A galvanometer of resistance `30Omega` is connected to a battery of emf 2 V with `1970Omega` 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 thrught the gaveanometer is given by
`l=(V)/(R_("eff"))=(2)/(1970+30)=(2)/(2000)`
`=10^(-3) A`
As we know this current provides full scale defection (i.e 20 div) in order to limit the deflection to 10 divisions the resistacne needed to connect such that the current reduces can be obtaned as
`theta =(nLAB)/(K )(i.e theta prop l)`
[ symbols have thier usual meaning ]
`rarr (theta_(1))/(theta_(2))=-(l_(1))/(l_(2))=2`
again `l=(V)/(R_("eff")+R_(5)) rarr R_(5)=(V)/(l)-R_("eff")`
`=(2)/(5xx10^(-4))-2000`
`=4xx10^(-3)-2000=2000 omega`
so the resistance of 1970 `omega` is to be relplaced by 1970+2000 =3970 `omega`