A) 50 cm
B) 80 cm
C) 40 cm
D) 70 cm
Correct Answer: A
Solution :
| [a] 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) |
| 2nd 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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