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A 1kg lead(density=11200kg/m^3) block floats in mercury(13600kg/m^3). Determine the submerged volume of the block,the % of the block submerged and whether the object will submerge fully in water(1k kg/m^3)?

1 Answer

Aug 10, 2015

Volume submerged $= 7.35 \times 10^{- 5} m^{3}$ $= 7.35xx10^(-5) m^3$
Percentage submerged $= 82.36 %$ $= 82.36%$
The object will be completely submerged in water.

Explanation:

$1 m^{3}$ $1 m^3$ needs to displace $11200$ $11200$ Kgm of mercury in order to float.

$11200$ $11200$ Kgm of mercury has a volume of
$XXXX$ $color(white)("XXXX")$ $\frac{11200}{13600} m^{3} = \frac{14}{17} m^{3}$ $11200/13600 m^3 = 14/17 m^3$

For each $m^{3}$ $m^3$ of lead floating in mercury
$XXXX$ $color(white)("XXXX")$ 14/17 m^3 = 82.36%# will be submerged

$1$ $1$ kgm of mercury has a volume of
$XXXX$ $color(white)("XXXX")$ $\frac{1}{11200} m^{3}$ $1/11200 m^3$

$82.36 %$ $82.36%$ of $\frac{1}{11200} m^{3}$ $1/11200 m^3$
$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $= 7.35 \times 10^{- 5} m^{3}$ $= 7.35xx10^(-5) m^3$

Since the density of water is less than the density of lead, it is impossible for water to support the block of lead.

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