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Elementary ideas of work, power, and energy

Chapter: Mechanics

Question 1 of 5 NDA MCQ

 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

Detailed Explanation

Correct Answer

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

📖 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.

🧪 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.

🔍 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.

⚡ 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.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

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

Question 2 of 5 NDA MCQ

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?

Detailed Explanation

Correct Answer

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

📖 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:

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

🧪 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.

📌 Hints / Properties Used:
  • Moment of Inertia of a disc: \(I = \frac{1}{2} m r^2\)
  • Moment of Inertia of a sphere: \(I = \frac{2}{5} m r^2\)
📊 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.

📖 Factual Verification & Reference:

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

Question 3 of 5 NDA MCQ

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

Detailed Explanation

Correct Answer

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

📖 Governing Law / Core Concept

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

\[ p = m \cdot v \]

🧪 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:

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

🔍 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.

⚡ 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.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

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

Question 4 of 5 NDA MCQ

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?

Detailed Explanation

Correct Answer

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

📖 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.

🧪 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.

🔍 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

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

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

Question 5 of 5 NDA MCQ

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

Detailed Explanation

Correct Answer

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

📖 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.

🧪 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).

🔍 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.

⚡ 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.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

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

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