To know the resistance G of a galvanometer by half deflection method, a battery of emf VE and resistance R is used to deflect the galvanometer by angle \theta . If a shunt of resistance S is needed to get half deflection then G, R and S are related by the equation :

Magnetics PYQ — JEE Main 2016 · 9 Apr · Shift 1 · Q72 with Solution

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Correct answer: B

Let the galvanometer resistance be $G$ , battery emf be $E$ , and external resistance in series be $R$ .

Then the current through the galvanometer initially is $I_g=\frac{E}{R+G}.$

Since deflection is proportional to current, deflection $\theta$ corresponds to current $I_g$ .

Now the galvanometer of resistance $G$ is shunted by resistance $S$ .

The equivalent resistance of the parallel combination is $\frac{GS}{G+S}.$

Hence total current from the battery becomes $I=\frac{E}{R+\frac{GS}{G+S}}.$

This current divides between $G$ and $S$ .

Current through galvanometer is given by current division: $I_g' = I\cdot \frac{S}{G+S}.$

So, $I_g' = \frac{E}{R+\frac{GS}{G+S}}\cdot \frac{S}{G+S}.$

Half deflection means current through galvanometer becomes half of the original current: $I_g' = \frac{I_g}{2}.$

So, $\frac{E}{R+\frac{GS}{G+S}}\cdot \frac{S}{G+S} = \frac{1}{2}\cdot \frac{E}{R+G}.$

Cancel $E$ : $\frac{1}{R+\frac{GS}{G+S}}\cdot \frac{S}{G+S} = \frac{1}{2(R+G)}.$

Now simplify the left side: $R+\frac{GS}{G+S} = \frac{R(G+S)+GS}{G+S}.$

Therefore,

= \frac{G+S}{R(G+S)+GS}\cdot \frac{S}{G+S} = \frac{S}{R(G+S)+GS}.$$ Hence, $$\frac{S}{R(G+S)+GS} = \frac{1}{2(R+G)}.$$ Cross-multiplying, $$2S(R+G)=R(G+S)+GS.$$ Expand RHS: $$2SR+2SG=RG+RS+GS.$$ Rearrange: $$SR+SG=RG.$$ Thus, $$S(R+G)=RG.$$ --- 4. **Matching with options** This corresponds to: **Option B:** $$S(R+G)=RG.$$ So the correct answer is **B**.