.

Accepted Answer

The diagram is shown below:(i) In equilibrium,Effort E = T3       (1)Tension T1 in the string passing over the pulley A is given as 2T1 = LT1 = L2\bold{\dfrac{L}{2}}2L​       (2)Tension T2 in the string passing over the pulley B is given as2T2 = T1T2= T12\dfrac{T_1}{2}2T1​​Substituting value of T1 from equation 2,T2 = L22\bold{\dfrac{L}{2^2}}22L​       (3)Tension T3 in the string passing over the pulley C is given as2T3 = T2T3 = T22\dfrac{T_2}{2}2T2​​Substituting value of T2 from equation 3,T3 = L23\bold{\dfrac{L}{2^3}}23L​       (4)In equilibrium, T3 = EFrom equation 4,Load L = 23 x T3       (5)Mechanical advantage M.A.=LEM.A.=23×T3T3⇒M.A.=23\text{Mechanical advantage M.A.} = \dfrac{L}{E} \[0.5em] M.A. = \dfrac{2^3 \times T_3}{T^3} \[0.5em] \Rightarrow M.A. = 2^3Mechanical advantage M.A.=EL​M.A.=T323×T3​​⇒M.A.=23(ii) As we know, one end of each string passing over a movable pulley is fixed, so the other end of string moves up twice the distance moved by the axle of the movable pulley.If the load L attached to the pulley A moves a distance d,then dL = dNow, the string connected to the axle of pulley B, moves up by a distance,2 times d = 2d.Then the string connected to the axle of the pulley C, moves up by a distance,2 times 2d = 22dThen the end of the string passing over the fixed pulley D, moves up by a distance,2 times 22d = 23d.Hence, dE = 23d.As we know,Velocity Ratio V.R.=Distance moved by effort Distance moved by load=dEdL\text{Velocity Ratio V.R.} = \dfrac{\text{Distance moved by effort }}{\text{Distance moved by load}} \[0.5em] = \dfrac{d_E}{d_L} \[0.5em]Velocity Ratio V.R.=Distance moved by loadDistance moved by effort ​=dL​dE​​(iii) Substituting the values in the formula we get,=23×dd=23Efficiency(η)=M.A.V.R.=2323=1= \dfrac {2^3 \times d}{d} \[0.5em] = 2^3 \[1.5em] \text{Efficiency} (\eta) = \dfrac{M.A.}{V.R.} \[0.5em] = \dfrac{2^3}{2^3} = 1 \[0.5em]=d23×d​=23Efficiency(η)=V.R.M.A.​=2323​=1

Question 13

Question 9A boy uses a single fixed pulley to lift a load of 50 kgf to some height. Another boy uses a single movable pulley to lift the same load to the same height. The ratio of the two effort is:1:22:11:1none of these

Accepted Answer

2:1Reason — We know that in case of a single fixed pulley, the effort applied is equal to the load.Hence,Ef=LE_f = L \[0.5em]Ef​=LIn the case of a single movable pulley the effort needed to lift a load is equal to half the load.Em=L2Em=50 kgf2⇒Em=25 kgfE_m = \dfrac{L}{2} \[1em] E_m = \dfrac{50 \text{ kgf}}{2} \[1em] \Rightarrow E_m = 25\text{ kgf} \[1em]Em​=2L​Em​=250 kgf​⇒Em​=25 kgfHence, the ratio of efforts applied by the respective pulley isEfEm=21\dfrac{E_f}{E_m} = \dfrac{2}{1} \[0.5em]Em​Ef​​=12​Therefore, Ef : Em = 2 : 1

Question 14

Question 12For a system of n pulleys, the weight of the lower block along with pulleys is w, then its efficiency is given by:η=1−wnE\eta = 1 - \dfrac{w}{nE}η=1−nEw​η=1+wnE\eta = 1 + \dfrac{w}{nE}η=1+nEw​η=wnE\eta = \dfrac{w}{nE}η=nEw​η=n−wE\eta = n - \dfrac{w}{E}η=n−Ew​

Accepted Answer

η=1−wnE\eta = 1 - \dfrac{w}{nE}η=1−nEw​Reason — For a system of n pulleys, let w be the total weight of the lower block along with the pulleys in it. In the balanced position,E = T and L + w = nTor L = nT - w = nE - wM.A. = LE=nE−wE=n−wE\dfrac{L}{E} = \dfrac{nE-w}{E} = n - \dfrac{w}{E}EL​=EnE−w​=n−Ew​Thus, the mechanical advantage is less than the ideal value n.The velocity ratio does not change, it remains n, i.e., V.R. = nHence, efficiency η=M.A.V.R.=n−wEn=1−wnE\eta = \dfrac{M.A.}{V.R.} = \dfrac{n - \dfrac{w}{E} }{n} = 1 - \dfrac{w}{nE}η=V.R.M.A.​=nn−Ew​​=1−nEw​.

Question 15

Question 2A fixed pulley is driven by a 100 kg mass falling at a rate of 8.0 m in 4.0 s. It lifts a load of 75.0 kgf.Calculate —(a) The power input to the pulley taking the force of gravity on 1 kg as 10 N.(b) the efficiency of the pulley, and(c) the height to which the load is raised in 4.0 s.

Accepted Answer

(a) As we know,Power=Force×velocity\text {Power} = \text {Force} \times \text {velocity} \[0.5em]Power=Force×velocityandVelocity=Change in distanceChange in timevelocity=84So,velocity=2m/sec\text {Velocity} = \dfrac{\text{Change in distance}}{\text{Change in time}} \[0.5em] \text {velocity} = \dfrac{8}{4} \[0.5em] So, \text {velocity} = 2 m/sec \[0.5em]Velocity=Change in timeChange in distance​velocity=48​So,velocity=2m/secGiven,Tension in pulley=100×10Force=Tension=1000N\text {Tension in pulley} = 100 \times 10 \[0.5em] \text {Force} = \text {Tension} = 1000 N \[0.5em]Tension in pulley=100×10Force=Tension=1000NSubstituting the values in the formula for power we get,Power=1000×2⇒Power=2000W\text {Power} = 1000 \times 2 \[0.5em] \Rightarrow \text {Power} = 2000 W \[0.5em]Power=1000×2⇒Power=2000W(b) We know,η=M.A.V.R.\eta = \dfrac{M.A.}{V.R.} \[0.5em]η=V.R.M.A.​Given,Load=75×10Load=750N\text {Load} = 75 \times 10 \[0.5em] \text {Load} = 750 N \[0.5em]Load=75×10Load=750NWe know,M.A.=LoadEffortM.A. = \dfrac{Load}{Effort}M.A.=EffortLoad​ Substituting the values for M.A. we get,M.A.=7501000⇒M.A.=0.75M.A. = \dfrac{750}{1000} \[0.5em] \Rightarrow M.A. = 0.75 \[0.5em]M.A.=1000750​⇒M.A.=0.75When the effort moves by a distance d downwards, the load moves by the same distance upwards. So we get,V.R. = 1Substituting the values of M.A. and V.R. in the formula for efficiency we get,η=0.751⇒η=0.75\eta = \dfrac{0.75}{1} \[0.5em] \Rightarrow \eta = 0.75 \[0.5em]η=10.75​⇒η=0.75Hence, efficiency = 75 %(c) As the load moves same distance upwards as much as effort moves downwards. So, the height achieved by the load is 8 m.

Question 16

Question 3A single fixed pulley and a movable pulley both are separately used to lift a load of 50 kgf to the same height. Compare the efforts applied in an ideal situation.

Accepted Answer

We know that in case of a single fixed pulley, the effort applied is equal to the load.Hence,Ef=LE_f = L \[0.5em]Ef​=LIn the case of a movable pulley the effort needed to lift a load is equal to half the load.Em=L2Em=50kgf2⇒Em=25kgfE_m = \dfrac{L}{2} \[0.5em] E_m = \dfrac{50 kgf}{2} \[0.5em] \Rightarrow E_m = 25kgf \[0.5em]Em​=2L​Em​=250kgf​⇒Em​=25kgfHence, the ratio of efforts applied by the respective pulley isEfEm=21\dfrac{E_f}{E_m} = \dfrac{2}{1} \[0.5em]Em​Ef​​=12​Therefore, Ef : Em = 2 : 1

Question 17

Question 4In a block and tackle system consisting of 3 pulleys, a load of 75 kgf is raised with an effort of 25 kgf.Find —(i) the mechanical advantage,(ii) the velocity ratio, and(iii) the efficiency.

Accepted Answer

(i) Given,Effort = 25 kgfLoad = 75 kgfAs we know that,M.A.=LoadEffortM.A. = \dfrac{\text{Load}}{\text{Effort}} \[0.5em]M.A.=EffortLoad​Substituting the values in the formula we get,M.A.=7525M.A.=31⇒M.A.=3M.A. = \dfrac{75}{25} \[0.5em] M.A. = \dfrac{3}{1} \[0.5em] \Rightarrow M.A. = 3 \[0.5em]M.A.=2575​M.A.=13​⇒M.A.=3(ii) As we know,V.R. = n = 3(iii) Also, η=M.A.V.R.η=33⇒η=1\eta = \dfrac{M.A.}{V.R.} \[0.5em] \eta = \dfrac{3}{3} \[0.5em] \Rightarrow\eta = 1 \[0.5em]η=V.R.M.A.​η=33​⇒η=1Therefore, efficiency = 100%

Question 18

Question 8Figure shows a system of four pulleys. The upper two pulleys are fixed and the lower two are movable.(a) Draw a string around the pulleys. Also show the point of application and direction in which the effort E is applied.(b) What is the velocity ratio of the system?(c) How are load and effort of the pulley system related?(d) What assumption do you make in arriving at your answer in part (c)?

Accepted Answer

(a) The diagram below shows the point of application and direction of the effort E:(b) As we know that in a system of 4 pulleys, if the load moves up through a distance d, each segment of the string is loosened by a length d so the effort end moves through a distance 4 times d.Velocity ratio=dEdLVelocity ratio=4dd⇒Velocity ratio=4\text{Velocity ratio} = \dfrac{d_E}{d_L} \[0.5em] \text{Velocity ratio} = \dfrac{4d}{d} \[0.5em] \Rightarrow \text{Velocity ratio}= 4 \[0.5em]Velocity ratio=dL​dE​​Velocity ratio=d4d​⇒Velocity ratio=4(c) As we know,M.A.=LoadEffortM.A. = \dfrac{\text {Load}}{\text {Effort}} \[0.5em]M.A.=EffortLoad​As M.A. = n = 4Substituting the values in the formula we get,LoadEffort=4⇒Load=4×Effort\dfrac{Load}{Effort} = 4 \[0.5em] \Rightarrow Load = 4 \times EffortEffortLoad​=4⇒Load=4×Effort(d) Assumptions made are —(i) There is no friction in the pulley bearings.(ii) Weight of lower pulleys is negligible.(iii) The effort is applied downwards.

Question 19

Question 10A block and tackle system has the velocity ratio 3. Draw a labelled diagram of the system indicating the points of application and the directions of load L and effort E. A man can exert a pull of 200 kgf.(a) What is the maximum load he can raise with this pulley system if its efficiency is 60%?(b) If the effort end moves a distance 60 cm, what distance does the load move?

Accepted Answer

Given,V.R = n (number of pulleys) = 3Hence, number of pulleys = 3Below is a labelled diagram of the system indicating the points of application and the directions of load L and effort E:(a) As we know,η=1−wnE\eta = 1 - \dfrac{w}{nE}η=1−nEw​Given,Effort = 200 kgfEfficiency = 60% = 0.6Substituting the values in the formula for efficiency we get,0.6=1−w3×200w600=1−0.6w=600×0.4⇒w=240 kgf0.6 = 1 - \dfrac{w}{3 \times 200} \[0.5em] \dfrac{w}{600} = 1 - 0.6 \[0.5em] w = 600 \times 0.4 \[0.5em] \Rightarrow w = 240 \text{ kgf}0.6=1−3×200w​600w​=1−0.6w=600×0.4⇒w=240 kgfAs we know,Load=(n×E)−wLoad=(3×200)−wLoad=600−240⇒Load=360kgf\text{Load} = (n \times E) - w \[0.5em] \text{Load} = (3 \times 200) - w \[0.5em] \text{Load} = 600 - 240 \[0.5em] \Rightarrow \text{Load} = 360kgfLoad=(n×E)−wLoad=(3×200)−wLoad=600−240⇒Load=360kgf(b) The velocity ratio of system is V.R = number of pulleys = 3As we know,V.R.=dEdLV.R. = \dfrac{d_E}{d_L} \[0.5em]V.R.=dL​dE​​Substituting the values in the formula for V.R. we get,3=dEdLdL=dE3dL=603⇒dL=20cm3 = \dfrac{d_E}{d_L} \[0.5em] d_L= \dfrac{d_E}{3} \[0.5em] d_L= \dfrac{60}{3} \[0.5em] \Rightarrow d_L= 20 cm \[0.5em]3=dL​dE​​dL​=3dE​​dL​=360​⇒dL​=20cm

Machines

Exercise 3A Long Questions

Exercise 3A Numericals

Exercise 3A Short Questions

Exercise 3B Long Questions

Exercise 3B Multiple Choice Type

Exercise 3B Numericals