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A wooden cube is floating in water when a mass $m= 0.2\;kg$ is kept on its top, When the mass is removed the cube rises by $2 \;cm$. The side of the cube is [density of water $=10^3 kg/m^3$]

\[\begin {array} {1 1} (a)\;6\;cm \\ (b)\;12\;cm \\ (c)\;10\;cm \\ (d)\;8\;cm \end {array}\]

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Since the cube rise by 2cm there is a decrease in upthrust due to the decrease in weight.
Volume of water $=V=2 \times 10^{-2} \times Area\; of\; base$
upthrust $=V \rho _w g= 2 \times 10^{-2} \times A \times10^3 \times g$
$\therefore mg = 2 \times 10^{-2} \times A \times 10^3 \times g$
$0.2= 2 \times 10^{-2} \times 10^3 \times A$
$A=10^{-2}\;m^2$
$\therefore$ Side of the cube $=10\;cm$
Hence c is the correct answer.

 

answered Oct 7, 2013 by meena.p
edited Feb 18, 2014 by meena.p
 

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