Question

In a metre 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:

• A. 80 cm
• B. 75 cm
• C. 67 cm
• D. 50 cm

Answer 1

Answer: (C)

67 cm

Answer 2

From question, $\frac{x}{y} = \frac{40}{100-40} = \frac{2}{3}$
$\Rightarrow x = \frac{2}{3}y$
Again, $\frac{3x}{y} = \frac{Z}{100-Z}$
or $\frac{3\times\frac{2y}{3}}{y} = \frac{Z}{100-Z}$
Solving we get $Z = 67$ cm
Therefore new position of null point
$? 67 \,cm$