| In a meter bridge, as shown in the figure, it is given that resistance \[Y=12.5\Omega \] and that the balance is obtained at a distance \[39.5cm\] from end \[A\] (by jockey J). After interchanging the resistances \[X\] and\[Y\], a new balance point is found at a distance \[{{l}_{2}}\] from end \[A\]. What are the values of \[X\]and \[{{l}_{2}}\]? [JEE Online 15-04-2018] |
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A) \[\text{19}\text{.15 }\Omega \] and \[\text{39}\text{.5 cm}\]
B) \[\text{8}\text{.16 }\Omega \text{ and 60}\text{.5 cm}\]
C) \[\text{19}\text{.15}\,\,\text{ }\!\!\Omega\!\!\text{ }\,\,\text{and}\,\,\text{60}\text{.5}\,\,\text{cm}\]
D) \[\text{8}\text{.16}\,\,\Omega \text{ and 39}\text{.5 cm}\]
Correct Answer: B
Solution :
| [b] \[\frac{X}{Y}=\] |
| \[\frac{39.5}{60.5}\] |
| \[X=\frac{39.5}{60.5}\times 12.5\] |
| \[=8.16\,\,\,\Omega \] |
| \[\frac{12.5}{8.16}=\frac{{{l}_{2}}}{100-{{l}_{2}}}\] |
| \[1.53(100+{{l}_{2}})={{l}_{2}}\] |
| \[153-1.53\,\,{{l}_{2}}={{l}_{2}}\] |
| \[\frac{153}{2.53}={{l}_{2}}\] |
| \[=60.5cm\] |
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