A ball of density sigma and radius r is dropped on the surface of a liquid of density rho from certain height if speed of ball does not change even on entering in the liquid
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your question is incomplete . A complete question is ----> A ball of density
and radius R is dropped on the surface of a liquid of density ρ from a certain height. if speed of ball does not change even on entering in liquid and viscosity of liquid is η, then the Height from which ball drop is ?
solve :- use formula,

where
terminal velocity of ball,
Let h is the height from which ball drops .
so, v² = u² + 2as
h = v²/2g
=
=
solve :- use formula,
where
Let h is the height from which ball drops .
so, v² = u² + 2as
h = v²/2g
=
=
Answered by
0
Answer:
your question is incomplete . A complete question is ----> A ball of density and radius R is dropped on the surface of a liquid of density ρ from a certain height. if speed of ball does not change even on entering in liquid and viscosity of liquid is η, then the Height from which ball drop is ?
solve :- use formula,
where terminal velocity of ball,
Let h is the height from which ball drops .
so, v² = u² + 2as
h = v²/2g
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