In Millikan’s oil drop experiment, an oil drop of mass 16 × {10^{ - 6}}kg is balanced by an e

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11.Dual Nature of Radiation and matter

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In Millikan’s oil drop experiment, an oil drop of mass $16 \times {10^{ - 6}}kg$ is balanced by an electric field of ${10^6}V/m.$ The charge in coulomb on the drop, assuming $g = 10\,m/{s^2}$ is

A

$6.2 \times {10^{ - 11}}$

B

$16 \times {10^{ - 9}}$

C

$16 \times {10^{ - 11}}$

D

$16 \times {10^{ - 13}}$

Solution

(c) $eE = mg$

==> $e = \frac{{mg}}{E}$ $ = \frac{{16 \times {{10}^{ – 6}} \times 10}}{{{{10}^6}}} = 16 \times {10^{ – 11}}C.$

Standard 12

Physics

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