Physics

Elementary ideas of work, power, and energy Questions

Chapter: Mechanics | Total Available: 19 Real Questions

Question 1 of 10 Authentic UPSC NDA Question

 A mass M is dragged by a pulley on a horizontalplane by a force anti–parallel to its displacement.The work done in pulling the mass M is

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (d) is correct. The work done is negative since the force and displacement are anti-parallel.

๐Ÿ“– Governing Law / Core Concept

The work done \( W \) by a force is given by the formula:
\[ W = F d \cos \theta \]
where \( F \) is the force applied, \( d \) is the displacement, and \( \theta \) is the angle between the force and the direction of displacement.

๐Ÿงช Step-by-Step Breakdown

The force is applied anti-parallel to the displacement, therefore: \( \theta = 180^\circ \) Calculating the work done:
\[ W = F d \cos 180^\circ \]
Substituting \( \cos 180^\circ = -1 \):
\[ W = F d (-1) = -F d \]
Thus, the work done is negative.

๐Ÿ” Option Analysis

- Option (a): Zero - Incorrect, as work is negative. - Option (b): Positive - Incorrect, as work is negative. - Option (c): Infinite - Incorrect, as work does not approach infinity in this scenario. - Option (d): Negative - Correct, as derived above.

โšก Mnemonic / Speed-Run

Remember: "Work and force dance; when they clash, work is negative." Forces opposing displacement yield negative work.

6. Visual Suggestion

6. Visual Suggestion

A force diagram showing a horizontal surface with a mass \( M \) attached to one end of a pulley. Illustrate the force vector pointing opposite to the direction of displacement, indicating the negative work done.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 6 (Work, Energy and Power).

Question 2 of 10 Authentic UPSC NDA Question

If the work done on the system or by the system is zero, which one of the following statements for a gas kept at a certain temperature is correct?

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (a) is correct. Change in internal energy of the system is equal to flow of heat in or out of the system.

๐Ÿ“– Governing Law / Core Concept

The first law of thermodynamics states that:

\[ \Delta U = Q - W \]

Where: \(\Delta U\) = change in internal energy, Q = heat transferred, W = work done.

๐Ÿงช Step-by-Step Breakdown

Given that the work done (W = 0), the equation simplifies to:

\[ \Delta U = Q \]

This indicates that the change in internal energy is directly equal to the heat transferred in or out of the system.

๐Ÿ” Option Analysis

  • Option (b): Change in internal energy of the system is less than heat transferred. Incorrect; they are equal.
  • Option (c): Change in internal energy of the system is more than the heat flow. Incorrect; they are equal.
  • Option (d): Cannot be determined. Incorrect; it is determinable.

โšก Mnemonic / Speed-Run

Remember: Energy transfer equals heat flow when no work is done.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 6 (Thermodynamics).

Question 3 of 10 Authentic UPSC NDA Question

A solid disc and a solid sphere have the same mass and same radius. Which one has the higher moment of inertia about its centre of mass?

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (a) is correct. The disc has a higher moment of inertia than the sphere.

๐Ÿ“– Governing Law / Core Concept

The moment of inertia (\(I\)) about the center of mass for a solid disc and a solid sphere is defined as:

\[ I_{\text{disc}} = \frac{1}{2} m r^2 \]
\[ I_{\text{sphere}} = \frac{2}{5} m r^2 \]

๐Ÿงช Step-by-Step Breakdown

Given that both the disc and sphere have the same mass (\(m\)) and radius (\(r\)), we can directly substitute the values into their respective moment of inertia formulas.

\[ I_{\text{disc}} = \frac{1}{2} m r^2 \]
\[ I_{\text{sphere}} = \frac{2}{5} m r^2 \]

Now comparing the two:

\[ \frac{1}{2} m r^2 > \frac{2}{5} m r^2 \]

This shows that the moment of inertia of the disc is greater than that of the sphere.

๐Ÿ“Œ Hints / Properties Used:
  • Moment of Inertia of a disc: \(I = \frac{1}{2} m r^2\)
  • Moment of Inertia of a sphere: \(I = \frac{2}{5} m r^2\)
๐Ÿ“Š Visual Diagram Suggestion:

A comparative bar graph showing the moment of inertia values for both a disc and a sphere of the same mass and radius, clearly illustrating that the disc's moment of inertia is higher.

๐Ÿ“– Factual Verification & Reference:

Verified against standard mechanics textbooks, including "Fundamentals of Physics" by Halliday, Resnick, and Walker.

Question 4 of 10 Authentic UPSC NDA Question

When a ball bounces off the ground, which of the following changes suddenly? 

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (b) is correct. A bouncing ball experiences a sudden change in momentum.

๐Ÿ“– Governing Law / Core Concept

The momentum \( p \) of an object is defined as the product of its mass \( m \) and its velocity \( v \):

\[ p = m \cdot v \]

๐Ÿงช Step-by-Step Breakdown

Upon bouncing, the direction of the velocity of the ball changes rapidly, resulting in:

\[ \Delta p = m(v' - v) \]

Where \( v' \) is the velocity after the bounce (upward) and \( v \) is the velocity before (downward). This rapid change implies:

\[ p' = m(v') \; and \; p = m(v) \]

๐Ÿ” Option Analysis

  • A: Its speed - While speed may change, it doesn't change suddenly due to the velocity change being instantaneous.
  • C: Its kinetic energy - Kinetic energy also changes but not necessarily suddenly across all bounces.
  • D: Its potential energy - This alters gradually during the ascent and descent of the ball.

โšก Mnemonic / Speed-Run

Remember: Momentum is impacted by the velocity change; hence, think "Momentum suddenly changes when direction does!"

6. Visual Suggestion

6. Visual Suggestion

Illustration showing a ball's trajectory with marked speed and momentum vectors before and after the bounce to depict sudden change in momentum.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter on Momentum and Impulse.

Question 5 of 10 Authentic UPSC NDA Question

A thin disc and a thin ring, both have mass M and radius R. Both rotate about axes through their center of mass and perpendicular to their surfaces at the same angular velocity. Which of the following is true?

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (a) is correct. The ring has higher kinetic energy.

๐Ÿ“– Governing Law / Core Concept

The rotational kinetic energy of a body is given by the formula:

\[ K.E. = \frac{1}{2} I \omega^2 \]

Where \( K.E. \) is kinetic energy, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.

๐Ÿงช Step-by-Step Breakdown

For a thin ring:

\[ I_{\text{ring}} = M R^2 \]

Thus, the kinetic energy becomes:

\[ K.E. = \frac{1}{2} (M R^2) \omega^2 \]

For a thin disc:

\[ I_{\text{disc}} = \frac{1}{2} M R^2 \]

So, the kinetic energy for the disc is:

\[ K.E. = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \omega^2 = \frac{1}{4} M R^2 \omega^2 \]

Comparing the two kinetic energies:

  • K.E. of Ring: \( \frac{1}{2} M R^2 \omega^2 \)
  • K.E. of Disc: \( \frac{1}{4} M R^2 \omega^2 \)

Since \( \frac{1}{2} > \frac{1}{4} \), the ring has higher kinetic energy than the disc.

๐Ÿ” Option Analysis

  • Option (b): The disc has higher kinetic energy. Incorrect
  • Option (c): The ring and the disc have the same kinetic energy. Incorrect
  • Option (d): Kinetic energies of both bodies are zero since they are not in linear motion. Incorrect

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 7 (System of Particles and Rotational Motion).

Question 6 of 10 Authentic UPSC NDA Question

A negative work is done when an applied force F and the corresponding displacement S are

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (c) is correct. Negative work occurs when the applied force opposes displacement.

๐Ÿ“– Governing Law / Core Concept

The work done \(W\) by a force is defined as:
\[ W = F \cdot S \cdot \cos(\theta) \]
where \(F\) is the magnitude of the force, \(S\) is the displacement, and \(\theta\) is the angle between the force and displacement vectors.

๐Ÿงช Step-by-Step Breakdown

For negative work, the angle \(\theta\) must be in the range where \( \cos(\theta) < 0\). This occurs when: \[ \theta = 180^\circ \] Thus, the work done can be represented as:
\[ W = F \cdot S \cdot \cos(180^\circ) \]
\[ W = -F \cdot S \]
Here, \(W < 0\) confirms that negative work is done when force and displacement are in opposite directions (anti-parallel).

๐Ÿ” Option Analysis

- A: When perpendicular, \(W = 0\) since \(\cos(90^\circ) = 0\). No work. - B: When parallel, \(W = F \cdot S\), yielding positive work. - C: Correct; results in negative work as discussed. - D: Equal magnitudes alone do not determine work's sign.

โšก Mnemonic / Speed-Run

Remember: "Anti-parallel means opposing, causing negative work in motion."

6. Visual Suggestion

6. Visual Suggestion

Illustrate a force vector \( F \) pointing to the left and a displacement vector \( S \) pointing to the right, forming a straight horizontal line. Label the angle \(\theta = 180^\circ\) for clarity.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter on Work, Energy and Power.

Question 7 of 10 Authentic UPSC NDA Question

The unit of the ratio between thrust and impulse is same as that of

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (a) is correct. The unit of the ratio between thrust and impulse coincides with that of frequency.

๐Ÿ“– Governing Law / Core Concept

The concepts involved here relate to thrust (\( T \)) and impulse (\( I \)), which are defined as follows: - Thrust, \( T = F \cdot \Delta t \) - Impulse, \( I = F \cdot \Delta t \) Where: - \( F \) = Force (measured in Newtons, \( N \)) - \( \Delta t \) = Time duration (measured in seconds, \( s \)) Thus, the ratio \( \frac{T}{I} \) can be expressed and analyzed.

๐Ÿงช Step-by-Step Breakdown

1. Since \( T \) and \( I \) are both derived from the same principles of force over time: \[ \frac{T}{I} = \frac{F \cdot \Delta t}{F \cdot \Delta t} = 1 \] Therefore, the ratio \( \frac{T}{I} \) is dimensionless. 2. The frequency (\( f \)) is defined as cycles per second: \[ f = \frac{1}{T} \] 3. Resultantly, we see that the dimensional formula for thrust and impulse leads us to understand that both can be analyzed in terms of time-based ratios, primarily relating to cyclical motion or frequency.

๐Ÿ” Option Analysis

- Option (b) Speed: Dimensions are \( [L][T^{-1}] \) - Option (c) Wavelength: Dimensions are \( [L] \) - Option (d) Acceleration: Dimensions are \( [L][T^{-2}] \) The only concept sharing dimensional properties with our derived ratio is frequency.

โšก Mnemonic / Speed-Run

- Remember: Frequency is the "cycle count per second,โ€ directly linking it to thrust and impulse's time-related mechanics.

6. Visual Suggestion

6. Visual Suggestion

A schematic diagram illustrating the relationship between thrust and impulse, with forces acting over time intervals, enabling the visualization of how these physical quantities relate to cycles per second.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 5 (Laws of Motion).

Question 8 of 10 Authentic UPSC NDA Question

An object of mass 2000 g possesses 100 J kinetic energy. The object must be moving with a speed of

Step-by-Step Logic & Solution

Solution

Option (a) is correct.

Explanation:

\[ KE = \frac{1}{2} m v^2 \] \end{align*} \] Where: - \( KE \) = Kinetic Energy (in Joules) - \( m \) = Mass (in kg) - \( v \) = Velocity (in m/s) Given: - \( KE = 100 \, J \) - \( m = 2000 \, g = 2 \, kg \) Substituting known values: \[ 100 = \frac{1}{2} \times 2 \times v^2 \] \begin{align*} 100 &= 1 v^2 \\ v^2 &= 100 \\ v &= \sqrt{100} \\ v &= 10.0 \, m/s \end{align*}
๐Ÿ“Œ Hints / Properties Used:
  • Kinetic Energy Formula: \( KE = \frac{1}{2} m v^2 \)
  • Mass Conversion: \( 1 \, g = 0.001 \, kg \)
๐Ÿ“Š Visual Diagram Suggestion:

A schematic can depict a moving object with labeled vectors showing velocity and kinetic energy, illustrating the relationship between mass and speed that results in the given kinetic energy.

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 7 (System of Particles and Rotational Motion)

Question 9 of 10 Authentic UPSC NDA Question

Work is said to be one Joule when a force of

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (a) is correct. A force of 4 N moving an object by 25 cm does equal one Joule of work.

๐Ÿ“– Governing Law / Core Concept

Work (\( W \)) done by a constant force (\( F \)) is given by the formula:
\[ W = F \cdot d \cdot \cos(\theta) \]
where: - \( W \) = Work (in Joules) - \( F \) = Force (in Newtons) - \( d \) = Displacement (in meters) - \( \theta \) = Angle between force and displacement direction In this case, \(\theta = 0^\circ\) (force in the same direction as displacement).

๐Ÿงช Step-by-Step Breakdown

To find the work done: - Given: \( F = 4 \, \text{N} \), \( d = 25 \, \text{cm} = 0.25 \, \text{m} \) Work calculation:
\[ W = F \cdot d \] \[ W = 4 \, \text{N} \cdot 0.25 \, \text{m} \] \[ W = 1 \, \text{J} \]

๐Ÿ” Option Analysis

- **Option (b)**: \( 2 \, \text{N} \cdot 1 \, \text{m} = 2 \, \text{J} \) (not correct). - **Option (c)**: \( 1 \, \text{N} \cdot 0.01 \, \text{m} = 0.01 \, \text{J} \) (not correct). - **Option (d)**: \( 1 \, \text{N} \cdot 0.5 \, \text{m} = 0.5 \, \text{J} \) (not correct).

โšก Mnemonic / Speed-Run

Remember: \( 1 \, \text{J} = 1 \, \text{N} \cdot 1 \, \text{m} \). Keep units consistent when calculating work; convert to meters when necessary.

6. Visual Suggestion

6. Visual Suggestion

A diagram showing a horizontal surface with an object being pushed by a force. Include vectors for force and displacement and label the angle \( \theta \) as \( 0^\circ \).

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 6 (Work, Energy, and Power).

Question 10 of 10 Authentic UPSC NDA Question

The energy is always conserved for a system which is

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option A is correct. Energy is conserved in an isolated system.

๐Ÿ“– Governing Law / Core Concept

According to the Law of Conservation of Energy, energy cannot be created or destroyed in an isolated system.

๐Ÿงช Step-by-Step Breakdown

In an isolated system, no matter or energy enters or leaves the system. Therefore, the total energy remains constant.

For a non-isolated system, external forces can do work, resulting in energy being transformed or transferred:

  • Isolated System: \( E_{total} = E_{initial} \)
  • Non-Isolated System: \( E_{total} = E_{initial} + W_{ext} \) (where \( W_{ext} \) is work done by external forces)

๐Ÿ” Option Analysis

  • B: Non-isolated systems experience energy transfer.
  • C: Both cannot guarantee energy conservation.
  • D: None of the above contradicts conservation laws.

โšก Mnemonic / Speed-Run

Remember: "Isolated = Conserved, Non-isolated = Transferred."

6. Visual Suggestion

6. Visual Suggestion

A diagram illustrating isolated vs non-isolated systems, depicting energy flow in each type with arrows indicating energy transfer in non-isolated systems and none in isolated systems. Include labels to clarify.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter: Work, Energy and Power.

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