A) 50 cm
B) 80 cm
C) 40cm
D) 70cm
Correct Answer: A
Solution :
Meter bridge is an arrangement which works on Wheatstones principle, so the balancing condition is \[\frac{R}{S}=\frac{{{l}_{1}}}{{{l}_{2}}}\] Where \[{{l}_{2}}=100-{{l}_{1}}\] 1st case: \[R=X,S=Y,{{l}_{1}}=20\,cm,\] \[{{l}_{2}}=100-20=80cm\] \[\therefore \] \[\frac{X}{Y}=\frac{20}{80}\] ?.(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) \[\frac{4X}{Y}=\frac{l}{100-l}\] \[\Rightarrow \] \[\frac{X}{Y}=\frac{l}{4(400-l)}\] ?(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, \[l=50\text{ }cm\]You need to login to perform this action.
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