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Q.

A small sphere of density $0.5 gm / {cm}^{3}$ is brought to a depth of 5m below the surface of water in a lake and released. Find the height to which the sphere will rise above the free surface of water. Ignore viscous effect of water and take $g = 10 m / s^{2}$

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a

10 m

b

8 m

c

5 m

d

$\sqrt{10} m$

answer is C.

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Detailed Solution

After the sphere is released, its acceleration is
$a = \frac{ρ_{w} vg - ρ_{s} vg}{ρ_{s} v} = (\frac{ρ_{w}}{ρ_{s}} - 1) g = (\frac{1}{0 .5} - 1) \times 10 m / s^{2} = 10 m / s^{2}$
At the surface, velocity of sphere, $v = \sqrt{2 . a . h} = \sqrt{2 \times 10 \times 5} m / s = 10 m / s$
Height above the surface of water $= \frac{v^{2}}{2 g} = \frac{{(10)}^{2}}{2 \times 10} m = 5 m$

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