question_answer

The galvanometer deflection, when key \[{{K}_{1}}\]is closed but \[{{K}_{2}}\] is open, equals \[{{\theta }_{0}},\] (see figure). On closing \[{{K}_{2}}\]also and adjusting \[{{R}_{2}}\] to \[5\Omega ,\]the deflection in galvanometer becomes\[\frac{{{\theta }_{0}}}{5}.\] The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] [JEE Main Online Paper Held On 12-Jan-2019 Morning]

A) \[25\Omega \]                                        

B) \[22\Omega \]

C) \[5\Omega \]                             

D)   \[12\Omega \]

Correct Answer: B

Solution :

Initially, let \[{{I}_{1}}\]be the current through G. then \[{{I}_{1}}=\frac{V}{220+G}.\] After the key \[{{K}_{2}}\]is closed, the circuit is shown as Apply KVL on loop 1, \[5{{I}_{3}}=G{{I}_{4}}\Rightarrow {{I}_{3}}=\frac{G{{I}_{4}}}{5}\] Also,\[{{I}_{3}}+{{I}_{4}}={{I}_{2}}\]\[\Rightarrow \]\[\left( \frac{G}{5}+1 \right){{I}_{4}}={{I}_{2}}\] \[\Rightarrow \]\[{{I}_{4}}=\frac{V}{\left( {{R}_{1}}+\frac{5G}{G+5} \right)\left( \frac{G+5}{5} \right)}\] For a galvanometer, \[I\propto \theta \] So,\[\frac{{{I}_{1}}}{{{I}_{4}}}=\frac{{{\theta }_{0}}}{{{\theta }_{0}}/5}\Rightarrow 5=\frac{V}{220+G}\frac{{{R}_{1}}(G+5)+5G}{5V}\] \[\Rightarrow \]\[25(220+G)=(220)(G+5)+5G\] \[\Rightarrow \]\[25(220)+20G=220G+1100\] \[\Rightarrow \]\[200G=4400\Rightarrow G=22\Omega \]