question_answer

With a resistance R connected in series with galvanometer of resistance \[100\,\,\Omega \], it acts as a voltmeter of range 0 - V. To double the range a resistance of \[1000\,\,\Omega \] is to be connected in series with R. Then the value of-R is:

A)  \[400\,\,\Omega \]

B)  \[900\,\,\Omega \]

C)  \[1100\,\,\Omega \]

D)  \[1000\,\,\Omega \]

Correct Answer: B

Solution :

Resistance added to galvanometer is                                 \[R=\frac{V}{{{i}_{g}}}-G\] or            \[R=\frac{V}{{{i}_{g}}}-100\]                 \[\Rightarrow \] \[R+100=\frac{V}{{{i}_{g}}}\] ... (i) and        \[R+1000=\frac{2V}{{{i}_{g}}}-100\] \[\Rightarrow \] \[R+1100=\frac{2V}{{{i}_{g}}}\] ... (ii) Dividing Eq. (ii) by Eq. (i), we get                 \[\frac{R+1100}{R+100}=2\] or            \[R+1100=2\,R+200\] or \[R=900\,\,\Omega \]