| The resistances in the two arms of the meter bridge are \[5\,\Omega \] and \[R\,\Omega \], respectively. When the resistance R is shunted with an equal resistance, the new balance point is at \[1.6\,{{l}_{1}}\]. [NEET 2014] |
| The resistance R, is |
 |
A) \[10\,\,\Omega \]
B) \[15\,\,\Omega \]
C) \[20\,\,\Omega \]
D) \[25\,\,\Omega \]
Correct Answer: B
Solution :
| [b] For first case, |
| \[\frac{5}{{{l}_{1}}}=\frac{R}{(100-{{l}_{1}})}\] (i) |
| Now, by shunting resistance R by an equal resistance R, new resistance in that arm become \[\frac{R}{2}\] |
| So \[\frac{5}{1.6{{l}_{1}}}\frac{R/2}{(100-{{l}_{1}})}\] (ii) |
| \[\Rightarrow \] From Eqs. (i) and (ii) |
| \[\frac{1.6}{1}=\frac{(100-1.6{{l}_{1}})}{100-{{l}_{1}}}\times 2\] |
| \[\Rightarrow \] \[160-1.6{{l}_{1}}=200-3.2{{l}_{1}}\] |
| \[1.6{{l}_{1}}=40\] |
| \[{{l}_{1}}=\frac{400}{1.6}=25m\] |
| From Eq.(i), \[\frac{5}{25}=\frac{R}{75}\Rightarrow R=15\Omega \] |
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