The value of universal gravitational constant G is
A. 6.67 × 10⁻¹¹ N m²/kg²
B. Both A and C
C. 6.67 × 10⁻¹¹ dyne cm²/g²
D. 9.8 m/s²
Answer: Option B
Solution (By JKSSB Mock Tests)
G = 6.67 × 10⁻¹¹ N m²/kg² in SI, and 6.67 × 10⁻⁸ dyne cm²/g² in CGS. Both are correct. Important constant, measured by Cavendish. g is acceleration due to gravity, different.
Explanation:
Bohr model: radius r_n = (4πε₀ħ²n²)/(m_e e²) = n² a₀, where a₀ is Bohr radius (≈0.529 Å). Thus r_n ∝ n². Energy E_n ∝ -1/n². Memory tip: 'Bohr radius: r ∝ n²; energy: E ∝ -1/n²'. This atomic physics formula is frequently tested in competitive exams. Always recall that n is principal quantum number; higher n means larger orbit, less tightly bound electron. This problem assesses understanding of quantization in early quantum theory.
Explanation:
Least count of screw gauge = Pitch / Number of circular scale divisions. Given pitch = 1 mm, divisions = 100, so LC = 1 mm / 100 = 0.01 mm. This represents the smallest measurement the instrument can accurately detect. Screw gauges measure small dimensions like wire diameter with high precision. Memory tip: Least count formula is universal for vernier and screw instruments: value per division on main scale divided by total circular divisions. This concept is frequently tested in practical physics sections of competitive exams.
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