\[\begin {array} {1 1} (a)\;5\;m \\ (b)\;6\;m \\ (c)\;10 \;m \\ (d)\;12\;m \end {array}\]

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Let V be the volume of ball

The ball attains a velocity

$v =\sqrt {2gh}$

$\quad= \sqrt {2 \times 9.8 \times 10}$

$\quad= \sqrt {196}\; m/s$

Inside water - density of ball $= 0.5 \times 10^3\;kg/m^3$

density of water $=1 \times 10^3\;kg/m^3$

Net retardation produced

$ \qquad=\large\frac{upthrust - weight}{mass}$

$\qquad =\large\frac{V \times 10^3 \times g - V \times 0.5 \times 10^3 \times g}{V \times 0.5 \times 10^3}$

$\qquad=g$

Since the effective retardation inside liquid $=g$ the ball will fall $10\; m$ inside water before raising up.

Hence c is the correct answer.

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