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A galvanometer of resistance 50Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be
Options
(a) 6050 Ω
(b) 4450 Ω
(c) 5050 Ω
(d) 5550 Ω
Correct Answer:
4450 Ω
Explanation:
I=3 / (50+2950) = 10⁻³ A
Current for 30 divisions=10⁻³ A
Current for 20 divisions=10⁻³ * 20/30
=2*10⁻³ / 3 Amperes
For the same deflection to obtain for 20 divisions let resistance added be R, therefore,
2*10⁻³ / 3 = 3 / (50+R)
(50+R) = 9 *10⁻³ / 2
(50+R) = 4500
Therefore R=4450Ω
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Topics: Alternating Current
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Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The amplitude of S.H.M. y=2 (sin 5πt+√2 cos 5πt) is
- When ₃Li⁷ nuclei are bombarded by protons, and the resultant nuclei are ₄Be⁸,
- Four indentical thin rods each of mass M and length l, form a square frame
- The binding energy of deutron is 2.2 MeV and that of ₂⁴He is 28 MeV.
- A small part of the rim of a fly wheel breaks off while it is rotating at a constant
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Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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