Physics - Current Electricity Question with Solution | TestHub

Let $G$ be the resistance of the galvanometer and $I_g$ be the current for full scale deflection.

For the first case, the range of the voltmeter is $V_1$:
$V_1=I_g(G+R)$

For the second case, the series resistance is $4R$ and the range is $V_2$:
$V_2=I_g(G+4R)$

Multiplying the first equation by $4$ gives:
$4V_1=4I_g(G+R)=4I_{g}G+4I_{g}R$

Subtracting the expression for $V_2$ from $4V_1$:
$4V_1-V_2=(4I_{g}G+4I_{g}R)-(I_{g}G+4I_{g}R)$

$4V_1-V_2=3I_{g}G$

Since the galvanometer resistance $G$ and the full scale deflection current $I_g$ are both positive quantities, $3I_{g}G>0$.

Therefore, $4V_1-V_2>0\Rightarrow V_2<4V_1$.

Answer: $V_2<4V_1$