# Frank Solutions for Chapter 1 Force, Work, Energy and Power Class 10 Physics ICSE

**Exercise**

**1. What do you mean by the turning effect of a force?**

**Answer **

Moment of a force = Force x perpendicular distance of the pivot from the force.

Its SI unit is newton-metre (Nm).

**2. Name the physical quantity whose SI unit is Nm.**

**Answer**

**3. What is meant by a torque?**

**Answer**

**4. A body is pivoted at a point. A force of 10 N is applied at a distance of 30 cm from the pivot. Find the moment of force about the pivot.**

**Answer**

= 10×30×10

^{-2}= 3 Nm

**5. Is moment of a force a scalar quantity?**

**Answer**

**6. Is it easier to turn a steering wheel of large diameter than that of small diameter? Why?**

**Answer**

**7. Define the centre of gravity of a body.**

**Answer**

**8. What is meant by equilibrium?**

**Answer**

**9. What is meant by the principle of moments?**

**Answer**

**10. What is the difference between mass and weight?**

**Answer**

Mass |
Weight |

1. It is the measure of quantity of matter contained in the body, at rest. |
1. It is the force with which the earth attracts a body. |

**11. The moment of a force of 20 N about a fixed point 0 is 10 Nm. Calculate the distance of the point 0 from the line of action of the force.**

**Answer**

∴ 10 = 20× perpendicular distance from the line of action

or, Perpendicular distance from the line of action = 0.5 m

**12. What is meant by the term 'center of gravity of a body'?**

**Answer**

**13. Where is the centre of gravity of a uniform ring situated?**

**Answer**

**14. State the factors on which the centre of gravity of a body depends.**

**Answer**

- Body's weight
- Body's shape

**15. Can the centre of gravity be situated outside the material of the body? Give an example.**

**Answer**

**17. What is meant by a centripetal force? Is it same as centrifugal force?**

**Answer**

It is not the same as the centrifugal force.

**18. Name the physical quantity whose unit is kgfm. Define it.**

**Answer**

Torque may be defines as the turning effect produced by a force on a rigid body about a point, pivot or fulcrum. It is measured by the product of force and the perpendicular distance of the pivot from the line of action of force.

**19. What is meant by the turning effect of force? Give two examples.**

**Answer**

Examples of turning effect of force:

(i) Turning a steering wheel

(ii) Tightening a cap

**20. What is meant by the state of equilibrium of a body?**

**Answer**

**21. What is the difference between static and dynamic equilibrium?**

**Answer**

_{net}= 0.

Static equilibrium is the situation where the object upon which the forces act is no moving.

The object is "static" hence the state of equilibrium gets its name.

Dynamic equilibrium is the situation where an object is in constant velocity motion.

This object can't experience an acceleration which means, F

_{net}>0

**22. State the conditions of a stable equilibrium of a body.**

**Answer**

Conditions for stable equilibrium:

- The body should have a broad base.
- Centre of gravity of the body should be as low as possible.
- Vertical line drawn from the centre of gravity should fall within the base of the support.

**23. A body is acted upon by two forces, each of magnitude F, but in opposite directions. State the effect of the forces when**

**(a) Both forces act at the same point of the body.**

(b) The two forces act at two different points of the body at a separation d.

(b) The two forces act at two different points of the body at a separation d.

**Answer**

**24. A wheel of diameter 3 m is shown in fig. 2 with axle at 0. A force F = 8 N is applied at Q in the direction shown in figure. Calculate the moment of force about:**

**(i) centre o, and**

(ii) point p.

(ii) point p.

**Answer**

∴ Moment of force at O = 8×1.5 = 12 Nm

(ii) Moment of force = applied force × perpendicular distance from the line of action

∴ Moment of force at P = 8× 3 = 24 Nm

**Answer**

In the given figure, forces 10 N and 100 N act clockwise and the forces 15 N and 4 N act anticlockwise moment of force = sum of clockwise moments - sum of anticlockwise moments

∴ Moment of force = (10× 5 + 100 × 0 + 4× 6) - (15× 4) = 74 - 60 = 14 Nm

Or, 14 Nm in clockwise direction.

**26 . State the principle of moments. A meter scale is pivoted at 30 cm mark and it is in equilibrium when a mass of 40 g is suspended from 10 cm mark. Calculate the mass of the ruler.**

**Answer**

∴ Sum of clockwise moments = Sum of anticlockwise moments

or, Mass of scale = 800/30 = 26.7 g

**27. A meter scale is provided at its mid-point when the various masses are suspended on it as shown in fig. 4. from which point will you suspend a 50 g mass in order to keep the ruler in equilibrium?**

**Answer**

80×30 = (40×10)+ (50×d)

∴ 2400 = 400 + 50d

or, 50d = 2000

or, d = 40 cm to the right of the mid - point.

**28. Fig. 5 shows a uniform meter scale weighing 200 gf. Provided at its centre. Two weights 300 gf and 500 gf are suspended from the ruler as shown in the diagram. Calculate the resultant torque of the ruler and hence calculate the distance from mid-point where a 100 gf should be suspended to balance the meter scale.**

**Answer**

Taking , moments about the mid - point,

Resultant torque = (300 × 40) - (500 × 20)

or, Resultant torque = 12000 - 10000 = 2000 gf - cm

Let a mass of 100 gf be suspended at a distance 'd' from the mid - point towards the right side, so as to balance the metre scale.

Then, in balanced condition :

sum of clockwise moments = sum of anticlockwise moments

(300 × 40) = ( 500 × 20) + (100 × d)

or, 12000 = 10000 + 100 d

or, 100 d = 2000

or d = 20 cm to the right of the mid - point.

**29. A meter scale is pivoted at its mid point and a 50 g mass suspended from the 20 cm mark. What mass balances the ruler when suspended from 65 cm mark?**

**Answer**

In balanced condition:

sum of clockwise moments = sum of anticlockwise moments

50×(50-20) = m × (65 - 50)

or, 50×30 = m × 15

or, m = 1500/15 = 100g

**30. What do you mean by the state of equilibrium? What are the conditions for stable equilibrium?**

**Answer**

Conditions for stable equilibrium:

- The body should have a broad base.
- Centre of gravity of the body should be as low as possible.
- Vertical line drawn from the centre of gravity should fall within the base of the support.

**31. Fig. 6 shows the dimensions of an acute angled triangle. By geometrical construction mark the C.G. of the triangle.**

**Answer**

The center of gravity is at the intersection of lines BE and AD. The distance a can be calculated as a = h/3.

**32. A right-angled triangle cardboard piece is placed as shown in fig. 7. Redraw the diagram showing the relative position of the vertices of the triangle when it is suspended by a pin from the hole A. Explain why the position changes?**

**Answer**

The position of vertices changes because the triangle is in equilibrium and the centre of gravity lies on the vertical line through the point of suspension as the weight acts along this same line.

**33. Give scientific reasons for the following:**

**(i) It is easier to push a boy standing on one leg than on both legs.**

(ii) When a man climbs a slope he bends forward.

(iii) There are chances of toppling when a truck takes a sharp turn especially when it is not fully loaded.

(iv) A man runs in the direction of train while getting down from a moving train.

(v) Passengers in a bus are pushed backward when it starts suddenly.

(ii) When a man climbs a slope he bends forward.

(iii) There are chances of toppling when a truck takes a sharp turn especially when it is not fully loaded.

(iv) A man runs in the direction of train while getting down from a moving train.

(v) Passengers in a bus are pushed backward when it starts suddenly.

**Answer**

A boy standing on both legs has his COG in balanced position and is thus in stable equilibrium but a boy standing on one leg has his COG in unbalanced position which makes him quite unstable and hence it is easier to push him.

**34. What are the different methods by which you can increase the stability of a body?**

**Answer**

**35. (i) What do you understand by the term couple of forces?**

**(ii) Calculate the moment of a couple shown in fig. 8.**

**Answer**

= 4×10 = 40 Nm

**36. Give scientific reasons for the following:**

**(i) Even though the Tower of Pisa is leaning through an angle it does not fall.**

(ii) While climbing a hill you will try to bend your body forward.

(iii) In a moving bus the standing passenger stands keeping both his legs apart.

(iv) In a doubled decker bus passengers are not allowed to stand in the upper deck.

(ii) While climbing a hill you will try to bend your body forward.

(iii) In a moving bus the standing passenger stands keeping both his legs apart.

(iv) In a doubled decker bus passengers are not allowed to stand in the upper deck.

**Answer**

**37. Three forces A, B and C are acting on a rigid body which can turn about O in fig.9. If all the three forces are applied simultaneously, in which direction will the body move? Explain.**

**Answer**

Net torque = Sum of clockwise moments - sum of anticlockwise moments

In the given figure, force 40 N is acting clockwise and forces 20 N and 60 N are acting anticlockwise

∴ Net torque = (40× 3) - [(20×2)+(60×1)]

= 120 - 100 =20 Nm

Hence, the body will move in clockwise direction about 'O'.

**38. Fig. 10 shows a uniform meter scale weighing 100 N pivoted at its centre. Two weights of 500 N and 300 N are hung from the ruler as shown in fig. 10.**

**(i) Calculate total clockwise and anticlockwise moments.**

(ii) Calculate difference in clockwise moment and anticlockwise moment.

(iii) Calculate the distance from O where a 100 N weight should be suspended to balance the meter scale.

(ii) Calculate difference in clockwise moment and anticlockwise moment.

(iii) Calculate the distance from O where a 100 N weight should be suspended to balance the meter scale.

**Answer**

(i) Then, total clockwise moment = 500×(50-40) = 500×10 = 5000 Nm

Total anticlockwise moment = 300 ×(80 - 50) = 300× 30 = 9000 Nm

Then, total clockwise moment = 5000 + 100 d

In balanced condition, sum of clockwise moments = sum of anticlockwise moments

∴ 5000 + 100d = 9000

or, 100 d = 4000

or, d = 40 cm

∴ the weight 100 N should be hung from the 40 cm mark so as to balance the scale.

**39. A meter scale is provided at 10 cm mark and is balanced by suspending 400 g from 0 cm mark (fig. 11). Calculate the mass of meter scale.**

**Answer**

In balanced condition, sum of clockwise moments = sum of anticlockwise moments

∴ 400×10 =m × 90

or, 4000 = 90 m

or, m = 44.4g

∴ the mass of the scale is 44.4 g.

**40. (a) Define the term work.**

**(b) Name the CGS and SI unit of work.**

**Answer**

It is equal to the product of force and the displacement of the point of application of the force in the direction of force.

The SI unit of work is 'joules' and the CGS unit is 'erg'.

**41. The following are some of the energy transformations.**

**A. Electrical to light**

B. Work to heat

C. Chemical to light

D. Electrical to sound

E. Mechanical to electrical

Identify the energy transformation that takes place in the following by inserting the corresponding letter in the shape provided.

(i) A candle flame

(ii) A torch is lighted

(iii) A microphone is used in a meeting

(iv) A cycle dynamo

(v) A piece of metal is being filed

B. Work to heat

C. Chemical to light

D. Electrical to sound

E. Mechanical to electrical

Identify the energy transformation that takes place in the following by inserting the corresponding letter in the shape provided.

(i) A candle flame

(ii) A torch is lighted

(iii) A microphone is used in a meeting

(iv) A cycle dynamo

(v) A piece of metal is being filed

**Answer**

(ii) A

(iii) D

(iv) E

(v) B

**42. A boy pulls a box up to 10 m with a force of 5 kgf. Calculate the work done by him.**

**Answer**

^{-2 }= 50 N

displacement = 10 m

Work done = force × displacement = 50 × 10 = 500 J

**43. Calculate the amount of work done by a child carrying a bag of 20 kg when he moves a distance of 40 m in**

**(a) Vertical direction, and**

(b) Horizontal direction

(b) Horizontal direction

**Answer**

displacement = 40 m

(a) In vertical direction work is done against gravity

Work done = mgh = 20× 9.8× 40 = 7840 J

Work done = Fs cosÎ¸ = (mg)s cos0° = 20× 9.8×40 = 7840 J

**44. (a) Define power and name its unit.**

**(b) A girl weighting 50 kg climbs up 60 steps each of 20 cm height in 5 minutes. Calculate the power developed.**

**Answer**

**(a)**The rate of doing work is called power .

SI unit of power = watt

CGS unit of power = erg per second

**(b)**Given, mass = 50 kg

Total height traveled = 60× 20 cm = 1200 cm = 12 m

g = 10 ms

^{-2 }

time taken = 5 min = 5×60s = 300 s

Work done = mgh = 50×10×12 = 6000 J

Power = Work done/time taken = 6000/300 = 20 watt

**45. Define energy and name its unit.**

**Answer**

The SI unit of work is 'joules' and the CGS unit is 'erg'.

**46. When an elevator starts to move down suddenly we experience 'weightlessness'. Explain.**

**Answer**

- Weight of the body acting downwards
- Normal reaction of the floor acting upwards
- The centrifugal force acting on the body, acting upwards.

Normal reaction is the force that is exerted by the elevator floor in response to the force with which the body presses itself against the floor.

The centrifugal force here is fictious force that acts on the body in the direction opposite to the acceleration of the reference frame, here it is, the elevator floor. It is given by ma, where m is the mass of the body & a is the acceleration of the elevator floor. Centrifugal force is directed opposite to the acceleration of the elevator floor.

Weightlessness is the condition of the zero apparent weight.

When the acceleration of the elevator is such that the upward centrifugal force Fc completely balances the downward weight Wt. of the body, the resultant normal reaction (N =Fc - Wt.) of the body is reduce to zero. That's when the body on the elevator floor will experience the state of weightlessness.

**47. A block of mass 20 kg is pulled up a slope (fig.12) with a constant speed by applying a force of 500 N parallel to the slope. A and B are initial and final positions of the block.**

**(a) Calculate the work done by the force in moving the block from A and B.**

(b) Calculate the potential energy gained by the block.

(b) Calculate the potential energy gained by the block.

**Answer**

Vertical displacement = Final position - initial position = BC = 4m

∴ work done = force × displacement = 500 × 4 = 2000 J

(b) P.E. gained = mgh = (mg)h = 500 ×4 = 2000 J

**48. How fast should a boy weighting 30 kg run so that his kinetic energy is 375 joule?**

**Answer**

**49. A girl of mass 50 kg runs up a flight of 40 steps in 20 seconds. If each step in 20 cm high, calculate the power developed by the girl. (g= 10 ms**

^{-2}).**Answer**

Total height traveled = 40×20 cm = 800 cm = 8 m

g = 10 ms

^{-2 }

time taken = 20s

Work done = mgh = 50 × 10× 8 = 4000 J

Power = Work done/time taken = 4000/20 = 200 watt

**50. A water pump raises 80 kg of water through a height of 20 m in 10 s. Calculate the power of the pump.**

**Answer**

Height = 20 m

g = 10 ms

^{-2 }

time taken = 10s

Work done = mgh = 80 × 10×20 = 16000 J

Power = Work done /time taken = 16000/10 = 1600 watt

**51. The truck has to apply a force of 3000 N to overcome friction while moving with a uniform speed of 36 kmhr**

^{-1}. What is the power developed by the truck?**Answer**

Here, force = 3000 N,

velocity = 36 kmhr

^{-1 }= 10 m/s

∴ power = 3000×10 = 30000 watt

**52. Define energy and state the unit of energy and the law of conservation of energy.**

**Answer**

The SI unit of work is 'joules' and the CGS unit is 'erg'.

According to the law of conservation of energy, energy can neither be created nor be destroyed but can be transformed from one form to another. In other words, energy can be transformed from one form to another but the total amount of all the energies remain the same.

**53. List any six forms of energy and write a short note on each of them.**

**Answer**

**Solar energy:**The energy radiated by the sun is called the solar energy. Inside the sun, energy is produced by nuclear fusion reaction. Solar energy cannot be used to do work directly, because it is too diffused and is not always uniformly available. However, a number of devices such as solar panels, solar cells etc. have been invented to make use of solar energy.**Heat energy:**The energy released on burning coal, oil, wood or gas is the heat energy. The stem possesses heat energy it has capacity to do work.**Light energy:**It is the form of energy in presence of which other objects are seen. The natural source of light energy is sun. Many other sources of heat energy also give light energy.**Chemical or fuel energy:**The energy possessed by fossil fuels such as coal, petroleum and natural gas is called chemical energy or fuel energy. These fuels are formed from the decayed remains of dead plants and animals that lived millions of years ago. In the interior of earth, due to high pressure and temperature the remains slowly changed into fossil fuels.**Hydro energy:**The energy possessed by fast moving water is called the hydro energy. This energy is used to generate electricity in hydroelectric power stations. For this, dams are built across the rivers high up in the hills to store water. Water is allowed to run down the pipes and the energy of running water is used to turn a turbine. The turbine drives generators to produce electrical energy.**Nuclear energy:**The energy released during the processes of nuclear fission and fusion is called nuclear (or atomic) energy. In both these processes, there is loss in mass which converts into energy in accordance with Einstein's mass-energy relation, E =mc^{2}.

**54. What do you understand by 'potential energy' and 'Kinetic energy'? Give three examples of each to illustrate your answer.**

**Answer**

Examples of potential energy:

- Water stored at a height in a reservoir.
- A stretched spring.
- A bent bow.

Examples of kinetic energy:

- Air in motion has kinetic energy.
- A swinging pendulum.
- Moving hands of a clock.

**55. A bullet is of mass 'm' g and is moving with a velocity 'v' m/s. Find the kinetic energy of the bullet when**

**(a) The mass is doubled,**

(b) The velocity is tripled.

(b) The velocity is tripled.

**Answer**

**56. A block of iron of mass 400 kg is used as a pile driver. If it is raised to a height of 20 meters, calculate the potential energy possessed by the iron block. (Assume g = 10 ms**

^{-2}).**Answer**

^{-2 }

P.E. = mgh = 400 × 10× 20 = 80000 J

**57. Identify the type of energy possessed by the body in each of the following:**

**(a) A coiled spring of the toy car**

(b) A hammer which is raised

(c) A stone shot from a catapult

(d) Water stored in the overhead tank

(e) A tadpole moving in water.

(b) A hammer which is raised

(c) A stone shot from a catapult

(d) Water stored in the overhead tank

(e) A tadpole moving in water.

**Answer**

(b) Potential energy

(c) Kinetic energy

(d) Potential energy

(e) Kinetic energy

**59. State the energy changes which take place when**

**(a) Water stored in the dam is used to turn turbine in dynamo.**

(b) An electric bulb glows when it is connected to a source of electric current.

(c) A piece of magnesium wire is burnt in a jar of oxygen.

(d) A stone dropped from the top of a cliff reaches the ground after some time.

(e) A toy car with a wound spring moves on ground.

(b) An electric bulb glows when it is connected to a source of electric current.

(c) A piece of magnesium wire is burnt in a jar of oxygen.

(d) A stone dropped from the top of a cliff reaches the ground after some time.

(e) A toy car with a wound spring moves on ground.

**Answer**

**60. Give one example of each of the following:**

**(a) Electrical energy changes of sound energy.**

(b) Chemical energy changes to heat energy.

(c) Chemical energy changes to electrical energy.

(d) Light energy changes to electrical energy.

(e) Electrical energy changes to heat energy.

(b) Chemical energy changes to heat energy.

(c) Chemical energy changes to electrical energy.

(d) Light energy changes to electrical energy.

(e) Electrical energy changes to heat energy.

**Answer**

(b) Candle flame

(c) Dry cell

(d) Solar cell

(e) Electric iron

**61.Find the kinetic energy of a car of mass 1000 kg traveling at 72 km/h.**

**Answer**

velocity V = 72 km/hr = 20 m/s

**62. A bullet of mass 25 g has a velocity of 600 ms**

^{-1}. What is the kinetic energy of the bullet? If it penetrates 50 cm into a target, find the resistive force offered by the target.**Answer**

**63. A lead bullet moving at 70 ms**

^{-1}is brought to rest on hitting a target. If 80% of its energy is converted into heat energy, find the rise in temperature (Sp. Heat cap. Of lead is 140 J/kgK).**Answer**

final velocity, v = 0 m/s

sp. heat capacity of lead, s = 140 J/kgK

Let Î”t be the change in temperature.

**64. If 60% of the potential energy available in a waterfall is converted into heat energy, find the height of the waterfall, when the temperature difference between the top and the bottom of the fall is 0.21°C (sp. Heat cap. Of water = 4200J/kg K).**

**Answer**

Temperature difference, Î”t = 0.21°C

Let 'h' be the height of the waterfall and 'm' be the mass of water.

Then, P.E. of the water = mgh

Given that, heat energy, H = 60% of P.E.

**65. Define the following:**

**(a) Simple machine**

**(b) Lever**

(c) Mechanical advantage

(c) Mechanical advantage

**(d) Velocity ratio**

(e) Efficiency

(e) Efficiency

**Answer**

**(a) Simple machine:**A machine is a device by which we can either overcome a large resistive force at some point by applying a small force at a convenient point and in a desired direction or by which we can obtain a gain in speed.

**(b) Lever:**A lever is a rigid, straight or bent bar which is capable of turning about a fixed axis.

**(c) Mechanical advantage (M.A.):**The ratio of the load to the effort is called the mechanical advantage of the machine.

**(d) Velocity ratio (V.R.):**The ratio of the velocity of effort to the velocity of load is called the velocity ratio of the machine.

It is also defined as the ratio of the displacement of effort to the displacement of load.

**(e) Efficiency:**Efficiency of a machine is the ratio of the useful work done by the machine to the work put into the machine by the effort. In other words, it is the ratio of the work output to the work input.

**66. Class I lever can have M.A. = 1, M.A. < 1 and M.A. > 1. Explain each giving examples.**

**Answer**

E.g. of class I lever with M.A. = 1: A pair of scissors used to cut a piece of cloth has blades longer than the handle (i.e. effort arm is shorter than the load arm), thus its M.A. = 1.

E.g. of class I lever with M.A. = 1: Shears used for cutting thin metal sheets have much longer handles as compared to the blades (i.e. effort arm is longer than the load arm), thus its M.A. = 1 and it serves as a force multiplier.

**67. Fig. shows a spade. It is being used to lift soil weighing 30N from the ground.**

**(a) Mark the direction of the least force on the handle necessary to keep the spade balanced. (b) Calculate the least force on the handle necessary to keep the spade balanced (the weight of the spade is negligible).(c) If the left hand was move towards the soil on the spade, would the force on the handle necessary to keep the soil balance the greater or less? Give a reason for your answer.(d) To which class of lever does the spade belong?**

**Answer**

**(a) **diagram showing the direction of application of least force on the handle :

**(b)**Given, Load, L = 30 N

Let, effort =E

effort - arm = 20 cm

load - arm = 20 + 10 = 30 cm

Now, load × load - arm = effort × effort - arm

∴ effort = (30×30)/20 = 45 N

**(c)**on moving the left hand towards the soil on the spade, the length of effort arm will increase and effort being inversely proportional to the length of effort arm, the force or effort necessary to keep the soil balanced would be less.

**(d)**A spade belongs to Class III lever.

**68. A crowbar 4 m long has its fulcrum 50 cm from one end. What minimum effort is required to displace a weight of 500 kgf? Calculate the M.A. of the crowbar.**

**Answer**

Given, load, L = 500 kgf

Length of crowbar = 4m = 400 cm

Therefore, load arm = 50 cm

Effort arm = 400 - 50 cm = 350 cm

Let effort = E

**69. A pair of nut crackers is 12 cm long. An effort of 10 gf is required to crack a nut which is passed at a point 3 cm from the finger. Calculate the load (Fig. 15).**

**Answer**

Given, load arm = 3 cm

effort arm = 12 cm

Effort, E = 10 gf

**70. (a) Fig. 16 represents an incomplete diagram of a simple string pulley system. Copy this diagram on a new page and complete it and mark where the effort must be applied to lift the load.**

**(b) What is the velocity ratio of this system?**

(c) If the pulley system is 80% efficient and the load is 720 N, then

(i) What effort must be applied to lift the load?

(ii) What work must be done must be done in lifting the load through a distance of 2 m using this machine?

(c) If the pulley system is 80% efficient and the load is 720 N, then

(i) What effort must be applied to lift the load?

(ii) What work must be done must be done in lifting the load through a distance of 2 m using this machine?

**Answer**

(c) (i) Given, Load, L = 720 N

Let, Effort = E

**72. (a) What makes a balance faulty?**

**(b) A faulty balance of equal arms but pans of unequal weight is used to find the weight of a body. By the method of double weighing the weights are found as 8 kg and 8.2 kg. Find the actual weight of the body.**

(c) The arms of a beam balance are 20 cm and 21 cm, but the pans are of equal weight. By the method of double weighing the weights are found to be 1000 g and 20 g. Find the actual weight of the body.

(d) A faulty balance of unequal arms and pans of unequal weights is used to find the true weight of a metal. By double weighing the weights are found to be 1210 g and 1000 g. Calculate the true weight of the metal.

(c) The arms of a beam balance are 20 cm and 21 cm, but the pans are of equal weight. By the method of double weighing the weights are found to be 1000 g and 20 g. Find the actual weight of the body.

(d) A faulty balance of unequal arms and pans of unequal weights is used to find the true weight of a metal. By double weighing the weights are found to be 1210 g and 1000 g. Calculate the true weight of the metal.

**Answer**