In a meter bridge experiment null point is obtained at $40\, cm$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$, then the new position of the null point from the same end, if one decides to balance a resistance of $3X$ against $Y$, will be close to .............. $cm$
JEE MAIN 2013, Diffcult
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From question, $\frac{x}{y}=\frac{40}{100-40}=\frac{2}{3}$
$\Rightarrow x=\frac{2}{3} y$
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 $Z=67\, \mathrm{cm}$
Therefore new position of null point
$\cong 67\, \mathrm{cm}$
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