A galvanometer of resistance 30 2 is connected to a battery of emf 2 V with 1970 Q resistance in series. A full scale deflection of 20 divisions is obtained in the galvanometer. To reduce the deflection to 10 divisions, the resistance in series required is

Options

A. `4030 Omega`
B. `4000 Omega`
C. `3970 Omega`
D. `2000 Omega`

Answer

C

Explanation

According to question, Net current through the galvanometer is given by
`l=V/R_(eff)=2/(1970+30)=2/(2000)`
= `10^-3`
As we know, this current provides full scale deflection (i.e. 20 div). In order to limit the deflection to 10 divisions, the resistance needed to connect such that the current reduces can be obtained as
`theta=(nlAB)/K ("i.e. "theta alpha l`)
[Symbols have their usual meanings.]
Righrarrow `theta_1/theta_2=l_1/l_2=2`
`Rightarrow l_2=l_1/2=10^-3/2=5xx10^-4`A
Again,`l=V/(R_(eff)+R_5)`
`Rightarrow R_5=V/l-R_("eff")`
`2/(5xx10^-4)-2000`
`=4xx10^3-2000=2000 Omega`
So, the resistance of `1970 Omega` is to be replaced by `1970+2000=3970 Omega.`