Question

Please solve q 206?

Answer 1

D. `2/3pir^4rhog`

Explanation:

on applying conservation of energy,
`KE_i.+PE_i=KE_f+PE_f`
=`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

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)"`