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In a metre bridge, the gaps are closed by two resistance `P` and `Q` and the balance point is obtained at `40 cm`. When `Q` is shunted by a resistance of `10 Omega`, the balance point shifts to `50 cm`. The values of `P` and `Q` are. .A. ` 10/3 Omega, 5 Omega`B. `20 Omega, 30 Omega`C. `10 Omega, 15 Omega`D. ` 5 Omega, 15/2 Omega` |
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Answer» Correct Answer - A For meter bridge to be balanced `P/Q =( 40)/(60) = 2/3` `:. p =2/3Q` When `Q` is shunted, i.e. a resistance of `10 Omega` is connected in parallel across `Q`, the net resistance becomes `(10 Q)/(10 + Q)` Now, the balance point shifts to `50 cm`, i.e. `p/(((10Q)/(10 + Q))) = 1` `:. 2/3 = (10)/(10 + Q)` `:. 20 + 2Q = 30` `:. Q = 5 Omega` and `P = (10)/(3) Omega`. |
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