Question

The galvanometer deflection, when key $K_1$ is closed but $K_2$ is open, equals ${θ}_0$ (see figure). On closing $K_2$ also and adjusting $R_2$ to $5Î\copyright$, the deflection in galvanometer becomes $\frac{{θ}_0}{5}$. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :

• A. $12Î\copyright$
• B. $25Î\copyright$
• C. $5Î\copyright$
• D. $22Î\copyright$

Answer 1

Answer: (D)

case I $i_g=\frac{E}{220+R_g}=C{θ}_0$ ....(i)
case II $i_g=\left(\frac{E}{220+\frac{5R_g}{5+R_g}}\right)×\frac{5}{\left(R_g+5\right)}=\frac{C{θ}_0}{5}$ ....(ii)
$⇒\frac{5E}{225R_g+1100}=\frac{C{θ}_0}{5}$
$\frac{E}{220+R_g}=Cθ$
$⇒\frac{225R_g+1100}{1100+5R_g}=5$
$⇒5500+25R_g=225R_g+1100$
$200R_g=4400$
$R_g=22Î\copyright$