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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2013
  • Hard
  • Numerical

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 .............. \(cm\)

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Answer & Solution

Correct Answer

(C) \(67\)

Step-by-step Solution

Detailed explanation

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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