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Q.

A wooden sphere is floating in water with 60% of its volume submerged in water. Now the sphere is taken to a depth of 5 m and released. What will be the acceleration of the sphere after release? Take g=10 m/s2

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a

7.5 m/s2

b

6.67 m/s2

c

8.42 m/s2

d

5 m/s2

answer is B.

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Detailed Solution

If V be volume of sphere, σ be the density of wood and P be the density of water, then by law of floatation. σvg=ρ(60100V).g⇒σ=35ρ∴ Acceleration a=ρvg−σvgσv=g(ρσ−1)⇒a=10[ρ3ρ5−1]m/s2=6.67 m/s2

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