Q198 / 225 · JEE Main 2004 · Physics

In a metre bridge experiment null point is obtained at $20 \mathrm{~cm}$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$, then where will be the new position of the null point from the same end, if one decides to balance a resistance of $4 X$ against Y?

  • Correct answer
    A
    $50 \mathrm{~cm}$
  • B
    $80 \mathrm{~cm}$
  • C
    $40 \mathrm{~cm}$
  • D
    $70 \mathrm{~cm}$

Solution

We have from meter bridge experiment, $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\ell_1}{\ell_2}$, where $\ell_2=\left(100-\ell_1\right) \mathrm{cm}$ In the first case, $X / Y=20 / 80$ In the second case $\frac{4 X}{Y}=\frac{\ell}{100-\ell}$ $\ell=50 \mathrm{~cm}$.

Hence, the correct answer is option: 'A'.

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