A number of capacitors, each of capacitance 1 $μ$ F and each one of which gets punctured if a potential difference just exceeding 500 volt is applied, are provided. Then an arrangement suitable for giving a capacitor of capacitance 2 $μ$ F across which 3000 volt may be applied requires at least

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answer is C.

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Detailed Solution

Number of capacitors required in series $= \frac{30000}{500} = 6$
The capacitance of series combination will be $(1 / 6) μF .$
In order to obtain the capacitance of 2 $μ$ F, we should use 12 such combinations in parallel to each other.
$∴$ Total number of capacitor required = 12 x 6 = 72.