NCERT Solutions for Class 9 Science Chapter 7: Work, Energy and Simple Machines – Free PDF Download

Chapter 7 of the new NCERT “Exploration” textbook for Class 9 Science, “Work, Energy and Simple Machines,” builds on the ideas of force and motion from Chapters 4 and 6 to introduce work, kinetic and potential energy, the work-energy theorem, conservation of mechanical energy, power, and simple machines (pulleys, inclined planes and levers). These solutions were compiled from the official NCERT textbook PDF (ncert.nic.in) and cross-checked against TiwariAcademy, LearnCBSE and Vedantu, with every numerical answer independently recomputed rather than copied.

Last Updated: September 23, 2026

NCERT Solutions for Class 9 Science Chapter 7: Work, Energy and Simple Machines

Revise, Reflect, Refine (NCERT Textbook, Page No. 137)

1. State whether True or False.
(i) Work is said to be done when a force is applied, even if the object does not move.
(ii) Lifting a bucket vertically upward results in positive work done on the bucket.
(iii) The SI unit for both work and energy is joule (J).
(iv) A motionless stretched rubber band has kinetic energy.
(v) Energy can change from one form to another.
(i) False. Work is done only when a force produces displacement; if the object does not move, no work is done.
(ii) True. The applied force and the displacement of the bucket are both directed upward, so the work done is positive.
(iii) True. Both work and energy are measured in the same SI unit, the joule (J).
(iv) False. A motionless object has zero velocity, so its kinetic energy is zero. A stretched rubber band stores elastic potential energy, not kinetic energy.
(v) True. This is the principle behind the law of conservation of energy — energy is never destroyed, only converted from one form to another.

2. Fill in the blanks.
(i) Work done = ________________ × ________________ (in the direction of force).
(ii) 1 joule of work is done when a force of ________________ newton displaces an object by 1 metre in the direction of the force.
(iii) The expression for kinetic energy of a body of mass m and velocity v is ________________.
(iv) The potential energy of an object of mass m at a small height h from the Earth’s surface is ________________.
(v) Power is defined as the ________________ at which work is done.
(i) Force, displacement
(ii) 1 (newton)
(iii) ½mv²
(iv) mgh
(v) rate

3. When a ball thrown upwards reaches its highest point, tick which of the following statement(s) are correct?
(i) The force acting on the ball is zero.
(ii) The acceleration of the ball is zero.
(iii) Its kinetic energy is zero.
(iv) Its potential energy is maximum.
Statements (iii) and (iv) are correct.
(i) is incorrect — the force of gravity (mg) continues to act on the ball throughout its flight, including at the highest point.
(ii) is incorrect — since gravity still acts, the acceleration remains g (≈ 9.8 m/s², downward) at the highest point.
(iii) is correct — the ball’s velocity becomes momentarily zero at the highest point, so KE = ½mv² = 0.
(iv) is correct — since the ball is at its maximum height, its gravitational potential energy (mgh) is maximum there.

4. For each of the following situations, identify the energy transformation that takes place: (i) a truck moving uphill, (ii) unwinding of a watch spring, (iii) photosynthesis in green leaves, (iv) water flowing from a dam, (v) burning of a matchstick, (vi) explosion of a firecracker, (vii) speaking into a microphone, (viii) a glowing electric bulb, and (ix) a solar panel.
(i) Truck moving uphill: kinetic energy → gravitational potential energy (as the truck climbs, it gains height and slows relative to the energy diverted into height gain).
(ii) Unwinding of a watch spring: elastic (stored) potential energy → kinetic energy (motion of the watch hands).
(iii) Photosynthesis in green leaves: light energy → chemical energy (stored in food).
(iv) Water flowing from a dam: gravitational potential energy → kinetic energy (and further to electrical energy when it drives a turbine, as in a hydroelectric dam).
(v) Burning of a matchstick: chemical energy → heat energy + light energy.
(vi) Explosion of a firecracker: chemical energy → heat energy + light energy + sound energy + kinetic energy.
(vii) Speaking into a microphone: sound energy → electrical energy.
(viii) A glowing electric bulb: electrical energy → light energy + heat energy.
(ix) A solar panel: light energy → electrical energy.

5. A student is slowly lifted straight up in an elevator from the ground level to the top floor of a building. Later, the same student climbs the staircase, all the way to the top. Given that the height of the building is h = 72.5 m, acceleration due to gravity is g = 10 m s⁻², and student’s mass is m = 50 kg.
(i) Find the gain in the potential energy if the student is lifted straight up to the top.
(ii) Find the gain in the potential energy when the student climbs the stairs to the same top.
(iii) What do you conclude about the dependence of the potential energy on the path taken?
(i) Gain in PE = mgh = 50 × 10 × 72.5 = 36,250 J.
(ii) Since the student reaches the same final height (72.5 m), the gain in potential energy is the same: 36,250 J.
(iii) Potential energy depends only on the mass and the vertical height gained (the initial and final positions), not on the path taken to get there. Whether the student rides the elevator or climbs the stairs, the gain in potential energy is identical.

6. A crane lifts a mass m to the 10th floor of a building in a certain time. It then raises the same mass to the 20th floor of the same building in double the time. How much more energy and power are required? Assume that the height of all floors is equal.
Let the height of one floor be d and the time to reach the 10th floor be t.
Energy to reach the 10th floor: E₁ = mg(10d) = 10mgd.
Energy to reach the 20th floor: E₂ = mg(20d) = 20mgd = 2E₁.
So twice as much energy is required to reach the 20th floor.
Power to the 10th floor: P₁ = E₁/t = 10mgd/t.
Power to the 20th floor (in time 2t): P₂ = E₂/2t = 20mgd/2t = 10mgd/t = P₁.
So the power required stays the same — the energy needed doubles, but since the crane also takes double the time, the rate of doing work (power) is unchanged.

7. Which factors determine the energy required to raise a flag from the ground to the top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change the amount of work done? If the speed at which the flag is raised is doubled, how does the power requirement change? Explain your answers.
The energy required equals the gain in the flag’s gravitational potential energy, E = mgh, so it depends on the mass of the flag (m), the height of the flagpole (h), and the acceleration due to gravity (g).
Raising the flag slowly or quickly does not change the work done — work depends only on force and displacement (here, on m, g and h), not on the time taken, so W = mgh either way.
If the speed is doubled, the time taken to raise the flag is halved. Since power = work/time, halving the time while keeping the work the same doubles the power required.

8. A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? Assume that the energy transfer to the scooter happens entirely due to fuel, and no other losses occur due to air resistance and friction.
Fuel used is proportional to the kinetic energy gained, KE = ½mv², so for the same final speed v, KE depends only on the total mass being accelerated.
Day 1 total mass = 60 + 100 = 160 kg → KE₁ = ½ × 160 × v² = 80v².
Day 2 total mass = 60 + 40 + 100 = 200 kg → KE₂ = ½ × 200 × v² = 100v².
Ratio of fuel used = KE₁ : KE₂ = 80v² : 100v² = 4 : 5.

9. On a seesaw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice that of the child. The seesaw however is balanced. Draw a figure which depicts this situation showing the distances from the fulcrum where the child and the adult are seated.
For the seesaw (a Class I lever) to balance, the moment on each side about the fulcrum must be equal: effort × effort arm = load × load arm.
Let the child’s weight be W (so the adult’s weight is 2W), the child’s distance from the fulcrum be d₁, and the adult’s distance be d₂.
Balance condition: W × d₁ = 2W × d₂, which gives d₁ = 2d₂.
So the lighter child must sit exactly twice as far from the fulcrum as the heavier adult — for example, if the adult sits 1 m from the fulcrum, the child must sit 2 m from the fulcrum on the opposite side.

10. A ball of mass 2 kg is thrown up with a velocity of 20 m s⁻¹.
(i) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion.
(ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance (assume g = 10 m s⁻²)?
(i) During the upward motion, gravity acts downward while the ball’s displacement is upward — force and displacement are opposite, so the work done by gravity is negative. During the downward motion, both gravity and displacement point downward, so the work done by gravity is positive.
(ii) Without air resistance, the ball would rise to h = u²/2g = (20)²/(2×10) = 20 m. Its initial kinetic energy is ½mv² = ½ × 2 × (20)² = 400 J, and at the actual maximum height of 19.4 m its potential energy is mgh = 2 × 10 × 19.4 = 388 J (kinetic energy is zero at the highest point). Since gravity’s effect is already accounted for through the PE, the “missing” energy was removed by air resistance: work done by air resistance = 388 − 400 = −12 J.

11. A 10.0 kg block is moving on a horizontal floor with negligible friction. As shown in Fig. 7.37, a variable force is applied on the block in its direction of motion from its position at 0 m till 4 m. If the block had a kinetic energy of 180 J when it was at 0 m, find the block’s speed (i) at 0 m, and (ii) at 4 m. Does the block have negative acceleration in any portion of its motion?
(i) Speed at 0 m: 180 = ½ × 10 × v² ⇒ v² = 36 ⇒ v = 6 m/s.
(ii) Work done = area under the force-displacement graph: from 0–1 m (triangle) = ½ × 1 × 50 = 25 J; from 1–3 m (rectangle) = 2 × 50 = 100 J; from 3–4 m (triangle) = ½ × 1 × 50 = 25 J. Total work = 25 + 100 + 25 = 150 J.
Final KE = 180 + 150 = 330 J ⇒ ½ × 10 × v² = 330 ⇒ v² = 66 ⇒ v = √66 ≈ 8.12 m/s.
The block does not have negative acceleration at any point, because the applied force always acts in the direction of motion (the force never becomes negative on the graph), so the block keeps speeding up throughout.

12. The gravitational attraction on the surface of the Moon is about 1/6th of that on the surface of the Earth. An astronaut can throw a ball up to a height of 8 m from the surface of the Earth. How far up will the ball thrown with the same upward velocity travel from the surface of the Moon?
Using h = u²/2g, height is inversely proportional to g for a fixed launch velocity u: h ∝ 1/g.
Since g on the Moon is 1/6th of g on Earth, the height reached on the Moon will be 6 times greater: h(Moon) = 6 × 8 = 48 m.

13. A 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of motion of the car starting from the instant the driver spots the traffic ahead is shown in Fig. 7.38.
(i) Describe how the car moves between positions A and B.
(ii) Calculate the kinetic energy of the car at A.
(iii) State the work done by the brakes in bringing the car to a halt between B and C.
(iv) What does the kinetic energy of the car transform into?
(i) Between A and B, the speed-time graph is a horizontal line at 35 m/s, showing the car moves with constant speed (uniform motion, zero acceleration) — this is the driver’s reaction-time interval before the brakes are applied.
(ii) KE at A = ½mv² = ½ × 1000 × (35)² = 500 × 1225 = 612,500 J.
(iii) Between B and C the brakes bring the car from 35 m/s to rest, so final KE = 0. Work done by the brakes = final KE − initial KE = 0 − 612,500 = −612,500 J. The negative sign shows the braking force opposes the car’s motion.
(iv) The car’s kinetic energy is converted mainly into heat energy (due to friction between the brake pads and wheels, and between the tyres and the road), with a small portion converted into sound energy.

14. The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. 7.39. At O, the velocity of the ball is 0 m s⁻¹ and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.
Since the track is frictionless, total mechanical energy is conserved. At O, KE = 0 and PE = 30 J, so total mechanical energy = 30 J everywhere on the track.
At P, PE = 20 J, so KE = 30 − 20 = 10 J. Using KE = ½mv²: 10 = ½ × 0.5 × v² ⇒ v² = 40 ⇒ v ≈ 6.32 m/s.
At Q, PE = 30 J, so KE = 30 − 30 = 0 J ⇒ v = 0 m/s (Q is a turning point on the track, like O).
At R, PE = 40 J, which is greater than the ball’s total mechanical energy (30 J). This would require negative kinetic energy, which is impossible — so the ball can never reach R. It turns back at Q before getting anywhere near R.

15. A coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by making a depression in the sand.
(i) Calculate the velocity of the coconut just before it hits the sand.
(ii) Assume that the average resistive force of sand is 3000 N and all of the coconut’s energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume g = 10 m s⁻².
(i) Using v² = 2gh = 2 × 10 × 10 = 200 ⇒ v = √200 = 10√2 ≈ 14.14 m/s.
(ii) Kinetic energy just before impact = mgh = 1.5 × 10 × 10 = 150 J. This energy is used entirely to do work against the sand’s resistive force: Force × depth = 150 J ⇒ 3000 × d = 150 ⇒ d = 0.05 m = 5 cm.

In-Text Questions (Think It Over, Pause and Ponder & Activities)

Think It Over — What will be the magnitude of velocity of the child at the bottom of the blue slide? (Page No. 116)
Using conservation of energy, mgh = ½mv², so v = √(2gh). The velocity at the bottom depends only on the height h of the slide and on g — not on the mass of the child or the shape of the slide (assuming friction is negligible).

Think It Over — Will two children of different masses reach the bottom of the same slide with the same velocity? (Page No. 116)
Yes. Since v = √(2gh), the mass cancels out of the equation, so both children reach the bottom of the same slide with the same speed regardless of their individual masses.

Think It Over — Which of the slides will result in the largest magnitude of velocity for the child at its bottom? (Page No. 116)
The slide with the greatest vertical height h gives the largest velocity at the bottom, since v = √(2gh) increases with height.

Pause and Ponder — A weightlifter is shown holding a barbell steady in her hands. Is she doing any work on the barbell while holding it steady? (Page No. 119)
No. Since the barbell has zero displacement while she holds it steady (s = 0), the work done is W = F × s = F × 0 = 0 J. She feels tired because her muscles are continuously contracting and relaxing internally, consuming the body’s chemical energy — but no scientific work is done on the barbell itself.

Pause and Ponder — Is the work done by friction on a stack of coins that travels on a rough surface positive, negative, or zero? (Page No. 119)
Negative. Friction acts opposite to the direction of motion of the coins, so the force and displacement point in opposite directions, making the work done by friction negative.

Pause and Ponder — When you pedal a bicycle on a flat road, your muscles supply energy. In what forms does this muscular energy appear as you ride? (Page No. 121)
The muscular (chemical) energy is converted mainly into the kinetic energy of the bicycle and rider, with some lost as thermal energy (heat) due to friction in the wheels and chain and due to air resistance, and a small amount as sound energy from the moving mechanical parts.

Pause and Ponder — Two objects A and B of mass m and 4m have the same kinetic energy. What is the ratio of the magnitude of velocities of A and B? (Page No. 123)
Equating kinetic energies: ½mv²A = ½(4m)v²B ⇒ v²A = 4v²B ⇒ vA/vB = 2/1. So the ratio of velocities vA : vB is 2 : 1 (the lighter object A moves twice as fast as the heavier object B).

Pause and Ponder — Does the kinetic energy of an object moving with constant velocity change with its position? (Page No. 123)
No. Kinetic energy KE = ½mv² depends only on mass and speed. If the velocity is constant, the kinetic energy stays the same regardless of the object’s position.

Pause and Ponder — Does the potential energy of an object near the surface of the Earth change if it moves with constant velocity in the horizontal direction? What if the object is gradually raised in the vertical direction? (Page No. 126)
In horizontal motion at constant velocity, the height h does not change, so potential energy U = mgh remains constant. If the object is gradually raised vertically, h increases, so its potential energy increases proportionally with height.

Pause and Ponder — For a ball dropped from height h, calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is mgh. (Page No. 129)
Just before hitting the ground, height h′ = 0, so PE = 0. Using v² = 2gh, KE = ½mv² = ½m(2gh) = mgh. So total mechanical energy = KE + PE = mgh + 0 = mgh — exactly equal to the initial mechanical energy at the top (where v = 0 and PE = mgh), confirming that mechanical energy is conserved.

Pause and Ponder — In a science-park exhibit where a ball is released from the highest point and rolls over a series of humps (A, B, C, D, E), describe how kinetic and potential energy change at points A, B and C. Why do later humps have progressively lower heights? Could this be due to friction? (Page No. 129)
At the higher points (like A and C), the ball has more potential energy and less kinetic energy; at the lowest point (B), it has less potential energy and more kinetic energy — energy keeps converting between the two forms. The heights of successive humps (C, D, E) get progressively lower because the ball’s total mechanical energy is gradually reduced by friction and air resistance, which convert part of the mechanical energy into heat, so less energy remains available to reach the same height as before.

Pause and Ponder — Explain why roads on hills are built to wind around in gentle slopes rather than going straight up. (Page No. 132)
A winding road acts as an inclined plane of longer length L for the same height h gained, so its mechanical advantage (MA = L/h) is much greater than a steep, straight road. A greater MA means the engine needs to exert a much smaller force to climb the same height, even though it travels a longer distance — the total work done (mgh) stays the same either way.

Pause and Ponder — To reach a higher floor, we find climbing an inclined ladder easier compared to climbing a vertical ladder. Explain why. (Page No. 132)
An inclined ladder behaves like an inclined plane: since its length L is greater than the height h it covers, its mechanical advantage (L/h) is greater than 1, so the force needed at each step is less than the climber’s full body weight. Total work done is the same as climbing a vertical ladder, but the effort at each moment is reduced.

Pause and Ponder — Why is it easier to open the lid of a can by using a spoon? (Page No. 135)
The spoon acts as a Class I lever, with the rim of the can as the fulcrum. Pressing down at the far end of the spoon creates a long effort arm compared to the short load arm at the lid. Since MA = effort arm/load arm is greater than 1, a small force applied at the end of the spoon produces a much larger force at the lid, popping it open.

Pause and Ponder — Why do you push an object closer to the scissors’ fulcrum when you want to cut something hard? (Page No. 135)
Scissors are a Class I lever, and the load arm is the distance from the fulcrum to the object being cut. Moving the object closer to the fulcrum shortens the load arm, which increases the mechanical advantage (MA = effort arm/load arm), so the same hand effort produces a much larger cutting force on the object.

Pause and Ponder — Many designs for perpetual motion machines have been proposed throughout history, but none actually work. Why do all real machines eventually slow down and stop? Explain in terms of work and energy. (Page No. 135)
Every real machine loses some of its energy to friction, air resistance and similar effects while it operates, and this lost energy is converted mainly into heat (and sometimes sound) that cannot be recovered and reused by the machine. Because usable energy keeps decreasing with every cycle, the machine eventually runs out of energy to do further work and stops. A perpetual motion machine would need to operate with zero energy loss, which never happens in the real world — this is why such machines are impossible.

Activity 7.1 — Let Us Investigate: Relationship between drop height and depression in sand (Page No. 125)
Aim: To study how the height from which a ball is dropped affects the depth of the depression it creates in sand.
Observation: The higher the ball is dropped from, the deeper the depression it makes in the sand — a ball dropped from 2 m creates a deeper depression than one dropped from 1 m.
Conclusion: Raising the ball to a greater height requires more work, so it possesses greater gravitational potential energy (mgh) at that height. When released, this PE converts into KE, and a ball with more KE on impact creates a deeper depression.

Activity 7.2 — Let Us Experiment: Conservation of mechanical energy in a simple pendulum (Page No. 127)
Aim: To demonstrate conservation of mechanical energy using an oscillating pendulum.
Observation: At the extreme position P, the bob is at maximum height with PE = mgh and KE = 0. At the lowest point Q, PE = 0 and KE is maximum. At the other extreme R, the bob almost regains its original height, with KE returning to nearly 0.
Conclusion: The sum KE + PE (total mechanical energy) stays constant throughout the swing. In practice, the pendulum’s swing gradually shrinks because friction at the support and air resistance slowly convert some mechanical energy into heat.

Activity 7.3 — Let Us Experiment: Does an inclined plane reduce the force needed to raise a load? (Page No. 131)
Aim: To find out whether an inclined plane reduces the effort needed to raise an object to a height.
Observation: The spring-balance reading needed to pull a cart up an inclined plank is smaller than the force needed to lift it straight up. As the plank is made less steep, the required force decreases further, but the distance the force must be applied over increases.
Conclusion: An inclined plane reduces the effort needed to raise a load, at the cost of applying that effort over a greater distance — the total work done (mgh) remains unchanged, consistent with conservation of energy.

Activity 7.4 — Let Us Investigate: Lifting a heavy object with a lever (Page No. 133)
Aim: To show how a lever can lift a heavier object using a much smaller force.
Observation: Placing a heavy stapler close to a pencil (acting as the fulcrum) and pressing down with a light eraser at the far end of a ruler can lift the stapler.
Conclusion: A much heavier load can be lifted with a lighter effort when the effort arm is made longer than the load arm, following the principle effort × effort arm = load × load arm, so MA = effort arm/load arm.

Activity 7.5 — Let Us Experiment: Verifying the law of levers with a beam balance (Page No. 133)
Aim: To verify the law of levers using a beam balance loaded with coins at different distances from the fulcrum.
Observation and Result: The beam balances whenever (number of coins on one side × its distance from the fulcrum) equals (number of coins on the other side × its distance from the fulcrum), i.e., effort × effort arm = load × load arm.
Conclusion: Increasing the effort arm allows a smaller effort to balance (or move) a larger load. This confirms MA = load/effort = effort arm/load arm for a lever.

Why This Chapter Matters

Work, Energy and Simple Machines gives students the tools to explain motion and forces in situations where Chapter 4 (kinematics) and Chapter 6 (laws of motion) alone become cumbersome — from a car braking to a stop, to a ball rolling down a hill, to a crane lifting building materials. The concepts of energy conservation and the work-energy theorem introduced here recur throughout higher-secondary physics, in topics ranging from rotational mechanics to thermodynamics and electricity, while the section on simple machines connects directly to everyday technology like pulleys, ramps and levers that students already use without thinking about the physics behind them. A solid grasp of this chapter is essential both for scoring well in CBSE Class 9 exams and for building intuition needed in Class 11 Physics.

More on This Chapter

Extra Questions | Revision Notes | Formulas Handbook

Frequently Asked Questions

What is the difference between the work-energy theorem and the law of conservation of mechanical energy?
The work-energy theorem (W = ΔE) is a general rule stating that the work done on an object equals the change in its energy — it holds even when friction or other non-conservative forces act. The law of conservation of mechanical energy is a special case that applies only when no force other than gravity (or another conservative force) does work on the object, in which case KE + PE remains exactly constant. Whenever friction or air resistance is present, some mechanical energy is lost as heat, and only the more general work-energy theorem still holds.

Why does a fixed pulley not provide any mechanical advantage?
A fixed pulley only changes the direction of the applied force (for example, letting you pull down instead of lift up) — it does not change the amount of force needed. Since the effort required equals the load being lifted, its mechanical advantage MA = load/effort is exactly 1. Movable pulleys or pulley systems are needed to actually reduce the effort required.

Does the mass of an object affect its speed at the bottom of a frictionless slide?
No. Using conservation of energy, mgh = ½mv², the mass m cancels from both sides, giving v = √(2gh). The final speed depends only on the height of the slide and on g, not on the mass of the object — this is confirmed by the “Think It Over” discussion on page 116 of the chapter.

Why can’t perpetual motion machines exist, according to this chapter?
Every real machine loses some energy to friction, air resistance or similar effects during operation, and this lost energy is converted into heat that cannot be recovered and reused. Since a machine’s usable energy keeps decreasing with each cycle, it will eventually run out of energy and stop unless more energy is supplied. A perpetual motion machine would require zero energy loss, which is impossible for any real, physical system.

What’s new in this “Exploration” edition chapter compared to the earlier NCERT “Work and Energy” chapter?
This revised chapter significantly expands the earlier “Work and Energy” chapter by adding an entire new section (7.6) on simple machines — covering pulleys, inclined planes, and all three classes of levers with hands-on activities and mechanical-advantage derivations. It also introduces the work-energy theorem more formally, derives the expression for kinetic energy directly from the kinematic equations, and links conservation of energy to real-world systems such as watermills, escape ramps and roller-coaster-style exhibits.

Written by Satish

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