A screw gauge has a pitch of 1 mm and 100 divisions on its circular scale. The least count of this instrument is:
A. 0.001 mm
B. 0.1 mm
C. 0.0001 mm
D. 0.01 mm
Answer: Option D
Solution (By JKExamLibrary)
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.
Explanation:
Diamagnetic materials have small negative susceptibility (χ ≈ -10⁻⁵), meaning they weakly repel magnetic fields. Paramagnetic: small positive χ; ferromagnetic: large positive χ. This classification is based on response to external magnetic fields. Memory aid: 'Diamagnetic: repelled by magnets; χ < 0'. This conceptual question tests magnetism fundamentals, crucial for competitive exams. Always link susceptibility sign to material behavior: negative χ ⇒ diamagnetic (weak repulsion), positive ⇒ paramagnetic/ferromagnetic (attraction).
Explanation:
Optical fibers: TIR confines light in core. Diamonds: high n gives small θ_c, causing multiple TIRs for sparkle. Memory aid: 'TIR applications: fibers (communication), diamonds (optics)'. Optics application frequently tested in competitive exams.
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