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JEE PHYSICS MCQ: CURRENT ELECTRICITY

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Application +4 | -1
A cell of emf 2 V2 \mathrm{~V} and internal resistance 0.1 Ω0.1 ~\Omega is connected to a 3.9 Ω3.9 ~\Omega external resistance. What will be the p.d. across the terminals of the cell ?
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Correct option: D
1.95 V1.95 \mathrm{~V}
Option A (E=ma0eE=\frac{m a_0}{e}) is correct because it represents the electric field required to produce the same acceleration on an electron as the acceleration of the rod.

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