A screw gauge has a pitch of 1 mm and 100 divisions on its circular scale. The least count of this instrument is:
A. 0.0001 mm
B. 0.01 mm
C. 0.001 mm
D. 0.1 mm
Answer: Option B
Solution (By JKSSB Mock Tests)
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:
In Simple Harmonic Motion (like a swinging pendulum), velocity is maximum when the particle passes through its equilibrium or mean position. Since Kinetic Energy = ½mv², KE is maximum at the mean position. At the extreme positions, the velocity is briefly zero, meaning kinetic energy is zero and potential energy is maximum.
Explanation:
Apparent weight = m(g+a) = 60×(10+2) = 720 N, which corresponds to 72 kg mass on scale (if scale calibrated in kg under g=10). Reading 72 kg. Downward acceleration would reduce.
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