A) 50 cm
B) 80 cm
C) 40 cm
D) 70 cm
Correct Answer: A
Solution :
Meter bridge is an arrangement which works on Wheatstone's bridge principle, so the balancing condition is \[\frac{R}{S}=\frac{{{l}_{1}}}{{{l}_{2}}}\] where, \[{{l}_{2}}=100-{{l}_{1}}\] 1st case \[R=X,S=Y,{{l}_{1}}=20\,cm,\] \[{{l}_{2}}=100-20=80\,cm\] \[\therefore \] \[\frac{X}{Y}=\frac{20}{80}\] ??(i) IInd case Let the position of null point is obtained at a distance \[l\] from same end. \[\therefore \]\[R=4X,S=Y,{{l}_{1}}=l,{{l}_{2}}=100-l\] So, from Eq. (i), \[\frac{4X}{Y}=\frac{l}{100-l}\] \[\Rightarrow \] \[\frac{X}{Y}=\frac{l}{4(100-l)}\] ?..(ii) Therefore, from Eqs. (i) and (ii), \[\frac{l}{4(100-l)}=\frac{20}{80}\] \[\Rightarrow \]\[\frac{l}{4(100-l)}=\frac{1}{4}\]\[\Rightarrow \]\[l=100-l\] \[\Rightarrow \]\[2l=100\] Hence. \[l=50\,cm\]
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