When a resistance of 100`Omega` is connected in series with a galvanometer of resistance R, then its range is V. To double its range, a resistance of 1000`Omega` is connected in series. Find the value of R.

Options

A. `700 Omega`
B. `800 Omega`
C. `900 Omega`
D. `100 Omega`

Answer

C

Explanation

When a resistance of `100 Omega` is connected in series current , i= `(V)/(100 + R) " " …. (i)`
when a resistance of `1000 Omega` is connected in series , the its range double
current , `I = (2V)/(1100 + R) " " …. (ii)`
From Eqs.(i) and (ii)
`(V)/(100 + R) = (2V) / (1100 +R)`
`R = 900 Omega`