First slide

Fluid statics

Question

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 $g = 10 m / s^{2}$

Moderate

A

7.5 $m / s^{2}$
B

6.67 $m / s^{2}$
C

8.42 $m / s^{2}$
D

5 $m / s^{2}$

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.
$σ v g = ρ (\frac{60}{100} V) . g \Rightarrow σ = \frac{3}{5} ρ$
$∴$ Acceleration $a = \frac{ρ v g - σ v g}{σ v} = g (\frac{ρ}{σ} - 1)$
$\Rightarrow a = 10 [\frac{ρ}{\frac{3 ρ}{5}} - 1] m / s^{2} = 6.67 m / s^{2}$

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