NEET Physics: Atoms and Nuclei — MCQ Test 18

Q6. Why is the nuclear density nearly constant across all nuclei?

Q7. What is the primary challenge in achieving controlled thermonuclear fusion?

Q8. A nucleus has a mass defect of \( 0.05 \, \text{u} \). What is its binding energy in MeV? (Given \( 1 \, \text{u} = 931.5 \, \text{MeV/c}^2 \))

Q9. What is the approximate radius of a nucleus with mass number 64? (Given \( R_0 = 1.2 \times 10^{-15} \, \text{m} \))

Q10. A nucleus with mass number 50 has a binding energy of \( 425 \, \text{MeV} \). What is its binding energy per nucleon?

NEET Physics — Atoms and Nuclei

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What is the nuclear density of a nucleus with mass \( 1.66 \times 10^{-27} \, \text{kg} \) and radius \( 1.5 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

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About this NEET MCQ Test

This page provides a free online MCQ test for NEET Physics preparation, focusing specifically on the chapter Atoms and Nuclei. It contains 10 carefully selected multiple-choice questions (MCQs) designed to test your conceptual understanding and exam readiness.

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Key concepts covered in this specific test include Stability, Extremely, Temperatures, Equivalent, Constant, and Challenge.

Taking these timed MCQ tests helps you simulate the real NEET exam environment. Detailed step-by-step explanations are provided for every question to help you learn from your mistakes and strengthen your fundamentals.

Preview of Questions in this Test

Q1. What is the nuclear density of a nucleus with mass \( 1.66 \times 10^{-27} \, \text{kg} \) and radius \( 1.5 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

  • A. 1.0 × 10¹⁷ kg/m³
  • B. 1.1 × 10¹⁷ kg/m³
  • C. 1.2 × 10¹⁷ kg/m³
  • D. 1.18 × 10¹⁷ kg/m³

Q2. Why is the binding energy per nucleon a useful measure of nuclear stability?

  • A. Measures nuclear radius
  • B. Indicates nuclear density
  • C. Shows Coulomb force strength
  • D. Average energy per nucleon

Q3. Why does nuclear fusion require extremely high temperatures?

  • A. To increase nuclear density
  • B. To reduce binding energy
  • C. To overcome electrostatic repulsion
  • D. To enhance nuclear force

Q4. What is the energy equivalent of \( 0.15 \, \text{g} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

  • A. 1.2 × 10¹³ J
  • B. 1.3 × 10¹³ J
  • C. 1.4 × 10¹³ J
  • D. 1.35 × 10¹³ J

Q5. What is the energy equivalent of \( 0.05 \, \text{g} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

  • A. 4.0 × 10¹² J
  • B. 4.5 × 10¹² J
  • C. 5.0 × 10¹² J
  • D. 5.5 × 10¹² J

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