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            \[R=\frac{V}{{{i}_{g}}}-100\]                 \[\Rightarrow \]               \[R+100=\frac{V}{{{i}_{g}}}\]                 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            \[R+1100=2R+200\]                 or            \[R=900\,\Omega \]