Question

# The terminal velocity of a liquid drop of radius $$r$$ falling through air is $$v$$. If two such drops are combined to form a bigger  drop, the terminal velocity with which the bigger drop falls through air is (Ignore any buoyant force due to air) :

A
2v
B
2v
C
34v
D
32v

Solution

## The correct option is C $$\sqrt[3]{4}v$$On merging the two drops, the volume will become 2 times and since $$V=\dfrac { 4 }{ 3 } \pi { r }^{ 3 }$$, radius will become $${ 2 }^{ \dfrac { 1 }{ 3 } }$$ times. The expression for terminal velocity as derived from Stoke's Law is $${ v }_{ t }=\dfrac { 2 }{ 9 } \dfrac { ({ \rho }_{ p }-{ \rho }_{ f }) }{ \mu } g{ r }^{ 2 }$$Since this is proportional to $${ r }^{ 2 }$$, new terminal velocity will be $${ 2 }^{ \dfrac { 2 }{ 3 } }=\sqrt [ 3 ]{ 4 }$$ times $$v$$Physics

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