practically not possible !! X~)
a balll of density Ï0 falls from rest from a point P onto the surface of a liquid of density Ï in time T. It enters the liquid, stops, moves up n returns to P in a total time 3T. Neglect the viscosity, surface tension and splashing. the ratio Ï/Ï0 is :::
A) 1.5 B) 2 C) 3 D) 4
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9 Answers
Ï/Ï0 = 2 .. in side the liquid u will get that T = T/(Ï/Ï0 - 1)
possible if this = 2 ..
force in the upward direction is given by
(Ï-Ï0)Vg
V is the volume
Force = ma
(Ï-Ï0)Vg = Ï0Va
a= (Ï/Ï0-1)g
Velocity when entering the fluid = gT
final displacement = 0
s= ut+1/2at2
s= 0, a=(Ï/Ï0-1)g and t= 3T, u = -gT
ut+1/2at2
2u=at
2gT=(Ï/Ï0-1)g (T) {Here 2T is removed because it is the time in air again!}
2=(Ï/Ï0-1)
Ï/Ï0 = 3
balll of density Ï0 falls from rest from a point P
MEANS p is in air, then how :
moves up n returns to P
something i want to share ... if the ball takes time T to fall to a point P then again from P to T while it was returning it takes time T .. . so inside the liq it takes 2T .. that implies while going down the liquid it will take time T so gT - (Ï/Ï0 - 1)gT = 0 ... and i think my answer is correct !!!
yes ankit.. i correct it already :)
may be just when you were typing this message :)