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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. `19 Omega`C. `20 Omega`D. `25 Omega` |
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Answer» Correct Answer - b Initially `(5)/(R )= (l_(1))/((100 -l_(1))`…(i) Finally `(5)/(R//2)= (1.6l_(1))/((100 -16l_(1)))` or `(5)/(R )= (1.6l_(1))/(2(100 -16l_(1))`….(ii) `:.(l_(1))/((100 -l_(1))) = (16l_(1))/(2(100 -16l_(1)))` On solving we get `l_(1) = 25 cm` From(i) , `(5)/(R )=(25)/((100 - 25)) = (1)/(3) or R = 15 Omega` |
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