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The resistance 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 `1.6 l_(1)`. The resistance `R` is A. `10 Omega`B. `15 Omega`C. `20 Omega`D. `25 Omega` |
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Answer» Correct Answer - B (b) Initially, `(5)/(l_(1)) = (R )/(100 - l_(1))` Finally, `(5)/(1.6 l_(1)) = (R )/(2 (100 - 1.6l_(1))` `implies (R )/(1.6 (100 - l_(1))) = (R )/(2 (100 = 1.6 l_(2)))` `implies 160 - 1.6 l_(1) = 200 - 3.2 l_(1)` `implies 1.6 l_(1) = 40` `implies l_(1) = 25` From Equaiton (i), `(5)/(25) = (R )/(75) implies R = 15 Omega` |
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