Current through the galvanometer I=(50+2950)3=10−3 A Current for 30 divisions =10−3 A Current for 20 divisions =3010−3×20 =32×10−3 A For the same deflection to obtain for 20 divisions, let resistance added be R ∴32×10−3=(50+1R)3or R=4450Ω
NEET UG 2008 — Chemistry Physical Chemistry
A galvanometer of resistance 50Ω is connected to a battery of 3 V along with a resistance of 2950Ω 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
Held on 30 Apr 2008 · Verified 9 Jul 2026.
5050Ω
5550Ω
6050Ω
4450Ω
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