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A ball of density $0.5 \times 10^3$ is dropped from a height of $10\; m$ and then enters water . Upto what depth will the ball go inside the water before it starts raising up.

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

Can you answer this question?
 
 

1 Answer

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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.

 

answered Oct 8, 2013 by meena.p
edited Feb 18, 2014 by meena.p
 

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