Physics

Everyday physics applications

Chapter: Modern Physics & Applications

Question 1 of 5 NDA MCQ

Which one of the following types of radiations has the smallest wavelength ?

Detailed Explanation

Correct Answer

โœ“ Option (d) is correct. X-rays have the smallest wavelength among the given options.

๐Ÿ“– Governing Law / Core Concept

The relationship between the wavelength (\( \lambda \)) and the frequency (\( f \)) of electromagnetic radiation is given by the equation:
\[ c = \lambda f \] \
where: - \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)), - \( \lambda \) is the wavelength in meters, - \( f \) is the frequency in hertz.

๐Ÿงช Step-by-Step Breakdown

For electromagnetic spectrum classification: - Microwaves: \( \lambda \approx 1 \, \text{mm} - 1 \, \text{m} \) - Infra-red: \( \lambda \approx 700 \, \text{nm} - 1 \, \text{mm} \) - Visible light: \( \lambda \approx 400 \, \text{nm} - 700 \, \text{nm} \) - X-rays: \( \lambda \approx 0.01 \, \text{nm} - 10 \, \text{nm} \) Observing the above ranges, it is clear that: \[ \lambda_{X-rays} < \lambda_{Visible} < \lambda_{Infra-red} < \lambda_{Microwaves} \]

๐Ÿ” Option Analysis

- **A: Microwaves** - Longer wavelengths, leading to lower frequencies. - **B: Infra-red** - Wavelength range still longer than X-rays. - **C: Visible light** - Wavelength is shorter than microwaves and infra-red but longer than X-rays. - **D: X-rays** - Shortest wavelengths in this range, therefore having the highest frequency.

โšก Mnemonic / Speed-Run

Remember the mnemonic: **MIV-X** for frequency order: **Microwaves** > **Infra-red** > **Visible** > **X-rays**.

6. Visual Suggestion

6. Visual Suggestion

Create a schematic chart comparing the electromagnetic spectrum, illustrating the wavelength ranges of microwaves, infrared, visible light, and X-rays. Include arrows leading to a diagram of the electromagnetic spectrum to emphasize the increasing frequency and decreasing wavelength.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 8 (Electromagnetic Waves)

Question 2 of 5 NDA MCQ

The energy, E, of a photon can be expressed as E = hf, where f is the frequency and h is Planck’s constant. The dimensions of h are the same as that of

Detailed Explanation

Correct Answer

โœ“ Option B is correct. The dimensions of Planck's constant h match those of angular momentum.

๐Ÿ“– Core Concept

The energy of a photon is given by the equation E = hf, where \( E \) is energy, \( h \) is Planck's constant, and \( f \) is frequency. The equation reveals that Planck's constant functions as a bridge between energy and frequency and has dimensions equivalent to angular momentum.

๐Ÿงช Step-by-Step Breakdown

The dimension analysis begins with:

\[ [E] = [h][f] \]

Given that:

\[ [E] = [ML^2T^{-2}] \]
\[ [f] = [T^{-1}] \]

Thus, the dimensions of Planck's constant \( h \) can be derived:

\[ [h] = [E][T] = [ML^2T^{-2}][T] = [ML^2T^{-1}] \]

This result corresponds explicitly with the dimensions of angular momentum, which is also expressed as \( [ML^2T^{-1}] \).

๐Ÿ” Option Analysis

  • Option A: Linear momentum has dimensions of \( [MLT^{-1}] \), which is distinct from those of \( h \).
  • Option C: Displacement dimensions are \( [L] \); these do not align with \( h \).
  • Option D: Torque dimensions are \( [ML^2T^{-2}] \); these are not equivalent to \( h \).

โšก Mnemonic / Speed-Run

To recall, remember that Planck's constant \( h \) relates energy to frequency, with dimensions linked to angular momentum denoted as \( [ML^2T^{-1}] \).

๐Ÿ“– Visual Suggestion

6. Visual Suggestion

Create a diagram illustrating the equation \( E = hf \) with dimensions labeled. Include angular momentum \( [ML^2T^{-1}] \) to visualize the relationship clearly.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against standard textbooks of Quantum Mechanics and NCERT Class 12 Physics, Chapter on Quantum Physics.

Question 3 of 5 NDA MCQ

Which one of the following radiations has longer wavelength?

Detailed Explanation

Correct Answer

โœ“ Option (c) is correct. Microwaves have longer wavelengths than Infra-red, Ultraviolet, and X-rays.

๐Ÿ“– Governing Law / Core Concept

The electromagnetic spectrum orders radiation types based on wavelength. The order from shortest to longest wavelength is:

  • X-rays
  • Ultraviolet
  • Infra-red
  • Microwaves

๐Ÿงช Step-by-Step Breakdown

Radiation types and their approximate wavelength ranges are:

  • X-rays: \(10^{-11} \text{ m} \) to \(10^{-8} \text{ m} \)
  • Ultraviolet: \(10^{-8} \text{ m} \) to \(10^{-7} \text{ m} \)
  • Infra-red: \(10^{-7} \text{ m} \) to \(10^{-4} \text{ m} \)
  • Microwaves: \(10^{-4} \text{ m} \) to \(10^{-1} \text{ m} \)

The relation shows that:

\[ \lambda_{\text{microwaves}} > \lambda_{\text{IR}} > \lambda_{\text{UV}} > \lambda_{\text{X-rays}} \]

๐Ÿ” Option Analysis

Options A, B, and D all represent wavelengths shorter than microwaves:

  • A: X-rays - shortest wavelength
  • B: Ultraviolet - longer than X-rays but shorter than IR
  • D: Infra-red - longer than UV but shorter than microwaves

โšก Mnemonic / Speed-Run

Remember the order: "X-U-IR-M" (X-rays, Ultraviolet, Infra-red, Microwaves) helps to quickly recall increasing wavelength.

6. Visual Suggestion

6. Visual Suggestion

A schematic diagram showing the electromagnetic spectrum, labeled from shortest to longest wavelength, with highlighted regions for X-rays, Ultraviolet, Infra-red, and Microwaves, making it easy to visualize the relationships between them.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class 12 Physics, Chapter on Electromagnetic Waves.

Question 4 of 5 NDA MCQ

Which of the following pairs of physical phenomenon and the discoverer is/are correctly matched?

(1) James Chadwick : Photoelectric effect

(2) Albert Einstein : Neutron

 (3) Marie Curie : Radium

 Select the correct answer using the code given below:

Detailed Explanation

Correct Answer

โœ“ Option D is correct. Marie Curie is matched correctly with Radium.

๐Ÿ“– Governing Law / Core Concept

The discovery of radiation in the late 19th century led to significant advancements in nuclear physics. Marie Curie's work focused on the isolation of radioactive isotopes, specifically Radium, a fundamental element in the understanding of radioactivity.

๐Ÿงช Step-by-Step Breakdown

1. **Pairings Evaluation**: - **(1) James Chadwick : Photoelectric effect**: Incorrect. Chadwick discovered the neutron (1932). - **(2) Albert Einstein : Neutron**: Incorrect. Einstein's contribution was in photoelectric effect explanation (1905). - **(3) Marie Curie : Radium**: Correct. Curie discovered radium in 1898. Given the above evaluations, only the third pairing is accurate.

๐Ÿ” Option Analysis

- **Option A (1, 2 and 3)**: Incorrect. Two pairs are wrong. - **Option B (1 and 2 only)**: Incorrect. Both listed are wrong. - **Option C (2 and 3 only)**: Incorrect. Only (3) is correct. - **Option D (3 only)**: Correct. Only Curie and Radium is a valid pairing.

โšก Mnemonic / Speed-Run

Remember "C for Curie, R for Radium." To differentiate correct from incorrect pairs, recall: Chadwick = Neutron, Einstein = Photoelectric.

6. Visual Suggestion

6. Visual Suggestion

A flowchart depicting the contributions of Chadwick, Einstein, and Curie to their respective areas, highlighting the correct pairing of Curie and Radium, with incorrect links shown in red.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class 12 Physics Chapter on Nuclear Physics and Radioactivity.

Question 5 of 5 NDA MCQ

Nuclear energy is generated by

Detailed Explanation

Correct Answer

โœ“ Option (a) is correct. Nuclear energy is primarily generated by nuclear fission, and its expression was proposed by Einstein.

๐Ÿ“– Governing Law / Core Concept

The reaction enabling nuclear energy is governed primarily by the principle of nuclear fission, which is expressed mathematically through Einstein's equation:
\[ E = mc^2 \]
where \( E \) is the energy released, \( m \) is the mass lost, and \( c \) is the speed of light in a vacuum (approximately \( 3 \times 10^8 \, \text{m/s} \)).

๐Ÿงช Step-by-Step Breakdown

In nuclear fission, a heavy nucleus (like Uranium-235) absorbs a neutron and becomes unstable. This instability leads to the nucleus splitting into smaller, stable nuclei, releasing energy and additional neutrons in the process: \[ \text{Uranium-235} + \text{neutron} \rightarrow \text{Barium} + \text{Krypton} + 3 \text{neutrons} + \text{energy} \] The energy released during this reaction can be quantified using \( E = mc^2 \), which denotes the conversion of mass to energy.

๐Ÿ” Option Analysis

- **Option (b)**: Incorrect. Rutherford's work primarily focused on the discovery of the nucleus, not on fission. - **Option (c)**: Incorrect. Fusion, the process of combining lighter nuclei, was not proposed by Bohr in relation to fission. - **Option (d)**: Incorrect. Heisenberg's contributions centered on quantum mechanics, without specific reference to fusion.

โšก Mnemonic / Speed-Run

Remember "Fission Flares" for nuclear fission and "E=mcยฒ" for energy-mass equivalence.

6. Visual Suggestion

A schematic diagram of nuclear fission depicting Uranium-235, the incoming neutron, and the resulting products of the reaction along with the energy release would greatly aid in understanding the process.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter on Nuclear Physics.

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