Please solve q 206?

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Please solve q 206?

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Answer 1 · 2018-04-08T16:15:59

D. $\frac{2}{3} π r^{4} ρ g$ $2/3pir^4rhog$

Explanation:

on applying conservation of energy,
$K E_{i} . + P E_{i} = K E_{f} + P E_{f}$ $KE_i.+PE_i=KE_f+PE_f$
= $0 + {mg}_{i} - {buoyant force}_{i} = 0 + {mg}_{f} - {buoyant force}_{f} + W D$ $0+"mg"_i-"buoyant force"_i=0+"mg"_f-"buoyant force"_f+WD$
= $[0.5 * 4/3 * 3.14 * r3 * g].r – [d * 2/3 * 3.14 * r3 * g].r+W.D.$
= $[d * 4/3 * 3.14 * r3 * g].r -[d * 2/3 * 3.14 * r3 * g].r = W.D$
= $W.D = 2/3(3.14*r^4rhog)= 2/3pir^4rhog$

Answer 2 · 2018-04-08T16:27:27

The answer is $"option (4)"$

Explanation:

The force (to overcome the buoyancy) necessary to push down the sphere is

$F=2/3xxpixxr^3xxg$

The distance is $d=r$

The work done is

$W=Fxxd=2/3pir^3rhogxxr=2/3pir^4rhog$

The answer is $"option (4)"$

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Please solve q 206?

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