IIT JEE , AIEEE Forum - Physics , Mechanics

cuty

a rubber ball of mass m and radius r is submerged into water of density p to depth h and relesed. then to whatheight will ball jump above the surface of water?

a)(p*4/3

r^3-m)h/m b)(m-p*4/3

r^3)h/m c)(p*4/3

r^3-m)h/2m

d)(p*4/3

r^3+m)h/m

rajat

it should be a

but the ball should be very small

taruntanuj007

See cuty the approach here seems very imporatnt so wat i wud suggest is that the work done in bringing the ball from the surface of the liquid to a height h below the surface that much work will result in its final PE right after all its conservation of energy!!!!!

so first calculate the equilibrium position and then i'm sure with my hints u'll be able to solve it if not nudge me!!!! dont forget to rate nme!!!!!!

cvramana

Use the fact that a buoyant force acts on the body and gravity. So find the resultant and use work energy theorem.

kghedriu

F resultant = mg- 4/3 pi r^3 gp

now, acceleration resultant can b found out by dividing with m....

.: a = g{1-4/3 pi* r^3 * p/m}..................(1)

dist = h

accn = given above frm (1)

initial vel = 0

final vel = v = ?

use v^2 - u^2 = 2as and substitute values from above....

now, max ht reached = v^2/2g

hence the solution..

the final answwer comes out to be:

h/m [ m- 4/3 p*

r^3/m] i.e option (b)

see the following figure......hey...m sorry...there is a small mistake in my solution...i didnot follow the sign convention......rest all, the procedure is correct........thanx stuart for correcting me... :)

pinnaka

Sorry Goutham u r solution is wrong. Why? u didn?t stick to your coordinate axis which u taken. In your solution u have assumed the downward direction to be positive and the upward to be negative. U didn?t stick to this axis in the latter part of your solution.

Actually the answer is a

Going by the diagram given by Goutham and the upward direction to be the positive one.

In the fluid
w_done = 1/2mv^2
4/3*rho*r^3gh ? mgh=1/2mv^2

Outside the body the 1/2mv^2 = mgx (the height reached)

4/3*rho*r^3gh ? mgh=mgx
On rearranging the terms u get the answer

as x =[4/3*rho*r^3 ? m]h/m

-----Hope this Helps
-----Stuart Anderson

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