A galvanometer of resistance 50 Ω 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
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
a
5050 Ω
b
5550 Ω
c
6050 Ω
d
4450 Ω
answer is D.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
initial current i=33000=10−3A10−3A→30divisions20div→10−3×23A For current 23×10−3 resistance supposed to be, 10−3×23=3R+50R+50=92×103=4500R=4450Ω