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In a metre bridge experiment null point is obtained at $40 \mathrm{~cm}$ from one end of the wire when resistance $\mathrm{X}$ is balanced against another resistance $\mathrm{Y}$. If $\mathrm{X} < \mathrm{Y}$, then the new position of the null point from the same end, if one decides to balance a resistance of $3 \mathrm{X}$ against $\mathrm{Y}$, will be close to :
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The correct answer is:
$67 \mathrm{~cm}$
$67 \mathrm{~cm}$
From question, $\frac{x}{y}=\frac{40}{100-40}=\frac{2}{3}$
$$
\Rightarrow \mathrm{x}=\frac{2}{3} \mathrm{y}
$$
Again, $\frac{3 x}{y}=\frac{Z}{100-Z}$
or $\frac{3 \times \frac{2 y}{3}}{y}=\frac{Z}{100-Z}$
Solving we get $\mathrm{Z}=67 \mathrm{~cm}$
Therefore new position of null point
$$
\cong 67 \mathrm{~cm}
$$
$$
\Rightarrow \mathrm{x}=\frac{2}{3} \mathrm{y}
$$
Again, $\frac{3 x}{y}=\frac{Z}{100-Z}$
or $\frac{3 \times \frac{2 y}{3}}{y}=\frac{Z}{100-Z}$
Solving we get $\mathrm{Z}=67 \mathrm{~cm}$
Therefore new position of null point
$$
\cong 67 \mathrm{~cm}
$$
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