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Two resistances are joined in parallel whose reusltant is `6//8 ohm`. One of the resistance wire is broken and the effective resistance becomes `2 Omega` . Then the resistance in ohm of the wire that got broken was
A. `3//5`
B. 2
C. `6//5`
D. 3

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Correct Answer - C
(C ) If resistance are `R_(1)` and `R_(2)` then `(R_(1) R_(2))/(R_(1) + R_(2)) = (6)/(8)` ……..(i)
Suppose `R_(2)` is broken then `R_(1) = 2 Omega ………(ii)`
On solving equations (i) and (ii) we get `R_(2) = 6//5 Omega`

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