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

Answer

C

Explanation

When a resistance of `100 Omega` is connected in series
circuit, `I=(I)/(100 R)` ...(i)
When a resistance of `1000 Omega` is connected in series to double its range
`I=(2V)/(1100+R)` ...(ii)
From `eq^(n)`.(i) and (ii) we have,
`( V)/(100+R)=(2V)/(1100 +R) :.R=900 Omega`.