Question

A battery of emf $E_0=12V$ is connected across a $4m$ long uniform wire having resistance $4\Omega /m$. The cells of small emfs ${ϵ}_1=2V$ and ${ϵ}_2=4V$ having internal resistance $2\Omega$ and $6\Omega$ respectively, are connected as shown in the figure. If galvanometer shows no deflection at the point $N$, the distance of point $N$ from the point $A$ is equal to

• A. $\frac{1}{6}m$
• B. $\frac{1}{3}m$
• C. $25\text{cm}$
• D. $50\text{m}$

Answer 1

The correct option is C $25\text{cm}$

$\text{Voltage drop on the balancing length of the wire}\Rightarrow i\rho x$

$\text{Here}i=\text{Current flowing in primary circuit}$

$\rho =\text{Resistance per unit length of the wire}$

$x=\text{Balancing length of wire AB when deflection in galvanometer is zero}$

$\frac{1}{R_{eq.}}\Rightarrow (\frac{1}{2}+\frac{1}{6})=\frac{4}{6}$

$E_{eq.}=\frac{(\frac{E_1}{r_1}-\frac{E_2}{r_2})}{{\frac{1}{R}}_{eq.}}=\left(\frac{\frac{2}{2}-\frac{4}{6}}{\frac{4}{6}}\right)=\frac{1}{2}\text{volt}$

$\text{when deflection is zero}\Rightarrow E\text{of secondary circuit}=\text{voltage drop across the wire AN}$

$\frac{1}{2}=\frac{12}{8+16}\times 4\times x$

$\Rightarrow x=\frac{1}{4}\text{m}.$$\Rightarrow x=25\text{cm}$