A number of capacitors each of capacitance $1\,\mu F$ and each one of which get punctured if a potential difference just exceeding $500\,volt$ is applied, are provide, then an arrangement suitable for givin a capacitance of $2\,\mu F$ across which $3000\,volt$ may be applied requires at least
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Number of capacitors required in series $=\frac{3000}{500}=6.$
The capacitance of series combination $=(1 / 6)\, \mu \mathrm{F}.$
To obtain a capacitor of $2\, \mu \mathrm{F},$ we should use $12$ such combinations.
$\therefore $Total number of capacitors required $=12 \times 6=72$
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