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A oil drop having a mass $4.8 \times 10^{10}\;g$ and charge $2.4 \times 10^{-18}\;C$ stands still between two charged horizontal plates separated by a distance of 1 cm. If now the polarity of the plates is changed, instantaneous acceleration of the drop is : $(g= 10 ms^{-2}$)

\[\begin {array} {1 1} (a)\;5\;ms^{-2} & \quad (b)\;10\;ms^{-2} \\ (c)\;15\;ms^{-2} & \quad (d)\;20\;ms^{-2} \end {array}\]

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$(b)\;10 ms^{-2}$
answered Nov 7, 2013 by pady_1

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