A spherical ball of radius r and relative density 0.5 is floating in equilibrium in the water with half of it immersed in water. The work done in pushing the ball down so that whole of it is just immersed in water is
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By applying Energy conservation theorem
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)
This should be the answer.
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)
This should be the answer.
Answered by
2
Answer:
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
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Explanation:
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