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Two resistances are connected in two gaps of a meter bridge. The balance point is `20 cm` from the zero end. A resistance of `15 Omega` is connected in series with the smaller of the two. The null point shifts to `40 cm`. The value of the smaller resistance in `Omega` isA. 3B. 6C. 9D. 12 |
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Answer» Correct Answer - C `P/Q=20/(100-20) or Q=4P`….i `:. `PltQ` Now,` (P+15)/Q=40/(100-40)` `or (P+15)/Q=2/3` Solving these two equations we get `P=9Omega` |
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