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A rod of 10 cm length is moving perpendicular to uniform magntic field of intensity 5×10⁻⁴Wbm⁻². If the acceleration of the rod is 5ms⁻¹, then the rate of increase of induced emf is
Options
(a) 25×10⁻⁴ Vs⁻¹
(b) 2.5×10⁻⁴ Vs⁻¹
(c) 2.0×10⁻⁴ Vs⁻¹
(d) 20×10⁻⁴ Vs⁻¹
Correct Answer:
2.5×10⁻⁴ Vs⁻¹
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
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Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The half life of radium is 1620 yr and its atomic weight is 226 kg per kilo mol.
- Two slits in Young’s experiment have widths in the ratio 1:25. The ratio of intensity
- A spherical drop of mercury having a potential of 2.5 v is obtained as a result
- A rod of weight W is supported by two parallel knife edges A and B is in equilibrium
- Consider a uniform square plate of side ɑ and mass m. The moment of inertia
Topics: Electromagnetic Induction (76)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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