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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 .............. \(cm\)
Pick an option to check your answer.
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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