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We have a galvanometer of resistance `25 Omega`. It is shunted by a `2.5 Omega` wire. The part of total current that flows through the galvanometer is given asA. `(I)/(I_(0)) = (1)/(11)`B. `(I)/(I_(0)) = (1)/(10)`C. `(I)/(I_(0)) = (3)/(11)`D. `(I)/(I_(0)) = (4)/(11)` |
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Answer» Correct Answer - A |
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