Current Electricity 4 Question 9
9. The resistance of the meter bridge $A B$ in given figure is $4 \Omega$. With a cell of emf $\varepsilon=0.5 V$ and rheostat resistance $R _h=2 \Omega$. The null point is obtained at some point $J$. When the cell is replaced by another one of $\operatorname{emf} \varepsilon=\varepsilon _2$, the same null point $J$ is found for $R _h=6 \Omega$. The emf $\varepsilon _2$ is
(2019 Main, 11 Jan I)
(a) $0.6 V$
(b) $0.3 V$
(c) $0.5 V$
(d) $0.4 V$
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Solution:
- Let length of null point ’ $J$ ’ be ’ $x$ ’ and length of the potentiometer wire be ’ $L$ ‘. In first case, current in the circuit
$$ I _1=\frac{6}{4+2}=1 A $$
$\therefore$ Potential gradient $=I \times R=\frac{1 \times 4}{L}$
$\Rightarrow$ Potential difference in part ’ $A J$ '
Given,
$$ \begin{aligned} = & \frac{1 \times 4}{L} \times x=\varepsilon _1 \\ \varepsilon _1 & =0.5=\frac{4 x}{L} \text { or } \frac{x}{L}=\frac{1}{8} \end{aligned} $$
In second case, current in the circuit
$$ I _2=\frac{6}{4+6}=0.6 A $$
$\therefore$ Potential gradient $=\frac{0.6 \times 4}{L}$
$\Rightarrow$ Potential difference in part ’ $A J$ '
$$ \begin{array}{rlrl} & =\frac{0.6 \times 4}{L} \times x=\varepsilon _2 \\ \Rightarrow \quad & \varepsilon _2 & =\frac{0.6 \times 4}{L} \times \frac{L}{8} \\ \Rightarrow \quad & \varepsilon _2 & =0.3 V \end{array} $$
