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Question

An oil drop is negatively charged and weights 5×104N .The drop is suspended in an electric field intensity of 2.6×104N/C .The number of electrons the oil drop is in x×1010. Then x is

A
8
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B
10
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C
16
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D
12
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Solution

The correct option is C 12
Electric force acting on the drop in upward direction is balanced by weight of the drop i.e. qE=W
Thus charge of drop q=WE=5×1042.6×104=1.92×108 C
Number of electrons n=qe=1.92×1081.6×1019=12×1010

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