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In metre bridge experiment, the null point is obtained at $20 \mathrm{~cm}$ from left end of the wire, when resistance $X$ is balanced against another resistance $Y(X Options:
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$50 \mathrm{~cm}$
The correct option is (D).
Concept: Meter bridge experiment is based on two facts:
(1) the resistance across a wire is directly proportional to its length
(2) balanced Wheatstone bridge has null deflection when the ratio of resistances is equal, and no current flows through the circuit.
The current through galvanometre is zero, when the ratio of resistances, i.e., $\frac{P}{Q}=\frac{R}{S}$.
As given in the problem, $P=X, Q=Y, R \propto 20 \mathrm{~cm}$ and $S \propto 80 \mathrm{~cm}$.
Therefore,
$\frac{X}{Y}=\frac{20}{80}=\frac{1}{4}$
Similarly, to balance a resistance $P=4 X$ against $Q=Y$, let us assume a length $l$ is required. Then, $R=l$
$\frac{4 X}{Y}=\frac{l}{100-l}$
On solving, we get $l=50 \mathrm{~cm}$.
Concept: Meter bridge experiment is based on two facts:
(1) the resistance across a wire is directly proportional to its length
(2) balanced Wheatstone bridge has null deflection when the ratio of resistances is equal, and no current flows through the circuit.
The current through galvanometre is zero, when the ratio of resistances, i.e., $\frac{P}{Q}=\frac{R}{S}$.
As given in the problem, $P=X, Q=Y, R \propto 20 \mathrm{~cm}$ and $S \propto 80 \mathrm{~cm}$.
Therefore,
$\frac{X}{Y}=\frac{20}{80}=\frac{1}{4}$
Similarly, to balance a resistance $P=4 X$ against $Q=Y$, let us assume a length $l$ is required. Then, $R=l$
$\frac{4 X}{Y}=\frac{l}{100-l}$
On solving, we get $l=50 \mathrm{~cm}$.
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