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A positively charged oil drop of charge 8×10^–15 C remains stationary in the electric field between two horizontal plates separated by a distance of 2cm and having potential difference 6V. Mass of the oil drop is (g=10 ms^{– 2})

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23 × 10^–16 kg

6 × 10^–14 kg

12 × 10^–14 kg

24 × 10^–14 kg

answer is A.

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Detailed Solution

$\begin{array}{l} mg = Eq = \frac{V}{d} q; m (10) = \frac{6 \times (8 \times 10^{- 15})}{2 \times 10^{- 2}} \\ m = 24 \times 10^{- 14} Kg \end{array}$

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