In Millikan's oil drop experiment, an oil drop carrying a charge Q is held stationary by a potential difference of 2400 V between the plates. To keep a drop of half the radius stationary the potential difference had to be made 600 volt. What is the charge on the second drop

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$\frac{3 Q}{2}$

$\frac{Q}{4}$

$\frac{Q}{2}$

answer is B.

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Detailed Solution

$\begin{aligned} \Rightarrow Q = 4 / 3 π R^{3} dgL / V \\ Q \propto R^{3} / V \\ Q_{2} / Q_{1} = \frac{R_{2}^{3} \cdot V_{1}}{R_{1}^{3} V_{2}} \\ Q_{2} = \frac{Q \times (1 / 2)^{3}}{2400 / 600} \\ Q_{2} = \frac{Q}{2} \end{aligned}$

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