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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Current Electricity
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In a meter 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 $Y$?

$\begin {array} {1 1} (A)\;50\: cm & \quad (B)\;80\: cm \\ (C)\;40\: cm & \quad (D)\;70\: cm \end {array}$

 

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We have from meter bridge experiment
$ \large\frac{R_1}{R_2}$ $ = \large\frac{l_1}{l_2}$ where $ l_2 = ( 100 - l_1 \: cm)$
In the first case, $ \large\frac{X}{Y}$ $ = \large\frac{20}{80}$
In the second case, $ \large\frac{4X}{Y} $ $ = \large\frac{l}{100-l}$
$ l = 50 \: cm$
Ans : (A)

 

answered Feb 16, 2014 by thanvigandhi_1
edited Mar 14, 2014 by thanvigandhi_1
 

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