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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Feb 2008 20:21:05 IST
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a rubber ball of mass m and radius r is submerged into water of density 
to depth h and released. then to what height will the ball jump up above the surface of water?
[ans: ( 4/3pi r3 -m)h/m]
plz explain the metd...
rates assured
(missing images r rho{density})
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FAILURE IS NOT FALLING IN LIFE BUT NOT RISING AGAIN AFTER FALLING!!!!!!
I LIKE WAVES NOT BECAUSE THEY RISE AND FALL..
BUT BECAUSE EVERYTIME THEY FALL THEY RISE AGAIN!!!!!!!
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19 Feb 2008 20:30:26 IST
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plzzz answer
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FAILURE IS NOT FALLING IN LIFE BUT NOT RISING AGAIN AFTER FALLING!!!!!!
I LIKE WAVES NOT BECAUSE THEY RISE AND FALL..
BUT BECAUSE EVERYTIME THEY FALL THEY RISE AGAIN!!!!!!!
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19 Feb 2008 20:33:07 IST
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4/3pir3 is the density of the rubber ball with respect to water
and rho4/3pir3multiplied by h/m with be the height if the water is not considered and we must subtract the term rhopir3multipled by h/m to get the correct height to which it will rise. ( h is height which the bob is submerged and it must be considered because the upward thrust + h = new height
therefore rho4/3pir3h/m-rho4/3pir3h/m is the answer
hence[ans: (rho4/3pi r3 -m)h/m]
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19 Feb 2008 20:42:53 IST
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just do this simple method............
u know the effective gravity felt by the ball at a depth h.
use tht value for g in v^2 = 2.g.h
now u got the velocity at the surface of the water level.
let the total height reached be H.
now use energy conservation principle .equate KE=PE.
THUS YU HAVE THE ANSWER...........
IF USEFUL PLZ RATE ME............
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