Selina Concise Solutions for ICSE Class 10 Physics Chapter 2 Work Energy And Power

Exercise 2(A)

Question 1. Define work. Is work a scalar or a vector?
Answer: We define work as occurring when a force acting on an object causes it to move. It is classified as a scalar quantity because it has no direction.
In simple words: Work is done when you push or pull something and it moves. It does not have a direction, so it is a scalar.

Exam Tip: Always remember that displacement is a necessary condition for work to be done. If there is no movement, the work done is zero.

Question 2. How is the work done by a force measured when (i) force is in direction of displacement, (ii) force is at an angle to the direction of displacement?
Answer:
(i) If the applied force matches the path of movement, the work is calculated as \( W = F \times S \).
(ii) If the force is applied at an angle relative to the movement path, the work is determined using the formula \( W = F \, S \cos\theta \).
In simple words: When pushing straight, multiply force by distance. If pushing at an angle, multiply that result by the cosine of the angle.

Exam Tip: Be sure to write the formula clearly for both cases and define what each symbol (\( F, S, \theta \)) represents.

Question 3. A force F acts on a body and displaces it by a distance S in a direction at an angle θ with the direction of force. (a) Write the expression for the work done by the force. (b) what should be the angle between the force and displacement to get the (i) zero work (ii) maximum work?
Answer:
(a) When a force acts at an angle to the movement direction, the work is given by \( W = F \, S \cos\theta \).
(b)
(i) To obtain zero work, the angle must be \( 90^\circ \) because \( \cos 90^\circ = 0 \). Thus, \( W = F \, S \cos 90^\circ = F \, S \times 0 = 0 \).
(ii) To get the greatest amount of work, the angle must be \( 0^\circ \) since \( \cos 0^\circ = 1 \). Therefore, \( W = F \, S \cos 0^\circ = F \, S \).
In simple words: Work is greatest when you push in the exact direction of motion. If you push perpendicular to the movement, no work is done.

Exam Tip: Mention the specific trigonometric values of cosine (\( \cos 90^\circ = 0 \) and \( \cos 0^\circ = 1 \)) to secure full marks.

Question 4. A body is acted upon by a force. State two condition when the work done is zero.
Answer: The two requirements for work to be zero are:
(i) There is no movement of the object (\( S = 0 \)).
(ii) The movement occurs perpendicular to the direction of the applied force (\( \theta = 90^\circ \)).
In simple words: No work is done if the object does not move at all, or if it moves sideways relative to the push.

Exam Tip: Clearly state both conditions: zero displacement and perpendicular force-displacement relationship.

Question 5. State the condition when the work done by a force is (i) positive, (ii) negative. Explain with the help of examples.
Answer:
(i) Work is positive when an object moves in the same direction as the force applied to it. In this scenario, \( W = F \times S \). For instance, when a porter lifts a package upwards, the lifting force and the movement are both pointing upwards.
(ii) Work is negative when the movement of an object is in the exact opposite direction of the applied force. Here, \( W = -F \times S \). For example, as an object slides along a floor, the friction force acts opposite to the direction of travel, resulting in negative work by friction.
In simple words: Positive work happens when the push and the movement are in the same direction. Negative work happens when they point in opposite directions, like braking a car.

Exam Tip: Use clear real-world examples like friction (for negative work) and lifting objects (for positive work) to illustrate these concepts in detail.

Question 6. A body is moved in a direction opposite to the direction of force acting on it. State whether the work is done by the force or work is done against the force.
Answer: In this scenario, work is performed against the applied force.
In simple words: If you move opposite to where a force is pushing, you are working against that force.

Exam Tip: Remember that moving against any force always implies a negative work value relative to that specific force.

Question 7. When a body moves in a circular path, how much work is done by the body? Give reason.
Answer: No work is performed when an object travels along a circular path. This is because the centripetal force pulls the object directly toward the center, while its movement at any given moment is tangential. Consequently, the force and displacement are perpendicular to each other at all times.
In simple words: Moving in a circle does no work because the pulling force points inward while the movement is sideways along the edge of the circle.

Exam Tip: State that the angle between centripetal force and instantaneous displacement is \( 90^\circ \), which mathematically results in \( \cos 90^\circ = 0 \).

Question 8. A satellite revolves around the earth in a circular orbit. What is the work done by the force of gravity? Give reason.
Answer: The gravitational force performs zero work on the satellite. This occurs because the pull of gravity, acting as a centripetal force, points perpendicular to the direction in which the satellite is moving at any instant.
In simple words: Gravity does not do any work on a circular satellite because it pulls the satellite at a right angle to its flight path.

Exam Tip: Specify that the gravitational force is perpendicular to the instantaneous displacement vector.

Question 9. In which of the following cases, is work being done? (i) A man pushing a wall. (ii) a coolie standing with a load of 12 kgf on his head. (iii) A boy climbing up a staircase.
Answer: Work is performed solely when the boy climbs up the stairs. In the first two scenarios, there is no displacement, so no work is done.
In simple words: Work only happens when something moves. The wall and the standing porter do not move, but the climbing boy does.

Exam Tip: Explicitly state why work is not done in the other cases (i.e., zero displacement).

Question 10. A coolie carrying a load on his head and moving on a frictionless horizontal platform does no work. Explain the reason.
Answer: As the porter carries the load horizontally, gravity acts straight down. Since the displacement is horizontal, it is perpendicular to the vertical gravitational force, resulting in zero work against gravity.
In simple words: Walking horizontally with a load on your head does no work against gravity because you are moving sideways while gravity pulls straight down.

Exam Tip: Clearly identify that the angle between the vertical gravitational force and the horizontal movement is \( 90^\circ \).

Question 11. The work done by a fielder when he takes a catch in a cricket match, is negative Explain.
Answer: The fielder exerts an opposing force on the incoming ball to slow it down. Because this stopping force acts opposite to the direction the ball is moving, the resulting work done by the fielder is negative.
In simple words: The player pushes back against the moving ball to stop it. Because they push in the opposite direction of the ball's motion, the work is negative.

Exam Tip: Mention that the angle between the retarding force applied by the fielder and the displacement of the ball is \( 180^\circ \).

Question 12. Give an example when work done by the force of gravity acting on a body is zero even though the body gets displaces from its initial position.
Answer: An example is a porter carrying luggage across a flat platform. Even though the luggage travels horizontally, the downward pull of gravity is at a right angle to this motion, meaning gravity does no work.
In simple words: When you walk on flat ground with a backpack, gravity does zero work because you move sideways while gravity pulls down.

Exam Tip: This is a classic board exam question; always relate the zero work to the perpendicular orientation of the force and displacement vectors.

Question 13. What are the S.I. and C.G.S units of work? How are they related? Establish the relationship.
Answer: The SI unit for work is the Joule (J), and its CGS unit is the erg.
To derive their relationship, we use:
\( 1 \text{ Joule} = 1 \text{ N} \times 1 \text{ m} \)
Since \( 1 \text{ N} = 10^5 \text{ dyne} \) and \( 1 \text{ m} = 10^2 \text{ cm} \):
\( 1 \text{ Joule} = 10^5 \text{ dyne} \times 10^2 \text{ cm} = 10^7 \text{ dyne} \cdot \text{ cm} = 10^7 \text{ erg} \)
Therefore, \( 1 \text{ Joule} = 10^7 \text{ erg} \).
In simple words: The main standard unit of work is the Joule, while the smaller CGS unit is the erg. One Joule contains ten million (or ten to the power of seven) ergs.

Exam Tip: Clearly show the intermediate conversions for both Force (\( 1\text{ N} = 10^5\text{ dyne} \)) and Displacement (\( 1\text{ m} = 10^2\text{ cm} \)) to get full marks on derivations.

Question 14. State and define the S.I. unit of work.
Answer: The SI unit of work is the Joule. One Joule of work is completed when an applied force of one Newton moves an object a distance of one meter along the line of that force.
In simple words: One Joule is the amount of work done when a force of one Newton pushes an object by one meter.

Exam Tip: Make sure to specify "in the direction of force" when defining the Joule, as displacement must align with the force.

Question 15. Express joule in terms of erg.
Answer: We can relate the Joule to the erg through the following steps:
\( 1 \text{ Joule} = 1 \text{ N} \times 1 \text{ m} \)
Given that \( 1 \text{ N} = 10^5 \text{ dyne} \) and \( 1 \text{ m} = 10^2 \text{ cm} \):
\( 1 \text{ Joule} = 10^5 \text{ dyne} \times 10^2 \text{ cm} = 10^7 \text{ erg} \)
So, \( 1 \text{ Joule} = 10^7 \text{ erg} \).
In simple words: One Joule is equal to ten to the power of seven ergs.

Exam Tip: Although similar to the previous derivation, keep the presentation neat and show each step of exponent addition clearly.

Question 16. A body of mass m falls down through a height h. Obtain an expression for the work done by the force of gravity.
Answer: Consider an object of mass \( m \) falling vertically downward through a distance \( h \), whether along a straight drop or down a slope. The gravitational force acting on this mass is \( F = mg \), pointing downwards. Since the downward displacement is \( S = h \), the work done by gravity is:
\( W = F \, S = mgh \)
In simple words: When an object falls, gravity pulls it down. The work done by gravity is found by multiplying its mass, gravity, and the height it falls.

Exam Tip: Note that this work is positive because both the gravitational force and the downward displacement are in the same direction.

Question 17. A boy of mass m climbs up a staircase of vertical height h. (a) What is the work done by the boy against the force of gravity? (b) What would have been the work done if he uses a lift in climbing the same vertical height?
Answer:
(a) When a boy of mass \( m \) goes up a height \( h \), gravity exerts a downward force \( F = mg \) while his displacement is upward (\( S = -h \)). The work done by gravity is \( W = -mgh \). Therefore, the work done *against* gravity is \( W = mgh \).
(b) The work done against gravity would remain \( mgh \) even if he uses an elevator, because the vertical displacement and the gravitational force do not depend on the path taken.
In simple words: Whether walking up stairs or riding an elevator, the effort against gravity is the same because the final height climbed is identical.

Exam Tip: State clearly that the work done against gravity depends only on the vertical height and not on the route chosen.

Question 18. Define the term energy and state its S.I. unit.
Answer: Energy represents an object's ability to perform work. It shares its SI unit with work, which is the Joule (J).
In simple words: Energy is the power or capacity to do any physical work. Its unit is the Joule.

Exam Tip: Remember that energy and work are directly related, meaning they are measured using the exact same unit.

Question 19. What physical quantity does the electron volt (eV) measure? How is it related to the S.I. unit of that quality?
Answer: The electron volt (eV) is a unit used to measure the energy of atomic and subatomic particles.
Its relationship to the Joule is:
\( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \)
In simple words: An electron volt is a tiny unit of energy used for very small particles like atoms.

Exam Tip: Write the exponent accurately as \( -19 \) when stating this conversion value to avoid simple errors.

Question 20. Complete the following sentence: 1 J = Calorie
Answer: \( 1 \text{ J} = 0.24 \text{ calorie} \)
In simple words: One Joule is equal to about 0.24 calories.

Exam Tip: Memorize both conversions (\( 1\text{ J} = 0.24\text{ cal} \) and \( 1\text{ cal} = 4.18\text{ J} \)) as they are frequently tested.

Question 21. Name the physical quantity which is measured in calorie. How is it related to the S.I. unit of the quality?
Answer: A calorie is a unit used to measure heat energy. Its relationship to the SI unit of energy (Joule) is expressed as:
\( 1 \text{ calorie} = 4.18 \text{ J} \) (or approximately \( 4.2 \text{ J} \)).
In simple words: Calories measure heat. One calorie is worth about 4.18 Joules of energy.

Exam Tip: Mention "heat energy" specifically rather than just "energy" to show a precise understanding.

Question 22. Define a kilowatt hour. How is it related to joule?
Answer: A kilowatt-hour (kWh) is defined as the total energy consumed (or work performed) by an appliance of power \( 1 \text{ kW} \) operating continuously for \( 1 \text{ hour} \).
The conversion to Joules is:
\( 1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J} \)
In simple words: A kilowatt-hour is the energy a 1000-watt machine uses in one hour. This equals 3.6 million Joules.

Exam Tip: Show the derivation steps (\( 1000\text{ W} \times 3600\text{ s} \)) if a multi-mark question asks you to establish the relationship.

Question 23. Define the term power. State its S.I. unit.
Answer: Power is defined as the rate at which work is performed or energy is transferred. Its SI unit is the watt (W).
In simple words: Power is how fast you can do work. It is measured in watts.

Exam Tip: State the mathematical definition (\( P = W/t \)) along with the unit to give a complete answer.

Question 24. State two factors on which power spent by a source depends. Explain your answer with examples.
Answer: The power output of any source depends on two primary elements:
(i) The total work performed by the source.
(ii) The duration required to complete this work.
For example, if Porter A lifts a suitcase to the top of a bus in 1 minute, while Porter B takes 2 minutes for the identical task, both perform the same total work. However, Porter A exerts twice the power of Porter B because the work is completed in half the time.
In simple words: Power depends on how much work you do and how fast you do it. Doing the same job quicker requires more power.

Exam Tip: Remember the formula \( \text{Power} = \frac{\text{Work}}{\text{Time}} \) to help explain why shorter time leads to higher power.

Question 25. Differentiate between work and power.
Answer:

In simple words: Work is the total effort made to move an object, whereas power measures how fast that effort is completed.
Exam Tip: Presenting comparisons in tabular format with clear points is the best way to earn full marks in board exams.

Question 26. Differentiate between energy and power.
Answer:

In simple words: Energy is the total capacity to perform tasks, while power is the speed at which that energy is used up.
Exam Tip: Remember that both work and energy share the same SI unit (Joule), while power is measured in watts.

Question 27. State and define the S.I. unit of power.
Answer: The SI unit of power is the watt (W). One watt of power is spent when one Joule of work is done in a time span of one second.
In simple words: One watt means doing one Joule of work every second.

Exam Tip: Write the equation \( 1 \text{ W} = \frac{1 \text{ J}}{1 \text{ s}} \) to explicitly define the unit.

Question 28. What is horse power (H.P)? How is it related to the S.I. unit of power?
Answer: Horsepower (H.P.) is an alternative unit of power commonly used in mechanical applications. It relates to the watt as:
\( 1 \text{ H.P.} = 746 \text{ W} \)
In simple words: Horsepower is an older unit for power often used for car engines. One horsepower is equal to 746 watts.

Exam Tip: Keep this conversion factor (\( 746 \text{ W} \)) memorized as it is very common in power numericals.

Question 29. Differentiate between watt and watt hour.
Answer: The watt (W) is a unit of power, whereas the watt-hour (Wh) is a unit of work or energy, since multiplying power by time gives work.
In simple words: Watt tells you how fast energy is being used, while watt-hour tells you the total amount of energy used over time.

Exam Tip: Be careful not to confuse units of power (watt) with units of energy (watt-hour).

Question 30. Name the quality which is measured in (a) kWh (b) kW (c) Wh (d) eV
Answer:
(a) Energy is measured in kilowatt-hours (kWh).
(b) Power is measured in kilowatts (kW).
(c) Energy is measured in watt-hours (Wh).
(d) Energy is measured in electron volts (eV).
In simple words: Kilowatts measure how much power a machine has, while kilowatt-hours, watt-hours, and electron-volts measure how much energy is spent.

Exam Tip: Notice that the letter 'h' (hour) indicates a time multiplication, converting a power unit (W or kW) into an energy unit.

Multiple Choice Type

Question 1. One horse power is equal to: (a) 1000 W (b) 500 W (c) 764 W (d) 746 W
Answer: (d) 746 W
In simple words: One horsepower is a unit used to measure power, and it is exactly equal to 746 watts.

Exam Tip: This conversion factor is very useful for solving electrical and mechanical energy problems.

Question 2. kWh is the unit of: (a) power (b) force (c) energy (d) none of these
Answer: (c) energy
In simple words: Kilowatt-hour measures the total amount of electrical energy consumed by appliances over time.

Exam Tip: Remember that watt is for power, but multiplying power by time (like hours) gives energy.

Numericals

Question 1. A body, when acted upon by a force of 10 kgf, gets displaced by 0.5 m. Calculate the work done by the force, when the displacement is (i) in the direction of force, (ii) at an angle of 60° with the force, and (iii) normal to the force. (g = 10 N kg-1)
Answer: First, convert force to Newtons:
\( F = 10 \text{ kgf} = 10 \times 10 \text{ N} = 100 \text{ N} \)
The displacement is given as \( S = 0.5 \text{ m} \).
(i) When displacement is in the same direction as the force:
\( W = F \times S = 100 \text{ N} \times 0.5 \text{ m} = 50 \text{ J} \)
(ii) When displacement is at a \( 60^\circ \) angle to the force:
\( W = F \, S \cos\theta \)

\( \implies W = 100 \times 0.5 \times \cos 60^\circ \)
Since \( \cos 60^\circ = 0.5 \):
\( W = 100 \times 0.5 \times 0.5 = 25 \text{ J} \)
(iii) When displacement is normal (perpendicular) to the force:
\( W = F \, S \cos 90^\circ \)

\( \implies W = 100 \times 0.5 \times 0 = 0 \text{ J} \)
In simple words: Pushing in the same direction gives 50 Joules of work. Pushing at a 60-degree angle gives 25 Joules. Pushing straight down while moving sideways does 0 Joules.

Exam Tip: Always convert force from kgf to Newtons by multiplying by \( g \) before starting calculations.

Question 2. A boy of mass kg runs upstairs and reaches the 8 m high floor in 5 s Calculate: the force of gravity acting on the boy. (i) the work done by him against gravity. (ii) the power spent by boy. (Take g = 10 m s-2)
Answer: We are given the mass of the boy as \( 40 \text{ kg} \), the vertical height as \( h = 8 \text{ m} \), and the time taken as \( t = 5 \text{ s} \).
The gravitational force acting on the boy is:
\( F = mg = 40 \text{ kg} \times 10 \text{ m s}^{-2} = 400 \text{ N} \)
(i) The work performed by the boy against gravity is:
\( W = F \times h = 400 \text{ N} \times 8 \text{ m} = 3200 \text{ J} \)
(ii) The power exerted by the boy is:
\( P = \frac{\text{Work done}}{\text{Time taken}} = \frac{3200 \text{ J}}{5 \text{ s}} = 640 \text{ W} \)
In simple words: Gravity pulls the boy with 400 Newtons of force. To climb 8 meters, he does 3200 Joules of work. Because he does this in 5 seconds, his power output is 640 watts.

Exam Tip: Write down every formula used (\( F=mg \), \( W=F \cdot h \), and \( P=W/t \)) step-by-step to earn full step-marks.

Question 3. It takes 20 s for a person A to climb up the stairs, while another person B does the same in 15 s. Compare the (i) Work done and (ii) power developed by the persons A and B.
Answer:
(i) The total work done against gravity is independent of the duration. Because both people climb the same staircase, they do equal work:
\( \frac{\text{Work done by A}}{\text{Work done by B}} = \frac{1}{1} = 1:1 \)
(ii) Power is inversely proportional to the time taken when work is constant (\( P \propto \frac{1}{t} \)). Therefore, the ratio of power is:
\( \frac{\text{Power developed by A}}{\text{Power developed by B}} = \frac{t_B}{t_A} = \frac{15 \text{ s}}{20 \text{ s}} = \frac{3}{4} = 3:4 \)
In simple words: Both people do the exact same amount of work to reach the top. However, because B climbs faster, B generates more power, giving a power ratio of 3 to 4.

Exam Tip: When comparing ratios, remember to invert the time ratio to find the power ratio, since less time means more power.

Question 4. A boy weighing 350 N runs up a flight of 30 steps, each 20 cm high in 1 minute, Calculate: (i) the work done and (ii) power spent.
Answer:
(i) First, calculate the total height climbed:
\( h = 30 \times 20 \text{ cm} = 600 \text{ cm} = 6 \text{ m} \)
The work performed in climbing is:
\( W = F \times h = 350 \text{ N} \times 6 \text{ m} = 2100 \text{ J} \)
(ii) Convert the time to seconds: \( 1 \text{ minute} = 60 \text{ s} \).
The power spent is:
\( P = \frac{W}{t} = \frac{2100 \text{ J}}{60 \text{ s}} = 35 \text{ W} \)
In simple words: The boy climbs a total of 6 meters, doing 2100 Joules of work. Spreading this work over 60 seconds gives a power output of 35 watts.

Exam Tip: Always convert time from minutes to seconds (\( 1 \text{ min} = 60 \text{ s} \)) before calculating power in watts.

Question 5. A man spends 6.4 KJ energy in displacing a body by 64 m in the direction in which he applies force, in 2.5 s Calculate: (i) the force applied and (ii) the power Spent (in H.P) by the man.
Answer:
(i) The energy spent is equivalent to the work done:
\( W = 6.4 \text{ kJ} = 6400 \text{ J} \)
Using \( W = F \times S \):
\( 6400 \text{ J} = F \times 64 \text{ m} \)

\( \implies F = \frac{6400}{64} = 100 \text{ N} \)
(ii) The power in watts is:
\( P = \frac{W}{t} = \frac{6400 \text{ J}}{2.5 \text{ s}} = 2560 \text{ W} \)
To convert to horsepower (H.P.):
\( P = \frac{2560 \text{ W}}{746 \text{ W/H.P.}} \approx 3.43 \text{ H.P.} \)
In simple words: The man uses a force of 100 Newtons to move the object. His power output is 2560 watts, which is about 3.43 horsepower.

Exam Tip: Make sure to convert kilojoules to Joules (\( 1\text{ kJ} = 10^3\text{ J} \)) as the very first step in your calculation.

Question 6. A weight lifter a load of 200 kgf to a height of 2.5 m in 5 s. Calculate: (i) the work done, and (ii) the power developed by him. Take g = 10 N kg-1
Answer:
(i) First, convert the force from kgf to Newtons:
\( F = 200 \text{ kgf} = 200 \times 10 \text{ N} = 2000 \text{ N} \)
The work done to lift this load is:
\( W = F \times S = 2000 \text{ N} \times 2.5 \text{ m} = 5000 \text{ J} \)
(ii) The power developed is:
\( P = \frac{W}{t} = \frac{5000 \text{ J}}{5 \text{ s}} = 1000 \text{ W} \) (or \( 1 \text{ kW} \))
In simple words: The weightlifter lifts a 2000-Newton force up by 2.5 meters, doing 5000 Joules of work. Doing this in 5 seconds means his power is 1000 watts.

Exam Tip: Always write down units like Joules (J) and Watts (W) for final values to avoid losing easy marks.

Question 7. A machine raises a load of 750 N through a height of 16 m in 5 s. calculate: (i) energy spent by machine, (ii) power at which the machine works.
Answer:
(i) The energy spent corresponds to the work done against gravity:
\( W = F \times S = 750 \text{ N} \times 16 \text{ m} = 12000 \text{ J} \) (or \( 12 \text{ kJ} \))
(ii) The power at which the machine operates is:
\( P = \frac{W}{t} = \frac{12000 \text{ J}}{5 \text{ s}} = 2400 \text{ W} \) (or \( 2.4 \text{ kW} \))
In simple words: The machine uses 12000 Joules of energy to lift the load. Since it works for 5 seconds, its power level is 2400 watts.

Exam Tip: Keep in mind that energy spent and work done are equivalent in these gravitational lifting problems.

Question 8. An electric heater of power 3 KW is used for 10 h. How much energy does it consume? Express your answer in (i) kWh, (ii) joule.
Answer:
Power consumption of the heater is calculated by multiplying its power by the operational time.
(i) Energy used = \( 3\text{ kW} \times 10\text{ h} = 30\text{ kWh} \)
(ii) Since \( 1\text{ kWh} = 3.6 \times 10^6\text{ J} \), the total energy in joules is:
\( 30 \times 3.6 \times 10^6\text{ J} = 1.08 \times 10^8\text{ J} \)
In simple words: Electrical energy is power times time. Convert kilowatts and hours to get kilowatt-hours, then multiply by \( 3.6 \times 10^6 \) to convert to joules.
Exam Tip: Keep units correct. Convert power to kW and time to hours for kWh first, then use the direct conversion factor to find joules.

Question 9. A boy of mass 40 kg runs up a flight of 15 steps each 15 cm high in 10 s. Find:
(i) the work done and
(ii) the power developed by him
Take g = 10 N kg -1
Answer:
First, determine the gravitational force acting on the boy:
\( F = mg = 40 \times 10 = 400\text{ N} \)
Now, find the total vertical height climbed:
\( S = 15 \times 15\text{ cm} = 225\text{ cm} = 2.25\text{ m} \)
(i) Work performed while climbing is the force multiplied by the displacement in the direction of the force:
\( W = F \times S = 400 \times 2.25 = 900\text{ J} \)
(ii) Power generated is calculated as:
\( P = \frac{\text{Work}}{\text{Time}} = \frac{900\text{ J}}{10\text{ s}} = 90\text{ W} \)
In simple words: The boy's weight is 400 N, and he climbs a height of 2.25 m. Multiplying these gives the work of 900 J, and dividing by 10 seconds gives the power of 90 W.

Exam Tip: Remember to convert the step height from centimeters to meters before calculating work, otherwise your energy and power values will be incorrect.

Question 10. A water pump raises 50 litres of water through a height of 25 m in 5 s. Calculate the power which the pump supplies. (Take g = 10 N kg-1 and density of water = 1000 kg m-3)
Answer:
We first compute the mass of the water:
\( \text{Volume of water} = 50\text{ L} = 50 \times 10^{-3}\text{ m}^3 \)
\( \text{Density of water} = 1000\text{ kg m}^{-3} \)
\( \text{Mass of water } (m) = \text{Volume} \times \text{Density} = 50 \times 10^{-3} \times 1000 = 50\text{ kg} \)
The work performed to lift this mass of water against gravity is:
\( W = mgh \)
Therefore, the rate of work done, which is power, is:
\( P = \frac{\text{Work done}}{\text{Time}} = \frac{mgh}{t} = \frac{50 \times 10 \times 25}{5} = 2500\text{ W} \)
In simple words: Since 50 liters of water weighs 50 kg, the pump does work equal to \( mgh \). Dividing this total work by the 5 seconds it takes gives the pump's power output of 2500 W.

Exam Tip: For water, 1 liter always has a mass of 1 kg. You can use this shortcut directly in exams unless you are specifically asked to derive the mass using volume and density.

Question 11. A man raises a box of mass 50 kg to a height of 2 m in 2 minutes, while another man raises the same box to the same height in 5 minutes. Compare:
(i) the work done and
(ii) the power developed by them.
Answer:
(i) Work performed to lift a 50 kg load to a height of 2 m against gravity is given by:
\( W = mgh \)
Because the mass and height are identical for both individuals, they perform an equal amount of work.
\( \frac{\text{Work done by A}}{\text{Work done by B}} = \frac{mgh}{mgh} = \frac{50 \times 10 \times 2}{50 \times 10 \times 2} = \frac{1}{1} = 1:1 \)
(ii) Man A performs the task in 2 minutes (120 s), while Man B takes 5 minutes (300 s).
Power generated by A:
\( P_A = \frac{\text{Work}}{\text{Time}} = \frac{mgh}{t_A} = \frac{50 \times 10 \times 2}{120} = \frac{25}{3}\text{ W} \)
Power generated by B:
\( P_B = \frac{\text{Work}}{\text{Time}} = \frac{mgh}{t_B} = \frac{50 \times 10 \times 2}{300} = \frac{10}{3}\text{ W} \)
Since power is inversely proportional to time (\( P \propto \frac{1}{t} \)) when work is constant:
\( \frac{\text{Power of A}}{\text{Power of B}} = \frac{25/3}{10/3} = \frac{25}{10} = \frac{5}{2} = 5:2 \)
In simple words: Both men lift the same weight to the same height, so they do the exact same amount of work. However, because the first man does it faster, he produces more power in a 5:2 ratio.

Exam Tip: For comparison questions, always express your final answer as a simplified ratio (like 1:1 or 5:2) and state the steps clearly to show how you got it.

Question 12. A pump is used to lift 500 kg of water from a depth of 80 m in 10 s. calculate:
(a) the work done by the pump
(b) the power a which the pump works,
(c) the power rating of the pump if its efficiency is 40% (Take g = 10 m s-2)
Answer:
(a) The work required to raise the water against gravity is:
\( W = mgh = 500 \times 10 \times 80 = 4 \times 10^5\text{ J} \)
(b) The output power (useful power) of the pump is calculated by:
\( P = \frac{\text{Work done}}{\text{Time taken}} = \frac{4 \times 10^5\text{ J}}{10\text{ s}} = 40,000\text{ W} = 40\text{ kW} \)
(c) Given that the efficiency is 40% (or 0.4):
\( \text{Efficiency} = \frac{\text{Useful Power Output}}{\text{Power Input}} \)
\( 0.4 = \frac{40\text{ kW}}{\text{Power Input}} \)

\( \implies \text{Power Input} = \frac{40\text{ kW}}{0.4} = 100\text{ kW} \)
In simple words: The total work done in lifting the water is \( 4 \times 10^5 \) J. Doing this in 10 seconds means the pump puts out 40 kW of useful power. Since the pump is only 40% efficient, it requires a higher total input power of 100 kW.

Exam Tip: Be careful to distinguish between the actual useful power output of a machine and its total input power rating, which is always larger due to energy losses.

Question 13. An ox can apply a maximum force of 1000 N. It is taking part in a cart race and is able to pull the cart at a constant speed of 30 m s-1 while making its best effort. Calculate the power developed by the ox.
Answer:
The values provided are force \( F = 1000\text{ N} \) and constant speed \( v = 30\text{ m s}^{-1} \).
Power is the product of force and speed:
\( P = F \times v \)
\( P = 1000 \times 30 = 30,000\text{ W} = 30\text{ kW} \)
In simple words: When an object moves at a steady speed, we can find the power by multiplying the force applied by its speed. Here, multiplying 1000 N by 30 m/s gives 30,000 W of power.

Exam Tip: The formula \( P = F \times v \) is highly useful when a force is applied to keep an object moving at a constant speed.

Question 14. If the power of a motor is 40 kw, at what speed can it raise a load of 20,000 N?
Answer:
We are given:
\( \text{Power } (P) = 40\text{ kW} = 40,000\text{ W} \)
\( \text{Force } (F) = 20,000\text{ N} \)
Using the relationship between power, force, and velocity:
\( P = F \times v \)
\( v = \frac{P}{F} = \frac{40,000\text{ W}}{20,000\text{ N}} = 2\text{ m s}^{-1} \)
In simple words: To find the speed, divide the power of the motor (40,000 W) by the load it is lifting (20,000 N). This gives a lifting speed of 2 meters per second.

Exam Tip: Always convert power from kilowatts (kW) to watts (W) before performing division to keep the SI units consistent.

Exercise 2(B)

Question 1. What are the two forms of mechanical energy?
Answer:
Mechanical energy consists of two main types:
(i) Kinetic energy
(ii) Potential energy
In simple words: Mechanical energy is the total energy of an object due to its motion or its position, which means it is made of kinetic and potential energy.

Exam Tip: Mechanical energy is the sum of kinetic energy and potential energy. Always list both when asked about its forms.

Question 2. Name the forms of energy which a wound-up watch spring possesses.
Answer:
A wound-up spring in a watch contains elastic potential energy.
In simple words: When you wind up a watch spring, you deform it, storing energy inside that is released slowly to keep the watch running.

Exam Tip: Be specific when naming potential energy - mention "elastic potential energy" rather than just "potential energy" to secure full marks.

Question 3. Name the type of energy (kinetic energy K or potential energy U) possessed in the following cases:
(a) A moving cricket ball
(b) A compressed spring
(c) A moving bus
(d) The bob of a simple pendulum at its extreme position.
(e) The bob of a simple pendulum at its mean position.
(f) A piece of stone places on the roof.
Answer:
The types of energy for each scenario are:
(a) Kinetic energy (K)
(b) Potential energy (U)
(c) Kinetic energy (K)
(d) Potential energy (U)
(e) Kinetic energy (K)
(f) Potential energy (U)
In simple words: Objects that are moving have kinetic energy, while objects that are stored up at a height or deformed (like a spring) have potential energy.

Exam Tip: Remember that extreme position means maximum height/deformation (potential energy) and mean position means maximum speed (kinetic energy).

Question 4. When an arrow is shot from a bow, it has kinetic energy in it. Explain briefly from where does it get its kinetic energy?
Answer:
Pulling back a bowstring requires physical work, which becomes stored as elastic potential energy within the bent frame of the bow. Once the string is let go, this stored energy instantly converts into the kinetic energy of the arrow, sending it forward.
In simple words: The energy you use to pull the bowstring is stored in the bow as elastic potential energy. Releasing the string transfers this stored energy to the arrow as kinetic energy, making it fly.

Exam Tip: Clearly mention the step-by-step conversion: work done -> elastic potential energy of the bow -> kinetic energy of the arrow.

Question 5. Define the term potential energy of a body. State its different forms and give one example of each.
Answer:
The potential energy of an object is the energy it holds because of its particular state of deformation or its position.
The primary types of potential energy include:
(i) Gravitational potential energy: This is the energy an object has because of its location relative to the Earth's center.
Example: A boulder resting on top of a hill.
(ii) Elastic potential energy: This is the energy locked within an object when its physical shape is altered.
Example: A stretched rubber band.
In simple words: Potential energy is stored energy. It can come from being up high (gravitational) or from being stretched or squeezed (elastic).

Exam Tip: For a complete definition, make sure to mention both "position" and "changed configuration" as the two sources of potential energy.

Question 6. A ball is placed on a compressed spring. What form of energy does the spring possess? On releasing the spring, the ball flies away. Give a reason.
Answer:
Because the spring is squeezed, it possesses elastic potential energy. Letting go of the spring transforms this stored energy into kinetic energy, which performs work on the ball to propel it into the air with kinetic energy of its own.

In simple words: Squeezing a spring stores elastic energy in it. When let go, that energy changes into movement energy (kinetic energy), pushing the ball and making it fly.

Exam Tip: Explain the sequence of energy transfer clearly: spring's elastic potential energy -> work done on the ball -> ball's kinetic energy.

Question 7. What is meant by the gravitational potential energy? Derive expression for it.
Answer:
Gravitational potential energy refers to the energy stored in an object because of its position relative to the Earth's center.
To find its expression, we calculate the work needed to lift a mass \( m \) to a height \( h \) above the ground against gravity:
The minimum upward force required to raise the mass without acceleration must equal its weight:
\( F = mg \)
The work performed in raising it through a height \( h \) is:
\( W = \text{Force} \times \text{displacement} = mg \times h = mgh \)
Since this work is stored as potential energy, we get:
\( U = mgh \)
In simple words: The energy an object gets when lifted up is equal to the work done against gravity. Since work is force times height, and force is weight (\( mg \)), the potential energy is \( mgh \).

Exam Tip: When deriving \( U = mgh \), always state that the lifting force is equal to the weight of the body and that no acceleration occurs.

Question 8. Write an expression for the potential energy of a body of mass m places at a height h above the earth’s surface.
Answer:
An object of mass \( m \) positioned at a height \( h \) above the ground has gravitational potential energy equal to the work required to lift it:
\( U = mgh \)
In simple words: The gravitational potential energy is calculated by multiplying the mass, gravity, and height together, written as \( mgh \).

Exam Tip: Simply state \( U = mgh \), and briefly define each variable: \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is height.

Question 9. Name the form of energy which a body may possess even when it is not in motion. Give an example to support your answer.
Answer:
An object at rest can still hold potential energy. For instance, a rock sitting on the edge of a high cliff has gravitational potential energy because of its elevated position.
In simple words: Even if something isn't moving, it can have stored energy (potential energy) just by being high up or under tension.

Exam Tip: Provide a clear, everyday example (like a raised weight or a stretched spring) to back up your answer when requested.

Question 10. What do you understand by the kinetic energy of a body?
Answer:
Kinetic energy is the energy that a physical body possesses due to its motion. Any object that is actively moving contains this type of energy.
In simple words: Kinetic energy is the energy of movement. If something is moving, it has kinetic energy.

Exam Tip: Ensure you use the phrase "by virtue of its state of motion" as it is a key scientific definition that examiners look for.

Question 11. A body of mass m is moving with a velocity v. Write the expression for its kinetic energy.
Answer:
The kinetic energy \( K \) of an object of mass \( m \) travelling at a velocity \( v \) is given by:
\( K = \frac{1}{2}mv^2 \)
In simple words: The energy of a moving object is half of its mass times its speed squared.

Exam Tip: State the formula clearly and define each symbol to ensure you receive full credit.

Question 12. State the work energy theorem.
Answer:
The work-energy theorem states that the net work done by external forces on an object is equal to the change in its kinetic energy.
In simple words: When you do work on an object to speed it up, the amount of work you do is exactly equal to how much its movement energy increases.

Exam Tip: Always relate the "work done by a force" directly to the "increase or change in kinetic energy" when stating this theorem.

Question 13. A body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity changes from u to v. How much work is being done by the force.
Answer:
Let a body of mass \( m \) move with an initial velocity \( u \). Under a constant force \( F \), its velocity becomes \( v \) over a displacement \( S \), producing an acceleration \( a \).
The work done by this force is:
\( W = F \times S \) ---(i)
Using the equation of motion:
\( v^2 = u^2 + 2aS \)
\( \implies S = \frac{v^2 - u^2}{2a} \)
Substituting \( F = ma \) and \( S \) into equation (i):
\( W = ma \times \left(\frac{v^2 - u^2}{2a}\right) \)
\( \implies W = \frac{1}{2}m(v^2 - u^2) \)
\( \implies W = K_f - K_i \)
Where \( K_i = \frac{1}{2}mu^2 \) is the starting kinetic energy, and \( K_f = \frac{1}{2}mv^2 \) is the final kinetic energy.
Hence, the work performed equals the gain in kinetic energy:
\( W = \frac{1}{2}m(v^2 - u^2) \)
In simple words: The work done on a moving object is equal to its final kinetic energy minus its initial kinetic energy, which simplifies to \( \frac{1}{2}m(v^2 - u^2) \).

Exam Tip: Use the third equation of motion (\( v^2 = u^2 + 2aS \)) to substitute for displacement when deriving this relation.

Question 14. A light mass and a heavy mass have equal momentum. Which will have more kinetic energy? (Hint : Kinetic energy K = P\(^2\)/2m where P is the momentum)
Answer:
The relation between kinetic energy \( K \) and momentum \( p \) is given by:
\( K = \frac{p^2}{2m} \)
Since both objects possess the identical momentum \( p \), their kinetic energy is inversely proportional to their mass:
\( K \propto \frac{1}{m} \)
As a result, the lighter object will have greater kinetic energy because a smaller mass yields a higher kinetic energy value for the same momentum.
In simple words: Since kinetic energy equals momentum squared divided by two times the mass, if momentum is the same, the lighter object always ends up with more kinetic energy.

Exam Tip: Clearly state the inverse relationship \( K \propto \frac{1}{m} \) when momentum is constant to secure full marks.

Question 15. Name the three forms of kinetic energy and give on example of each.
Answer:
The three distinct types of kinetic energy are:
(i) Translational kinetic energy - for example, a body in free fall.
(ii) Rotational kinetic energy - for example, a spinning top.
(iii) Vibrational kinetic energy - for example, atoms in a solid oscillating about their fixed positions.
In simple words: Things can move in a straight line (translational), spin around (rotational), or shake back and forth (vibrational). Each movement has its own type of kinetic energy.

Exam Tip: Provide a distinct, clear example for each of the three types of kinetic energy to ensure full credit.

Question 16. Differentiate between the potential energy (U) and the kinetic energy (K)
Answer:
The differences between potential energy and kinetic energy are highlighted below:

In simple words: Potential energy is stored energy from position or shape, whereas kinetic energy is the energy of active motion.
Exam Tip: Tabulating differences with distinct heads (definition, forms, examples) is highly recommended for scoring maximum marks.

Question 17. Complete the following sentences:
(a) The kinetic energy of a body is the energy by virtue of its………….
(b) The potential energy of a body is the energy by virtue of its ……………….
Answer:
The completed statements are:
(a) motion
(b) position (or configuration)
In simple words: Movement gives an object kinetic energy, whereas its location or setup gives it potential energy.

Exam Tip: Fill in the blanks with precise technical terms like "motion" and "position" to secure full marks.

Question 18. Is it possible that no transfer of energy may take place even when a force is applied to a body?
Answer:
Yes, if the applied force is perpendicular to the direction of movement, zero energy is transferred.
In simple words: If you push sideways on an object moving forward, your force does no work on it, meaning no energy is transferred.

Exam Tip: Remember that when the angle \( \theta \) between force and displacement is \( 90^\circ \), work done \( W = F S \cos(90^\circ) = 0 \).

Question 19. Name the form of mechanical energy, which is put to use.
Answer:
Kinetic energy is the type of mechanical energy that is actively utilized.
In simple words: Kinetic energy is the energy of motion that actually does work, like a moving hammer hitting a nail.

Exam Tip: While potential energy is stored, it must convert to kinetic energy to perform active physical work.

Question 20. In what way does the temperature of water at the bottom of a waterfall differ from the temperature at the top? Explain the reason.
Answer:
The temperature of the water is slightly higher at the bottom than at the top. This happens because the stored potential energy of the water at the top transforms into kinetic energy as it falls, and upon hitting the ground, this kinetic energy is partially converted into thermal energy.
In simple words: Water at the bottom of a waterfall is warmer because its energy of motion turns into heat when it crashes into the ground.

Exam Tip: Ensure you write down the complete chain of energy conversions: potential energy -> kinetic energy -> heat energy.

Question 21. Name six different forms of energy?
Answer:
Six distinct types of energy are:
1. Solar energy
2. Thermal (heat) energy
3. Light energy
4. Chemical energy
5. Hydroelectric energy
6. Nuclear energy
In simple words: Energy exists in many types, such as sunlight, heat, light, chemical bonds, moving water, and atomic nuclei.

Exam Tip: List the forms clearly using a numbered list to make it easy for the examiner to award full marks.

Question 22. Energy can exist in several forms and may change from one form to another. For each of the following, state the energy changes that occur in:
(a) the unwinding of a watch spring
(b) a loaded truck when started and set in motion.
(c) a car going uphill
(d) photosynthesis in green leaves
(e) Charging of a battery,
(f) respiration,
(g) burning of a match stick
(h) explosion of crackers.
Answer:
The energy transformations for each process are:
(a) Elastic potential energy of the spring is converted into kinetic energy.
(b) Chemical energy from fuel converts into mechanical (kinetic) energy.
(c) Kinetic energy is transformed into gravitational potential energy.
(d) Light energy is converted into chemical energy.
(e) Electrical energy is stored as chemical energy.
(f) Chemical energy is converted into heat (thermal) energy.
(g) Chemical energy is transformed into heat and light energy.
(h) Chemical energy converts into heat, light, and sound energy.
In simple words: Energy cannot be created or destroyed; it only changes from one form (like chemical or potential) to another (like movement, light, or heat).

Exam Tip: Always identify the starting form of energy and state clearly what new forms of energy it turns into.

Question 23. State the energy changes in the following cases while in use:
(a) loudspeaker
(b) a steam engine
(c) microphone
(d) washing machine
(e) an electric bulb
(f) burning coal
(g) a solar cell
(h) bio-gas burner
(i) an electric cell in a circuit
(j) a petrol engine of a running car
(k) an electric toaster
(l) a photovoltaic cell
(m) an electromagnet.
Answer:
The energy conversions during operation are as follows:
(a) Electrical energy converts to acoustic (sound) energy.
(b) Heat energy is transformed into mechanical energy.
(c) Sound energy changes into electrical energy.
(d) Electrical energy converts to mechanical energy.
(e) Electrical energy transforms into light and heat energy.
(f) Chemical energy converts into thermal (heat) energy.
(g) Light energy is converted into electrical energy.
(h) Chemical energy converts into heat energy.
(i) Chemical energy is transformed into electrical energy.
(j) Chemical energy converts to mechanical energy.
(k) Electrical energy converts into thermal (heat) energy.
(l) Light energy transforms into electrical energy.
(m) Electrical energy is converted to magnetic energy.
In simple words: Devices are built to change one form of energy that we feed them into a different, useful form of energy.

Exam Tip: Memorize standard electrical appliances and their conversions, as they are frequently asked in short-answer questions.

Multiple Choice Type

Question 1. A body at a height possesses:
(a) kinetic energy
(b) potential energy
(c) solar energy
(d) heat energy
Answer: (b) potential energy
In simple words: When an object is lifted up high, it stores energy due to its position, which is known as potential energy.

Exam Tip: Any body placed at an elevation relative to the ground will have gravitational potential energy.

Question 2. In an electric cell which in use, the change in energy is from:
(a) electrical to mechanical
(b) electrical to chemical
(c) chemical to mechanical
(d) chemical to electrical
Answer: (d) chemical to electrical
In simple words: Inside a battery, chemical reactions happen to generate an electric current when you connect it to a circuit.

Exam Tip: Remember that during charging, electrical energy turns into chemical energy, but when in use (discharging), it converts from chemical to electrical.

Numericals

Question 1. Two bodies of equal masses are placed at heights h and 2h. Find the ration of their gravitational potential energies.
Answer:
Let the mass of both objects be \( m \).
The height of the first object is \( h_1 = h \) and the second object is \( h_2 = 2h \).
The potential energy of the first object is:
\( U_1 = mgh_1 = mgh \)
The potential energy of the second object is:
\( U_2 = mgh_2 = mg(2h) = 2mgh \)
Finding the ratio of their potential energies:
\( \frac{U_1}{U_2} = \frac{mgh}{2mgh} = \frac{1}{2} = 1:2 \)
In simple words: Since potential energy is directly proportional to height, doubling the height of an equal mass doubles its potential energy, resulting in a 1:2 ratio.

Exam Tip: For equal masses, the ratio of potential energies is directly equal to the ratio of their heights.

Question 2. Find the gravitational potential energy of 1 kg mass kept at a height of 5 m above the ground if g = 10 m s-2.
Answer:
Given:
\( \text{Mass } (m) = 1\text{ kg} \)
\( \text{Height } (h) = 5\text{ m} \)
\( \text{Acceleration due to gravity } (g) = 10\text{ m s}^{-2} \)
The formula for gravitational potential energy is:
\( U = mgh \)
\( U = 1 \times 10 \times 5 = 50\text{ J} \)
In simple words: Multiply the mass of 1 kg by gravity (10) and height (5 m) to find the stored energy, which is 50 joules.

Exam Tip: Ensure you include the correct unit (Joules, J) in your final numerical answer.

Question 3. A box of weight 150 kgf has gravitational potential energy stored in it equal to 14700 J. Find the height of the box above the ground. (Take g = 9.8 N kg-1)
Answer:
Given:
\( \text{Gravitational potential energy } (U) = 14,700\text{ J} \)
\( \text{Force of gravity (weight) } (F = mg) = 150\text{ kgf} = 150 \times 9.8\text{ N kg}^{-1} = 1470\text{ N} \)
Since potential energy is:
\( U = mgh \)
\( 14,700 = 1470 \times h \)
\( h = \frac{14,700}{1470} = 10\text{ m} \)
In simple words: The weight of the box in Newtons is 1470 N. Since potential energy is weight times height, we divide 14,700 J by 1470 N to get a height of 10 meters.

Exam Tip: Pay attention to the force units. Converting \( \text{kgf} \) to Newtons by multiplying with the given value of \( g \) is a crucial first step.

Question 4. A body of mass 5 kg falls from a height of 10 m to 4 m. Calculate: (i) the loss in potential energy of the body, (ii) the total energy possessed by the body at any instant? (Take g = 10 m s-2)
Answer:
(i) The mass of the object is \( 5\text{ kg} \).
The initial potential energy at \( 10\text{ m} \) height is calculated as:
\( U_1 = mgh = 5 \times 10 \times 10 = 500\text{ J} \)
The potential energy at \( 4\text{ m} \) height is given by:
\( U_2 = mgh = 5 \times 10 \times 4 = 200\text{ J} \)
Therefore, the reduction in potential energy is:
\( (500 - 200)\text{ J} = 300\text{ J} \)
(ii) During a free fall, the total mechanical energy of an object stays conserved at every point. This total energy is the sum of its potential and kinetic energies.
At the highest point of \( 10\text{ m} \), the kinetic energy is \( 0\text{ J} \).
Thus, the overall energy is:
\( \text{P.E.} + \text{K.E.} = 500\text{ J} + 0\text{ J} = 500\text{ J} \)
In simple words: The energy lost when the object drops is the difference between its start and end potential energies, which is 300 J. Because energy is conserved, the total energy is always 500 J.

Exam Tip: Remember that total energy remains constant during free fall. Just calculate the potential energy at the highest point where kinetic energy is zero to easily find the total energy.

Question 5. Calculate the height through which a body of mass 0.5 kg is lifted if the energy spent in doing so is 1.0 J. Take g = 10 m s-2
Answer:
We are given the mass of the object, \( m = 0.5\text{ kg} \), and the work done (potential energy gained), \( U = 1.0\text{ J} \).
The equation for gravitational potential energy is:
\( U = mgh \)
\( 1 = 0.5 \times 10 \times h \)
\( 1 = 5h \)

\( \implies h = \frac{1}{5} = 0.2\text{ m} \)
Hence, the object is raised to a height of \( 0.2\text{ m} \).
In simple words: To find the height, divide the energy spent by the weight of the object (mass multiplied by gravity).
Exam Tip: Ensure all quantities are in SI units before performing calculations. Double-check your basic division to avoid simple calculation errors like writing 1/5 as 0.5 instead of 0.2.

Question 6. A boy weighing 25 kgf climbs up from the first floor at height 3 m above the ground to the third floor at height 9m above the ground. What will be the increase in his gravitational potential energy? (Take g = 10 N kg -1)
Answer:
The downward force of gravity acting on the boy is:
\( F = mg = 25\text{ kg} \times 10\text{ N/kg} = 250\text{ N} \)
The rise in gravitational potential energy is given by:
\( \Delta U = mg(h_2 - h_1) \)
\( \Delta U = 250 \times (9 - 3) \)

\( \implies \Delta U = 250 \times 6 = 1500\text{ J} \)
Therefore, his potential energy increases by \( 1500\text{ J} \).
In simple words: The increase in energy is calculated by multiplying the boy's weight by the difference in height between the two floors.
Exam Tip: Note that \( 25\text{ kgf} \) is equivalent to a gravitational force of \( 250\text{ N} \) when \( g = 10\text{ N kg}^{-1} \). Always multiply mass by g to get the force in Newtons.

Question 7. A vessel containing 50 kg of water is placed at a height 15 m above the ground. Assuming the gravitational potential energy at ground to be zero, what will be the gravitational potential energy of water in the vessel? (g = 10 m s-2)
Answer:
The mass of the water is \( m = 50\text{ kg} \) and its height is \( h = 15\text{ m} \).
Using the formula for gravitational potential energy:
\( U = mgh \)

\( \implies U = 50 \times 10 \times 15 \)

\( \implies U = 7500\text{ J} \)
The potential energy of the water is \( 7500\text{ J} \).
In simple words: Multiply the water's mass by gravity and height to find the energy it stores.
Exam Tip: State the reference point clearly in your steps. Here, because potential energy at ground level is assumed to be zero, the total energy is exactly \( mgh \).

Question 8. A man of mass 50 kg climbs up a ladder of height 10 m. Calculate: (i) the work done by the man, (ii) the increase in his potential energy. (g = 9.8 m s-2)
Answer:
(i) The work performed by the man as he climbs is equivalent to \( mgh \):
\( W = 50\text{ kg} \times 9.8\text{ m s}^{-2} \times 10\text{ m} = 4900\text{ J} \)
(ii) The gain in his gravitational potential energy is evaluated relative to the ground:
\( \Delta U = mg(h_2 - h_1) \)

\( \implies \Delta U = 50 \times 9.8 \times (10 - 0) = 4900\text{ J} \)
Thus, both the work done and the energy increase are equal to \( 4900\text{ J} \).
In simple words: The work the man does to climb the ladder is completely stored as his potential energy, so both values are 4900 J.
Exam Tip: Pay attention to the value of \( g \). In this problem, it is \( 9.8\text{ m s}^{-2} \), not \( 10\text{ m s}^{-2} \). Using the wrong value will lead to a loss of marks.

Question 9. A block A, whose weight is 200 N, is pulled up a slope of length 5 m by means of a constant force F (= 150 N) as illustrated in Fig 2.13
(a) what is the work done by the force F in moving the block A, 5 m along the slope?
(b) By how much has the potential energy of the block A increased?
(c) Account for the difference in work done by the force and the increase in potential energy of the block.

Answer:
Given that the pulling force \( F = 150\text{ N} \) and the weight of the block \( W = mg = 200\text{ N} \).
(a) The work done by this force to move the block \( 5\text{ m} \) up the inclined plane:
\( \text{Work done} = \text{Force} \times \text{displacement in the force's direction} \)

\( \implies W_{\text{force}} = 150 \times 5 = 750\text{ J} \).
(b) The increase in gravitational potential energy of the block depends on the vertical height raised, \( h = 3\text{ m} \):
\( \Delta U = mgh = \text{Weight} \times \text{height} \)

\( \implies \Delta U = 200 \times 3 = 600\text{ J} \).
(c) The discrepancy between the work done (\( 750\text{ J} \)) and the gained potential energy (\( 600\text{ J} \)) is \( 150\text{ J} \). This remaining energy is consumed in overcoming frictional resistance between the slope and the block, which is released as thermal energy (heat).
In simple words: (a) Moving the block up the slope takes 750 J of work. (b) The actual height gained adds 600 J of potential energy. (c) The extra 150 J is lost as heat due to friction.
Exam Tip: Note that potential energy depends solely on the vertical height \( 3\text{ m} \), whereas work done by the pulling force depends on the actual path length \( 5\text{ m} \). This is a very common exam conceptual question.

Question 10. Find the kinetic energy of a body of mass 1 kg moving with a uniform velocity of 10 m s-1.
Answer:
We have the mass of the body, \( m = 1\text{ kg} \), and its constant speed, \( v = 10\text{ m/s} \).
The formula for kinetic energy is:
\( K = \frac{1}{2}mv^2 \)

\( \implies K = \frac{1}{2} \times 1 \times (10)^2 \)

\( \implies K = \frac{1}{2} \times 100 = 50\text{ J} \).
Thus, the body's kinetic energy is \( 50\text{ J} \).
In simple words: Put the mass and velocity into the kinetic energy formula to get 50 Joules.
Exam Tip: Always square the velocity first before multiplying by mass and halving it to prevent arithmetic slip-ups.

Question 11. If the speed of a car is halved, how does its kinetic energy change?
Answer:
Since kinetic energy is directly proportional to the square of the object's velocity (\( K \propto v^2 \)), reducing the speed of the car to half of its initial value (with constant mass) will reduce its kinetic energy to \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \) of the original value.
In simple words: Since kinetic energy depends on the square of speed, cutting speed in half makes the kinetic energy four times smaller.
Exam Tip: State the proportionality relation \( K \propto v^2 \) clearly in your answer to secure full explanation marks.

Question 12. Two bodies of equal masses are moving with uniform velocities v and 2v. Find the ratio of their kinetic energies.
Answer:
Let the first body's velocity be \( v_1 = v \) and the second body's velocity be \( v_2 = 2v \).
Both bodies have equal masses (\( m_1 = m_2 \)), which means their kinetic energy scales with the square of their velocities (\( K \propto v^2 \)).
The ratio of their kinetic energies is given by:
\( \frac{K_1}{K_2} = \frac{v_1^2}{v_2^2} \)

\( \implies \frac{K_1}{K_2} = \frac{v^2}{(2v)^2} = \frac{v^2}{4v^2} = \frac{1}{4} \)
Hence, the ratio of their kinetic energies is \( 1:4 \).
In simple words: Since the second body travels twice as fast, squaring this speed factor makes its kinetic energy four times larger, giving a 1:4 ratio.
Exam Tip: Write the ratio clearly with colons (\( 1:4 \)) as requested by the question, and don't forget to cancel out the common variables.

Question 13. A car is running at a speed of 15 km h-1 while another similar car is moving at a speed of 30 km h-1. Find the ration of their kinetic energies.
Answer:
Let the velocity of the first car be \( v_1 = 15\text{ km/h} \) and the second similar car be \( v_2 = 30\text{ km/h} \).
Because the cars are identical, their masses are identical, and kinetic energy is proportional to the square of the speed.
The ratio is:
\( \frac{K_1}{K_2} = \frac{v_1^2}{v_2^2} \)

\( \implies \frac{K_1}{K_2} = \frac{15^2}{30^2} = \frac{15 \times 15}{30 \times 30} = \frac{1}{2 \times 2} = \frac{1}{4} \)
So, the ratio of their kinetic energies is \( 1:4 \).
In simple words: Since both cars have the same weight and the second car goes twice as fast, its kinetic energy is four times higher.
Exam Tip: There is no need to convert the speeds from km/h to m/s when finding a ratio, as the conversion factors cancel out anyway. This saves valuable time in exams.

Question 14. A bullet of mass 0.5 kg slows down from a speed of 5 m s-1 to that of 3 m s-1. Calculate the change in kinetic energy of the ball.
Answer:
The mass of the object is \( m = 0.5\text{ kg} \).
Its starting speed is \( u = 5\text{ m/s} \), giving an initial kinetic energy:
\( K_i = \frac{1}{2} m u^2 = \frac{1}{2} \times 0.5 \times 5^2 = \frac{1}{2} \times 12.5 = 6.25\text{ J} \).
The final speed is \( v = 3\text{ m/s} \), giving a final kinetic energy:
\( K_f = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.5 \times 3^2 = \frac{1}{2} \times 4.5 = 2.25\text{ J} \).
The change in kinetic energy is calculated as:
\( \Delta K = K_f - K_i = 2.25\text{ J} - 6.25\text{ J} = -4\text{ J} \).
The negative sign signifies a reduction in kinetic energy of \( 4\text{ J} \).
In simple words: The object's kinetic energy drops from 6.25 J to 2.25 J, which is a decrease of 4 J.
Exam Tip: Remember that "change" in physics is always calculated as "final minus initial". A negative sign represents a loss or decrease in energy.

Question 15. A cannon ball of mass 500 g is fired with a speed of 15 m s-1. Find: (i) its kinetic energy and (ii) its momentum.
Answer:
Converting mass to kilograms: \( m = 500\text{ g} = 0.5\text{ kg} \).
The speed of the cannonball is \( v = 15\text{ m/s} \).
(i) The kinetic energy is:
\( K = \frac{1}{2}mv^2 \)

\( \implies K = \frac{1}{2} \times 0.5 \times (15)^2 = 56.25\text{ J} \).
(ii) The linear momentum is:
\( p = m \times v \)

\( \implies p = 0.5 \times 15 = 7.5\text{ kg m s}^{-1} \).
In simple words: (i) Its kinetic energy is 56.25 J. (ii) Its momentum is 7.5 kg m/s.
Exam Tip: Make sure to convert mass from grams to kilograms first, otherwise your final answers for kinetic energy and momentum will be off by a factor of 1000.

Question 16. A bullet of mass 50 g is moving with a velocity of 500 m s-1. It penetrated 10 cm into a still target and comes to rest. Calculate: (a) the kinetic energy possessed by the bullet, (b) the average retarding force offered by the target.
Answer:
First, convert all values to standard SI units:
Mass \( m = 50\text{ g} = 0.05\text{ kg} \)
Velocity \( v = 500\text{ m/s} \)
Stopping distance \( d = 10\text{ cm} = 0.1\text{ m} \)
(a) The kinetic energy of the bullet is:
\( K = \frac{1}{2} m v^2 \)

\( \implies K = \frac{1}{2} \times 0.05 \times (500)^2 = 6250\text{ J} \).
(b) The work performed by the bullet against the target is equal to its initial kinetic energy:
\( W = \text{Force} \times \text{distance} \)

\( \implies 6250 = F \times 0.1 \)

\( \implies F = \frac{6250}{0.1} = 62500\text{ N} \).
Therefore, the average retarding force is \( 62500\text{ N} \).
In simple words: (a) The moving bullet has 6250 J of energy. (b) To stop it in just 10 cm, the target must exert a huge stopping force of 62,500 Newtons.
Exam Tip: Remember that the work-energy theorem states that work done to stop an object equals its kinetic energy. Use this relation directly to solve resistive force problems.

Question 17. A body of mass 10 kg is moving with a velocity 20 m s-1. If the mass of the body is doubled and its velocity is halved, find the ratio of the initial kinetic energy to the final kinetic energy.
Answer:
Let the original mass be \( m_1 = 10\text{ kg} \) with speed \( v_1 = 20\text{ m/s} \).
The final mass is doubled, \( m_2 = 20\text{ kg} \), and the speed is halved, \( v_2 = 10\text{ m/s} \).
Calculating the initial kinetic energy:
\( K_1 = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} \times 10 \times 20^2 = 2000\text{ J} \).
Calculating the final kinetic energy:
\( K_2 = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} \times 20 \times 10^2 = 1000\text{ J} \).
Finding the ratio of initial to final kinetic energy:
\( \frac{K_1}{K_2} = \frac{2000}{1000} = \frac{2}{1} = 2:1 \).
In simple words: Doubling the mass makes kinetic energy twice as big, but halving the speed makes it four times smaller. Combined, the energy is halved, making the ratio 2:1.
Exam Tip: You can also solve this conceptually without values: \( K_1 = \frac{1}{2}mv^2 \), and \( K_2 = \frac{1}{2}(2m)(\frac{v}{2})^2 = \frac{1}{4}mv^2 \), so the ratio is indeed 2:1. Show both methods to guarantee full marks.

Question 18. A truck weighing 1000 kgf changes its speed from 36 km h-1 to 72 km h-1 in 2minutes. Calculate: (i) the work done by the engine and (ii) its power/ (g = 10 m s-2)
Answer:
Convert speeds to meters per second:
Initial speed, \( u = 36\text{ km/h} = 36 \times \frac{5}{18} = 10\text{ m/s} \)
Final speed, \( v = 72\text{ km/h} = 72 \times \frac{5}{18} = 20\text{ m/s} \)
Mass of the truck is \( m = 1000\text{ kg} \), and time interval \( t = 2\text{ minutes} = 120\text{ seconds} \).
(i) The work performed by the engine equals the gain in kinetic energy:
\( W = \frac{1}{2} m (v^2 - u^2) \)

\( \implies W = \frac{1}{2} \times 1000 \times (20^2 - 10^2) \)

\( \implies W = 500 \times (400 - 100) = 500 \times 300 = 150000\text{ J} = 1.5 \times 10^5\text{ J} \).
(ii) The power of the engine is computed as:
\( P = \frac{\text{Work done}}{\text{time taken}} \)

\( \implies P = \frac{1.5 \times 10^5\text{ J}}{120\text{ s}} = 1250\text{ W} = 1.25 \times 10^3\text{ W} \).
In simple words: (i) The engine does 150,000 Joules of work to speed up the truck. (ii) Doing this over 2 minutes requires a power output of 1250 Watts.
Exam Tip: Remember that \( 1\text{ km/h} \) is \( \frac{5}{18}\text{ m/s} \). Always convert minutes to seconds when calculating power in Watts.

Question 19. A body of mass 60 kg has the momentum 3000 kg m s-1. Calculate: (i) the kinetic energy and (ii) the speed of the body.
Answer:
We are given mass \( m = 60\text{ kg} \) and momentum \( p = 3000\text{ kg m s}^{-1} \).
(i) The kinetic energy can be expressed in terms of momentum:
\( K = \frac{p^2}{2m} \)

\( \implies K = \frac{3000^2}{2 \times 60} = \frac{9,000,000}{120} = 75000\text{ J} = 7.5 \times 10^4\text{ J} \).
(ii) Using the momentum formula:
\( p = m \times v \)

\( \implies 3000 = 60 \times v \)

\( \implies v = \frac{3000}{60} = 50\text{ m/s} \).
The velocity of the body is \( 50\text{ m/s} \).
In simple words: (i) The body possesses 75,000 J of kinetic energy. (ii) Dividing momentum by mass shows its speed is 50 m/s.
Exam Tip: The formula \( K = \frac{p^2}{2m} \) is extremely helpful and saves you from finding velocity first to compute kinetic energy. Memorize it.

Question 20. How much work is needed to be done on a ball of mass 50 g to give it s momentum of 500 g cm s-1?
Answer:
Convert all values into standard SI units:
Mass of the ball, \( m = 50\text{ g} = 0.05\text{ kg} \).
Momentum, \( p = 500\text{ g cm s}^{-1} = 500 \times 10^{-5}\text{ kg m s}^{-1} = 0.005\text{ kg m s}^{-1} \).
The work required is equivalent to the final kinetic energy:
\( W = K = \frac{p^2}{2m} \)

\( \implies W = \frac{(0.005)^2}{2 \times 0.05} = \frac{0.000025}{0.1} = 0.00025\text{ J} = 2.5 \times 10^{-4}\text{ J} \).
In simple words: The work needed is 0.00025 Joules to speed up the ball to the required momentum.
Exam Tip: Take extra care when converting CGS units (g cm/s) to SI units (kg m/s). Keep in mind that \( 1\text{ g cm s}^{-1} = 10^{-5}\text{ kg m s}^{-1} \).

Question 21. How much energy is gained by a box of mass 20 kg when a man (a) carrying the box waits for 5 minutes for a bus? (b) runs carrying the box with a speed of 3 m s-1 to catch the bus? (c) Raises the box by 0.5 m in order to place it inside the bus? (g = 10 m s-2)
Answer:
The mass of the box is \( m = 20\text{ kg} \).
(a) Since there is no physical displacement of the box while the man stands waiting, no work is performed, meaning the energy gained is \( 0\text{ J} \).
(b) When running, the box gains kinetic energy due to its motion:
\( K = \frac{1}{2} m v^2 \)

\( \implies K = \frac{1}{2} \times 20 \times 3^2 = 10 \times 9 = 90\text{ J} \).
(c) Raising the box increases its gravitational potential energy:
\( \Delta U = mgh \)

\( \implies \Delta U = 20 \times 10 \times 0.5 = 100\text{ J} \).
In simple words: (a) No energy is gained because the box does not move anywhere. (b) Running gives the box 90 J of motion energy. (c) Lifting it up stores 100 J of potential energy in it.
Exam Tip: In part (a), explain clearly that despite the effort felt by the man, the physics definition of work requires a non-zero displacement in the direction of the force to perform work.

Question 22. A spring is kept compressed by a small trolley of mass 0.5 kg lying on a smooth horizontal surface as shown in the adjacent fig. 2.14 when the trolley is released, it is found to move at a speed v = 2 m s-1. What potential energy did the spring possess when compressed?

Answer:
The mass of the trolley is \( m = 0.5\text{ kg} \), and its speed is \( v = 2\text{ m/s} \).
By the conservation of mechanical energy, the entire elastic potential energy stored in the compressed spring transforms into the kinetic energy of the trolley when it is let go.
\( \text{Potential energy of compressed spring} = \text{Kinetic energy of moving trolley} \)

\( \implies U = \frac{1}{2}mv^2 \)

\( \implies U = \frac{1}{2} \times 0.5 \times 2^2 \)

\( \implies U = \frac{1}{2} \times 0.5 \times 4 = 1\text{ J} \).
Therefore, the potential energy of the compressed spring was \( 1.0\text{ J} \).
In simple words: All the energy stored in the squeezed spring is turned into the movement energy of the trolley, which equals 1 Joule.
Exam Tip: State the Law of Conservation of Energy explicitly in your reasoning to justify equating the potential energy of the spring to the kinetic energy of the trolley.

Exercise 2(C)

Question 1. State two characteristic which a source of energy must have.
Answer:
An ideal source of energy should possess the following key features:
(i) It must deliver a sufficient quantity of useful energy at a constant, reliable rate for an extended duration.
(ii) It must be highly secure, user-friendly, easily accessible, and cost-effective.
In simple words: A good energy source needs to last a long time, provide steady power, be safe to use, and not cost too much.
Exam Tip: Learn these basic definitions and criteria by heart, as they are frequently asked direct 2-mark questions.

Question 2. Name the two groups in which various sources of energy are classified. State on what basis are they classified.
Answer:
Energy sources are broadly divided into two main categories:
1. Renewable (also termed non-conventional) energy sources.
2. Non-renewable (also termed conventional) energy sources.
This categorization is established primarily on the natural availability and replenishment rate of these energy resources.
In simple words: Energy sources are divided into renewable (which do not run out) and non-renewable (which do run out), based on how easily they replenish.
Exam Tip: Make sure to mention both names for each category (e.g., "renewable or non-conventional") to show complete knowledge of the syllabus terms.

Question 3. What is meant by the renewable and non-renewable sources of energy? Distinguish between them giving two examples of weach.
Answer:
The definitions and primary differences between renewable and non-renewable energy sources are detailed in the comparative table below:

In simple words: Renewable energy comes from sources that never run out, like the sun. Non-renewable energy comes from sources that will eventually disappear, like coal.
Exam Tip: When distinguishing between two terms, always format your response in a clear tabular layout to make it easier for the examiner to award full marks.

Question 4. Select the renewable and non-renewable sources of energy from the following: (a) Coal (b) Wood (c) Water (d) Diesel (e) Wind (f) Oil
Answer:
Based on the availability and nature of regeneration, the given sources are classified as follows:

In simple words: Wood, water, and wind can be replenished over time, while coal, diesel, and oil will eventually run out completely.
Exam Tip: Ensure you understand why wood is categorized as renewable (trees can be replanted), even though its large-scale use is discouraged.

Question 5. Why is the use of wood as a fuel not advisable although wood is a renewable source of energy?
Answer:
Even though wood is renewable, relying on it as a fuel source is not recommended for several reasons:
1. Trees require a long duration (typically over 15 years) to mature completely, making the replenishment of wood a very slow process.
2. Deforestation on a massive scale disrupts the ecological balance, damages habitats, and intensifies global warming.
Therefore, we should avoid burning wood as fuel.
In simple words: Trees take many years to grow back, so cutting them down quickly hurts the environment and causes forest loss.
Exam Tip: Highlight two distinct dimensions in your answer: the slow growth cycle of trees and the negative environmental impact of deforestation.

Question 6. Name five renewable and three non-renewable sources of energy.
Answer:
The requested examples of energy resources are listed below:
Five renewable sources:
1. Solar energy (the Sun)
2. Wind energy
3. Hydro energy (flowing water)
4. Biomass energy
5. Tidal energy
Three non-renewable sources:
1. Coal
2. Petroleum (crude oil)
3. Natural gas
In simple words: Renewable options include sun, wind, water, biomass, and tides. Non-renewable options include coal, petroleum, and natural gas.
Exam Tip: Standard examples are easy to write, but using scientific terms like "hydro energy" or "solar energy" instead of just "flowing water" or "sun" looks more professional.

Question 7. What is (i) tidal, (ii) ocean and (iii) geo thermal energy? Explain in brief.
Answer:
(i) Tidal energy: This is the energy derived from the rise and fall of ocean tides caused by gravitational forces. By building dams across narrow coastal inlets, we can capture this energy to generate electricity. However, it is limited because most locations do not experience tides large enough to support commercial power plants.
(ii) Ocean energy: The oceans store energy in two distinct ways:
(a) Ocean thermal energy: This exploits the temperature difference between the warm surface water and the cold deep ocean water. An Ocean Thermal Energy Conversion (OTEC) power plant can harness this thermal gradient to produce electrical power.
(b) Oceanic wave energy: This is the kinetic energy of strong sea waves. While experimental systems exist, it has not yet been widely adopted for commercial use.
(iii) Geothermal energy: This refers to the thermal energy trapped inside the Earth's crust. Deep hot rocks heat up underground water reservoirs, turning them into high-pressure steam. By drilling deep wells, we can extract this steam to spin turbines and drive electric generators.
In simple words: Tidal energy comes from ocean tides, ocean energy comes from temperature differences or waves, and geothermal energy comes from heat deep inside the Earth.
Exam Tip: Note the spelling and abbreviation of OTEC (Ocean Thermal Energy Conversion). Mentioning this term and OTEC power plants shows thorough preparation.

Question 8. What is the main source of energy for earth?
Answer:
The primary and ultimate source of energy for our planet is the Sun.
In simple words: The Sun is the main source of energy for almost everything on Earth.
Exam Tip: This is a direct factual question. Keep your answer brief and direct.

Question 9. What is solar energy? How is the solar energy used to generate electricity in a solar power plant?
Answer:
Solar energy is the radiant light and heat emitted by the Sun.
In a thermal solar power plant, this heat energy is utilized to produce electricity through these steps:
1. Large curved concave mirrors concentrate sunlight onto black-painted pipes located at their focal points.
2. The intense heat boils the water inside these pipes, turning it into high-pressure steam.
3. This steam is directed to spin a steam turbine, which is linked to a generator that produces electricity.
In simple words: Solar energy is heat and light from the sun. Solar plants focus this heat to boil water, create steam, and spin turbines to make electricity.
Exam Tip: Explain the mechanism step-by-step: focus of sunlight, steam generation, turbine rotation, and power production to get full marks.

Question 10. What is a solar cell? State two uses of solar cells. State whether a solar cell produces a.c. or d.c. Give one disadvantage of using a solar cell.
Answer:
A solar cell (or photovoltaic cell) is a semiconductor device that directly converts light energy into electricity via the photovoltaic effect. It is typically manufactured from materials like silicon and gallium.
Two major uses or benefits of solar cells:
1. They function with virtually no maintenance and have an exceptionally long lifespan with zero fuel or operational costs.
2. They are highly practical for supplying power to remote, off-grid locations where standard electricity cables cannot be run.
Type of current produced:
A solar cell generates direct current (d.c.).
One disadvantage:
The setup and purchase cost of solar cell panels is quite expensive initially.
In simple words: A solar cell turns sunlight directly into DC electricity. They are great for remote areas and need no fuel, but they cost a lot to buy at first.
Exam Tip: When asked about the current type, write "d.c." or "direct current" clearly. Ensure you mention "photovoltaic effect" as it is a key technical term.

Question 11. State two advantages and two disadvantages of producing electricity from solar energy.
Answer:
Two advantages of generating electricity from solar energy:
1. It is a completely clean and environmentally-friendly source of energy that produces no harmful emissions.
2. After installation, the running and maintenance costs are practically zero.
Two disadvantages of solar energy:
1. The upfront purchase and installation costs of solar panels are very high.
2. They produce direct current (d.c.), which must be converted to alternating current (a.c.) to power standard home appliances.
In simple words: Solar panels provide clean energy with almost no running costs, but they are expensive to buy and make DC electricity which needs converting for home use.
Exam Tip: When listing advantages and disadvantages, write them as clear, separate bullet points. Mentioning the need to convert DC to AC is a great technical detail.

Question 12. What is wind energy? How is wind energy used to produce electricity? How much electric power is generated in India using the wing energy?
Answer: The kinetic energy from large, flowing masses of air is known as wind energy. We can harness this energy to generate electricity by using wind turbines to spin the shaft of an electrical generator. Currently, India produces upwards of 1025 MW of electrical power through this wind-based method.
In simple words: Wind energy is the power of blowing air. Windmills capture this power and spin generators to make electricity.

Exam Tip: When defining wind energy, remember to state that it is the kinetic energy of moving air. Always state the correct unit (MW) for power output in India.

Question 13. State two advantages and two disadvantages of using wind energy for generating electricity.
Answer:
Advantages of utilizing wind power:
1. It is completely clean and does not produce environmental pollution.
2. It serves as an inexhaustible, renewable resource of energy.
Disadvantages of utilizing wind power:
1. Setting up wind farms requires a significant financial investment.
2. A vast expanse of land is required to set up these farms.
In simple words: Wind power is great because it never runs out and is clean. However, wind farms cost a lot of money to build and take up a lot of space.

Exam Tip: Make sure to write distinct points for advantages and disadvantages, using clear headings or numbered lists to make your answer easy to read.

Question 14. What is hydro energy? Explain the principle of generating electricity from hydro energy. How much hydroelectric power is generated in India?
Answer: Hydroelectric or water energy refers to the kinetic energy found in running water. The underlying principle of a hydroelectric station is to collect river water from high altitudes inside a tall dam or reservoir. This stored water is then released to fall onto a turbine positioned near the dam's base. As the turbine rotates, its central shaft turns the armature of an attached electrical generator or dynamo, producing power. Right now, hydro energy contributes about 23% of the overall electricity produced in India.
In simple words: Hydro energy uses the movement of falling water to spin a turbine. This spinning turbine runs a generator to create electricity.

Exam Tip: Be sure to explain the flow of energy: stored water has potential energy, which turns into kinetic energy as it falls to rotate the turbine.

Question 15. State two advantage and two disadvantages of producing hydroelectricity.
Answer:
Advantages of generating hydroelectricity:
1. It is a clean process that releases no harmful pollutants into the environment.
2. It utilizes a renewable energy source that does not deplete.
Disadvantages of generating hydroelectricity:
1. Building large dams across rivers floods nearby land, which destroys local plant and animal habitats.
2. The natural ecosystem downstream can be severely disrupted.
In simple words: Hydroelectricity is clean and renewable, but building dams can ruin nature and hurt local wildlife.

Exam Tip: For environmental disadvantages, focus on habitat destruction and the downstream ecological impact.

Question 16. What is nuclear energy? Explain the principle of producing electricity using the nuclear energy.
Answer: Nuclear energy is the immense energy released when a heavy atomic nucleus is hit by slow-moving neutrons, causing it to split into two lighter nuclei of comparable size. This splitting process is known as nuclear fission. During fission, the combined mass of the resulting products is slightly less than the initial mass of the reactants, and this missing mass is transformed into energy.
Principle: In a nuclear reactor, the heat generated by the controlled fission of uranium-235 is taken up by a circulating coolant. This coolant moves through a heat exchanger to warm up water and turn it into steam. The pressurized steam then drives a turbine, which spins the generator's armature within a magnetic field to produce electric current.
In simple words: Nuclear energy comes from splitting heavy atoms apart. The heat from this process boils water into steam, which spins a generator to make electricity.

Exam Tip: Clearly mention "nuclear fission" and the conversion of "lost mass into energy" (\( E = mc^2 \)) when defining nuclear energy.

Question 17. What percentage of total electrical power generated in india is obtained from nuclear power plant? Name two places in india where electricity is generated from nuclear power plants.
Answer: Currently, nuclear power plants contribute roughly 3% of the overall electricity generated across India. Two notable locations in India that house nuclear power stations are Tarapur in Maharashtra and Narora in Uttar Pradesh.
In simple words: Nuclear energy makes up only about 3% of India's electricity. Two places that have these plants are Tarapur and Narora.

Exam Tip: Remember the names of the states (Maharashtra and Uttar Pradesh) along with the power plant locations (Tarapur and Narora) to secure full marks.

Question 18. State two advantages and two disadvantages of using nuclear energy for producing electricity.
Answer:
Advantages of using nuclear power:
1. A tiny quantity of nuclear fuel is capable of producing an incredibly large amount of energy.
2. After the fuel is initially loaded into the reactor, it can keep producing power continuously for many years.
Disadvantages of using nuclear power:
1. It cannot be considered fully clean because the reaction generates dangerous radioactive radiation.
2. The radioactive waste produced presents a serious long-term hazard to the environment.
In simple words: Nuclear power is useful because a little fuel lasts for years and makes tons of energy. But, it creates dangerous radiation and radioactive waste.

Exam Tip: Focus on the contrast between the high energy yield/long fuel lifespan and the critical hazards of radioactive waste and radiation leaks.

Question 19. State the energy transformation on the following:
(i) Electricity is obtained from solar energy.
(ii) electricity is obtained from wind energy
(iii) Electricity is obtained from hydro energy.
(iv) Electricity is obtained from nuclear energy.
Answer:
(i) Light energy converts to electrical energy.
(ii) Mechanical energy converts to electrical energy.
(iii) Mechanical energy converts to electrical energy.
(iv) Nuclear energy (or thermal energy) converts to electrical energy.
In simple words: Different power sources convert stored or moving energy into electric current.

Exam Tip: Remember that wind and running water both possess kinetic energy, which is a form of mechanical energy.

Question 20. State four ways for the judicious use of energy.
Answer:
Four strategies to use energy wisely:
(a) Non-renewable fossil fuels like coal, oil, and natural gas should only be used as a last resort when alternative green options are unavailable.
(b) We must actively minimize any unnecessary consumption or waste of energy.
(c) Shared or community-based energy solutions should be encouraged to optimize resource consumption.
(d) Deforestation must be strictly prohibited, and active afforestation programs should be promoted to plant more trees.
In simple words: We can save energy by using fossil fuels only when needed, stopping waste, sharing resources, and planting more trees.

Exam Tip: When discussing the wise use of energy, cover both individual habits (stopping waste) and broader ecological solutions (planting trees, using public transport).

Multiple Choice Type

Question 1. The ultimate source of energy is:
(a) wood
(b) wind
(c) water
(d) sun
Answer: (d) sun
In simple words: Nearly all forms of energy on Earth, like wind, water, and plants, originally get their power from the Sun.

Exam Tip: The Sun drives the water cycle, wind currents, and plant growth (photosynthesis), making it the primary source of almost all energy.

Question 2. Renewable source of energy is:
(a) Coal
(b) fossil fuels
(c) natural gas
(d) sun
Answer: (d) sun
In simple words: The sun is a renewable resource because its light and heat will never run out.

Exam Tip: Coal, natural gas, and other fossil fuels are non-renewable because they take millions of years to form and will eventually be completely used up.

Exercise 2(D)

Question 1. State the law of conservation of energy.
Answer: The law of conservation of energy states that energy cannot be created or destroyed. It can only be transformed from one form to another.
In simple words: You cannot make new energy or make energy disappear. You can only change it into a different kind of energy.

Exam Tip: State both parts of the law: energy cannot be created or destroyed, and it can only be converted from one form to another.

Question 2. What do you understand by the conservation of mechanical energy? State the condition under which the mechanical energy is conserved.
Answer: The conservation of mechanical energy means that when potential and kinetic energy transition into each other, their combined sum stays constant (\( K + U = \text{constant} \)), provided no resistive forces act on the system.
This principle is only strictly true when there is no friction or air resistance. Consequently, total mechanical energy is perfectly conserved in a vacuum where such resistive forces do not exist.
In simple words: The total amount of potential and kinetic energy combined stays the same as long as there is no friction to slow things down.

Exam Tip: Always specify the essential condition: the absence of frictional or resistive forces (like air resistance).

Question 3. Name two examples in which the mechanical energy of a system remains constant.
Answer: The oscillation of an ideal simple pendulum and the descent of a freely falling body under gravity (assuming air resistance is ignored).
In simple words: A swinging pendulum and an object falling without any air resistance are two systems where total energy stays constant.

Exam Tip: If air resistance is mentioned, state that it must be neglected for mechanical energy to remain completely constant.

Question 4. A body is thrown vertically upwards. Its velocity keeps on decreasing. What happens to its kinetic energy as its velocity becomes zero?
Answer: As the body rises and slows down, its kinetic energy is entirely converted into gravitational potential energy by the time it comes to a temporary stop at its peak.
In simple words: When you throw something up, it slows down because its energy of motion turns into stored energy. When it stops at the top, all that energy is stored potential energy.

Exam Tip: At the highest point, kinetic energy becomes zero, and the potential energy reaches its maximum value.

Question 5. A body falls freely under gravity from rest. Name the kind of energy it will possess.
(a) at the point from where it falls.
(b) while falling.
(c) on reaching the ground.
Answer:
(a) Only potential energy is present.
(b) It possesses both potential energy and kinetic energy.
(c) Only kinetic energy remains just before hitting the ground.
In simple words: At the top, a resting object has only stored energy. As it falls, some stored energy turns into motion energy. When it hits the ground, all of it has turned into motion energy.

Exam Tip: Remember that during free fall, potential energy is continuously converted into kinetic energy, but the sum of both remains constant at any point.

Question 6. Show that the sum of kinetic energy and potential energy (i.e., total mechanical energy) is always conserves in the case of a freely falling body under gravity (with air resistance neglected) from a height h by finding it when (i) the body is at the top, (ii) the body has fallen a distance x, (iii) the body has reached the ground.
Answer: Let a body of mass \( m \) drop freely under gravity from an initial height \( h \) above the earth (starting from position A). We will determine the total mechanical energy (sum of kinetic energy \( K \) and potential energy \( U \)) at three separate locations: position A (at the maximum height \( h \)), position B (after descending a distance \( x \)), and position C (upon touching the ground).

(i) At position A (at height \( h \) above the earth):
The object starts from rest, so its initial velocity is zero (\( u = 0 \)).
Thus, the kinetic energy is:
\( K = 0 \)
The gravitational potential energy at this altitude is:
\( U = mgh \)
So, the total mechanical energy at point A is:
\( E_A = K + U = 0 + mgh = mgh \) - (1)

(ii) At position B (after dropping a distance \( x \)):
Let the speed of the body at point B be \( v_1 \). Here, the initial speed \( u = 0 \), distance traveled \( S = x \), and acceleration \( a = g \).
Using the equation of motion \( v^2 = u^2 + 2aS \):
\( v_1^2 = 0^2 + 2gx = 2gx \)
Now, we calculate the kinetic energy at this point:
\( K = \frac{1}{2} m v_1^2 = \frac{1}{2} m (2gx) = mgx \)
The height of the body above the ground is now \( h - x \), so its potential energy is:
\( U = mg(h - x) \)
Thus, the total mechanical energy at point B is:
\( E_B = K + U = mgx + mg(h - x) = mgx + mgh - mgx = mgh \) - (2)

(iii) At position C (on the ground):
Let the velocity of the body when it reaches the ground be \( v \). Here, \( u = 0 \), vertical distance \( S = h \), and acceleration \( a = g \).
Using the equation of motion \( v^2 = u^2 + 2aS \):
\( v^2 = 0^2 + 2gh = 2gh \)
Therefore, the kinetic energy just before hitting the ground is:
\( K = \frac{1}{2} m v^2 = \frac{1}{2} m (2gh) = mgh \)
Since the height above the ground is zero (\( h = 0 \)), the potential energy is:
\( U = 0 \)
Thus, the total mechanical energy at point C is:
\( E_C = K + U = mgh + 0 = mgh \) - (3)

By comparing equations (1), (2), and (3), we find that the total mechanical energy remains constant at \( mgh \) throughout the entire motion, showing that energy is conserved.
In simple words: At the top, the falling object has only potential energy. In the middle, it has a mix of potential and kinetic energy. At the bottom, it has only kinetic energy. In all three cases, the total energy adds up to the exact same value.

Exam Tip: In derivations, clearly define variables like \( m \), \( h \), and \( x \). Write down the three distinct positions (top, intermediate, and ground) with equations to score full marks.

Question 7. A pendulum is oscillating on either side of its rest position. Explain the energy changes that takes place in the oscillating pendulum. How does the mechanical energy remains constant in it? Draw the necessary diagram.
Answer: As the pendulum bob travels from its central rest position A towards the maximum point B, its speed decreases, converting its kinetic energy into potential energy. At the peak point B, the bob pauses briefly, meaning its potential energy reaches a maximum while kinetic energy drops to zero. As it swings back down from B to A, this stored potential energy is converted back into kinetic energy, and this cycle continues to repeat.
Consequently, the bob possesses purely potential energy at its highest limits (points B and C), and strictly kinetic energy at its lowest resting spot (point A). At any point in between, the bob has a combination of both kinetic and potential energy. The total mechanical energy (their sum) stays constant throughout the entire oscillation, neglecting air friction.

In simple words: At the highest points on the left and right, the swing stops for a split second, so all its energy is potential. At the very bottom, it moves fastest, meaning all its energy is kinetic. In between, it has some of both.

Exam Tip: Always label the potential and kinetic energies at both extreme positions and the central position on your diagram to ensure full marks.

Question 8. A pendulum with bob of mass m is oscillating on either side from its resting position A between the extremes B and C at a vertical height h and A. what is the kinetic energy K and potential energy U when the pendulum is at position (i) A, (ii) B and (iii) C?
Answer:
(i) At position A: The potential energy is zero and the kinetic energy reaches its maximum value. Thus, \( K = mgh \) and \( U = 0 \).
(ii) At position B: The kinetic energy drops to zero while the potential energy reaches its maximum value. Thus, \( K = 0 \) and \( U = mgh \).
(iii) At position C: Similar to position B, the kinetic energy is zero and potential energy is at its peak. Thus, \( K = 0 \) and \( U = mgh \).
In simple words: At the bottom point A, the pendulum has maximum kinetic energy but zero potential energy. At both high sides B and C, it has zero kinetic energy and maximum potential energy.

Exam Tip: Clearly state that total energy is always \( K + U = mgh \) at every single position of the pendulum.

Question 9. What do you mean by degradation of energy? Explain it by taking two examples of your daily life.
Answer: Energy degradation refers to the gradual conversion of useful energy into non-usable forms, such as heat lost due to friction.
Examples:
1. While cooking food over a flame, a substantial amount of the heat produced by the fuel escapes into the surrounding air, becoming unavailable for any useful work.
2. When running household electrical appliances, a significant portion of the electrical energy is wasted as heat rather than being converted into the desired form of work.
In simple words: Energy degradation means useful energy gets wasted as heat that we cannot use anymore, like when a machine gets warm while running.

Exam Tip: Remember that energy is never destroyed during degradation; it simply turns into a disordered, non-usable form (mostly low-grade heat).

Numericals

Question 1. A ball of mass 0.20 kg is thrown vertically upwards with an initial velocity of 20 m s-1. Calculate the maximum potential energy it gains as it goes up.
Answer: The maximum potential energy gained by the ball is equal to its initial kinetic energy at the throw:
Potential energy at maximum height = Initial kinetic energy
\( U = \frac{1}{2} m v^2 \)
\( U = \frac{1}{2} \times 0.20 \text{ kg} \times 20 \text{ m/s} \times 20 \text{ m/s} \)
\( U = 40 \text{ J} \)
Thus, the ball gains a maximum potential energy of 40 J.
In simple words: Since energy is conserved, all the motion energy of the ball at the start gets converted into stored potential energy at the highest point.

Exam Tip: Use the conservation of energy principle to equate the kinetic energy at the bottom directly to the potential energy at the top, saving calculation time.

Question 2. A stone of mass 500g is thrown vertically upwards with a velocity of 15 m s-1. Calculate: (a) the potential energy at the greatest height, (b) the kinetic energy on reaching the ground, (c) the total energy at its half-way point.
Answer: First, convert the mass to kilograms:
\( m = 500 \text{ g} = 0.5 \text{ kg} \)

(a) Potential energy at the greatest height:
By the conservation of energy, the potential energy at the peak is equal to the initial kinetic energy:
\( \text{Potential Energy (P.E.)} = \frac{1}{2} m v^2 \)
\( \text{P.E.} = \frac{1}{2} \times 0.5 \text{ kg} \times (15 \text{ m/s})^2 \)
\( \text{P.E.} = 0.25 \times 225 = 56.25 \text{ J} \)

(b) Kinetic energy on reaching the ground:
When the stone hits the ground, all its potential energy is converted back into kinetic energy:
\( \text{Kinetic Energy (K.E.)} = \text{P.E. at peak} = 56.25 \text{ J} \)

(c) Total energy at the half-way point:
According to the conservation of mechanical energy, the total energy (\( K + U \)) remains constant at every point in the flight:
\( \text{Total Energy} = 56.25 \text{ J} \)
In simple words: The total energy never changes. It starts as 56.25 J of movement energy, turns into 56.25 J of stored energy at the top, and stays at 56.25 J total even halfway up.

Exam Tip: Remember that total energy is always conserved and remains the same at all heights, including the half-way point.

Question 3. A metal ball of mass 2 kg is allowed to fall freely from rest from a height of 5m above the ground. (Take g = 10 m s-2) (a) Calculate the potential energy possessed by the ball when initially at rest. (b) what is the kinetic energy of the ball just before it hits the ground? (c) what happens to the mechanical energy after the ball hits the ground and comes to rest?
Answer: Given: Mass \( m = 2 \text{ kg} \), height \( h = 5 \text{ m} \), and \( g = 10 \text{ m/s}^2 \).

(a) Potential energy at rest:
\( \text{P.E.} = mgh \)
\( \text{P.E.} = 2 \text{ kg} \times 10 \text{ m/s}^2 \times 5 \text{ m} = 100 \text{ J} \)

(b) Kinetic energy just before hitting the ground:
By the conservation of energy, the kinetic energy right before impact equals the initial potential energy:
\( \text{K.E.} = 100 \text{ J} \)

(c) Transformation of mechanical energy after impact:
When the ball hits the ground and comes to rest, its mechanical energy is transformed into non-mechanical forms, primarily heat and sound energy.
In simple words: The ball starts with 100 J of stored energy, which turns into 100 J of speed energy as it falls. When it crashes, that energy turns into a thud sound and a little bit of warmth.

Exam Tip: For part (c), always specify both heat and sound energy as the main products of the collision.

Question 4. The diagram given below shows a ski jump. A skier weighing 60 kgf stands at A at the top of ski jump. He moves from A to B and takes off for his jump at B. (a) Calculate the change in the gravitational potential energy of the skier between A and B. (b) If 75% of the energy in part (a) becomes kinetic energy at B. Calculate the speed at which the skier arrives at B. (Take g=10 m s-2)
Answer: Given: Mass of the skier \( m = 60 \text{ kg} \), starting height \( h_1 = 75 \text{ m} \), final height \( h_2 = 15 \text{ m} \), and \( g = 10 \text{ m/s}^2 \).

(a) Change in gravitational potential energy:
The decrease in gravitational potential energy as the skier goes from A to B is:
\( \Delta U = mg(h_1 - h_2) \)
\( \Delta U = 60 \text{ kg} \times 10 \text{ m/s}^2 \times (75 \text{ m} - 15 \text{ m}) \)
\( \Delta U = 60 \times 10 \times 60 = 36000 \text{ J} = 3.6 \times 10^4 \text{ J} \)
Therefore, the change in potential energy is \( 3.6 \times 10^4 \text{ J} \).

(b) Speed of the skier at point B:
Given that 75% of the lost potential energy is converted into kinetic energy at B:
\( \text{Kinetic Energy (K.E.) at B} = \frac{75}{100} \times 3.6 \times 10^4 \text{ J} \)
\( \text{K.E.} = 0.75 \times 36000 = 27000 \text{ J} \)

Since \( \text{K.E.} = \frac{1}{2} m v^2 \):
\( 27000 = \frac{1}{2} \times 60 \times v^2 \)
\( 27000 = 30 \times v^2 \)
\( v^2 = \frac{27000}{30} = 900 \)
\( v = \sqrt{900} = 30 \text{ m/s} \)
So, the skier reaches point B at a speed of 30 m/s.
In simple words: First, find the drop in height (60 m) to get the lost potential energy (36,000 J). Then, take 75% of that to find the kinetic energy (27,000 J) and use it to solve for velocity (30 m/s).

Exam Tip: Carefully calculate the difference in heights (\( h_1 - h_2 \)) first before finding the change in potential energy.

Question 5. A hydroelectric power station takes its water from a lake whose water level if 50 m above the turbine. Assuming an overall efficiency of 40%, calculate the mass of water which must flow through the turbine each second to produce power output of 1 MV.
Answer: Let \( m \) be the mass of water flowing through the turbine every second. The height of the water level \( h = 50 \text{ m} \), and the acceleration due to gravity \( g = 10 \text{ m/s}^2 \).
The total potential energy available per second is:
\( \text{P.E. per second} = mgh = m \times 10 \times 50 = 500m \)

Since the efficiency of the power station is 40%, the useful power output is:
\( \text{Useful Power Output} = 40\% \text{ of } \text{P.E. per second} \)
\( \text{Power Output} = 0.40 \times 500m = 200m \)

The desired power output is 1 MW (which is \( 1 \times 10^6 \text{ W} \)):
\( 200m = 1 \times 10^6 \)
\( m = \frac{10^6}{200} = 5000 \text{ kg} \)
Thus, 5000 kg of water must flow through the turbine every second.
In simple words: Only 40% of the falling water's energy is turned into electricity. To get 1,000,000 Watts of power, we need 5,000 kilograms of water to rush through every second.

Exam Tip: Remember that power is work done per second, so mass per second \( m \) can be solved directly by setting the useful energy output per second equal to the electrical power output.

Question 6. The bob of a simple pendulum is imparted a velocity 5 m s-1 when it is at its mean position. To what maximum vertical height will it rise on reaching to its extreme position if 60% of its energy is lost in overcome friction of air?
Answer: Let \( m \) be the mass of the bob and \( H \) be the maximum height it reaches. The velocity at the mean position is \( v = 5 \text{ m/s} \).
Since 60% of the energy is lost to air resistance, only 40% is converted into gravitational potential energy at the extreme height:
\( \text{Potential Energy at highest point} = 40\% \text{ of } \text{Initial Kinetic Energy} \)
\( mgH = 0.40 \times \left(\frac{1}{2} m v^2\right) \)

Dividing both sides by the mass \( m \):
\( gH = 0.40 \times 0.5 \times v^2 \)
\( gH = 0.2 \times v^2 \)

Substitute \( g = 10 \text{ m/s}^2 \) and \( v = 5 \text{ m/s} \):
\( 10 \times H = 0.2 \times 5^2 \)
\( 10 \times H = 0.2 \times 25 \)
\( 10 \times H = 5 \)
\( H = \frac{5}{10} = 0.5 \text{ m} \)
Thus, the pendulum bob will rise to a maximum height of 0.5 m.
In simple words: If 60% of the energy is lost, only 40% of the starting speed energy turns into height. Solving the formula gives a height of 0.5 meters.

Exam Tip: Don't worry about the mass of the bob since it cancels out on both sides of the energy conservation equation.