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An unknown resistance `R_(1)` is connected is series with a resistance of `10 Omega`. This combination is connected to one gap of a meter bridge, while other gap is connected to another resistance `R_(2)`. The balance point is at `50 cm` Now , when the `10 Omega` resistance is removed, the balanced point shifts to `40 cm` Then the value of `R_(1)` is.A. `10 Omega`B. `20 Omega`C. `40 Omega`D. `60 Omega` |
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Answer» Correct Answer - b For first case `(R_(1) + 10)/(R_(2)) = (50)/(100 - 50) = 1` or `R_(2) = R_(1) + 10`…..(i) When `10 Omega ` resistance the balance point will shift toward lower length `:.` New balancing length `= 50 - 10 = 40 cm` for second case `(R_(1))/(R_(2)) = (40)/(100 - 40) = (40)/(60) = (2)/(3) :. R_(2) = (3)/(2)R_(1)` From (i) `(3)/(2)R_(1) = R_(1) = 10` or `R_(1) = 20 Omega ` |
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