The process of emission of electrons from a metal surface by heating is MCQ with Answer and Explanation
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Question #1
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The kinetic energy of a body is increased by 300%. The percentage increase in its momentum is:
Correct Answer: Option D
Explanation:
KE ∝ p² (since KE = p²/(2m)). If KE increases by 300%, new KE = 4× original KE. Thus p_new² / p_original² = 4 ⇒ p_new / p_original = 2 ⇒ momentum doubles, i.e., 100% increase. Memory tip: 'KE ∝ p²; so %Δp = ½ %ΔKE for small changes, but for large changes use ratio'. This proportional reasoning problem tests energy-momentum relationship, frequently appearing in competitive exams. Always derive from fundamental relations when percentages are large; linear approximations fail here.
KE ∝ p² (since KE = p²/(2m)). If KE increases by 300%, new KE = 4× original KE. Thus p_new² / p_original² = 4 ⇒ p_new / p_original = 2 ⇒ momentum doubles, i.e., 100% increase. Memory tip: 'KE ∝ p²; so %Δp = ½ %ΔKE for small changes, but for large changes use ratio'. This proportional reasoning problem tests energy-momentum relationship, frequently appearing in competitive exams. Always derive from fundamental relations when percentages are large; linear approximations fail here.
Question #2
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According to Kepler's Third Law of Planetary Motion, the square of the time period of revolution of a planet around the Sun is directly proportional to:
Correct Answer: Option B
Explanation:
Kepler's Third Law (Law of Periods) states that the square of the time period (T^2) of any planet is proportional to the cube of the semi-major axis (r^3) of its orbit. Mathematically, T^2 is proportional to r^3. This law helps in calculating orbital periods or distances of planets and satellites.
Kepler's Third Law (Law of Periods) states that the square of the time period (T^2) of any planet is proportional to the cube of the semi-major axis (r^3) of its orbit. Mathematically, T^2 is proportional to r^3. This law helps in calculating orbital periods or distances of planets and satellites.
Question #3
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The refractive index of a medium is 1.5. The speed of light in this medium is:
Correct Answer: Option D
Explanation:
Refractive index n = c/v, where c = 3×10⁸ m/s (vacuum speed). Thus v = c/n = 3×10⁸ / 1.5 = 2×10⁸ m/s. This direct calculation tests optics fundamentals. Memory aid: 'Higher n ⇒ slower light; v = c/n'. Competitive exams frequently test this with common refractive indices. Always use c = 3×10⁸ m/s unless specified; verify units (m/s). This problem assesses basic formula application skills essential for wave optics sections.
Refractive index n = c/v, where c = 3×10⁸ m/s (vacuum speed). Thus v = c/n = 3×10⁸ / 1.5 = 2×10⁸ m/s. This direct calculation tests optics fundamentals. Memory aid: 'Higher n ⇒ slower light; v = c/n'. Competitive exams frequently test this with common refractive indices. Always use c = 3×10⁸ m/s unless specified; verify units (m/s). This problem assesses basic formula application skills essential for wave optics sections.
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