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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: fluids
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Nithy (400)

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Olaaa!! Perrrfect answer. 66  [101 rates]

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the Q is:

an object of density d dropped frm rest at h height into liq of density D in tall container. neglect dissipativ effect and no spilling of liq

time taken by object to reach starting point ( frm which its dropped)

D = 2d

???????????

have no clue.  what is it supposed to mean ???????

got this frm YGFile of BT..............


ANS== 4[ ] (2h/g)

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luv_u_physics (25)

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Olaaa!! Perrrfect answer. 5  [5 rates]

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yes de ans is 4 (root)2h/g
 
let me explain in detail...
 
luk here height is - h
so velocity at th surface of water must be (root)2gh................1
 
now wen de object reachez de surface it goes down due to inertia....
 
force acting on body:
 
a: Mg downwards
b: mg upward (m is de mass f water displaced )
so
 
Mg - mg = Ma
 
here M = Vd
m = VD = 2Vd
 
thus, Vdg - 2vdg = Ma or
 
-Vdg = Ma
 
-Mg = Ma
 
hence a= -g.(ie upward)........................2
 
v^2 - u^2 = 2as
here v=0
hence
 
2gh = 2gs
 
s=h
 
ie it will go down a dist of h.....time take for this is root2h/g
 
therefore total time taken while going down is 2 root 2h/g.......3
 
then the object will go up coz of the acc g in upward directin......from 2 n will reach the surface with same vel oe root2gh n the will go up to height h
 
total time taken to reach height h from for depth h below the surface of water is 2 root 2h/g..................4
 
from 3 n 4
 
total time taken is 4 root 2h/g
 
 
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