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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).
Related Questions: - A simple pendulum has a time period T₁ when on the earth’s surface and T₂ When taken
- Two spherical nuclei have mass numbers 216 and 64 with their radii R₁ and R₂,
- A wire of length 1 m is moving at a speed of 2 ms⁻¹ perpendicular to its length
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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
- A simple pendulum has a time period T₁ when on the earth’s surface and T₂ When taken
- Two spherical nuclei have mass numbers 216 and 64 with their radii R₁ and R₂,
- A wire of length 1 m is moving at a speed of 2 ms⁻¹ perpendicular to its length
- 300j work is done in sliding a 2kg block up an inclined plane of height 10m
- Shear modulus is zero for
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.