CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
2v
loader
C
34v
loader
D
32v
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image