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Work, Energy and Power PYQs – Physics, SSC & Railway

196 Work, Energy and Power (Physics) questions asked in recent SSC and Railway exams, newest first, with exam, date & shift, answers and solutions – page 2 of 2.

📚 196 questionsSSC 12 · Railway 184
12
RRB NTPC UG CBT-1 2025-26 – 9 May 2026 Shift 1 · Work and its sign

Q101. In which of the following cases is the work done negative?

  1. (a) Force and displacement are between 0° and 90°
  2. (b) Force and displacement are between 90° and 180°
  3. (c) Force and displacement are in the same direction
  4. (d) Force and displacement are mutually perpendicular
Answer & solution

Answer: (b) Force and displacement are between 90° and 180°

Work Done Formula:W = F × d × cos θ
where θ = angle between force and displacement
Analysis:
• 0° < θ < 90° → cos θ is positive → W is positive
• θ = 90° → cos 90° = 0 → W = 0
• 90° < θ < 180° → cos θ is negative →W is negative
• θ = 0° (same direction) → cos 0° = 1 → W is positive
When angle is obtuse (90° to 180°), work done is negative.
Answer: Option 2 → Force and displacement are between 90° and 180°
RRB NTPC UG CBT-1 2025-26 – 9 May 2026 Shift 3 · Conservation and transformation of energy

Q102. The conversion of chemical energy into electrical energy and then into light and thermal energy occurs in which of the following cases?

  1. (a) When a fan rotates
  2. (b) When a candle burns
  3. (c) When water is boiled
  4. (d) When a flashlight is turned on
Answer & solution

Answer: (d) When a flashlight is turned on

Energy Conversion in Flashlight:
Chemical → Electrical → Light + Heat
Battery stores chemical energy
Chemical reactions produce electrical energy
Bulb/LED converts electrical energy to light and heat
Other options:
Fan: Electrical → Mechanical (rotation)
Candle: Chemical → Light + Heat (direct)
Boiling water: Electrical/Heat → Thermal
Answer: Option 4 → When a flashlight is turned on
SSC GD 2026 – 5 May 2026 Shift 2 · Conservation and transformation of energy

Q103. During free fall (ignoring air resistance), just before an object hits the ground, how are its potential energy and kinetic energy related?

  1. (a) Potential energy is maximum, kinetic energy is minimum
  2. (b) Potential energy is minimum, kinetic energy is maximum
  3. (c) Both kinetic and potential energies are zero
  4. (d) Some energy is lost due to conversion
Answer & solution

Answer: (b) Potential energy is minimum, kinetic energy is maximum

Correct Answer: Potential energy is minimum, kinetic energy is maximum — Option (b)
During free fall, total mechanical energy (potential energy + kinetic energy) is conserved (ignoring air resistance). As an object falls, its height decreases — so its gravitational potential energy (PE = mgh ) continuously decreases. Simultaneously, its velocity increases, so its kinetic energy (KE = ½mv²) continuously increases. Just before hitting the ground, height is nearly zero → PE is minimum (nearly zero), while velocity is maximum → KE is maximum. All of the initial PE has been converted to KE.
Option (a) Potential energy is maximum, kinetic energy is minimum — Incorrect: This describes the initial state (before falling), not just before impact.
Option (c) Both KE and PE are zero — Incorrect: Just before impact, velocity is maximum so KE is maximum; PE is near zero but KE is not zero.
Option (d) Some energy is lost due to conversion — Incorrect: In ideal free fall (no air resistance), energy is conserved; no energy is lost.
RRB NTPC GL CBT-1 2025-26 – 25 March 2026 Shift 1 · Conservation and transformation of energy

Q104. Which two forms of energy together make up the total mechanical energy of an object?

  1. (a) Light and heat energy
  2. (b) Potential and kinetic energy
  3. (c) Heat and sound energy
  4. (d) Electrical and chemical energy
Answer & solution

Answer: (b) Potential and kinetic energy

Mechanical Energy Components
Mechanical energy is the energy possessed by an object due to its motion or position.
Two forms:
1.Potential Energy (PE)- Energy due to position or configuration
2.Kinetic Energy (KE)- Energy due to motion
Total Mechanical Energy = PE + KE
Example: A swinging pendulum has both PE (at height) and KE (during motion).
Answer: Option 2 → Potential and kinetic energy
RRB NTPC GL CBT-1 2025-26 – 22 March 2026 Shift 3 · Potential energy

Q105. Which of the following is a clear example of potential energy?

  1. (a) A falling rock
  2. (b) A burning candle
  3. (c) A stretched rubber band
  4. (d) A moving car
Answer & solution

Answer: (c) A stretched rubber band

Topic: Potential Energy
Potential energy isstored energydue to position or state.
Analysis of options:
1. Falling rock → Kinetic energy (motion)
2. Burning candle → Chemical/Heat energy
3. Stretched rubber band →Elastic potential energy(stored)
4. Moving car → Kinetic energy (motion)
Stretched rubber band has stored energy ready to be released.
Answer: Option 3 → A stretched rubber band
RRB NTPC GL CBT-1 2025-26 – 19 March 2026 Shift 1 · Conservation and transformation of energy

Q106. At time t = 0, a body of mass 1 Kg starts to fall freely from rest from a height of 100 m. Compared to the kinetic energy at 3 seconds, the kinetic energy at 4 seconds ________.

  1. (a) Is larger
  2. (b) Is smaller
  3. (c) Remains same
  4. (d) Is not determinable
Answer & solution

Answer: (a) Is larger

Given: m = 1 kg, h = 100 m, free fall from rest
Kinetic Energy formula: KE = ½mv²
For free fall: v = gt (velocity increases with time)
At t = 3s: v₃ = 10 × 3 = 30 m/s
KE₃ = ½ × 1 × (30)² = 450 J
At t = 4s: v₄ = 10 × 4 = 40 m/s
KE₄ = ½ × 1 × (40)² = 800 J
Since 800 J > 450 J, KE at 4s is larger
Answer: Option 1 → Is larger
RRB NTPC GL CBT-1 2025-26 – 19 March 2026 Shift 3 · Potential energy

Q107. When a spring is compressed from its equilibrium position, its potential energy will ___________________.

  1. (a) first decrease then increase
  2. (b) decrease
  3. (c) increase
  4. (d) remain unchanged
Answer & solution

Answer: (c) increase

Topic:Elastic Potential Energy in Springs
Formula: PE =½kx²
where k = spring constant, x = displacement from equilibrium
At equilibrium position: x = 0, so PE = 0
When compressed: x increases (displacement from equilibrium)
Since PE ∝ x², potential energyincreases
Note: PE increases whether spring is compressed OR stretched
Answer: Option 3 → increase
RRB NTPC GL CBT-1 2025-26 – 17 March 2026 Shift 2 · Potential energy

Q108. Raising an object from the ground results in the storage of which type of energy?

  1. (a) Heat energy
  2. (b) Electrical energy
  3. (c) Kinetic energy
  4. (d) Potential energy
Answer & solution

Answer: (d) Potential energy

When an object is raised above the ground, work is done against gravity.
This work gets stored asGravitational Potential Energy (PE).
Formula: PE = mgh
where m = mass, g = gravity, h = height
The higher the object, the more potential energy it stores.
Answer: Option 4 → Potential energy
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 2 · Numericals on work, energy and power

Q109. A body of mass 10 kg is thrown vertically upwards with a velocity of 10 m/s. How much potential energy will be possessed by the body when it reaches the maximum height? (Take g = 10 m/s²)

  1. (a) 250 J
  2. (b) 2.5 J
  3. (c) 500 J
  4. (d) 25 J
Answer & solution

Answer: (c) 500 J

Given
mass (m) = 10 kg, initial velocity (u) = (10 m)/(s, acceleration due to gravity (g)) = (10 m)/(s²)
At maximum height, velocity v = (0 m)/(s)
Use the formula for maximum height (h)
v² = u² - 2gh
0 = 10² - 2 × 10 × h
= > 0 = 100 - 20h
= > 20h = 100
= > h = (100)/(20) = 5 m
Calculate potential energy (PE) at max height
PE = mgh
PE = 10 × 10 × 5 = 500 Joules
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 3 · Numericals on work, energy and power

Q110. A car has a mass of 120 kg. How much work must be done to raise its speed from 72 km/h to 108 km/h?

  1. (a) 2.0 × 10⁴ J
  2. (b) 1.0 × 10⁴ J
  3. (c) 3.0 × 10⁴ J
  4. (d) 4.0 × 10⁴ J
Answer & solution

Answer: (c) 3.0 × 10⁴ J

To find the work done to increase the speed, calculate the change in kinetic energy.
Initial speed u = 72 km/h = 72 × (1000/3600) = 20 m/s.
Final speed v = 108 km/h = 108 × (1000/3600) = 30 m/s.
Mass m = 120 kg.
Kinetic energy change = 0.5 × m × (v² − u²).
= 0.5 × 120 × (30² − 20²).
= 60 × (900 − 400) = 60 × 500 = 30000 J = 3.0 × 10⁴ J.
None of the given options exactly match 3.0 × 10⁴ J except option 3, but calculation shows 3.0 × 10⁴ J is correct.
Hence, correct option is Option 3.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 1 · Numericals on work, energy and power

Q111. A constant force of 21 N moves a body through 3 m in its own direction. Work done is _____.

  1. (a) 7 J
  2. (b) 24 J
  3. (c) 18 J
  4. (d) 63 J
Answer & solution

Answer: (d) 63 J

Work done (W) = Force (F) × Distance (d)
Given, F = 21 N, d = 3 m
So, W = 21 × 3 = 63 Joules
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 2 · Numericals on work, energy and power

Q112. A girl weighing 420 N climbs a rope to a height of 6 m and takes 20 seconds to do it. What is her power output?

  1. (a) 105 W
  2. (b) 90 W
  3. (c) 126 W
  4. (d) 155 W
Answer & solution

Answer: (c) 126 W

Power is the rate of doing work or energy transfer.
Work done = force × distance moved in the direction of the force.
Here, force = weight = 420 N, distance = height = 6 m.
Work done = 420 N × 6 m = 2520 J (joules).
Time taken = 20 seconds.
Power = Work done / time = 2520 J / 20 s = 126 W.
But among options given, 126 W is Option 3, so check carefully:
Calculation confirms power output = 126 W.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 1 · Numericals on work, energy and power

Q113. An object has a mass of 12 kg and moves with a constant speed of 8 m/s. How much kinetic energy does it have?

  1. (a) 256 J
  2. (b) 768 J
  3. (c) 384 J
  4. (d) 512 J
Answer & solution

Answer: (c) 384 J

Kinetic energy (KE) is calculated by the formula KE = 1/2 × mass × velocity².
Given: mass (m) = 12 kg, velocity (v) = 8 m/s.
Calculate KE = 1/2 × 12 × (8)²
= 6 × 64
= 384 Joules.
Therefore, the kinetic energy of the object is 384 J.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 1 · Numericals on work, energy and power

Q114. An object with a mass of 15 kg is placed 8 m above the ground. How much potential energy does the object have? (Using g = 9.8 m/s²)

  1. (a) 1500 J
  2. (b) 1176 J
  3. (c) 980 J
  4. (d) 784 J
Answer & solution

Answer: (b) 1176 J

Potential energy (PE) is given by the formula
PE = m × g × h
Here, mass (m) = 15 kg,
gravitational acceleration (g) = (9.8 m)/(s²,)
height (h) = 8 m
Calculate PE = 15 × 9.8 × 8
PE = 15 × 78.4
PE = 1176 Joules
Therefore, potential energy of the object is 1176 J.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 3 · Numericals on work, energy and power

Q115. A body undergoes a displacement of 50 m under the action of a constant force of 5 N in the same direction as that of the force. What will be the work done under the action of this force?

  1. (a) - 250 J
  2. (b) + 25 J
  3. (c) - 25 J
  4. (d) + 250 J
Answer & solution

Answer: (d) + 250 J

Work done by a force is given by the formula: work = force × displacement × cos θ
Here, force = 5 N, displacement = 50 m, and the force and displacement are in the same direction, so θ = 0°
cos 0° = 1
Therefore, work done = 5 × 50 × 1 = 250 J
Since force and displacement are in the same direction, work done is positive.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 2 · Numericals on work, energy and power

Q116. A porter lifts a 10 kg luggage from the ground and places it on his head at a height of 2.5 m. What is the work done by the porter on the luggage? (Using g =10 m/s²)

  1. (a) 250 J
  2. (b) 100 J
  3. (c) 25 J
  4. (d) 500 J
Answer & solution

Answer: (a) 250 J

Work done = Force × Displacement in the direction of force.
Force here is the weight of the luggage = mass × gravity = 10 kg × 10 m/s² = 100 N.
Displacement vertically upwards = 2.5 m.
So, Work done = 100 N × 2.5 m = 250 J.
Therefore, the porter does 250 joules of work on the luggage.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 2 · Work and its sign

Q117. When standing still with a heavy load on your head, you expend energy and get tired. Why is zero work done on the load?

  1. (a) The force applied is vertical.
  2. (b) There is no displacement of the load.
  3. (c) The time elapsed is irrelevant.
  4. (d) Force is not being applied.
Answer & solution

Answer: (b) There is no displacement of the load.

Work is defined as force applied multiplied by displacement in the direction of the force.
When standing still with a heavy load on your head, the load does not move; hence, displacement is zero.
Since displacement is zero, the work done on the load is zero, despite the force acting to support it.
Energy expenditure comes from muscle effort, not work done on the load by physics definition.
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 3 · Potential energy

Q118. The gravitational potential energy of an object at a height is the work done to lift it from the ground to that height _____________.

  1. (a) In the direction of gravity
  2. (b) Along a curved path
  3. (c) Against gravity
  4. (d) With constant velocity
Answer & solution

Answer: (c) Against gravity

The gravitational potential energy (GPE) is the work done to move an object against the force of gravity.
It requires energy to lift an object from ground level to a height.
Hence, the work done, and thus the GPE, is calculated against the gravitational force.
Therefore, the correct option is 'Against gravity'.
Answer completed.
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 3 · Numericals on work, energy and power

Q119. The kinetic energy of an object of mass m moving with a velocity of 5 m/s is 25 J. If its velocity is doubled to 10 m/s, what will be its kinetic energy?

  1. (a) 200 J
  2. (b) 100 J
  3. (c) 125 J
  4. (d) 50 J
Answer & solution

Answer: (b) 100 J

Kinetic energy (KE) is given by the formula: KE = (1/2) m v²
Initially, KE = 25 J, velocity v = 5 m/s.
When velocity is doubled, v = 10 m/s.
New kinetic energy = (1/2) m (10)² = (1/2) m × 100
Since (1/2) m × 25 = 25 J, (1/2) m = 25/25 = 1 J
Therefore, new KE = 1 × 100 = 100 J
Hence, the kinetic energy quadruples when velocity is doubled.
Final answer: Option 2
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 1 · Work and its sign

Q120. What is the work done on an object if its displacement is zero, even when a force is applied?

  1. (a) Zero
  2. (b) Positive
  3. (c) Negative
  4. (d) Infinite
Answer & solution

Answer: (a) Zero

Work done is calculated by the formula: Work = Force × Displacement × cos(θ).
If displacement is zero, regardless of the force applied, work done = Force × 0 × cos(θ) = 0.
Hence, no work is done on the object if it does not move.
Correct Option: 1
Final answer: Option 1
RRB JE CBT-1 2025-26 – 25 February 2026 Shift 3 · Kinetic energy

Q121. A student claims that two cars of equal mass, one moving at 40 km/h and another at 80 km/h, have the same kinetic energy if their engines provide the same power. What is the error in this reasoning?

  1. (a) Power supplied determines energy, regardless of speed.
  2. (b) The cars have equal energy if the engines are identical.
  3. (c) Kinetic energy depends on the square of speed, not just on power supplied.
  4. (d) Both cars have equal kinetic energy since both cover the same distance.
Answer & solution

Answer: (c) Kinetic energy depends on the square of speed, not just on power supplied.

Kinetic energy (KE) = 1/2 m v², so it depends on the square of the speed.
Even if engine power is same, KE is not the same for different speeds.
Power relates to rate of doing work, not directly to KE at a given instant.
Therefore, two cars at different speeds have different kinetic energies.
Hence, the error is ignoring that kinetic energy depends on speed squared.
Final answer: Option 3
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 2 · Potential energy

Q122. An object is placed on a shelf 2 m above the ground. If the shelf is raised to 4 m, what happens to the potential energy of the object?

  1. (a) The potential energy triples
  2. (b) The potential energy halves
  3. (c) The potential energy remains the same
  4. (d) The potential energy doubles
Answer & solution

Answer: (d) The potential energy doubles

Potential energy (PE) is given by the formula PE = mgh , where
m = mass of the object,
g = acceleration due to gravity,
h = height from the ground.
When height increases from 2 m to 4 m, height doubles.
Since PE is directly proportional to height, the potential energy also doubles.
Therefore, when the shelf is raised from 2 m to 4 m, potential energy doubles.
Final answer: Option 4
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 3 · Potential energy

Q123. What happens to the potential energy of an object if its mass is halved while its height above the ground remains constant?

  1. (a) The potential energy becomes zero
  2. (b) The potential energy is halved
  3. (c) The potential energy remains unchanged
  4. (d) The potential energy is doubled
Answer & solution

Answer: (b) The potential energy is halved

Potential energy (PE) is given by the formula: PE = mgh, where:
m = mass of the object,
g = acceleration due to gravity,
h = height above the ground.
If the mass is halved, m becomes m/2, while g and h remain constant.
So, new potential energy = (m/2) × g × h = (1/2) × original potential energy.
Therefore, the potential energy is halved when the mass is halved, keeping height constant.
Final answer: Option 2
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 1 · Work and its sign

Q124. What happens to the work done if the displacement becomes zero despite a force being applied?

  1. (a) The work done remains constant.
  2. (b) The work done decreases.
  3. (c) The work done increases.
  4. (d) The work done is zero.
Answer & solution

Answer: (d) The work done is zero.

Work done is calculated by the formula: Work = Force × Displacement × cos(θ).
If displacement is zero, then Work = Force × 0 × cos(θ) = 0.
Thus, no matter how much force is applied, if there is no displacement, the work done is zero.
Therefore, the correct answer is option 4.
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 3 · Conservation and transformation of energy

Q125. Which statement accurately describes how energy transforms during the free fall of an object under gravity?

  1. (a) Potential energy and kinetic energy both decrease.
  2. (b) Potential energy is converted into heat energy during the fall.
  3. (c) Both potential and kinetic energies remain constant.
  4. (d) Potential energy decreases while kinetic energy increases, maintaining constant total mechanical energy.
Answer & solution

Answer: (d) Potential energy decreases while kinetic energy increases, maintaining constant total mechanical energy.

Option 4 is correct.
During free fall under gravity, an object's potential energy decreases as it moves downwards.
This lost potential energy converts into kinetic energy, increasing the object's speed.
The total mechanical energy (potential + kinetic) remains constant if we ignore air resistance.
Hence, the energy transforms from potential to kinetic, conserving total mechanical energy throughout the fall.
This is a direct consequence of the law of conservation of energy in a gravitational field.
Final answer: Option 4
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 3 · Conservation and transformation of energy

Q126. A ball is placed at the edge of a table and left undisturbed. Which form of energy does the ball primarily possess due to its position, and what can happen if it is pushed off the table?

  1. (a) The ball possesses chemical energy and will release it when it falls.
  2. (b) The ball possesses kinetic energy and will remain at rest unless a force acts on it.
  3. (c) The ball possesses potential energy and will convert it to kinetic energy as it falls.
  4. (d) The ball possesses elastic potential energy due to its shape.
Answer & solution

Answer: (c) The ball possesses potential energy and will convert it to kinetic energy as it falls.

The ball placed at the edge of the table is at a height from the ground.
Due to its position, it possesses gravitational potential energy.
When the ball is pushed off the table, this potential energy converts into kinetic energy as it falls.
Hence, the correct energy form is potential energy, which becomes kinetic energy upon the fall.
Final answer: Option 3
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 1 · Numericals on work, energy and power

Q127. A cricket ball of mass 0.15 kg is thrown horizontally at 30 m/s from the top of a building. What is the ball's kinetic energy as it leaves the hand?

  1. (a) 135 J
  2. (b) 90 J
  3. (c) 45.5 J
  4. (d) 67.5 J
Answer & solution

Answer: (d) 67.5 J

Mass of the cricket ball, m = 0.15 kg
Velocity, v = 30 m/s
Kinetic energy (KE) formula: KE =½m v²
Substitute given values: KE = 1/2 × 0.15 × (30)²
KE = 0.075 × 900 = 67.5 J
Therefore, kinetic energy as it leaves the hand is 67.5 joules.
Final answer: Option 4
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 2 · Work and its sign

Q128. A student drags a suitcase on a flat airport floor. If the force applied and the distance moved both double, how does the work done change?

  1. (a) Work done halves.
  2. (b) Work done remains unchanged.
  3. (c) Work done becomes four times greater.
  4. (d) Work done doubles.
Answer & solution

Answer: (c) Work done becomes four times greater.

Work done is calculated by the formula: Work = Force × Distance.
If both force and distance are doubled, the work done becomes:
Work_new = 2 × Force × 2 × Distance = 4 × (Force × Distance).
Thus, the work done becomes four times the original amount.
Final answer: Option 3
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 1 · Work and its sign

Q129. When a man pushes a wall but the wall does not move, what is the work done by the man on the wall?

  1. (a) Maximum work is done as the man uses full effort.
  2. (b) Zero work is done as there is no displacement of the wall.
  3. (c) Positive work is done as force is applied.
  4. (d) Negative work is done as the wall resists the push.
Answer & solution

Answer: (b) Zero work is done as there is no displacement of the wall.

Work done is defined as force applied multiplied by displacement in the direction of the force.
Here, the man applies force on the wall but the wall does not move.
Since displacement is zero, no work is done on the wall.
Therefore, work done by the man on the wall is zero.
Final answer: Option 2
SSC MTS 2025 – 6 February 2026 Shift 3 · Kinetic energy

Q130. The kinetic energy of an object increases with its _____.

  1. (a) speed
  2. (b) friction
  3. (c) volume
  4. (d) density
Answer & solution

Answer: (a) speed

Correct Answer: 1 (speed)
Kinetic Energy (KE) = 1/2 x mass x velocity^2
KE = (1/2)mv^2
As speed (velocity) increases, kinetic energy increases proportionally to the square of the speed. For example, if speed doubles, KE becomes 4 times.
Option 2 (friction) is incorrect - friction is a force that opposes motion and actually dissipates kinetic energy as heat.
Option 3 (volume) is incorrect - volume does not directly determine kinetic energy.
Option 4 (density) is incorrect - density is not a direct factor in the kinetic energy formula.
SSC CHSL Tier-1 2025 – 30 November 2025 Shift 3 · Numericals on work, energy and power

Q131. A force of 10 N acts on an object. The object is displaced by 2 metres in the direction of the force. The force acts on the object all through the displacement. What is the work done?

  1. (a) 5 J
  2. (b) 10 J
  3. (c) 15 J
  4. (d) 20 J
Answer & solution

Answer: (d) 20 J

Correct Answer: Option 4 – 20 J
Work Done = Force × Displacement × cos(theta)
Since force and displacement are in the same direction, theta = 0 degrees, cos(0) = 1.
Work Done = 10 N × 2 m × 1 = 20 J
Why other options are wrong:
Option 1 (5 J): This would result if Force × Displacement = 5, which is incorrect.
Option 2 (10 J): This would be Force × Displacement only if one value were halved.
Option 3 (15 J): Incorrect; Work = F × d = 10 × 2 = 20 J, not 15 J.
SSC CHSL Tier-1 2025 – 25 November 2025 Shift 1 · Numericals on work, energy and power

Q132. A girl with 50 kg mass climbs up a rope through a height of 5 metres. She takes 25 s to accomplish the task. What is the power (in watts) expended by the girl?

  1. (a) 49
  2. (b) 98
  3. (c) 68.6
  4. (d) 88.2
Answer & solution

Answer: (b) 98

Power = Work done / Time = mgh / t
Given: m = 50 kg, g = 9.8 m/s², h = 5 m, t = 25 s
Power = (50 × 9.8 × 5) / 25 = 2450 / 25 = 98 watts
Hence, the power expended by the girl is 98 W.
SSC CHSL Tier-1 2025 – 24 November 2025 Shift 3 · Conservation and transformation of energy

Q133. Mechanical energy is mainly composed of how many types?

  1. (a) 2
  2. (b) 3
  3. (c) 4
  4. (d) 5
Answer & solution

Answer: (a) 2

Correct Answer: 2 (Option 1)
Mechanical energy is mainly composed of two types: (1) Kinetic Energy (KE) — the energy possessed by a body due to its motion (KE = ½mv²); and (2) Potential Energy (PE) — the energy stored in a body due to its position or configuration (e.g., gravitational PE = mgh ). Total mechanical energy = KE + PE, and in the absence of non-conservative forces, it is conserved (Law of Conservation of Mechanical Energy).
3, 4, or 5 types are incorrect. Mechanical energy is specifically classified into only two types — kinetic and potential energy.
SSC CHSL Tier-1 2025 – 22 November 2025 Shift 3 · Conservation and transformation of energy

Q134. During the free fall of an object, as it falls, what happens to its potential energy and kinetic energy?

  1. (a) Both potential energy and kinetic energy decrease
  2. (b) Both potential energy and kinetic energy increase
  3. (c) Potential energy decreases while kinetic energy increases
  4. (d) Potential energy increases while kinetic energy decreases
Answer & solution

Answer: (c) Potential energy decreases while kinetic energy increases

Correct Answer: Potential energy decreases while kinetic energy increases (Option 3)
During free fall, an object falls under the influence of gravity with no other forces acting on it. As the object falls:
- Its height decreases, so its gravitational Potential Energy (PE = mgh ) decreases.
- Its velocity increases due to gravitational acceleration, so its Kinetic Energy (KE = 1/2 mv2) increases.
The total mechanical energy (PE + KE) remains constant (conservation of energy), as the decrease in PE exactly equals the increase in KE.
Why other options are wrong:
Option 1: Both cannot decrease simultaneously as total energy must be conserved.
Option 2: PE cannot increase while the object is falling (height is decreasing).
Option 4: PE increases only when the object moves upward, not during free fall.
SSC CHSL Tier-1 2025 – 13 November 2025 Shift 1 · Conservation and transformation of energy

Q135. Which of the following is correct regarding law of conservation of energy?
I. Energy can only be converted from one form to another.
II. Energy can be easily destroyed.

  1. (a) Only I
  2. (b) Only II
  3. (c) Both I and II
  4. (d) Neither I nor II
Answer & solution

Answer: (a) Only I

Correct Answer: Only I
Statement I is correct: According to the Law of Conservation of Energy, energy can neither be created nor destroyed; it can only be converted from one form to another. For example, electrical energy is converted to light and heat in a bulb.
Statement II is incorrect: Energy cannot be destroyed. This directly contradicts the Law of Conservation of Energy. The total energy of an isolated system remains constant.
Why 'Only II' is wrong: Statement II is false.
Why 'Both I and II' is wrong: Statement II is incorrect.
Why 'Neither' is wrong: Statement I correctly describes the law.
SSC CHSL Tier-1 2025 – 12 November 2025 Shift 2 · Potential energy

Q136. What is gravitational potential energy?

  1. (a) Energy possessed by an object due to its motion
  2. (b) Energy stored in an object due to its position above the ground
  3. (c) Energy that is released during a chemical reaction
  4. (d) Energy transferred through heat
Answer & solution

Answer: (b) Energy stored in an object due to its position above the ground

Gravitational potential energy (GPE) is the energy stored in an object because of its height/position above the ground. It is given by GPE = mgh . Energy due to motion is kinetic energy; energy released in a chemical reaction is chemical energy; energy transferred through heat is thermal/heat energy.
SSC CHSL Tier-1 2025 – 12 November 2025 Shift 2 · Potential energy

Q137. What is gravitational potential energy?

  1. (a) Energy possessed by an object due to its motion
  2. (b) Energy stored in an object due to its position above the ground
  3. (c) Energy that is released during a chemical reaction
  4. (d) Energy transferred through heat
Answer & solution

Answer: (b) Energy stored in an object due to its position above the ground

Correct Answer: Energy stored in an object due to its position above the ground
Gravitational Potential Energy (GPE) is the energy stored in an object due to its height (position) above a reference point (usually the ground). Formula: GPE = mgh , where m = mass, g = acceleration due to gravity, h = height.
Why other options are wrong:
- Energy due to motion: This is kinetic energy (KE = ?mv?).
- Energy released during chemical reaction: This is chemical energy.
- Energy transferred through heat: This is thermal/heat energy.
SSC Selection Post 2025 (Phase XIII) – 26 July 2025 Shift 2 · Kinetic energy

Q138. What happens to the kinetic energy of a moving object if its velocity is tripled?

  1. (a) It becomes 6 times
  2. (b) It becomes 9 times
  3. (c) It becomes 3 times
  4. (d) It becomes 12 times
Answer & solution

Answer: (b) It becomes 9 times

Correct Answer: Option 2 – It becomes 9 times
Kinetic Energy (KE) = ½mv². If velocity v is tripled (v becomes 3v): New KE = ½m(3v)² = ½m × 9v² = 9 × (½mv²) = 9 × original KE. Therefore, kinetic energy becomes 9 times the original. This shows that KE is proportional to the square of velocity; doubling velocity quadruples KE, tripling velocity increases KE nine times.
Option 1 (6 times): Incorrect; KE ∝ v², not v. Tripling v means KE × 3² = × 9, not × 6.
Option 3 (3 times): Incorrect; if KE were proportional to v (not v²), it would triple. But KE ∝ v², so it increases 9 times.
Option 4 (12 times): Incorrect; 12 has no mathematical basis for this calculation.
RRB NTPC GL CBT-1 2024-25 – 6 June 2025 Shift 3 · Units of work, energy and power

Q139. What is the SI unit of energy?

  1. (a) Joule
  2. (b) Watt
  3. (c) Kelvin
  4. (d) Newton
Answer & solution

Answer: (a) Joule

RRB Technician Grade III 2024-25 – 30 December 2024 Shift 1 · Units of work, energy and power

Q140. 1 kilowatt is equal to _________.

  1. (a) 1000 J s
  2. (b) 1000 J s⁻²
  3. (c) 1000 J
  4. (d) 1000 J s⁻¹
Answer & solution

Answer: (d) 1000 J s⁻¹

Correct Answer: Option 4 — 1000 J s⁻¹
Power is the rate of doing work, so
P = W/t, giving the unit joule per second, which is named the watt (W). Therefore 1 W = 1 J/s = 1 J s⁻¹.
The kilowatt is simply one thousand watts:
1 kW = 1000 W = 1000 J s⁻¹
So a 1 kW heater converts 1000 joules of electrical energy into heat every second. Note carefully that the kilowatt measures power (a rate), while the kilowatt-hour (kWh) measures energy — the energy used by a 1 kW appliance running for one hour, equal to 3.6 × 10⁶ J. This is the "unit" on an electricity bill.
❌ Why the other options are incorrect
✗ 1000 J s — joule multiplied by second is not power at all; it is the unit of "action", as in Planck's constant.
✗ 1000 J s⁻² — joule per second squared would be the *rate of change of power*, not power itself.
✗ 1000 J — the joule alone is the unit of work and energy; without dividing by time it says nothing about how fast the work is done.
📌 SSC & Railway Exam Facts
• Power P = W/t = Energy/Time; SI unit watt (W) = J s⁻¹.
• 1 kW = 1000 W; 1 MW = 10⁶ W; 1 horsepower (hp) = 746 W.
• 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J — the commercial unit of electrical energy.
• Electrical power: P = VI = I²R = V²/R.
• The watt is named after James Watt; the joule after James Prescott Joule.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 1 · Potential energy

Q141. When a body of mass m is raised to a height H against gravity, the potential energy of the body is:

  1. (a) decreased by an amount mgH/2
  2. (b) decreased by an amount mgH
  3. (c) increased by an amount mgH
  4. (d) increased by an amount mgH/2
Answer & solution

Answer: (c) increased by an amount mgH

Correct Answer: Option 3 — increased by an amount mgH
To lift a body of mass m slowly through a height H, you must apply an upward force just equal to its weight, F = mg. The work you do against gravity is
W = Force × displacement = mg × H = mgH
This work is not lost — it is stored in the body as gravitational potential energy because of its new position. Hence the potential energy increases by mgH. If the body is released, this stored energy converts back into kinetic energy, and just before hitting the ground ½mv² = mgH, giving v = √(2gH).
Raising a body always increases its potential energy; lowering it decreases the PE. Only the change in height matters, so the actual path taken — straight up, along a ramp or a staircase — makes no difference. Gravity is therefore called a conservative force.
❌ Why the other options are incorrect
✗ decreased by an amount mgH/2 — wrong in both respects: PE rises, not falls, and there is no factor of ½ (that ½ belongs to kinetic energy, ½mv²).
✗ decreased by an amount mgH — the magnitude is right but the sign is reversed; PE falls only when a body comes down.
✗ increased by an amount mgH/2 — the direction of change is right, but the work done against gravity is the full mgH, since the force mg acts over the whole height H.
📌 SSC & Railway Exam Facts
• Gravitational PE = mgh; SI unit joule (J).
• KE = ½mv²; Mechanical energy = KE + PE, conserved when friction is absent.
• Speed on reaching the ground from height h: v = √(2gh).
• PE depends only on the change in height, not on the path — gravity is a conservative force.
• Work done is zero when a body is carried horizontally, since the height does not change.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 1 · Potential energy

Q142. The gravitational potential energy of an object of mass 'm' raised through a height 'h' from the earth's surface is given by ________.

  1. (a) m g v
  2. (b) m g h
  3. (c) m a v
  4. (d) m a
Answer & solution

Answer: (b) m g h

Correct Answer: Option 2 — m g h
The gravitational potential energy of a body is the energy it possesses because of its position above the ground. To raise a body of mass m through a height h, a force equal to its weight mg must act over the distance h, so the work done is
W = mg × h = mgh
This work is stored in the body, hence
Gravitational PE = m g h
Here g is the acceleration due to gravity, about 9.8 m s⁻² on the Earth's surface. The unit works out correctly: kg × m s⁻² × m = N m = joule. Potential energy is measured from a chosen reference level (usually the ground, where PE is taken as zero), and only the difference in PE between two positions is physically meaningful.
❌ Why the other options are incorrect
✗ m g v — this mixes velocity into a position-based quantity; mass × g × velocity has the unit watt, i.e. power, not energy.
✗ m a v — again mass × acceleration × velocity gives power (J/s), not potential energy.
✗ m a — mass × acceleration is simply force (newton); multiplying by a distance is still needed to obtain energy.
📌 SSC & Railway Exam Facts
• Gravitational PE = mgh; Kinetic energy KE = ½mv²; both in joules.
• g = 9.8 m s⁻² on Earth (often taken as 10), about 1.6 m s⁻² on the Moon.
• Energy is a scalar; it can be transformed but never created or destroyed (law of conservation of energy).
• A body falling from height h hits the ground with v = √(2gh).
• Other kinds of potential energy: elastic PE in a stretched spring, chemical PE in fuel, electrical PE in a charged capacitor.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 2 · Numericals on work, energy and power

Q143. A body undergoes a displacement of 2m on application of a constant force of 10N; the work done will be:

  1. (a) 10 J
  2. (b) 2 J
  3. (c) 20 J
  4. (d) 0 J
Answer & solution

Answer: (c) 20 J

Correct Answer: Option 3 — 20 J
📖 Given
• Constant force applied, F = 10 N
• Displacement produced, s = 2 m
• Force and displacement are along the same direction
✍️ Formula
W = F × s
🔢 Solution
• W = 10 × 2
• W = 20 J
So the force transfers 20 joules of energy to the body while pushing it through 2 metres.
❌ Why the other options are incorrect
✗ 10 J — this is just the value of the force in newtons; you get it only if you forget to multiply by the displacement (or take s = 1 m).
✗ 2 J — this is only the displacement in metres; it ignores the force completely.
✗ 0 J — work is zero only when the displacement is zero, or when the force acts at 90° to the displacement. Here the body actually moves 2 m along the force.
📌 SSC & Railway Exam Facts
• Work = Force × Displacement in the direction of the force; in general W = F s cos θ.
• SI unit of work is the joule (J); 1 J = 1 N × 1 m = 1 kg m² s⁻².
• Work is a scalar quantity — it has magnitude but no direction.
• Work done is zero when θ = 90°, e.g. a coolie carrying a load on his head walking on level ground, or the centripetal force in circular motion.
• Work done against gravity in lifting a body of mass m through height h is W = mgh.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 2 · Power and its units (watt, horsepower)

Q144. Which of the following quantities determine the rate of work performed by a machine?

  1. (a) Power
  2. (b) Total force exerted
  3. (c) Total energy consumed
  4. (d) Rate of change of velocity
Answer & solution

Answer: (a) Power

Correct Answer: Option 1 — Power
Power is defined as the rate of doing work, or equivalently the rate of transfer of energy. It tells us how *fast* a machine finishes a job, not how much work the job needs.
P = W/t = Energy/Time
Its SI unit is the watt (W), and 1 watt = 1 joule per second. Two machines may do exactly the same 1000 J of work, but the one that finishes in 10 s has a power of 100 W while the one taking 20 s has only 50 W. This is why a motor is rated in watts or horsepower, never in joules.
❌ Why the other options are incorrect
✗ Total force exerted — force alone tells nothing about time or even about distance; a force that moves nothing does zero work.
✗ Total energy consumed — this is the total work done, the whole job. Divided by time it gives power, but by itself it has no "rate" in it.
✗ Rate of change of velocity — that is the definition of acceleration, measured in m/s², a quantity of motion and not of work.
📌 SSC & Railway Exam Facts
• Power P = W/t; also P = F × v for a constant force.
• 1 W = 1 J/s; 1 kW = 1000 W; 1 horsepower (hp) = 746 W.
• Commercial unit of electrical energy is the kilowatt-hour (kWh); 1 kWh = 3.6 × 10⁶ J.
• Power is a scalar quantity; the watt is named after James Watt.
• Electrical power: P = VI = I²R = V²/R.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 2 · Units of work, energy and power

Q145. Work has the same unit as that of _______.

  1. (a) force
  2. (b) energy
  3. (c) acceleration
  4. (d) momentum
Answer & solution

Answer: (b) energy

Correct Answer: Option 2 — energy
Work and energy are measured in exactly the same unit — the joule (J). This is not a coincidence: energy is defined as the capacity to do work, so whatever amount of work a body can perform is precisely the amount of energy it carries.
1 joule = 1 newton × 1 metre = 1 kg m² s⁻²
When a force does 20 J of work on a body, the body gains 20 J of energy — this is the work–energy theorem, which states that the net work done equals the change in kinetic energy: W = ½mv² − ½mu². Because of this equality, work, kinetic energy, potential energy and heat are all quoted in joules.
❌ Why the other options are incorrect
✗ force — measured in the newton (N); force is what *causes* work, but N and J differ by one factor of metre (J = N m).
✗ acceleration — measured in m/s²; it describes how quickly velocity changes and is a purely kinematic quantity.
✗ momentum — measured in kg m/s (or N s); it is the product of mass and velocity, a vector, while work is a scalar.
📌 SSC & Railway Exam Facts
• SI unit of work and energy: joule (J); CGS unit: erg; 1 J = 10⁷ erg.
• 1 calorie = 4.18 J (often taken as 4.2 J); 1 kWh = 3.6 × 10⁶ J.
• Kinetic energy = ½mv²; Potential energy = mgh.
• Work–energy theorem: W(net) = ΔKE.
• The joule is named after James Prescott Joule.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 2 · Conservation and transformation of energy

Q146. Mechanical energy of a body includes ______.

  1. (a) kinetic energy only
  2. (b) potential energy only
  3. (c) heat energy only
  4. (d) both potential energy and kinetic energy
Answer & solution

Answer: (d) both potential energy and kinetic energy

Correct Answer: Option 4 — both potential energy and kinetic energy
The mechanical energy of a body is the sum of its kinetic energy and its potential energy:
Mechanical energy = KE + PE = ½mv² + mgh
Kinetic energy is the energy a body has because of its motion, and potential energy is the energy it has because of its position or configuration (height, a stretched spring, a bent bow). Together they describe the total energy a body possesses purely on account of mechanics.
The law of conservation of mechanical energy says that in the absence of friction and air resistance this sum stays constant. For a ball falling freely, PE keeps decreasing and KE keeps increasing by exactly the same amount, so KE + PE remains the same at every point of the fall.
❌ Why the other options are incorrect
✗ kinetic energy only — a stone held at the top of a tower is at rest, with zero KE, yet it clearly possesses mechanical energy in the form of PE.
✗ potential energy only — a bullet flying horizontally at constant height has constant PE but enormous KE, which is also mechanical energy.
✗ heat energy only — heat is the energy of random molecular motion; it is a thermal, not a mechanical, form of energy.
📌 SSC & Railway Exam Facts
• KE = ½mv²; Gravitational PE = mgh; both in joules.
• KE = p²/2m, where p = mv is the momentum.
• In free fall, mechanical energy is conserved; at the ground PE = 0 and KE is maximum.
• Doubling the speed makes KE four times larger, since KE ∝ v².
• Other energy forms: heat, light, sound, chemical, electrical, nuclear; energy can be transformed but never created or destroyed.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 2 · Numericals on work, energy and power

Q147. A boy of weight 600 N climbs up a rope through a height of 10 m. He takes 20 s to accomplish this task. What is the power expended by him?

  1. (a) 150 W
  2. (b) 200 W
  3. (c) 300 W
  4. (d) 100 W
Answer & solution

Answer: (c) 300 W

Correct Answer: Option 3 — 300 W
📖 Given
• Weight of the boy, W = 600 N (this is already the force mg)
• Height climbed, h = 10 m
• Time taken, t = 20 s
✍️ Formula
Work done against gravity: W(done) = Weight × h
Power: P = W(done)/t
🔢 Solution
• W(done) = 600 × 10
• W(done) = 6000 J
• P = 6000/20
• P = 300 W
The boy converts energy at the rate of 300 joules every second while climbing.
❌ Why the other options are incorrect
✗ 150 W — you get this by taking t = 40 s, or by wrongly halving the work done.
✗ 200 W — this comes from dividing 6000 J by 30 s, or from using a height of about 6.7 m.
✗ 100 W — this is 6000/60, i.e. mistaking the time for one minute instead of 20 s.
📌 SSC & Railway Exam Facts
• Power = Work/Time; SI unit watt (W), and 1 W = 1 J/s.
• Work done against gravity: W = mgh; here the weight 600 N is mg itself, so do not multiply by g again.
• If the boy's mass were asked: m = W/g = 600/10 = 60 kg.
• 1 horsepower = 746 W; an average human can sustain roughly 0.1 hp.
• Energy consumed = Power × Time; the commercial unit is the kilowatt-hour (kWh).
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 3 · Units of work, energy and power

Q148. 1 kW power is equal to _______. (kW stands for kilowatt.)

  1. (a) 1 J/s
  2. (b) 1000 J/s
  3. (c) 100 J/s
  4. (d) 10 J/s
Answer & solution

Answer: (b) 1000 J/s

Correct Answer: Option 2 — 1000 J/s
Power is the rate of doing work, so P = W/t. Its SI unit is therefore the joule per second, which is given the special name watt (W):
1 W = 1 J/s
The prefix kilo means one thousand, so
1 kW = 1000 W = 1000 J/s
A 1 kW electric heater therefore converts 1000 joules of electrical energy into heat every second. Keep the difference clear: kilowatt (kW) measures power, a rate, while the kilowatt-hour (kWh) measures energy — the energy used by a 1 kW appliance in one hour, equal to 3.6 × 10⁶ J. The kWh is the "unit" printed on an electricity bill.
❌ Why the other options are incorrect
✗ 1 J/s — this is just 1 watt, one-thousandth of a kilowatt; the prefix kilo has been ignored.
✗ 100 J/s — this is 100 W, roughly the rating of an old filament bulb; the factor is ten times too small.
✗ 10 J/s — this is only 10 W, a hundred times smaller than a kilowatt.
📌 SSC & Railway Exam Facts
• Power P = W/t = Energy/Time; SI unit watt (W) = J s⁻¹.
• 1 kW = 1000 W; 1 MW = 10⁶ W; 1 GW = 10⁹ W; 1 horsepower (hp) = 746 W.
• 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J — the commercial unit of electrical energy.
• Electrical power: P = VI = I²R = V²/R.
• Power is a scalar quantity; the watt is named after James Watt.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 3 · Work and its sign

Q149. Which of the following can increase the work done by a force on an object?

  1. (a) Reducing the displacement.
  2. (b) Reducing the mass.
  3. (c) Reducing the force.
  4. (d) Increasing the displacement.
Answer & solution

Answer: (d) Increasing the displacement.

Correct Answer: Option 4 — Increasing the displacement.
The work done by a constant force is
W = F × s (more generally W = F s cos θ)
Work is therefore directly proportional to both the force and the displacement produced in the direction of the force. Increasing either of them increases the work.
So if the displacement is made larger while the force stays the same, the work done rises in the same proportion. Pushing a box with a force of 10 N through 2 m does 20 J of work; pushing it through 5 m with the same force does 50 J. Notice that mass does not appear in the formula at all — the work done by a given force over a given displacement is the same whatever the body's mass.
❌ Why the other options are incorrect
✗ Reducing the displacement. — this decreases the work, since W ∝ s; if s becomes zero, the work done is zero.
✗ Reducing the mass. — mass is not part of W = Fs; changing it may change the acceleration, but not the work done by a stated force over a stated displacement.
✗ Reducing the force. — this also decreases the work, because W ∝ F.
📌 SSC & Railway Exam Facts
• W = F s cos θ; work is maximum at θ = 0° and zero at θ = 90°.
• SI unit of work is the joule (J); 1 J = 1 N m = 1 kg m² s⁻²; CGS unit erg, 1 J = 10⁷ erg.
• Work is a scalar quantity; it can be positive, negative or zero.
• Work done against gravity in lifting a body: W = mgh.
• Work–energy theorem: net work done = change in kinetic energy.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 3 · Potential energy

Q150. In which of the following situations is potential energy being stored?

  1. (a) Moving a pencil across a table
  2. (b) Lighting a matchstick
  3. (c) Pushing a vehicle on the plane road
  4. (d) Stretching a rubber band
Answer & solution

Answer: (d) Stretching a rubber band

Correct Answer: Option 4 — Stretching a rubber band
Potential energy is the energy a body possesses because of its position or its configuration (shape). When you stretch a rubber band, the work you do against the elastic restoring force changes the shape of the band and is stored inside it as elastic potential energy.
Nothing is lost: release the band and that stored energy is at once converted into the kinetic energy of the flying band. The same idea explains a stretched bow, a compressed spring (PE = ½kx²) and a wound-up watch spring.
For this to count as *stored* energy, the change must be recoverable — the system must be able to give the energy back. That is exactly what an elastic body does, and what a burning or a sliding body does not.
❌ Why the other options are incorrect
✗ Moving a pencil across a table — the work done here goes into kinetic energy and is then lost as heat by friction; nothing is stored.
✗ Lighting a matchstick — this releases stored chemical energy as heat and light; it is the opposite of storing energy.
✗ Pushing a vehicle on the plane road — the road is level, so there is no gain in height and no change of shape; the work done is spent against friction and in giving the vehicle kinetic energy.
📌 SSC & Railway Exam Facts
• Gravitational PE = mgh; Elastic PE of a spring = ½kx²; both in joules.
• KE = ½mv²; Mechanical energy = KE + PE.
• Energy is a scalar; it can be transformed but never created or destroyed.
• Examples of stored PE: stretched bow, compressed spring, water in a dam, charged capacitor, food and fuel (chemical PE).
• Work–energy theorem: net work done = change in kinetic energy.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 1 · Kinetic energy

Q151. The kinetic energy of a moving body of mass ‘m’ and velocity ‘v’ is __________.

  1. (a) directly proportional to its velocity but inversely proportional to its mass
  2. (b) directly proportional to its mass but inversely proportional to its velocity
  3. (c) directly proportional to its mass and square of its velocity
  4. (d) inversely proportional to its mass and square of its velocity
Answer & solution

Answer: (c) directly proportional to its mass and square of its velocity

Correct Answer: Option 3 — directly proportional to its mass and square of its velocity
📖 Given
• Mass of the body = m
• Velocity of the body = v
✍️ Formula
KE = ½mv²
🔢 Solution
• From KE = ½mv², with v kept constant, KE ∝ m — doubling the mass doubles the kinetic energy.
• With m kept constant, KE ∝ v² — doubling the velocity makes the kinetic energy 4 times, and tripling it makes the KE 9 times.
• So the kinetic energy is directly proportional to the mass and to the square of the velocity.
• Derivation: a body starting from rest reaches velocity v after a displacement s, where v² = 2as. Work done, W = F × s = ma × (v²/2a) = ½mv², which is stored as kinetic energy.
In plain language: a heavier body moving at the same speed carries more energy, but speed matters much more than mass, because it enters as a square.
❌ Why the other options are incorrect
✗ directly proportional to its velocity but inversely proportional to its mass — both parts are wrong; KE rises with mass and rises with v², not with v.
✗ directly proportional to its mass but inversely proportional to its velocity — the mass part is right, but KE increases with velocity, it never decreases.
✗ inversely proportional to its mass and square of its velocity — this would mean a fast heavy truck has almost no energy, which is clearly false.
📌 SSC & Railway Exam Facts
• KE = ½mv²; in terms of momentum, KE = p²/2m and p = √(2mKE).
• Potential energy near the Earth's surface, PE = mgh.
• SI unit of energy is the joule (J); 1 J = 1 N m, 1 J = 10⁷ erg, 1 kWh = 3.6 × 10⁶ J.
• Work–energy theorem: net work done on a body = change in its kinetic energy.
• If the speed of a vehicle is doubled, its braking distance becomes four times — a direct consequence of KE ∝ v².
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 1 · Numericals on work, energy and power

Q152. A porter lifts a luggage weighing 10 kg from the ground and puts it on his head 1 m above the ground. Then the work done by him on the luggage (g is acceleration due to gravity in SI unit) is _____________.

  1. (a) 15g J
  2. (b) 18g J
  3. (c) 20g J
  4. (d) 10g J
Answer & solution

Answer: (d) 10g J

Correct Answer: Option 4 — 10g J
📖 Given
• Mass of the luggage, m = 10 kg
• Height through which it is raised, h = 1 m
• Acceleration due to gravity = g (in SI unit)
✍️ Formula
W = F × s = mgh
🔢 Solution
• The porter must apply an upward force equal to the weight of the luggage, F = mg = 10 × g = 10g N.
• The luggage is displaced upward through s = h = 1 m, in the same direction as the applied force, so θ = 0°.
• W = F × s = (10g) × 1
• W = 10g J
• If we take g = 9.8 m s⁻², this equals 98 J; with g = 10 m s⁻² it is 100 J.
In plain words, the work done is exactly the gain in gravitational potential energy of the luggage, because it has simply been raised 1 m higher.
❌ Why the other options are incorrect
✗ 15g J — this would need m × h = 15, for example a 15 kg load raised 1 m; the given data do not give 15.
✗ 18g J — no combination of the given values 10 kg and 1 m produces 18; it is an arbitrary number.
✗ 20g J — this is the error of doubling the answer, as if the height were 2 m instead of 1 m.
📌 SSC & Railway Exam Facts
• W = F s cos θ; work is positive for θ = 0°, negative for θ = 180° and zero for θ = 90°.
• Work done against gravity = mgh = gain in potential energy.
• A porter walking on level ground with a load on his head does zero work against gravity, since force (vertical) and displacement (horizontal) are perpendicular.
• SI unit of work and energy: joule (J); 1 J = 1 N m.
• Power = Work/Time, SI unit watt (W); 1 W = 1 J/s and 1 hp = 746 W.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 2 · Work and its sign

Q153. A constant force F is applied on an object. The work done by the force is positive if the object:

  1. (a) Is displaced opposite to the direction of the force
  2. (b) is displaced perpendicular to the force
  3. (c) is not displaced
  4. (d) is displaced in the direction of the force
Answer & solution

Answer: (d) is displaced in the direction of the force

Correct Answer: Option 4 — is displaced in the direction of the force
Work done by a constant force is defined as W = F s cos θ, where F is the force, s the displacement and θ the angle between them. The sign of the work depends entirely on cos θ.
• When the displacement is along the force, θ = 0° and cos 0° = +1, so W = +Fs — the work is positive. Here the force adds energy to the body and its kinetic energy increases.
• When the displacement is opposite to the force, θ = 180° and cos 180° = −1, so W = −Fs — the work is negative (for example friction, or gravity on a body thrown upward).
• When the displacement is perpendicular to the force, θ = 90° and cos 90° = 0, so W = 0.
So the work done is positive only in the first case.
❌ Why the other options are incorrect
✗ Is displaced opposite to the direction of the force — here θ = 180°, so the work is negative, not positive.
✗ is displaced perpendicular to the force — θ = 90°, cos 90° = 0, so zero work is done, as in uniform circular motion.
✗ is not displaced — with s = 0 the work is zero, however large the force; pushing hard against a fixed wall does no work.
📌 SSC & Railway Exam Facts
• SI unit of work and energy is the joule (J); 1 J = 1 N m. CGS unit is the erg; 1 J = 10⁷ erg.
• W = F s cos θ — work is a scalar quantity though F and s are vectors.
• Work–energy theorem: net work done = change in kinetic energy.
• Power = Work/Time, SI unit watt (W); 1 horsepower = 746 W.
• A coolie carrying a load on his head and walking on level ground does zero work against gravity, since force and displacement are perpendicular.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 2 · Conservation and transformation of energy

Q154. Which of the following statements is NOT true?

  1. (a) Energy can neither be created nor destroyed.
  2. (b) The total energy before and after the transformation always remains constant.
  3. (c) The total energy before and after the transformation is not constant.
  4. (d) Energy can only be transformed from one form to another.
Answer & solution

Answer: (c) The total energy before and after the transformation is not constant.

Correct Answer: Option 3 — The total energy before and after the transformation is not constant.
This statement is NOT true because it directly contradicts the Law of Conservation of Energy, one of the most fundamental laws of physics.
The law states that energy can neither be created nor destroyed; it can only be transformed from one form into another, and the total energy of an isolated system always remains constant. So the total energy before a transformation is exactly equal to the total energy after it.
A falling stone is a good example: at the top it has maximum potential energy and zero kinetic energy; as it falls, PE steadily changes into kinetic energy, but at every instant PE + KE = constant. Similarly, an electric bulb converts electrical energy into light and heat, and the total remains unchanged.
Options 1, 2 and 4 are three different ways of stating this same law, so they are all true. Option 3 denies it and is therefore the required answer.
❌ Why the other options are incorrect
✗ Energy can neither be created nor destroyed. — this is the exact statement of the law of conservation of energy, hence true.
✗ The total energy before and after the transformation always remains constant. — true; this is the quantitative form of the same law.
✗ Energy can only be transformed from one form to another. — true; transformation, not creation or destruction, is what actually happens.
📌 SSC & Railway Exam Facts
• Law of conservation of energy — formulated from the work of Julius Robert Mayer, James Prescott Joule and Hermann von Helmholtz.
• For a freely falling body KE + PE = constant at every point (mechanical energy is conserved).
• Common conversions: electric motor — electrical → mechanical; microphone — sound → electrical; solar cell — light → electrical; dynamo/generator — mechanical → electrical.
• Einstein's mass–energy relation E = mc² extends the law to nuclear reactions.
• SI unit of energy is the joule (J); 1 kWh = 3.6 × 10⁶ J (the commercial 'unit' of electricity).
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 2 · Potential energy

Q155. When we stretch a spring, the form of energy stored in the spring is _________.

  1. (a) potential energy
  2. (b) electric energy
  3. (c) kinetic energy
  4. (d) gravitational energy
Answer & solution

Answer: (a) potential energy

Correct Answer: Option 1 — potential energy
When a spring is stretched or compressed, we do work against the restoring force of the spring. This work is not lost — it is stored inside the spring because of the change in its configuration (shape). Energy stored in a body due to its position or configuration is called potential energy, and in this particular case it is called elastic potential energy.
For a spring obeying Hooke's law (F = kx), the stored energy is U = ½kx², where k is the spring constant and x the extension or compression. Doubling the stretch therefore stores four times the energy.
The moment the spring is released, this stored potential energy converts back into kinetic energy — which is exactly how a wind-up toy, a spring balance and the mainspring of a watch work.
❌ Why the other options are incorrect
✗ electric energy — arises from the motion or position of charges; stretching a spring involves no charge at all.
✗ kinetic energy — energy due to motion, given by ½mv². A stretched spring that is being held is at rest, so it has no kinetic energy; KE appears only after release.
✗ gravitational energy — the potential energy due to height, U = mgh. Here the energy comes from a change of shape, not from a change of height.
📌 SSC & Railway Exam Facts
• Elastic PE of a spring = ½kx²; Gravitational PE = mgh; Kinetic energy = ½mv².
• Hooke's law: F = kx — within the elastic limit, stress ∝ strain, given by Robert Hooke.
• SI unit of spring constant k is N m⁻¹; a stiffer spring has a larger k.
• Energy in a stretched bow, catapult or rubber band is also elastic potential energy.
• Mechanical energy = KE + PE, and it stays constant when only conservative forces act.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 3 · Work and its sign

Q156. If a constant force of 2 Newton is applied on a body such that displacement of the body in the direction of the applied force is zero, then the work done by the force is _____.

  1. (a) 1 joules
  2. (b) 0
  3. (c) 2 joules
  4. (d) 20 joules
Answer & solution

Answer: (b) 0

Correct Answer: Option 2 — 0
📖 Given
• Constant force applied, F = 2 N
• Displacement in the direction of the force, s = 0
✍️ Formula
W = F × s (or W = F s cos θ in general)
🔢 Solution
• W = F × s
• W = 2 × 0
• W = 0 joule
• In physics, work is done only when the point of application of the force actually moves along the force. If that displacement is zero, the work is zero no matter how large the force is.
• The same result follows from W = F s cos θ — with s = 0, the whole product vanishes.
In plain language: a man pushing hard against a stationary wall gets tired, but from the point of view of physics he does no work at all, because the wall does not move.
❌ Why the other options are incorrect
✗ 1 joules — this would need a displacement of 0.5 m; here the displacement along the force is stated to be zero.
✗ 2 joules — this is the mistake of simply copying the force value (2 N) as if s = 1 m; the given s is 0.
✗ 20 joules — this comes from multiplying 2 N by a stray 10 m; no such displacement exists in the question.
📌 SSC & Railway Exam Facts
• W = F s cos θ — work is zero when s = 0, when F = 0, or when θ = 90°.
• SI unit of work is the joule (J); 1 J = 1 N m; CGS unit is the erg, 1 J = 10⁷ erg.
• Work is a scalar quantity even though force and displacement are vectors.
• The centripetal force in uniform circular motion does zero work, since it is always perpendicular to the motion.
• Work–energy theorem: net work done = change in kinetic energy; zero work means no change in speed.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 3 · Numericals on work, energy and power

Q157. A book is pushed with a constant force of 1 N. It displaces 1 m in the direction of the force. The work done by the force is:

  1. (a) 0
  2. (b) 2 J
  3. (c) −1 J
  4. (d) +1 J
Answer & solution

Answer: (d) +1 J

Correct Answer: Option 4 — +1 J
📖 Given
• Constant force applied, F = 1 N
• Displacement, s = 1 m, and it is in the direction of the force, so θ = 0°
✍️ Formula
W = F s cos θ
🔢 Solution
• Since the displacement is along the force, θ = 0° and cos 0° = +1.
• W = F × s × cos 0°
• W = 1 × 1 × 1
• W = +1 J
• The positive sign shows that the force adds energy to the book — its kinetic energy increases.
In plain language, 1 joule is defined as the work done when a force of 1 newton moves a body through 1 metre in its own direction. This question is simply the definition of the joule.
❌ Why the other options are incorrect
✗ 0 — zero work would need either s = 0 or the displacement perpendicular to the force; here the book clearly moves along the force.
✗ 2 J — this is the error of adding 1 + 1 instead of multiplying, or of assuming a 2 m displacement.
✗ −1 J — the magnitude is right but the sign is wrong; negative work occurs when the displacement is opposite to the force, as with friction.
📌 SSC & Railway Exam Facts
• 1 joule = 1 newton × 1 metre — the definition of the SI unit of work and energy.
• W = F s cos θ: positive for θ = 0°, negative for θ = 180°, zero for θ = 90°.
• Friction and air resistance always do negative work on a moving body.
• 1 J = 10⁷ erg; 1 calorie = 4.2 J; 1 kWh = 3.6 × 10⁶ J.
• Power = W/t, SI unit watt (W); 1 W = 1 J/s, 1 hp = 746 W.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 3 · Kinetic energy

Q158. The kinetic energy E possessed by an object of mass m and moving with a uniform velocity v is _______.

  1. (a) 1/3 m v²
  2. (b) m v⁻²
  3. (c) m v²
  4. (d) 1/2 m v²
Answer & solution

Answer: (d) 1/2 m v²

Correct Answer: Option 4 — 1/2 m v²
📖 Given
• Mass of the object = m
• Uniform velocity of the object = v
• To find: the expression for its kinetic energy E
✍️ Formula
E = ½mv²
🔢 Solution
• Suppose a body of mass m starts from rest and is accelerated uniformly to a velocity v over a displacement s by a force F.
• From the third equation of motion, v² = u² + 2as with u = 0, we get s = v²/2a.
• Work done by the force, W = F × s = (ma) × (v²/2a).
• The acceleration a cancels, leaving W = ½mv².
• This work is stored in the body as its kinetic energy, so E = ½mv².
In plain language, the kinetic energy is half the mass times the square of the speed — the factor ½ is not optional, it comes straight out of the derivation.
❌ Why the other options are incorrect
✗ 1/3 m v² — the coefficient is wrong; the derivation gives ½, never ⅓.
✗ m v⁻² — here v is in the denominator, which would mean a faster body has less energy — physically impossible.
✗ m v² — the ½ factor has been dropped; this expression is twice the actual kinetic energy.
📌 SSC & Railway Exam Facts
• KE = ½mv²; also KE = p²/2m and p = √(2m·KE), where p = mv is the momentum.
• KE ∝ v² — doubling the speed makes the kinetic energy four times.
• Potential energy PE = mgh; total mechanical energy = KE + PE is conserved in the absence of friction.
• SI unit of energy: joule (J); 1 J = 1 N m = 10⁷ erg; 1 eV = 1.6 × 10⁻¹⁹ J.
• Kinetic energy is a scalar and is always positive, whatever the direction of motion.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 1 · Numericals on work, energy and power

Q159. The potential energy possessed by an object of mass 8 kg when it is at a height of 5 m above the ground will be (g is acceleration due to gravity in SI unit):

  1. (a) 20g J
  2. (b) 100g J
  3. (c) 40g J
  4. (d) 200g J
Answer & solution

Answer: (c) 40g J

Correct Answer: Option 3 — 40g J
📖 Given
• Mass of the object, m = 8 kg
• Height above the ground, h = 5 m
• Acceleration due to gravity = g (in SI unit, m/s²)
✍️ Formula
Gravitational potential energy, PE = mgh
🔢 Solution
• PE = m × g × h
• PE = 8 × g × 5
• PE = 40g J
So the object stores 40g joule of energy purely because of its position. Taking g = 10 m/s², this is 400 J; taking g = 9.8 m/s², it is 392 J. This whole amount would be converted into kinetic energy by the time the object reached the ground.
❌ Why the other options are incorrect
✗ 20g J — this is what you get if you use half the product, i.e. wrongly applying ½mgh (the ½ belongs to kinetic energy ½mv², not to potential energy).
✗ 100g J — this comes from wrongly using the mass and height as 8 + 5 = 13 or from a stray multiplication; it does not follow from mgh.
✗ 200g J — this is 5 times the correct value; it comes from multiplying 8 × 5 and then again by 5.
📌 SSC & Railway Exam Facts
• PE = mgh; KE = ½mv²; both are measured in joule (J).
• 1 joule = 1 N m = 1 kg m² s⁻²; the CGS unit is the erg, 1 J = 10⁷ erg.
• At the ground (h = 0) the potential energy is zero; PE is measured with respect to a chosen reference level.
• Total mechanical energy = KE + PE stays constant for a freely falling body.
• g = 9.8 m/s² on Earth; PE does not depend on the path taken, only on the height.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 2 · Work and its sign

Q160. Work done can be defined as ______.

  1. (a) force × displacement
  2. (b) force / displacement
  3. (c) force − displacement
  4. (d) force + displacement
Answer & solution

Answer: (a) force × displacement

Correct Answer: Option 1 — force × displacement
Work is said to be done when a force acting on a body displaces it along the direction of that force. So work is the product of the force and the displacement produced in the direction of the force:
W = F × s
The SI unit of work is the joule (J), where 1 J = 1 N × 1 m. Work is a scalar quantity — it has magnitude but no direction. If the force makes an angle θ with the displacement, the general form is W = F s cos θ.
This is why a man pushing a heavy wall that does not move does zero work in physics — the displacement s is zero, however tired he may get.
❌ Why the other options are incorrect
✗ force / displacement — force divided by displacement gives a quantity measured in N/m, which is the spring constant (stiffness), not work.
✗ force − displacement — force is measured in newton and displacement in metre; quantities with different units can never be subtracted.
✗ force + displacement — same objection: only like physical quantities can be added together.
📌 SSC & Railway Exam Facts
• W = F s cos θ — work is positive at θ = 0°, zero at θ = 90° and negative at θ = 180°.
• SI unit of work and energy is the joule (J); the CGS unit is the erg; 1 J = 10⁷ erg.
• 1 joule of work is done when a force of 1 N moves a body through 1 m in its own direction.
• Work done against friction or by a body being lifted down by gravity is negative work.
• Power = Work / Time; SI unit watt (W), and 1 W = 1 J/s.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 2 · Conservation and transformation of energy

Q161. Which of the following is part of mechanical energy?

  1. (a) Only Light energy
  2. (b) Potential energy
  3. (c) Only Heat energy
  4. (d) Only Chemical energy
Answer & solution

Answer: (b) Potential energy

Correct Answer: Option 2 — Potential energy
Mechanical energy is the energy a body possesses because of its motion or because of its position/configuration. It therefore has exactly two parts:
• Kinetic energy (KE = ½mv²) — energy due to motion.
• Potential energy (PE = mgh) — energy due to position or state of strain.
Total mechanical energy = KE + PE. For a freely falling body (no friction), this total stays constant — that is the law of conservation of mechanical energy. As a stone falls, PE keeps converting into KE, but the sum never changes.
❌ Why the other options are incorrect
✗ Only Light energy — light (radiant) energy travels as electromagnetic waves; it is a separate form, not a part of mechanical energy.
✗ Only Heat energy — heat is the internal energy of random molecular motion; it is thermal energy, and mechanical energy lost to friction turns into it.
✗ Only Chemical energy — this is energy stored in the bonds of a fuel or food and is released in chemical reactions, not a form of mechanical energy.
📌 SSC & Railway Exam Facts
• Mechanical energy = Kinetic energy + Potential energy.
• KE = ½mv² and gravitational PE = mgh; both are measured in joule (J).
• A body at the highest point of its flight has maximum PE and minimum KE.
• Energy can neither be created nor destroyed — the law of conservation of energy.
• A stretched spring or bow stores elastic potential energy.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 3 · Work and its sign

Q162. The displacement of a body is perpendicular to the direction in which the force is applied; the work done in this case will be:

  1. (a) Positive
  2. (b) Zero
  3. (c) Infinite
  4. (d) Negative
Answer & solution

Answer: (b) Zero

Correct Answer: Option 2 — Zero
📖 Given
• A force F acts on a body
• The displacement s is perpendicular to the force, so the angle between them is θ = 90°
✍️ Formula
W = F s cos θ
🔢 Solution
• W = F × s × cos 90°
• cos 90° = 0
• W = F × s × 0
• W = 0 (zero)
So when the force and the displacement are at right angles, no work is done at all, however large the force and the displacement may be. The force in that case only changes the direction of motion, not the speed, so the body's kinetic energy stays the same.
Classic examples: a coolie carrying a load on his head while walking horizontally (the force is vertical, the displacement horizontal) and the Moon revolving around the Earth in a circular orbit, where the gravitational pull is always perpendicular to the motion.
❌ Why the other options are incorrect
✗ Positive — work is positive only when the angle is less than 90° (cos θ > 0), for example a body being pushed along the direction of the force.
✗ Infinite — work can never be infinite here; W = Fs cos θ is a finite product, and cos θ never exceeds 1.
✗ Negative — work is negative when the angle is more than 90° (cos θ < 0), for example work done against friction or by gravity on a body being lifted up.
📌 SSC & Railway Exam Facts
• W = F s cos θ: θ = 0° → maximum positive work, θ = 90° → zero work, θ = 180° → negative work.
• SI unit of work and energy is the joule (J); 1 J = 1 N m; CGS unit is the erg, 1 J = 10⁷ erg.
• In uniform circular motion the centripetal force does zero work, since it is always perpendicular to the velocity.
• Work is a scalar quantity, even though force and displacement are both vectors.
• Work–energy theorem: work done by the net force = change in kinetic energy.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 1 · Numericals on work, energy and power

Q163. The work done by a person pushing a brick parallel to the ground with a force of 2 N is 3 J. The displacement caused by the force is equal to ______.

  1. (a) 1.5 m
  2. (b) 5 m
  3. (c) 6 m
  4. (d) 0.66 m
Answer & solution

Answer: (a) 1.5 m

Correct Answer: Option 1 — 1.5 m
📖 Given
• Force applied, F = 2 N
• Work done, W = 3 J
• The force is parallel to the ground, i.e. along the displacement, so θ = 0° and cos θ = 1
✍️ Formula
W = F s cos θ, so for θ = 0°, s = W/F
🔢 Solution
• s = W/F
• s = 3/2
• s = 1.5 m
So the brick slides forward through one and a half metres while the person spends 3 joules of energy on it.
❌ Why the other options are incorrect
✗ 5 m — this comes from adding 2 and 3 instead of dividing; work and force cannot be added.
✗ 6 m — this is W × F = 3 × 2; multiplying gives a quantity with units J N, which is meaningless here.
✗ 0.66 m — this is F/W = 2/3, the ratio taken upside down; displacement is work divided by force.
📌 SSC & Railway Exam Facts
• W = F s cos θ; SI unit of work is the joule (J), and 1 J = 1 N × 1 m.
• Work is done only when the force produces a displacement in its own direction.
• θ = 0° → positive work, θ = 90° → zero work, θ = 180° → negative work.
• Work and energy are scalar quantities.
• Power = work/time, SI unit watt (W); 1 W = 1 J/s.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 1 · Numericals on work, energy and power

Q164. If a 5 N force applied to an object displaces it by 3 m from it's initial position, then what is the work done by the force?

  1. (a) 50 J
  2. (b) 20 J
  3. (c) 5/3 J
  4. (d) 15 J
Answer & solution

Answer: (d) 15 J

Correct Answer: Option 4 — 15 J
📖 Given
• Force applied, F = 5 N
• Displacement produced, s = 3 m
• The displacement is along the direction of the force, so θ = 0° and cos θ = 1
✍️ Formula
W = F s cos θ = F × s (when θ = 0°)
🔢 Solution
• W = 5 × 3 × cos 0°
• W = 5 × 3 × 1
• W = 15 J
So the force transfers 15 joules of energy to the object while moving it through 3 metres. The work is positive, because force and displacement point the same way.
❌ Why the other options are incorrect
✗ 50 J — this treats the displacement as 10 m instead of 3 m; 5 × 10 = 50.
✗ 20 J — this comes from using a force of 5 N over 4 m, or from a slip such as 5 + 15; neither matches the given data.
✗ 5/3 J — this is F/s = 5/3, obtained by dividing instead of multiplying; work is force times displacement.
📌 SSC & Railway Exam Facts
• W = F s cos θ; SI unit joule (J), 1 J = 1 N m; CGS unit erg, 1 J = 10⁷ erg.
• 1 joule of work is done when a force of 1 newton moves a body 1 metre in its own direction.
• Work is a scalar quantity.
• Work–energy theorem: net work done = change in kinetic energy.
• 1 calorie = 4.2 J; 1 kWh = 3.6 × 10⁶ J.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 2 · Units of work, energy and power

Q165. Out of the following, which is a unit used for measuring the amount of work done on a body:

  1. (a) Joule
  2. (b) Kilogram
  3. (c) Newton
  4. (d) Meter
Answer & solution

Answer: (a) Joule

Correct Answer: Option 1 — Joule
Work is said to be done when a force acting on a body produces a displacement in it. Numerically, W = F × s, where F is the force applied and s is the displacement in the direction of the force.
Putting the SI units into this formula, 1 unit of work = 1 newton × 1 metre = 1 N m. This unit is given the special name joule (J), in honour of the English physicist James Prescott Joule.
So 1 joule is the work done when a force of 1 newton moves a body through 1 metre in the direction of the force. Since work and energy are the same physical quantity, energy is also measured in joules.
❌ Why the other options are incorrect
✗ Kilogram — the SI unit of mass, a measure of the quantity of matter in a body, not of work.
✗ Newton — the SI unit of force; 1 N = 1 kg m/s². Force multiplied by displacement gives work.
✗ Meter — the SI unit of length or displacement, one of the two factors of work, not work itself.
📌 SSC & Railway Exam Facts
• W = F s cos θ; SI unit joule (J), CGS unit erg, and 1 J = 10⁷ erg.
• Work and energy are scalar quantities even though force and displacement are vectors.
• 1 calorie = 4.2 J; 1 kWh = 3.6 × 10⁶ J.
• Power = work/time, SI unit watt (W); 1 W = 1 J/s and 1 horsepower = 746 W.
• Zero work is done when displacement is zero or when the force is perpendicular to the displacement.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 3 · Work and its sign

Q166. A girl pushes a trolley with a constant force F through a distance s, the work done will be:

  1. (a) F + s
  2. (b) F × s
  3. (c) F/s
  4. (d) F − s
Answer & solution

Answer: (b) F × s

Correct Answer: Option 2 — F × s
📖 Given
• Constant force applied, F
• Distance moved in the direction of the force, s
• The force and the displacement are in the same direction (θ = 0°)
✍️ Formula
W = F s cos θ
🔢 Solution
• Here the trolley moves in the direction of the push, so θ = 0°
• cos 0° = 1
• W = F × s × 1
• W = F × s
So the work done is simply the product of the force and the distance moved along it — if the girl pushes with 20 N through 5 m, she does 20 × 5 = 100 J of work.
❌ Why the other options are incorrect
✗ F + s — force and distance are different physical quantities with different units (N and m); they can never be added.
✗ F/s — dividing force by distance gives N/m, which is the unit of a spring constant, not of work.
✗ F − s — subtracting two unlike quantities is meaningless; the result would not even have the unit of work.
📌 SSC & Railway Exam Facts
• W = F s cos θ; SI unit joule (J), and 1 J = 1 N × 1 m.
• Work is a scalar quantity, although force and displacement are both vectors.
• θ = 0° → positive work, θ = 180° → negative work, θ = 90° → zero work.
• CGS unit of work is the erg; 1 J = 10⁷ erg.
• Power = W/t, SI unit watt (W); 1 horsepower = 746 W.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 3 · Kinetic energy

Q167. The kind of energy possessed by a moving body by virtue of its motion is ________.

  1. (a) kinetic energy
  2. (b) potential energy
  3. (c) electromagnetic energy
  4. (d) nuclear energy
Answer & solution

Answer: (a) kinetic energy

Correct Answer: Option 1 — kinetic energy
The energy possessed by a body because of its motion is called kinetic energy (KE). A moving body can do work on anything that comes in its way — a moving hammer drives a nail, flowing water turns a turbine, and a moving bullet pierces a target.
Its value is given by KE = ½mv², where m is the mass and v the velocity of the body. Since KE depends on the square of the velocity, doubling the speed makes the kinetic energy four times larger, while doubling the mass only doubles it.
Kinetic energy is a scalar quantity, it is always positive or zero, and it is measured in joules (J).
❌ Why the other options are incorrect
✗ potential energy — this is the energy of position or configuration, PE = mgh, as in water stored in a dam or a stretched spring; it does not need motion.
✗ electromagnetic energy — the energy carried by electromagnetic waves such as light, radio waves and X-rays, not the energy of a moving body.
✗ nuclear energy — the energy released from the nucleus of an atom during fission or fusion, as in a nuclear reactor or the sun.
📌 SSC & Railway Exam Facts
• KE = ½mv² and p = mv, so KE = p²/2m.
• Work–energy theorem: work done by the net force = change in kinetic energy.
• Mechanical energy = KE + PE, and it stays constant when only gravity acts.
• SI unit of energy is the joule (J); 1 calorie = 4.2 J; 1 kWh = 3.6 × 10⁶ J.
• A body at rest has zero kinetic energy but may still possess potential energy.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 3 · Numericals on work, energy and power

Q168. A worker lifts a rice bag of 10 kg from the ground and puts it on the truck 2 m above the ground. Calculate the work done by him on the rice bag. (Given, g= 10 m s⁻² )

  1. (a) 100 J
  2. (b) 200 J
  3. (c) 59 J
  4. (d) 153 J
Answer & solution

Answer: (b) 200 J

Correct Answer: Option 2 — 200 J
📖 Given
• Mass of the rice bag, m = 10 kg
• Height to which it is raised, h = 2 m
• Acceleration due to gravity, g = 10 m s⁻²
✍️ Formula
The worker lifts the bag against gravity, so W = F × h = mgh
🔢 Solution
• Force needed = weight of the bag = m × g
• F = 10 × 10 = 100 N
• W = F × h
• W = 100 × 2
• W = 200 J
The 200 joules of work done by the worker is stored in the bag as gravitational potential energy, which it would release if it fell back to the ground.
❌ Why the other options are incorrect
✗ 100 J — this is only mg, the force in newtons; the student stopped before multiplying by the height h = 2 m.
✗ 59 J — no combination of 10 kg, 2 m and g = 10 m s⁻² produces this value; it is a distractor.
✗ 153 J — again unrelated to the data; mgh must come out as a clean multiple of 100 here.
📌 SSC & Railway Exam Facts
• Work done against gravity, W = mgh, which equals the gain in potential energy.
• W = F s cos θ; here the force and the displacement are both vertical, so cos 0° = 1.
• SI unit of work and energy is the joule (J); 1 J = 1 N m.
• Potential energy depends only on the height, not on the path taken to reach it.
• If the worker had merely carried the bag horizontally, the work done against gravity would be zero (θ = 90°).
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 3 · Work and its sign

Q169. The work done by a constant force acting on a body is _________.

  1. (a) always positive
  2. (b) either positive or negative
  3. (c) always zero
  4. (d) always negative
Answer & solution

Answer: (b) either positive or negative

Correct Answer: Option 2 — either positive or negative
The work done by a force is W = F s cos θ, where θ is the angle between the force and the displacement. The sign of the work is decided entirely by cos θ, so a constant force can do work of either sign:
• θ = 0° → cos 0° = +1 → W = +Fs, the work is positive. Example: gravity acting on a falling stone.
• θ = 180° → cos 180° = −1 → W = −Fs, the work is negative. Example: friction on a sliding box, or gravity on a stone thrown upward.
So the very same constant force — gravity — does positive work in one situation and negative work in another. Hence the work done by a constant force is either positive or negative, and cannot be fixed to one sign.
❌ Why the other options are incorrect
✗ always positive — false, because a force acting opposite to the displacement does negative work, as brakes do on a moving train.
✗ always zero — the work is zero only in the special case θ = 90°, as for a coolie carrying a load horizontally on his head.
✗ always negative — false, because a force acting along the displacement does positive work, as when you push a trolley forward.
📌 SSC & Railway Exam Facts
• W = F s cos θ; SI unit joule (J); work is a scalar quantity.
• Positive work adds energy to a body; negative work removes energy from it.
• In uniform circular motion, the centripetal force does zero work (θ = 90°).
• Work–energy theorem: net work done = change in kinetic energy.
• 1 J = 1 N m = 10⁷ erg; 1 calorie = 4.2 J.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 3 · Numericals on work, energy and power

Q170. A ball of mass 0.5 kg is moving with a uniform velocity of 6 m/s. The kinetic energy of the ball is:

  1. (a) 6 J
  2. (b) 15 J
  3. (c) 12 J
  4. (d) 9 J
Answer & solution

Answer: (d) 9 J

Correct Answer: Option 4 — 9 J
📖 Given
• Mass of the ball, m = 0.5 kg
• Uniform velocity, v = 6 m/s
✍️ Formula
KE = ½mv²
🔢 Solution
• KE = ½ × 0.5 × (6)²
• v² = 6 × 6 = 36
• KE = ½ × 0.5 × 36
• KE = 0.25 × 36
• KE = 9 J
So the moving ball carries 9 joules of energy, and it is exactly this much work that must be done against it to bring it to rest.
❌ Why the other options are incorrect
✗ 6 J — this is just the numerical value of the velocity; the mass and the factor ½ have been left out completely.
✗ 15 J — this needs v² = 60, since ½ × 0.5 × 60 = 15; but squaring 6 gives 36, not 60.
✗ 12 J — this needs v² = 48, since ½ × 0.5 × 48 = 12; again the square of 6 is 36, so 12 J cannot arise.
📌 SSC & Railway Exam Facts
• KE = ½mv² — kinetic energy depends on the square of the velocity.
• Doubling the speed makes the kinetic energy four times; doubling the mass only doubles it.
• KE = p²/2m, where p = mv is the momentum.
• Work–energy theorem: net work done on a body = change in its kinetic energy.
• SI unit of kinetic energy is the joule (J); it is a scalar and is never negative.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 1 · Units of work, energy and power

Q171. Which of the following pairs has the same unit?

  1. (a) (i) Force and (ii) Work
  2. (b) (i) Force and (ii) Inertia
  3. (c) (i) Work and (ii) Energy
  4. (d) (i) Energy and (ii) Inertia
Answer & solution

Answer: (c) (i) Work and (ii) Energy

Correct Answer: Option 3 — (i) Work and (ii) Energy
Work and energy are measured in the same unit, the joule (J), because energy is defined as the capacity to do work. When a force of 1 N moves a body through 1 m along its own direction, the work done is 1 J, and exactly 1 J of energy is transferred to the body. So 1 J = 1 N m.
This is also why the work–energy theorem can be written at all: net work done = change in kinetic energy. Both are scalar quantities, and their CGS unit is the erg, with 1 J = 10⁷ erg.
❌ Why the other options are incorrect
✗ (i) Force and (ii) Work — force is measured in newton (N) and work in joule (J); they differ by a factor of length, since J = N m.
✗ (i) Force and (ii) Inertia — inertia is measured by mass, so its unit is the kilogram (kg), while force is in newton (N).
✗ (i) Energy and (ii) Inertia — energy is in joule (J) and inertia (mass) in kilogram (kg); two completely different physical quantities.
📌 SSC & Railway Exam Facts
• Work = Energy → joule (J); 1 J = 1 N m = 10⁷ erg.
• Work–energy theorem: W(net) = ½mv² − ½mu².
• Other quantities sharing units: work, energy, heat and torque all reduce to N m (torque is written as N m, never as joule).
• Inertia ∝ mass, unit kilogram (kg); force unit newton (N).
• 1 kWh = 3.6 × 10⁶ J, 1 calorie = 4.2 J, 1 eV = 1.6 × 10⁻¹⁹ J.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 1 · Kinetic energy

Q172. What kind of energy is possessed by a moving bullet?

  1. (a) Kinetic energy
  2. (b) Light energy
  3. (c) Electrical energy
  4. (d) Potential energy
Answer & solution

Answer: (a) Kinetic energy

Correct Answer: Option 1 — Kinetic energy
The energy possessed by a body by virtue of its motion is called kinetic energy. A bullet fired from a gun is moving very fast, so it carries kinetic energy, given by
K = ½mv²
Notice that kinetic energy depends on the square of the speed. That is why a bullet of small mass but very high speed carries enough energy to pierce a thick plank — doubling the speed makes the energy four times larger. When the bullet strikes a target, this kinetic energy is spent as work done against friction, appearing as heat and deformation.
❌ Why the other options are incorrect
✗ Light energy — the energy carried by electromagnetic radiation; a moving bullet does not shine, and light plays no part in its motion.
✗ Electrical energy — the energy associated with the flow of charge through a circuit; a bullet carries no current.
✗ Potential energy — the energy stored because of position or configuration, such as gravitational PE = mgh or the elastic PE of a stretched spring. It is the stored form, not the form due to motion.
📌 SSC & Railway Exam Facts
• Kinetic energy K = ½mv²; K ∝ v²; SI unit joule (J); it is a scalar and always positive.
• Relation with momentum: K = p²/2m and p = √(2mK).
• Potential energy (gravitational) = mgh; a stretched bow, a wound spring and water in a dam store PE.
• Law of conservation of energy: energy can neither be created nor destroyed, only transformed.
• A falling body converts PE → KE; total mechanical energy stays constant when friction is absent.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 2 · Conservation and transformation of energy

Q173. Which of the following correctly describes the sum of kinetic energy and potential energy of an object?

  1. (a) Total electrical energy
  2. (b) Total mechanical energy
  3. (c) Total chemical energy
  4. (d) Total thermal energy
Answer & solution

Answer: (b) Total mechanical energy

Correct Answer: Option 2 — Total mechanical energy
The sum of the kinetic energy (KE) and the potential energy (PE) of a body is called its total mechanical energy.
• Kinetic energy is the energy of motion: KE = ½mv².
• Potential energy is the energy of position or configuration: PE = mgh for a body raised to height h.
• Therefore Total mechanical energy, E = ½mv² + mgh.
For a freely falling body, as it comes down its PE keeps decreasing and its KE keeps increasing, but the sum stays constant. This is the law of conservation of mechanical energy, valid when only gravity acts and friction or air resistance is absent.
❌ Why the other options are incorrect
✗ Total electrical energy — energy carried by moving charges, W = VIt; it is not made of KE and PE.
✗ Total chemical energy — energy stored in the bonds of a substance, released in food, fuels and cells.
✗ Total thermal energy — the internal heat energy of a body due to the random motion of its molecules.
📌 SSC & Railway Exam Facts
• SI unit of every form of energy is the joule (J); 1 J = 1 N m.
• 1 kWh (Board of Trade Unit) = 3.6 × 10⁶ J.
• Energy can be transformed but never created or destroyed — law of conservation of energy.
• At the top of its flight a ball has maximum PE and zero KE; just before hitting the ground it has maximum KE.
• Power is the rate of doing work: P = W/t, SI unit watt (W).
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 2 · Conservation and transformation of energy

Q174. Which of the following energies reflect mechanical energy when combined?

  1. (a) Chemical energy and electrical energy
  2. (b) Light energy and chemical energy
  3. (c) Potential energy and kinetic energy
  4. (d) Heat energy and light energy
Answer & solution

Answer: (c) Potential energy and kinetic energy

Correct Answer: Option 3 — Potential energy and kinetic energy
Mechanical energy is the energy a body possesses because of its motion or because of its position. It is therefore made up of exactly two parts:
• Kinetic energy (KE = ½mv²) — energy due to motion. A running train, flowing water and a moving bullet all have it.
• Potential energy (PE = mgh) — energy due to position or configuration. Water stored in a dam, a stretched rubber band and a book on a shelf all have it.
So Mechanical energy = KE + PE. In a swinging pendulum the two keep changing into each other while their sum stays constant — a neat demonstration of the conservation of mechanical energy.
❌ Why the other options are incorrect
✗ Chemical energy and electrical energy — these are separate forms; chemical energy is stored in bonds, electrical energy is carried by moving charges.
✗ Light energy and chemical energy — light is radiant energy; their combination is used in photosynthesis, not in mechanics.
✗ Heat energy and light energy — both are non-mechanical forms released, for example, by a burning candle.
📌 SSC & Railway Exam Facts
• KE = ½mv² — doubling the speed makes the kinetic energy four times.
• PE = mgh depends on mass, g and height, not on the path taken.
• In a hydroelectric power plant, PE of stored water → KE of falling water → electrical energy.
• Work–energy theorem: work done by the net force = change in kinetic energy.
• SI unit of work and energy is the joule (J); the CGS unit is the erg (1 J = 10⁷ erg).
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 2 · Work and its sign

Q175. In which of the following conditions is the work done considered negative?

  1. (a) When the force is opposite to the direction of displacement.
  2. (b) When the displacement is zero.
  3. (c) When the force is in the direction of displacement.
  4. (d) When the force is perpendicular to the direction of displacement.
Answer & solution

Answer: (a) When the force is opposite to the direction of displacement.

Correct Answer: Option 1 — When the force is opposite to the direction of displacement.
Work done by a force is W = F s cos θ, where θ is the angle between the force and the displacement. The sign of the work depends entirely on cos θ:
• θ = 0° → cos 0° = +1 → W = +Fs, work is positive.
• θ = 180° → cos 180° = −1 → W = −Fs, work is negative.
• θ = 90° → cos 90° = 0 → W = 0, work is zero.
So when the force acts opposite to the displacement, the work done by it is negative. Everyday examples are the work done by friction on a sliding box and the work done by gravity on a ball thrown upwards.
❌ Why the other options are incorrect
✗ When the displacement is zero — then W = F × 0 = zero, not negative. Pushing a wall that does not move is an example.
✗ When the force is in the direction of displacement — this gives maximum positive work, W = +Fs.
✗ When the force is perpendicular to the displacement — cos 90° = 0, so the work is zero; a coolie carrying a load on his head does no work against gravity.
📌 SSC & Railway Exam Facts
• Work, W = F s cos θ; SI unit joule (J), 1 J = 1 N × 1 m.
• Work is a scalar quantity, although both force and displacement are vectors.
• Negative work removes energy from a body; that is how brakes stop a train.
• In uniform circular motion the centripetal force is always perpendicular to motion, so its work is zero.
• 1 kilojoule = 1000 J; 1 calorie = 4.2 J.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 3 · Numericals on work, energy and power

Q176. A force of 20 N acts on a body in a direction opposite to the displacement of the body. If under the action of this force the displacement is 2m, the work done will be:

  1. (a) −40 J
  2. (b) 40 J
  3. (c) 0 J
  4. (d) −20 J
Answer & solution

Answer: (a) −40 J

Correct Answer: Option 1 — −40 J
📖 Given
• Force, F = 20 N
• Displacement, s = 2 m
• The force acts opposite to the displacement, so the angle θ = 180°
✍️ Formula
W = F s cos θ
🔢 Solution
• W = 20 × 2 × cos 180°
• cos 180° = −1
• W = 20 × 2 × (−1)
• W = −40 J
The minus sign means the force is taking energy away from the body instead of giving energy to it — exactly what friction or a brake does to a moving object.
❌ Why the other options are incorrect
✗ 40 J — this is the value you get if you forget the direction and take θ = 0°, i.e. treat the force as helping the motion.
✗ 0 J — work would be zero only if the displacement were zero or the force were perpendicular to the displacement (θ = 90°).
✗ −20 J — this uses only the force and drops the displacement; you must multiply F by s.
📌 SSC & Railway Exam Facts
• W = F s cos θ; SI unit of work is the joule (J), and 1 J = 1 N × 1 m.
• Work is positive at θ = 0°, negative at θ = 180° and zero at θ = 90°.
• Work and energy are scalar quantities even though force and displacement are vectors.
• CGS unit of work is the erg; 1 J = 10⁷ erg.
• Power = work/time, SI unit watt (W); 1 horsepower = 746 W.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 3 · Power and its units (watt, horsepower)

Q177. The rate of doing work can be defined as ______.

  1. (a) force
  2. (b) power
  3. (c) work
  4. (d) energy
Answer & solution

Answer: (b) power

Correct Answer: Option 2 — power
Power is defined as the rate of doing work, that is, the work done per unit time:
P = W/t
It can also be written as the rate of transfer of energy, P = E/t. Its SI unit is the watt (W), where 1 watt = 1 joule per second.
Power tells us how fast a job is done, not how big the job is. Two cranes may both lift the same load to the same height — so both do the same work — but the crane that finishes sooner has the greater power.
When a constant force F moves a body with a steady velocity v, power can also be written as P = F × v.
❌ Why the other options are incorrect
✗ force — force is a push or pull, F = ma, measured in newton (N); it has nothing to do with time.
✗ work — work is the product of force and displacement in its direction, W = Fs cos θ, measured in joule (J); it is the job itself, not the rate of doing it.
✗ energy — energy is the capacity to do work, also measured in joule (J); dividing energy by time gives power.
📌 SSC & Railway Exam Facts
• Power P = W/t = F v; SI unit watt (W), named after James Watt.
• 1 kilowatt = 1000 W; 1 horsepower (hp) = 746 W.
• Commercial unit of electrical energy is the kilowatt-hour (kWh); 1 kWh = 3.6 × 10⁶ J.
• Electric power P = VI = I²R = V²/R.
• Power is a scalar quantity.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 3 · Conservation and transformation of energy

Q178. Which of the following statements related to the law of conservation of energy is true?

  1. (a) Energy cannot be converted from one form to another.
  2. (b) Energy can be created or destroyed.
  3. (c) The total energy before and after a transformation process are different.
  4. (d) The law of conservation of energy is valid in all situations and for all kinds of transformations.
Answer & solution

Answer: (d) The law of conservation of energy is valid in all situations and for all kinds of transformations.

Correct Answer: Option 4 — The law of conservation of energy is valid in all situations and for all kinds of transformations.
The law of conservation of energy states that energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy before and after the transformation always remains the same.
This law has been tested in mechanics, heat, light, sound, electricity, chemistry and nuclear physics, and no exception has ever been found. That is why the correct statement is that it is valid in all situations and for every kind of transformation.
Everyday examples: in a bulb, electrical energy → light + heat; in a falling body, potential energy → kinetic energy; in a loudspeaker, electrical energy → sound energy.
❌ Why the other options are incorrect
✗ Energy cannot be converted from one form to another. — this is exactly the opposite of the truth; every machine works by converting one form of energy into another.
✗ Energy can be created or destroyed. — the law says precisely that energy can be neither created nor destroyed.
✗ The total energy before and after a transformation process are different. — the total energy always stays the same; what looks like a loss is usually energy turned into heat or sound.
📌 SSC & Railway Exam Facts
• Energy can neither be created nor destroyed, only transformed — total energy is always conserved.
• Mechanical energy = KE + PE; for a freely falling body the sum stays constant.
• In nuclear reactions mass and energy are linked by E = mc² (Einstein).
• Hydroelectric plant: PE of water → KE → electrical energy.
• SI unit of energy is the joule (J); 1 calorie = 4.2 J.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 3 · Work and its sign

Q179. A book is lying on a table and Rahul moves the book in the direction of the force for some distance D. The work done by the force is:

  1. (a) undetermined
  2. (b) zero
  3. (c) positive
  4. (d) negative
Answer & solution

Answer: (c) positive

Correct Answer: Option 3 — positive
Work done by a force is W = F s cos θ, where θ is the angle between the force and the displacement.
Here the book is moved in the direction of the force through a distance D, so θ = 0° and cos 0° = +1. Therefore
W = F × D × 1 = +FD
which is a positive quantity. Positive work means the force is giving energy to the book, and that energy shows up as the motion of the book across the table.
This is the standard case of positive work: a horse pulling a cart forward, a person pushing a trolley ahead, or gravity acting on a freely falling stone.
❌ Why the other options are incorrect
✗ undetermined — the work is perfectly well defined; F, D and θ = 0° are all known in principle, so W = FD.
✗ zero — work would be zero only if the displacement were zero or the force acted perpendicular to the displacement (θ = 90°).
✗ negative — negative work needs the force to act opposite to the displacement (θ = 180°), as friction does; here the force and displacement point the same way.
📌 SSC & Railway Exam Facts
• W = F s cos θ; SI unit joule (J), 1 J = 1 N m.
• θ = 0° → positive work, θ = 90° → zero work, θ = 180° → negative work.
• Work done against friction is always negative.
• A coolie carrying a load on his head on level ground does zero work against gravity.
• Work–energy theorem: net work done = change in kinetic energy.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 1 · Conservation and transformation of energy

Q180. A ball is dropped from the top of a tower. At the midpoint of the motion of the ball, the energy possessed by the ball is:

  1. (a) Potential
  2. (b) Kinetic
  3. (c) Sound
  4. (d) Both Potential and Kinetic
Answer & solution

Answer: (d) Both Potential and Kinetic

Correct Answer: Option 4 — Both Potential and Kinetic
At the top of the tower the ball is at rest, so it has only potential energy, P.E. = mgh. As soon as it is dropped it starts moving, and part of that potential energy is steadily converted into kinetic energy, K.E. = ½mv².
At the midpoint the ball has already fallen a height h/2, so its potential energy is mg(h/2), and the energy lost has appeared as kinetic energy, K.E. = mg(h/2) as well. Hence at the midpoint the ball possesses both potential and kinetic energy, each exactly half of the initial energy. By the law of conservation of energy the total mechanical energy (K.E. + P.E.) stays equal to mgh throughout the fall.
❌ Why the other options are incorrect
✗ Potential — the ball has potential energy only at the top, where it is still at rest; at the midpoint it is already moving, so it must also have kinetic energy.
✗ Kinetic — the ball possesses only kinetic energy just before striking the ground, where its height (and hence its potential energy) becomes zero.
✗ Sound — sound energy is produced only at the moment of impact with the ground; a ball falling freely through air produces no significant sound energy.
📌 SSC & Railway Exam Facts
• P.E. = mgh, K.E. = ½mv²; both are measured in joule (J).
• Law of conservation of energy — energy can neither be created nor destroyed, only converted from one form into another.
• For a freely falling body, K.E. + P.E. = mgh = constant at every point of the fall.
• Just before hitting the ground the whole energy is kinetic, so v = √(2gh).
• Free fall is independent of mass — a heavy and a light ball dropped together reach the ground at the same time in a vacuum.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 2 · Commercial unit of energy (kWh)

Q181. What is the energy consumed when 1 watt of power is used for 1 hour?

  1. (a) 1 watt hour
  2. (b) 1 kilowatt hour
  3. (c) 60 watt hour
  4. (d) 6 kilowatt hour
Answer & solution

Answer: (a) 1 watt hour

Correct Answer: Option 1 — 1 watt hour
📖 Given
• Power, P = 1 W
• Time, t = 1 hour
✍️ Formula
Energy consumed, E = Power × Time
🔢 Solution
• E = 1 W × 1 h
• E = 1 watt hour (1 Wh)
• In pure SI units, E = 1 W × 3600 s = 3600 J
In plain words, an appliance that draws 1 watt continuously for 1 hour uses up 1 watt hour of electrical energy — this is exactly how the "units" on an electricity bill are built up.
❌ Why the other options are incorrect
✗ 1 kilowatt hour — this is 1000 times bigger; it needs a 1000 W appliance running for 1 hour, not a 1 W one.
✗ 60 watt hour — comes from multiplying by 60 (minutes in an hour) even though the time was already taken in hours.
✗ 6 kilowatt hour — no step of the calculation produces this figure; it is 6000 times the correct answer.
📌 SSC & Railway Exam Facts
• 1 kWh = 1 "unit" of electricity = 3.6 × 10⁶ J.
• Electric power P = VI = I²R = V²/R; SI unit watt (W).
• 1 W = 1 J/s and 1 horsepower (hp) = 746 W.
• The joule (J) is the SI unit of energy; the kilowatt hour is its commercial unit.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 2 · Conservation and transformation of energy

Q182. When an object of mass 'm' falls freely to the ground from a height 'h', at what point is the object's kinetic energy maximum?

  1. (a) At one-fourth way through the fall.
  2. (b) At the start of the fall.
  3. (c) When it is about to hit the ground.
  4. (d) At halfway through the fall.
Answer & solution

Answer: (c) When it is about to hit the ground.

Correct Answer: Option 3 — When it is about to hit the ground.
During a free fall the total mechanical energy is conserved: KE + PE = constant. At the top the body is at rest, so KE = 0 and PE = mgh is maximum. As it falls, the height keeps decreasing and the speed keeps increasing, so potential energy is steadily converted into kinetic energy.
Just before touching the ground the height is almost zero, so PE ≈ 0 and the whole of the initial potential energy has become kinetic energy:
KE(max) = mgh, with speed v = √(2gh).
So the kinetic energy is largest at the instant the body is about to strike the ground, which is exactly why a fall from a greater height hurts more.
❌ Why the other options are incorrect
✗ At one-fourth way through the fall. — after falling h/4 the body has gained only mgh/4 of kinetic energy.
✗ At the start of the fall. — here the velocity is zero, so KE = ½mv² = 0; this is where potential energy is maximum instead.
✗ At halfway through the fall. — at h/2 the kinetic energy is mgh/2, exactly half of its maximum value; at that point KE = PE.
📌 SSC & Railway Exam Facts
• KE = ½mv², PE = mgh; SI unit of both is the joule (J).
• Law of conservation of energy — energy is neither created nor destroyed, only transformed.
• Free-fall speed on hitting the ground, v = √(2gh), is independent of mass.
• KE = p²/2m, where p = mv is the momentum.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 2 · Work and its sign

Q183. The work done (in Joule) by a constant force of 10 N on an object that is displaced by an amount D meter in the direction of the force is:

  1. (a) +10D
  2. (b) −10D
  3. (c) D
  4. (d) 0
Answer & solution

Answer: (a) +10D

Correct Answer: Option 1 — +10D
📖 Given
• Constant force, F = 10 N
• Displacement, s = D metre
• Displacement is in the direction of the force, so the angle θ = 0°
✍️ Formula
W = F × s × cos θ
🔢 Solution
• cos 0° = 1
• W = 10 × D × 1
• W = +10D joule
Because the force and the displacement point the same way, the force is helping the motion, so the work done on the object is positive.
❌ Why the other options are incorrect
✗ −10D — this is the answer only if the force acted opposite to the displacement (θ = 180°, cos 180° = −1), like friction or a braking force.
✗ D — here the 10 N has been dropped; work is force × displacement, not displacement alone.
✗ 0 — work is zero only when the displacement is zero or when the force is perpendicular to the displacement (θ = 90°), neither of which is the case here.
📌 SSC & Railway Exam Facts
• W = F s cos θ; SI unit joule (J), 1 J = 1 N m; work is a scalar.
• θ = 0° ⇒ positive work; θ = 90° ⇒ zero work; θ = 180° ⇒ negative work.
• A coolie carrying a load on his head on level ground does zero work against gravity (θ = 90°).
• Power = Work/Time; SI unit watt (W); 1 W = 1 J/s.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 3 · Work and its sign

Q184. A cart is moving with a uniform velocity and a retarding force F is applied in the opposite direction to stop the cart. The work done by the force is:

  1. (a) positive
  2. (b) undetermined
  3. (c) negative
  4. (d) zero
Answer & solution

Answer: (c) negative

Correct Answer: Option 3 — negative
Work done by a force is W = F s cos θ, where θ is the angle between the force and the displacement. Here the cart moves forward while the retarding force acts backwards, so θ = 180° and cos 180° = −1. Therefore W = −F s, which is negative work.
Negative work means the force is opposing the motion and is taking energy away from the body. That is exactly what happens: the cart's kinetic energy falls from ½mv² to zero, and the magnitude of the negative work equals the kinetic energy destroyed (work–energy theorem). Braking forces and friction are the standard examples of forces that always do negative work on a moving body.
❌ Why the other options are incorrect
✗ positive — positive work needs the force and displacement to point the same way (θ = 0°), as when an engine drives the cart forward. A retarding force does the opposite.
✗ undetermined — nothing is unknown here; the direction of the force relative to the displacement is clearly stated, so the sign of the work is fixed.
✗ zero — work is zero only when the displacement is zero (like pushing an immovable wall) or when the force is perpendicular to the displacement (θ = 90°). The cart does move, and the force is along the line of motion.
📌 SSC & Railway Exam Facts
• W = F s cos θ; SI unit joule (J), 1 J = 1 N m; work is a scalar.
• θ = 0° ⇒ positive work, θ = 90° ⇒ zero work, θ = 180° ⇒ negative work.
• Work–energy theorem: net work done = change in kinetic energy.
• Friction and air resistance always do negative work on a moving body.
• Kinetic energy K = ½mv²; a body at rest has K = 0.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 3 · Work and its sign

Q185. Ramu is working hard to push a huge rock by applying 1N force, but the rock does not move at all. The net work done by Ramu is:

  1. (a) 0 J
  2. (b) −1 J
  3. (c) 1000 J
  4. (d) +1 J
Answer & solution

Answer: (a) 0 J

Correct Answer: Option 1 — 0 J
📖 Given
• Force applied by Ramu, F = 1 N
• The rock does not move, so displacement s = 0 m
✍️ Formula
W = F × s × cos θ
🔢 Solution
• W = 1 × 0 × cos θ
• W = 0 J
In physics, work is done only when the point of application of the force actually moves. Ramu certainly gets tired, because his muscles keep using up chemical energy, but since the rock has no displacement, the work done on the rock is exactly zero. Tiredness is a biological effect, not a measure of physical work.
❌ Why the other options are incorrect
✗ −1 J — this would need a displacement of 1 m against the force; but here there is no displacement at all, so no negative work either.
✗ 1000 J — a large number chosen to match the idea of "working hard"; effort is not work. With s = 0, no amount of force can produce energy transfer.
✗ +1 J — this comes from simply multiplying 1 N × 1 and forgetting that the displacement is zero, not 1 m.
📌 SSC & Railway Exam Facts
• W = F s cos θ; SI unit joule (J), 1 J = 1 N m, CGS unit erg, 1 J = 10⁷ erg.
• Zero work cases: s = 0 (pushing a wall), F = 0, or θ = 90° (a coolie carrying a load on level ground).
• Work is a scalar quantity but can be positive, negative or zero.
• Power = W/t; SI unit watt (W), 1 W = 1 J/s, 1 hp = 746 W.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 3 · Work and its sign

Q186. An object is displaced horizontally because of the applied force. What is revealed by this action?

  1. (a) The force increases the mass of the object.
  2. (b) The force decreases the mass of the object.
  3. (c) Work is done by the force.
  4. (d) The force increases the inertia of the object.
Answer & solution

Answer: (c) Work is done by the force.

Correct Answer: Option 3 — Work is done by the force.
In physics, work is said to be done when a force acts on a body and the body gets displaced along the direction of the force. Both conditions are satisfied here: a force is applied and the object is actually displaced horizontally. Hence work is done by the force, and its amount is W = F × s (with θ = 0°, so cos θ = 1).
Work is the transfer of energy. The energy spent by the agent applying the force reappears as the kinetic energy of the object (and as heat if friction is present). If the object had not moved, the work would have been zero, no matter how large the force.
❌ Why the other options are incorrect
✗ The force increases the mass of the object. — mass is the amount of matter in a body; it does not change when a force is applied (in ordinary, non-relativistic physics).
✗ The force decreases the mass of the object. — equally wrong for the same reason; mass is constant, and a force only changes the body's state of motion.
✗ The force increases the inertia of the object. — inertia is measured by mass. Since the mass is unchanged, the inertia is unchanged too; the force changes velocity, not inertia.
📌 SSC & Railway Exam Facts
• Two conditions for work: a force must act and there must be a displacement along that force.
• W = F s cos θ; SI unit joule (J); 1 J = 1 N m; work is a scalar.
• Work–energy theorem: net work done = change in kinetic energy = ½mv² − ½mu².
• Mass is a measure of inertia and is the same everywhere; weight W = mg changes with place.
• 1 kWh = 3.6 × 10⁶ J; 1 calorie = 4.2 J.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 1 · Numericals on work, energy and power

Q187. The energy possessed by a ball of mass 5 kg when it is raised to a height of 5 m above the ground is (take g = 10 m/ s² ):

  1. (a) 2 J
  2. (b) 250 J
  3. (c) 2.5 J
  4. (d) 10 J
Answer & solution

Answer: (b) 250 J

Correct Answer: Option 2 — 250 J
📖 Given
• Mass of the ball, m = 5 kg
• Height above the ground, h = 5 m
• Acceleration due to gravity, g = 10 m/s²
• To find: the energy possessed by the ball at that height (gravitational potential energy)
✍️ Formula
Potential energy, E(p) = mgh
🔢 Solution
• E(p) = 5 × 10 × 5
• E(p) = 50 × 5
• E(p) = 250 J
In plain language: the work done against gravity in lifting the 5 kg ball through 5 m is 250 joule, and exactly that much energy is stored in the ball as potential energy because of its raised position.
❌ Why the other options are incorrect
✗ 2 J — this looks like 10/5 or a stray division; it ignores the formula mgh altogether.
✗ 2.5 J — obtained by dividing instead of multiplying (5 × 5 ÷ 10); potential energy grows with height, it does not shrink.
✗ 10 J — this is just the numerical value of g, mistaken for an energy.
📌 SSC & Railway Exam Facts
• Potential energy E(p) = mgh; kinetic energy E(k) = ½mv²; SI unit of both is the joule (J).
• 1 J = 1 N m = 1 kg m² s⁻²; the joule is named after James Prescott Joule.
• If this ball were dropped, it would hit the ground with v = √(2gh) = √100 = 10 m/s, its 250 J of PE having become KE.
• Law of conservation of energy: energy can neither be created nor destroyed, only transformed from one form to another.
• Work W = F × s × cos θ; work done is zero when the force is perpendicular to the displacement (θ = 90°).
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 1 · Numericals on work, energy and power

Q188. An object with a mass of 10 kg is moving with an initial velocity of 3 m/s and its final velocity becomes 5 m/s. Then the change in its kinetic energy is:

  1. (a) 50 J
  2. (b) 80 J
  3. (c) 160 J
  4. (d) 40 J
Answer & solution

Answer: (b) 80 J

Correct Answer: Option 2 — 80 J
📖 Given
• Mass of the object, m = 10 kg
• Initial velocity, u = 3 m/s
• Final velocity, v = 5 m/s
• To find: change in kinetic energy, ΔE(k)
✍️ Formula
Kinetic energy, E(k) = ½mv², so ΔE(k) = ½mv² − ½mu² = ½m(v² − u²)
🔢 Solution
• Initial KE = ½ × 10 × (3)² = 5 × 9 = 45 J
• Final KE = ½ × 10 × (5)² = 5 × 25 = 125 J
• ΔE(k) = 125 − 45
• ΔE(k) = 80 J
In plain language: the object's store of motion energy rises from 45 J to 125 J, so 80 J of work must have been done on it — this is the work–energy theorem in action.
❌ Why the other options are incorrect
✗ 50 J — obtained by using the difference of the velocities squared wrongly, e.g. ½ × 10 × (5 − 3)² × 2.5; the velocities must be squared separately.
✗ 160 J — this is double the correct value, from forgetting the factor ½ in ½m(v² − u²).
✗ 40 J — obtained by taking ½ × 10 × (5 − 3)² = 5 × 4 = 20, then doubling, or similar mis-handling; (v − u)² is not the same as v² − u².
📌 SSC & Railway Exam Facts
• E(k) = ½mv²; SI unit joule (J); kinetic energy is a scalar and is always positive.
• Work–energy theorem: work done by the net force = change in kinetic energy.
• KE is proportional to the square of the speed — doubling the speed makes the kinetic energy four times.
• Relation with momentum: E(k) = p²/2m and p = √(2mE(k)).
• Potential energy E(p) = mgh; the total mechanical energy E(k) + E(p) stays constant when only gravity acts.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 1 · Numericals on work, energy and power

Q189. An object A with a mass m and velocity v has a kinetic energy of 100 J. Then the kinetic energy of an object B with a mass of 2 m and velocity v is:

  1. (a) 50 J
  2. (b) 200 J
  3. (c) 400 J
  4. (d) 300 J
Answer & solution

Answer: (b) 200 J

Correct Answer: Option 2 — 200 J
📖 Given
• Object A: mass = m, velocity = v, kinetic energy E(A) = 100 J
• Object B: mass = 2m, velocity = v (the same velocity)
• To find: kinetic energy of B, E(B)
✍️ Formula
Kinetic energy, E(k) = ½mv² — at a fixed velocity, E(k) ∝ m
🔢 Solution
• For A: E(A) = ½ × m × v² = 100 J
• For B: E(B) = ½ × (2m) × v²
• E(B) = 2 × (½ × m × v²)
• E(B) = 2 × 100
• E(B) = 200 J
In plain language: the velocity is unchanged, so only the mass matters — and since the mass is doubled, the kinetic energy is also simply doubled to 200 J.
❌ Why the other options are incorrect
✗ 50 J — this is half of A's energy; it would follow only if the mass were halved, not doubled.
✗ 400 J — this is four times; the KE becomes four times when the velocity is doubled, not the mass.
✗ 300 J — three times the original; this would need a mass of 3m, which is not what is given.
📌 SSC & Railway Exam Facts
• E(k) = ½mv²: E(k) ∝ m at constant v, and E(k) ∝ v² at constant m.
• Doubling the velocity makes the kinetic energy 4 times; tripling it makes the energy 9 times.
• In terms of momentum, E(k) = p²/2m; here B has twice the momentum of A but only twice the energy.
• SI unit of energy is the joule (J); 1 J = 1 N m; 1 kWh = 3.6 × 10⁶ J.
• Work–energy theorem: the work done by the net force equals the change in kinetic energy.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 1 · Commercial unit of energy (kWh)

Q190. What is the value of 1 kWh in Joule(J)?

  1. (a) 3.6 × 10⁶
  2. (b) 3.6
  3. (c) 3.6 × 10⁻⁵
  4. (d) 3.6 × 10⁵
Answer & solution

Answer: (a) 3.6 × 10⁶

Correct Answer: Option 1 — 3.6 × 10⁶
📖 Given
• Energy to convert = 1 kWh (kilowatt-hour)
• 1 kilowatt = 1000 watt = 10³ W
• 1 hour = 60 × 60 = 3600 s
• 1 watt = 1 joule/second, so 1 W s = 1 J
✍️ Formula
Energy = Power × Time, i.e. 1 kWh = 1 kW × 1 h
🔢 Solution
• 1 kWh = 1000 W × 3600 s
• 1 kWh = 3 600 000 W s
• 1 kWh = 3 600 000 J
• 1 kWh = 3.6 × 10⁶ J
In plain language: an appliance of 1000 watt running for one full hour consumes 3.6 million joule of energy — and that is what the electricity meter counts as one unit.
❌ Why the other options are incorrect
✗ 3.6 — this is the value in megajoules (MJ), not in joules; the factor of 10⁶ is missing.
✗ 3.6 × 10⁻⁵ — a negative power of ten would make 1 kWh a tiny amount of energy, which is clearly wrong.
✗ 3.6 × 10⁵ — ten times too small; this comes from using 360 s or 100 W by mistake.
📌 SSC & Railway Exam Facts
• 1 kWh = 1 unit of electricity = 3.6 × 10⁶ J; this is the commercial unit of electrical energy.
• The joule (J) is the SI unit of energy; 1 J = 1 N m = 1 W s.
• Power P = W/t; SI unit watt (W), and 1 horsepower = 746 W.
• Electrical energy consumed = P × t, and electrical power P = VI = I²R = V²/R.
• A 100 W bulb burning for 10 hours consumes 1 kWh, i.e. exactly one unit.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 2 · Numericals on work, energy and power

Q191. If an object with a mass of 0.1 kg moves with a uniform velocity of 2 m s⁻¹, its kinetic energy is ______.

  1. (a) 0.1 joule
  2. (b) 0.2 joule
  3. (c) 0.05 joule
  4. (d) 0.4 joule
Answer & solution

Answer: (b) 0.2 joule

Correct Answer: Option 2 — 0.2 joule
📖 Given
• Mass, m = 0.1 kg
• Velocity, v = 2 m s⁻¹
✍️ Formula
K.E. = ½mv²
🔢 Solution
• K.E. = ½ × 0.1 × (2)²
• K.E. = ½ × 0.1 × 4
• K.E. = 0.05 × 4
• K.E. = 0.2 joule
So a light 100 g body moving at 2 m s⁻¹ carries only 0.2 J of energy of motion.
❌ Why the other options are incorrect
✗ 0.1 joule — this is ½mv (forgetting to square the velocity).
✗ 0.05 joule — this is just ½m, the velocity has not been used at all.
✗ 0.4 joule — this is mv², i.e. the factor ½ has been left out.
📌 SSC & Railway Exam Facts
• Kinetic energy is the energy of a body due to its motion; SI unit joule (J).
• K.E. is directly proportional to mass and to the square of velocity — doubling v makes K.E. four times.
• Potential energy of a raised body = mgh.
• 1 joule = 1 newton-metre = 10⁷ erg; 1 kWh = 3.6 × 10⁶ J.
• Relation with momentum: K.E. = p²/2m, where p = mv.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 2 · Numericals on work, energy and power

Q192. What is the value of power if a man does 50 J of work in 10 s ?

  1. (a) 500 W
  2. (b) 5 W
  3. (c) 0.5 W
  4. (d) 50 W
Answer & solution

Answer: (b) 5 W

Correct Answer: Option 2 — 5 W
📖 Given
• Work done, W = 50 J
• Time taken, t = 10 s
✍️ Formula
P = W/t
🔢 Solution
• P = 50/10
• P = 5 W
In plain words, the man converts 5 joule of energy every second, which is his rate of doing work.
❌ Why the other options are incorrect
✗ 500 W — this is 50 multiplied by 10 instead of divided; multiplying work by time gives no physical quantity here.
✗ 0.5 W — this is 50 divided by 100; the time has been wrongly taken as 100 s.
✗ 50 W — this is the work done, not the power; power is obtained only after dividing by the time.
📌 SSC & Railway Exam Facts
• Power is the rate of doing work; SI unit watt (W), a scalar quantity.
• 1 watt = 1 joule/second; 1 horsepower (hp) = 746 W.
• 1 kilowatt-hour (kWh) = 3.6 × 10⁶ J — this is the commercial unit of electrical energy (1 'unit').
• The unit watt is named after the Scottish engineer James Watt.
• For electricity, P = VI = I²R = V²/R.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 2 · Kinetic energy

Q193. Select the INCORRECT statement from the following options.

  1. (a) The kinetic energy is the energy possessed by an object due to its motion.
  2. (b) The kinetic energy of an object increases with the increase of its mass.
  3. (c) The kinetic energy is the energy possessed by an object due to its position.
  4. (d) The kinetic energy of an object increases with the increase of its velocity.
Answer & solution

Answer: (c) The kinetic energy is the energy possessed by an object due to its position.

Correct Answer: Option 3 — The kinetic energy is the energy possessed by an object due to its position.
This statement is wrong because energy possessed by a body because of its position or configuration is called potential energy, not kinetic energy. A stone lying on a roof, a stretched spring and water stored in a dam all have potential energy (P.E. = mgh).
Kinetic energy is the energy a body has because it is moving, and is given by K.E. = ½mv². Since m and v² both appear in the formula, the kinetic energy grows when either the mass or the velocity grows.
❌ Why the other options are incorrect
✗ The kinetic energy is the energy possessed by an object due to its motion — this is the correct definition of kinetic energy, so it cannot be the answer.
✗ The kinetic energy of an object increases with the increase of its mass — true, because K.E. ∝ m at a fixed speed.
✗ The kinetic energy of an object increases with the increase of its velocity — true, and in fact K.E. ∝ v², so it rises even faster.
📌 SSC & Railway Exam Facts
• K.E. = ½mv² and P.E. = mgh; both are measured in joule (J).
• If velocity is doubled, kinetic energy becomes 4 times; if it is tripled, it becomes 9 times.
• Mechanical energy = K.E. + P.E., and it stays constant for a freely falling body (law of conservation of energy).
• A body of mass m falling from height h has K.E. = mgh just before hitting the ground.
• Work-energy theorem: work done by the net force = change in kinetic energy.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 3 · Potential energy

Q194. What kind of energy is acquired by a stretched rubber band?

  1. (a) Light energy
  2. (b) Thermal energy
  3. (c) Kinetic energy
  4. (d) Potential energy
Answer & solution

Answer: (d) Potential energy

Correct Answer: Option 4 — Potential energy
When a rubber band is stretched, work has to be done against the restoring force of the band. This work is stored inside the band because of the change in its shape (configuration). Energy stored in a body due to its position or configuration is called potential energy, and the form stored in a stretched or compressed elastic body is called elastic potential energy.
The moment the band is released, this stored potential energy changes into kinetic energy and the band — or the stone placed in a catapult — flies off. The same idea explains a wound spring in a clock, a stretched bow and water stored in a dam.
❌ Why the other options are incorrect
✗ Light energy — a stretched rubber band does not glow; light energy comes from luminous sources such as the Sun, a bulb or a flame.
✗ Thermal energy — heat energy is due to the random motion of molecules; only a negligible amount appears here, so it is not the energy acquired by stretching.
✗ Kinetic energy — this is the energy of a body in motion; the stretched band is held at rest, so it has no kinetic energy until it is released.
📌 SSC & Railway Exam Facts
• Potential energy = mgh for a body raised to a height h; SI unit joule (J).
• Elastic potential energy of a spring = ½kx², where k is the spring constant.
• Kinetic energy = ½mv²; a stretched bow, a wound spring and a dam all store potential energy.
• Law of conservation of energy — energy can neither be created nor destroyed, only changed from one form into another.
• 1 joule = 1 N m; the commercial unit of energy is 1 kWh = 3.6 × 10⁶ J.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 3 · Conservation and transformation of energy

Q195. In an electric bulb, the electrical energy is converted into ______ energy.

  1. (a) Sound
  2. (b) Light
  3. (c) Wind
  4. (d) Potential
Answer & solution

Answer: (b) Light

Correct Answer: Option 2 — Light
In an electric bulb the current passing through the thin tungsten filament heats it to about 2500 °C, and the white-hot filament begins to glow. Thus the electrical energy supplied by the supply line is converted into light energy, together with a large amount of heat.
This is the heating and lighting effect of electric current, governed by Joule's law of heating, H = I²Rt. An ordinary incandescent bulb is very inefficient — only about 5 % of the electrical energy becomes light and the rest is wasted as heat; that is why LED and CFL lamps have replaced it.
❌ Why the other options are incorrect
✗ Sound — a bulb produces no sound; electrical energy becomes sound energy in a loudspeaker, an electric bell or a buzzer.
✗ Wind — moving air carries mechanical energy; a bulb creates no air movement. It is an electric fan that turns electrical energy into mechanical energy.
✗ Potential — potential energy depends on position or configuration; a glowing filament stores no such energy.
📌 SSC & Railway Exam Facts
• The filament of a bulb is made of tungsten (melting point 3380 °C) and the bulb is filled with inert gases such as argon and nitrogen.
• Joule's law of heating: H = I²Rt — the electric fuse works on this principle.
• Energy conversions: loudspeaker electrical → sound; electric motor electrical → mechanical; electric cell chemical → electrical.
• Power of a bulb, P = VI = V²/R; SI unit watt (W).
• 1 unit of electricity = 1 kWh = 3.6 × 10⁶ J.
RRB Technician Grade III 2024-25 – 20 December 2024 Shift 3 · Numericals on work, energy and power

Q196. If a machine does 500 J of work in 4 seconds, what is its power output?

  1. (a) 125 watt
  2. (b) 2000 watt
  3. (c) 250 watt
  4. (d) 8000 watt
Answer & solution

Answer: (a) 125 watt

Correct Answer: Option 1 — 125 watt
📖 Given
• Work done by the machine, W = 500 J
• Time taken, t = 4 s
• To find: power output, P
✍️ Formula
Power = Work done / Time taken, i.e. P = W/t
🔢 Solution
• P = 500/4
• P = 125 W (125 watt)
In plain language: the machine delivers 125 joules of energy every second, and that rate of doing work is exactly what we call its power.
❌ Why the other options are incorrect
✗ 2000 watt — obtained by multiplying 500 × 4 instead of dividing; that gives the wrong quantity altogether.
✗ 250 watt — this is 500 ÷ 2, i.e. taking t = 2 s instead of 4 s.
✗ 8000 watt — this comes from multiplying 2000 by 4 again, compounding the earlier mistake.
📌 SSC & Railway Exam Facts
• P = W/t; SI unit watt (W), where 1 W = 1 J/s, named after James Watt.
• 1 horsepower (hp) = 746 W; 1 kilowatt = 1000 W.
• Commercial unit of electrical energy: 1 kWh = 1 unit = 3.6 × 10⁶ J.
• Work W = F × s × cos θ; SI unit joule (J), and 1 J = 1 N m.
• Electrical power can also be written as P = VI = I²R = V²/R.
12

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