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

Options

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

Answer

C

Explanation

When a resistance of `100 Omega` is connected in series
current, `i=(V)/(100+R)` ….(i)
When a resistance of `100 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)`
`rArr " " R = 900 Omega`