Current Electricity Ques 102
- 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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Answer:
Correct Answer: 102.(b)
Solution:
Formula:
- 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} \quad …….(i) \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 ’ $A J$ '
$ =\frac{0.6 \times 4}{L} \times x=\varepsilon _2 $
$\Rightarrow \quad \varepsilon _2 =\frac{0.6 \times 4}{L} \times \frac{L}{8} \quad $ [using Eq. (i)]
$\Rightarrow \quad \varepsilon _2 =0.3$ $V$
BETA
