KCET 2017 Physics Solved Paper

Given, galvanometer resistance, RG=50Ω; battery, V = 3 V; resistance, R = 2950 Ω; deflection = 30 divisions. Now,
Thus, current when deflection is 30 divisions =10−3A
⇒ Current when deflection is 1 division =

10−3

30

⇒ Current when deflection is 20 division =

10−3

30

×20
Therefore, in order to have a deflection of 20 division the resistance should be
‌

2

3

×10−3=‌

3

(50+R)

⇒(50+R)=‌

3×3

2×10−3

50+R=4.5×1000
R=4500−50=4450
Hence, additional resistance required =4450−2950=1500