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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
- Two capacitors having capacitances C₁ and C₂ are charged with 120V and 200V batteries
- A bullet of mass m moving with velocity v strikes a block of mass M at rests
- A uniform force of (3i + j) newton acfs on a particle of mass 2kg
- If two vectors 2i+3j+k and -4i-6j-λk are parallel to each other, then the value of λ is
- The magnifying power of a telescope is 9. When it is adjusted for parallel rays
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.