A resistance of `4 Omega` and a wire of length 5 meters and resistance `5 Omega` are joined in series and connected to a cell of e.m.f. `10 V` and internal resistance `1 Omega`. A parallel combination of two identical cells is balanced across `300 cm` of wire. The e.m.f. `E` of each cell is A. `1.5 V`B. `3.0 V`C. `0.67 V`D. `1.33 V`

A resistance of `4 Omega` and a wire of length 5 meters and resistance

1.	A resistance of `4 Omega` and a wire of length 5 meters and resistance `5 Omega` are joined in series and connected to a cell of e.m.f. `10 V` and internal resistance `1 Omega`. A parallel combination of two identical cells is balanced across `300 cm` of wire. The e.m.f. `E` of each cell is A. `1.5 V`B. `3.0 V`C. `0.67 V`D. `1.33 V`
Answer» Correct Answer - B (b) `E = xl = (V)/(l) = (iR)/(L) xx l implies E = (e)/((R + R_(h) + r)) xx (R )/(L) xx l` `implies E = (10)/((5 + 4 + 1)) xx (5)/(5) xx 3 = 3 V`

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