In Millikan's oil drop experiment, an oil drop of density 8 times of air is held stationary by applying a field E. The field required to hold another drop of same radius and carrying same charge but density is 22 times the density of air is

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Solution

The correct option is C 3E
Given in first condition,
Let volume be $v$ and density be $8 a$ where $a$ is density of air,
$v 8 a g - v a g = q E$
$↓$
Buoyant force

$\Rightarrow 7 v a g = q e$ ----------(1)

For second case,
$22 v a g - v a g = q E^{1}$
$\Rightarrow 21 v a g = q E^{1}$ ------------------ [From (I)]
$\Rightarrow E^{1} = 3 E$
So, the answer is option (C).

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In Millikan's oil drop experiment, an oil drop of density 8 times of air is held stationary by applying a field E. The field required to hold another drop of same radius and carrying same charge but density is 22 times the density of air is

The correct option is C 3EGiven in first condition,Let volume be v and density be 8a where a is density of air,v8ag−vag=qE ↓Buoyant force ⇒7vag=qe----------(1)For second case,22vag−vag=qE1⇒21vag=qE1 ------------------ [From (I)]⇒E1=3ESo, the answer is option (C).