Question

In a metre bridge experiment null point is obtained at $20\,cm$ 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 $Y$ ?

• A. 50 cm
• B. 80 cm
• C. 40 cm
• D. 70 cm

Answer 1

Answer: (A)

50 cm

Answer 2

Using $l_2 = ( 100 - l_1)$, we get, $\frac{R_1}{R_2} = \frac{l_1}{l_2}$ as $\frac{X}{Y} = \frac{20}{80} =$in case 1 and $\frac{4X}{Y}=\frac{l}{100-l}$ in case 2 i.e. $4\times \frac{20}{80}=\frac{l}{100-l}$ $100\,-\,l \,=\, l\, or\, l =50cm$