The escape velocity from Earth's surface is approximately 11.2 km/s. If a planet has twice Earth's radius and same density, its escape velocity will be: MCQ with Answer and Explanation
The escape velocity from Earth's surface is approximately 11.2 km/s. If a planet has twice Earth's radius and same density, its escape velocity will be:
A. 22.4 km/s
B. 5.6 km/s
C. 15.8 km/s
D. 11.2 km/s
Answer: Option A
Solution (By JKExamLibrary)
Escape velocity v_e = √(2GM/R). Mass M = density ρ × volume = ρ×(4/3)πR³. Thus v_e = √(2Gρ×4/3πR³ / R) = √(8GρπR²/3) ∝ R√ρ. Given same density and R_planet = 2R_earth, v_e ∝ R, so v_e_planet = 2 × 11.2 = 22.4 km/s. This derivation shows escape velocity scales linearly with radius for constant density. Memory tip: v_e ∝ √(M/R) and M ∝ R³ for constant ρ, so v_e ∝ R. Such proportional reasoning questions test conceptual grasp of gravitation formulas in competitive exams without heavy calculation.
Explanation:
Photoelectric equation: work function φ = hν₀, where ν₀ is threshold frequency. Thus ν₀ = φ/h = 2.3 eV / (4.14×10⁻¹⁵ eV·s) ≈ 5.555×10¹⁴ Hz ≈ 5.6×10¹⁴ Hz. This direct application tests photoelectric effect fundamentals. Memory aid: 'ν₀ = φ/h; higher work function ⇒ higher threshold frequency'. Competitive exams frequently provide h in eV·s for such calculations. Always ensure units match: eV for φ and h to avoid conversion errors. This problem assesses numerical proficiency in modern physics applications.
Explanation:
Sir C. V. Raman was awarded the Nobel Prize in Physics in 1930 for discovering the 'Raman Effect'. The Raman effect involves the inelastic scattering of photons by molecules, resulting in a change in wavelength. This discovery proved that light has particle-like characteristics. National Science Day in India (Feb 28) commemorates this discovery.
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