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A spherical ball of radius r and relative density 0.5 is floating in equilibrium in water with half of it immersed in water.The work done in pushing the ball down so that the whole of it is immersed in water is(d=density of water) choose: (a)5/12(3.14*r4dg) (b)2/3(3.14*r4dg) and please tell how?

A spherical ball of radius r and relative density 0.5 is floating in equilibrium in water with half of it immersed in water.The work done in pushing the ball down so that the whole of it is immersed in water is(d=density of water)


choose:


(a)5/12(3.14*r4dg)   (b)2/3(3.14*r4dg)


and please tell how?

Grade:12

4 Answers

vivek kumar
57 Points
8 years ago

answer is b

this is bcuz when it will be immersed buoyancy= 4/33.14*r3.and weight acting downward =2/3*3.14 *r3and external force is f

,by this f= 2/3 *3.14*r3. and work done = b option

Vikas TU
14149 Points
8 years ago

Energy conservation applying –

K.Ei + P.Ei = K.Ef + P.Ef

0 + mgi – Buoyant forcei = 0 + mgf – Buoyant forcef + W.D

 [0.5 x 4/3 x 3.14 x r3 x g].r – [d x 2/3 x 3.14 x r3 x g].r =

[0.5 x 4/3 x 3.14 x r3 x g].r – [d x 4/3 x 3.14 x r3 x g].r+ W.D

[d x 4/3 x 3.14 x r3 x g].r -[d x 2/3 x 3.14 x r3 x g].r = W.D

W.D = 2/3(3.14*r4dg) 

B should be the answer.

FITJEE
43 Points
8 years ago

THIS CAN BE DONE BY ENERGY CONSERVATION ...

BY APLLYING E- CONSERVATION WE CAN CALCULATE IT.

SO YOUR ANSWER WILL BE ---- (b)

Karan Madan
33 Points
4 years ago
I Think, The Force Required Would Slowly Increase As We Go Down.. As The Volume Of The Ball Inside The Water Continuously Increases.. U Can Easily Find The Ans Using Integration.. Ans Is Equal To 5/12πr^4dg

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