Current Electricity 4 Question 4
4. In a meter bridge, the wire of length $1 m$ has a non-uniform cross-section such that the variation $\frac{d R}{d l}$ of its resistance $R$ with length $l$ is $\frac{d R}{d l} \propto \frac{1}{\sqrt{l}}$.
Two equal resistance are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point $P$. What is the length $A P$ ?
(2019 Main, 12 Jan I)
(a) $0.3 m$
(b) $0.25 m$
(c) $0.2 m$
(d) $0.35 m$
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Solution:
- As, galvanometer shows zero deflection.
This means, the meter bridge is balanced.
$$ \begin{aligned} \frac{R^{\prime}}{R _{A P}} & =\frac{R^{\prime}}{R _{P B}} \\ \Rightarrow \quad R _{A P} & =R _{P B} \end{aligned} $$
Now, for meter bridge wire
$$ \frac{d R}{d l}=\frac{k}{\sqrt{l}} $$
where, ’ $k$ ’ is the constant of proportionality.
$$ \Rightarrow \quad d R=\frac{k}{\sqrt{l}} d l $$
Integrating both sides, we get
$$ \begin{array}{lc} \Rightarrow & R=\int \frac{k}{\sqrt{l}} d l \\ \text { So, } & R _{A P}=\int _0^{l} \frac{k}{\sqrt{l}} d l=\left.k(2 \sqrt{l})\right| _0 ^{l}=2 k \sqrt{l} \\ \text { and } & R _{P B}=\int _l^{1} \frac{k}{\sqrt{l}} d l=\left.2 k(\sqrt{l})\right| _l ^{1} \\ & =2 k(\sqrt{1}-\sqrt{l})=2 k(1-\sqrt{l}) \end{array} $$
Substituting values of $R _{A P}$ and $R _{P B}$ in eq. (i), we get
$$ \begin{array}{rlrl} R _{A P} & =R _{P B} \\ \Rightarrow & & 2 k \sqrt{l} & =2 k(1-\sqrt{l}) \\ \Rightarrow & & \sqrt{l} & =\frac{1}{2} \text { or } \quad l=\frac{1}{4}=0.25 m \end{array} $$
