A battery of emf $10 \mathrm{~V}$ is connected to a uniform wire $A B$ of $1 \mathrm{~m}$ length and having…
A battery of emf $10 \mathrm{~V}$ is connected to a uniform wire $A B$ of $1 \mathrm{~m}$ length and having a resistance of $10 \Omega$ in series with a $10 \Omega$ resistor as shown in the figure. Two cells of emf $2 \mathrm{~V}$ and $3 \mathrm{~V}$ having internal resistance $2 \Omega$ and $3 \Omega$, respectively are connected as shown in the figure. If the galvanometer shows null deflection at point $J$ on the wire, then the distance of point $J$ from the point $B$ is.
48 cm
50 cm
52 cm
54 cm
Solution
Given $r=10 \Omega, R=10 \Omega, E=10 \mathrm{~V}$ and $L=100 \mathrm{~cm}$.
Now, voltage drop on the wire, $E^{\prime}=\frac{E}{(r+R)} \times R$
$
\Rightarrow \quad E^{\prime}=\frac{10}{20} \times 10=5 R
$
So, potential gradient on wire,
Effective emf of combination in secondary circuit,
$
\begin{array}{rlrl}
& & \frac{V}{r_{\text {eff }}} & =\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_2}{r_2} \\
\Rightarrow \text { Here, } \quad & \frac{1}{r_{\text {eff }}} & =\frac{1}{2}+\frac{1}{3}=\frac{5}{6} \\
\Rightarrow \quad & V & =\frac{6}{5}(1+1)=\frac{12}{5}
\end{array}
$
Now, at balancing point (from point $A$ ) is
$
\begin{aligned}
& l=V / x=\frac{12}{5} \times \frac{100}{5} \\
& l=48 \mathrm{~cm}
\end{aligned}
$
So, length from point $B$ is
$
100-48=52 \mathrm{~cm}
$
