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Electrically charged drops of mercury fall from an altitude h into a spherical metal vessel of radius R. There is a small opening in the upper part of the vessel. The mass of each drop is m, and the charge on the drop is Q. What will be the number n of the last drop that can still enter the sphere?

Text Solution

Verified by Experts

The correct Answer is:
`n=(4pi in_(0) mg(h-R)R)/Q^(2)`
Let `n^(th)` number last drop that can entre
`(KQ^(2))/Rxxn=mg (h-R) :. N=(4pi in_(0) mg(h-R)R)/Q^(2)`
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