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Unit of reduction factor is
Options
(a) ampere
(b) ohm
(c) tesla
(d) weber
Correct Answer:
ampere
Explanation:
Reduction factor K = i/tan θ
.·. i = K tanθ
The unit of current (i) is ampere. So, the unit of reduction factor (K) is equivalent to that of current (i).
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Question Type: Memory
(964)
Difficulty Level: Easy
(1008)
Topics: Physical World and Measurement
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The binding energy per nucleon of ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06MeV
- Perfectly black body radiates the energy 18 J/s at 300K. Another ordinary body of e=0.8
- The specific charge of a proton is 9.6×10⁷ C/kg. The specific charge of an alpha particle
- A cycle wheel of radius 4 m completes one revolution in two seconds. Then acceleration
- If the electric field lines is flowing along axis of a cylinder, then the flux
Question Type: Memory (964)
Difficulty Level: Easy (1008)
Topics: Physical World and Measurement (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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Reduction factor of say Tangent Galvanometer is actually numerically equal to the current in ampere needed to produce a deflection of 45° when plane of coil lies in magnetic meridian
K=Itan(theta). I has unit ampere and (theta) is the ratio b/w opposite side and adjacent side so length by length. therefore tan (theta) has no unit. So unit of K is Ampere.