Please solve q 206?

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2 Answers

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#

Apr 8, 2018

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)"#