Question

The resistance of the meter bridge AB in the given figure is $4\mathrm{\Omega }$ with a cell of $emf=0.5V$ and rheostat resistance $R_h=2\mathrm{\Omega }$ the null point is obtained at some point J. When the cell is replaced by another one of emf $\epsilon ={\epsilon }_2$ the same null point J is found for $R_h=6\mathrm{\Omega }$. The emf ${\epsilon }_2$is;

• A. $0.3V$
• B. $0.5V$
• C. $0.4V$
• D. $0.6V$

Answer 1

The correct option is A

$0.3V$

Step 1: Given Data:

Resistance of meter bridge AB is $\left(R\right)=4\Omega$

EMF of cell $\left(emf\right)=0.5V$

Rheostat resistance $\left(R_h\right)=2\mathrm{\Omega }$

J is a null point

Step 2: Formula Used:

Ohm's law,

$V=IR$

Where, $V$ is the potential difference, $R$ is the resistance

Step 3: Calculation of AJ:

As given, $R_h=2\mathrm{\Omega }$

Let $i_1$ be current when emf of cell is ${\epsilon }_1=0.5V$

$$\begin{gathered} i_1=\frac{V}{R+R_h} \\ \begin{array}{l}{i}_1=\frac{6}{4+2}=1A\end{array} \end{gathered}$$

Using-$\frac{\epsilon }{AJ}=\frac{i_1\times R}{AB}$

$$\begin{gathered} \begin{array}{l}\epsilon =\frac{1\times 4}{AB}\times AJ\end{array} \\ \begin{array}{l}AJ=\frac{AB}{4}\times \epsilon \end{array}\dots \dots \dots \dots \dots \left(1\right) \end{gathered}$$

Similarly, when $R_h=6\mathrm{\Omega }$, let $i_2$ be current when emf of cell is ${\epsilon }_2$

$$\begin{gathered} \begin{array}{l}{i}_2=\frac{6}{4+6}=0.6A\end{array} \\ \begin{array}{l}{\epsilon }_2=\frac{0.6\times 4}{AB}\times AJ\end{array} \\ \begin{array}{l}AJ=\frac{AB\times {\epsilon }_2}{0.6\times 4}\end{array}\dots \dots \dots \dots \dots \dots \left(2\right) \end{gathered}$$

Step 4: Calculating the emf ${\epsilon }_2$

Solving equation $\left(1\right)\mathrm{and}\left(2\right)$:

$$\begin{gathered} \begin{array}{l}\frac{{\epsilon }_2\times AB}{0.6\times 4}=\frac{\epsilon \times AB}{4}\end{array} \\ \begin{array}{l}{\epsilon }_2=0.6\times \epsilon =0.6\times 0.5=0.3V\end{array} \end{gathered}$$

Hence, option A is the correct answer.